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https://chemistry.stackexchange.com/questions/166446/question-regarding-aibn-in-toluene
|
Question regarding AIBN in toluene
|
[](https://i.stack.imgur.com/zzUs5.png)
Hi guys, I am trying to perform bromination of this compound using AIBN as a radical initiator. However, it seems like AIBN is only sold commercially as 0.2 M in toluene.
Is it possible to use AIBN directly as it (0.2 M in toluene) is or should I try to remove the toluene first before performing this reaction? (Not sure if toluene will be brominated instead of thymidine)
Thank you!
| 0
|
[
[
"\nBefore AIBN was developed as a radical initiator, the world used benzoyl peroxide [example here](https://pubs.acs.org/doi/10.1021/ja01172a016). If none of the AIBN family of initiators are available then you can use benzoyl peroxide under similar conditions.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/166440/labs-without-equipment-virtual-chem-lab
|
Labs without equipment - virtual chem lab?
|
I am an "international online" high school student with very spotty access to chemistry lab materials needed for some chemistry assignments I am being given. I would like to know if there exists online virtual chemistry labs, appropriate for a high school student, that will interactively/graphically simulate an actual lab experiment. For example, I'd like to be able to use an online interface to declare which chemicals, in what quantities, I am placing into a specified beaker, and perhaps heated at a certain temperature for X duration. This hypothetical chemistry simulation tool might show me the color changing of my solution, which I can then use to answer my homework questions.
I googled and found this example [virtual chem lab](https://chemcollective.org/vlabs), but it seems to only offer pre-determined labs that I can't adapt to my specific homework assignments. I guess I need something more flexible?
Below is a sample real lab-homework question I am currently expected to perform, even though I have no chemicals or equipment:
* Label one of the test tubes Fe2+. Add 4 milliliters of iron(II) sulfate to the test tube.
* Label the other test tube Fe3+. Add 4 milliliters of iron(III) nitrate to the test tube.
* Add 4 milliliters of potassium thiocyanate to each test tube.
* Observe the contents of the test tubes, noting any evidence of a chemical reaction. Record your observations in the table. If there is no evidence of a reaction, write “no reaction.”
Likewise, here is another example homework question:
* Reuse the same test tubes from part A, labeled Fe2+ and Fe3+. Be sure they’re clean.
* Add 4 milliliters of iron(II) sulfate to the test tube labeled Fe2+.
* Add 4 milliliters of iron(III) nitrate to the test tube labeled Fe3+.
* Add 4 milliliters of potassium iodide to each test tube.
* Add 1 milliliter of the prepared starch solution to each test tube.
* Record your observations, noting any evidence of a chemical reaction. If there is no evidence of a reaction, write “no reaction.”
If this hypothetical chem-sim tool doesn't exist, suitable for my needs, is there some other tactic I can use to find answers in the absence of the correct materials?
| 0
|
[] |
https://chemistry.stackexchange.com/questions/166439/what-would-be-the-products-when-2-iodobutane-is-heated-with-tertiary-butoxide
|
What would be the product(s) when 2-iodobutane is heated with tertiary butoxide?
|
The following question from *Black Book Organic Chemistry IIT JEE Advanced Level Papers* [1]:
>
> Choose the correct option(s) among the following about **[P]**:
>
>
> [](https://i.stack.imgur.com/pmXSu.png)
>
>
> A) Two C−H bonds in **[P]** are involved in hyperconjugation.
>
> B) **[P]** cannot show stereoisomerism.
>
> C) Hydrogenation of **[P]** gives mainly *n*-butane.
>
> D) Monochlorination of **[P]** gives 3
>
>
>
What is the product when 2-iodobutane is heated with tertiary butoxide? Is it 1-butene, 2-butene, or both?
Bulky bases generally favor the Hoffmann product, but my teacher told that in this specific case having much less hindrance Zaitsev product dominates. I am not able to completely digest it how this happened and is this correct or not.
Answer key: C, D.
### Reference
1. Ashish Mishra; Dr Ramu Petakamsetty. *Black Book Organic Chemistry IIT JEE Advanced Level Papers*; Blue Rose Publishers, **2022**. ISBN 978-93-5611-183-7.
| 5
|
[
[
"\nAs described in most introductory organic chemistry textbooks (my usual reference is Loudon), E2 elimination to form an alkene typically produces the more substituted alkene as the major product. This outcome is the Saytzeff/Zaitsev product (roman spelling varies depending on source).\n\n\nThe primary exception is if a sterically hindered (\"bulky\") base (the standard example is t-butoxide) is used, in which case the most accessible protons are removed, leading to a terminal alkene. Similarly, if the leaving group is quite large (trialkyl amines are typical examples), the terminal alkene is favored even for small bases. This outcome is the Hofmann product.\n\n\nAn exception to the exception is the case of large halide leaving groups. It has been observed that the percentage of 2-alkene (ie more substituted) product in eliminations using t-butoxide as base increase when one changes the leaving group from chloride to bromide and from bromide to iodide. For the 2-halobutanes specifically, the ratio of 1-butene to 2-butene (*cis* or *trans*) products is approximately 2:1 for 2-chlorobutane, 1:1 for 2-bromobutane and 1:2 for 2-iodobutane[1]. Competing explanations were put forth for these observations (sometimes referred to as the \"Ingold-Brown controversy\"), and I'm not aware of a consensus even today.\n\n\n[1] Brown, HC and Klimisch, RL (1966) *J Amer Chem Soc* **88:** 1425\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/166436/electron-affinities-of-the-alkali-and-alkali-earth-metals
|
Electron affinities of the alkali and alkali earth metals
|
According to my book, the former has a higher EA than the latter group of elements because alkali metals will attain $ns^2$ configuration whereas an alkaline earth metal in the same period will attain $ns^2np^1$. Fully and half filled configurations are more stable but according to [this](https://studylib.net/doc/7445445/stability-of-half-full-shells-and-other-myths) article, there should be unfavorable interactions due to pairing of electrons in same orbital in alkali metals. Please explain what should be correct order and why?
| -1
|
[
[
"\nA look at any electron affinity table reveals that in factvalkali metals have higher electron affinities than the alkaline earth metals that respectively follow them. We can say the same for nonmetallic hydrogen versus helium, these also being $s$-block elements. In fact helium, beryllium and magnesium seem to have no electron affinity at all! While the effect of pairing the valence electrons in a Group 1 element is unfavorable, having to add an electron to a higher subshell as the Group 2 elements have to do is worse.\n\n\nThis effect of subshell occupation shows up not only with electron affinities but also with actual compounds of the metals. Most alkali metals [can form compounds as anions](https://en.wikipedia.org/wiki/Alkalide), while alkaline earth metals are not known to have this possibility.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/166433/evaporation-rate-of-hydrochloric-acid
|
Evaporation Rate of Hydrochloric acid
|
I'm currently working in a Chemical Plant as an intern, and I was given a task to improve the efficiency of the blower, which is already in place right now, by designing a new blower for the appropriate flow rates.
We have a filling station in the plant for HCl liquid, which fills the HCl (30-32% conc. at 20 m$^3$/hr) into tankers for transportation purposes. The problem is when HCl is filled up in a tank, it releases fumes, which are harmful for the people present near the filling station.
We have to recommend a blower, so that most of the generated fumes can be treated at an appropriate flow rate. But we don't know how to proceed with this, as we wouldn't know about the evaporation rate of HCl while it is flowing inside the tank.
Some technical information:
1. Temperature, Pressure can be assumed as 30 $^{\circ}$C, 1 atm.
2. Flow rate of HCl inside tank: 20 m$^3$/hr.
3. Capacity of Tank: 25 Metric tonnes
| 1
|
[
[
"\nWhen HCl is put into a tank, it releases fumes... And this is not an equilibrium situation (just like pouring a carbonated beverage into a glass can be done rapidly or slowly, releasing various amounts of CO2).\n\n\nYou may be able to get some numbers useful for calculations (if you also specify a range) from the two references below (1 and 2). No. 1 has a nomograph that shows the concentration of HCl in the liquid and vapor, and No.2, which is a fairly large handbook of safe handling for HCl, has a different way of expressing the information:\n\n\n[](https://i.stack.imgur.com/X3zsJ.png)\n\n\n[](https://i.stack.imgur.com/6dHlA.png)\n\n\nKnowing your volumes and tank flow rates and other estimated required flow rates, along with the amount of HCl in the vapor could help decide what kind of pump you need. And don't forget to inquire of your HCl supplier, who might have significant, relatively private, but not actually confidential, information about how other customers succeed (or fail) with their methods of handling this task. Sometimes calculations don't take everything into account, but a successful solution is worthwhile, even if if is \"supported\" by calculations.\n\n\nRef 1: <https://pubs.acs.org/doi/10.1021/i460001a002>\n\n\nRef 2: <https://www.jsia.gr.jp/data/handling_02e.pdf>\n\n\n",
"1"
],
[
"\nThis is a job for a competent chemical engineer not an intern; find one and suggest hiring a consultant. I am dismayed that HCl gas is released to the atmosphere and think that OSHA should get more involved with your company. Since the air volumes displaced in the delivery tank and the receiving tank are the same [and should be readily determined] a vapor recovery system should be imperative in handling such materials. Look into it. Learn all you can about the overall process. If that fails, get a better internship.\n\n\nSome things to consider: 30 degrees Celsius seems a very high temperature to be handling concentrated hydrochloric acid; what is going on here? Also work with the same units; convert the masses to volumes; learn about head spaces etc. in the shipping.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/166428/why-does-equilibrium-of-ice-and-water-only-exist-at-273k-at-normal-pressure
|
Why does equilibrium of ice and water only exist at 273K at normal pressure?
|
I am a high school student and I am a little confused in the concept of spontaneity of a reaction and how equilibrium works for a reaction, I got some confusions:
1. Let's take example of freezing of water: In my textbook its written that freezing point of water at 1 atm pressure is 0 degrees and at this temperature water and ice remains in "dynamic equilibrium". So it means this phase transition from solid to liquid is a "reversible process" and we know at equilibrium the "change in Gibbs free energy is 0". Now if I heat it "∆G would become negative" and what happens is more ice melts but the temperature doesn't changes and stopped heating so a new equilibrium got established, so does it mean here equilibrium constant is changing(as more water is forming and more ice is melting) without increase in temperature? But I used to think it only depends on temperature. So what's going on. please explain it in simple language so that a high school student can understand?
2. Suppose I have ice at room temperature in a closed beaker. What we observe in our daily life is whole ice melts into water but shouldn't the "equilibrium" still exist because its a reversible reaction ,so just like evaporation in a closed beaker the water reaches an equilibrium state with its vapors at all temperatures, ice should also do the same with water but it doesn't. Why?
| 0
|
[
[
"\nTo write down the equilibrium constant expression, you use concentrations, not volumes or mass. To illustrate this with an example you can check in your kitchen, a saturated salt solution does not get saltier by adding more salt (you have pure undissolved solid salt in either case). It does not get less salty if you remove part of the undissolved salt.\n\n\nThe equilibrium between liquid and solid water is a bit special because both species are pure. Adding more ice or adding more water will not change the \"concentrations\" or the freezing point of water.\n\n\n\n> \n> 1. [...] more ice melts but the temperature doesn't changes and stopped heating so a new equilibrium got established [...]\n> \n> \n> \n\n\nNo, you still have pure water and pure ice, so the equilibrium constant is still the same, and the concentrations are still the same - it is not a new equilibrium. On the other hand, if you pour salt into the water, the liquid water concentration drops (no longer pure) and the solid water concentration remains constant, so then you are out of equilibrium, and ice will melt and the temperature will decrease. With pure water, when the temperature is higher or lower than the freezing point, ice will melt or water will freeze until we are at the freezing temperature (or we run out of ice or water).\n\n\n\n> \n> [... from OP's comments] how the temperature will decrease??? suppose if we have ice and water at 100 degrees and I add ideal solute to the water ,the concentration of water would decrease and it will only decrease the no. of water molecules going back to the solid state but the no of solid molecules going to liquid state remains the same so equilibrium will shift forward but here also \"the equilibrium constant\" should remain the same if we think this way.... but it doesn't if we think of its formula why?\n> \n> \n> \n\n\nIn ice, water has 4 hydrogen bonds. In liquid water, it has less. So transferring a water molecule from ice to water is an endothermic process, and it will cool the system down (thermal energy turns into potential energy).\n\n\n\n> \n> 2. [...] so just like evaporation in a closed beaker the water reaches an equilibrium state with its vapors at all temperatures, ice should also do the same with water but it doesn't.\n> \n> \n> \n\n\nThe difference between pure liquid water and a gas mixture of water and air (\"humid\" air) is that the water in humid air is not pure. You can have air with higher or lower humidity. The water in a half-full closed bottle will evaporate until the humidity reaches its equilibrium value. On the other hand, liquid and solid water are both pure, and are at equilibrium (at normal pressure) only if the temperature is equal to the normal freezing point.\n\n\n\n> \n> What we observe in our daily life is whole ice melts into water but shouldn't the \"equilibrium\" still exist because its a reversible reaction\n> \n> \n> \n\n\nBecause the ambient temperature in our lives is usually higher than the freezing point of water, and because our containers are not perfectly insulated, we keep transferring heat to the container. By the way, even when ice melts, at the molecular level, some liquid water molecules still attach to the solid. The process of water molecules going from solid to liquid just happens to be faster.\n\n\n\n> \n> [from OP's comment] does that mean the equilibrium between ice and water exists at all temperatures?? so why do textbooks says and also the phase diagrams says that it exists \"only\" at freezing point\n> \n> \n> \n\n\nIf water and ice come in contact, there will be a forward and a reverse process (freezing and melting) at the particular level. When the two phases have the same temperature and these two process have the same rate, it is called equilibrium. Because the rates are the same at equilibrium, there is no net change, i.e. the amount of ice will neither increase nor decrease. The temperature where this happens is called the freezing point or melting point, even though at this point, there is not bulk freezing or melting.\n\n\n\n> \n> [...from OP's comments] can u please explain\"dynamically\" what's going on when u add \"ideal solute\" to the water ?? I understand that the concentration of water would decrease and more ice will melt....but if we will think how the activity of the molecules actually affected on adding solute things get difficult? like if i add an ideal solute to the liquid it will not change the bonding because its ideal so what I think is the no. of water molecules going back to solid state should still remain the same??\n> \n> \n> \n\n\nAn ideal solute will not interact different with water than a water molecule, but it can't become part of the ice lattice (it does not fit). So instead of - say - 100 water molecules bumping into the ice surface, now 99 water molecules and one other molecule bump into it, decreasing the rate of attachment of water molecules to the ice.\n\n\n",
"2"
],
[
"\nFor the first question, consider a simple reaction for example take Haber's process in which nitrogen and hydrogen combine to form ammonia. Now, at a particular temperature (T), there will be some equilibrium constant and correspondingly, a particular value of $ ∆{{G}\\_{0}} $. What it simply means is that if you have a mixture of hydrogen, nitrogen and ammonia at a particular temperature, then equilibrium will only be achieved at this specific composition. Some other composition is also possible at the same temperature but it will simply not be at the same temperature but it will not be at equilibrium and will have to tendency to go towards equilibrium and this rate of *movement towards equilibrium* will depend upon other reaction conditions. So basically in your example,when you provide heat to the mixture, the temperature of mixture suddenly increases leading to change in value of $ ∆{{G}\\_{0}} $. As a result, melting of ice occurs which decreases temperature. This melting of ice happens till the temperature of mixture reaches the freezing point of water. Talking about equilibrium constant, it is not simple to describe since in high school,liquids and solids have no role in reaction quotient. But gibbs free energy change is the important thing. In real life, the freezing of water or melting of ice is so fast that it almost seems as if the externally supplied heat almost immediately was converted into latent heat\n\n\nFor your second question, as I said, equilibrium constant is a bit difficult to describe but it will definitely be something. So basically, some amount of ice must be left at room temperature but the equilibrium constant is *extremely shifted* to the right because of which it is in an extremely small amount not percievable practically. (It might be so less that it is less than even a few molecules, but since matter is quantised, it would not be possible to have fractional number of molecules.)\n\n\nEven when you were talking about water- water vapour equilibrium, if you have taken very small amount of water in very large container, then there will be negligible water left as liquid at equilibrium.\n\n\n",
"0"
],
[
"\nFirst some key points that are important to keep in mind when thinking about thermodynamic equilibrium: (1) thermodynamic equilibrium states describe systems that are unchanging, static; (2) an equilibrium constant is relevant only when a system can exist in either of multiple states (as you suggest); (3) the equilibrium constant for the melting of water is just that, a constant equal to the ratio (equal to 1) of the activities of water and ice at the melting point (activities are approximated by concentrations in some cases).\n\n\nWhat you are describing (melting of ice) occurs when the system is pushed out of its original equilibrium state by adding heat at a finite rate. Melting, in other words, is not an equilibrium state, it is a *transition* between different states occurring when you perturb the initial equilibrium state by adding heat.\n\n\nTo explain how melting affects the thermodynamic equilibrium between ice and water, consider the following process: (1) you start with ice and water at equilibrium at the melting point (surroundings and the system are at that point); (2) you isolate the system (by wrapping it with insulation) and transfer some energy into the system as heat; (3) the system returns to equilibrium by redistributing the input energy and forming a new state; (4) you remove the insulation and place the system into contact again with surroundings (at the melting point). Importantly, during steps 2 and 3 the system is not in an equilibrium state. In steps 2/3, if not too much heat is added (so as to melt all of the ice), then a new equilibrium can arise between water and ice, where both are at the melting point. If too much heat is added, then the transition of ice to a liquid will be complete. With sufficient heat the temperature of the system will rise above the melting temperature, with ice no longer present (but then at step 4 the system will be cooled again).\n\n\nTo help wrap your head around the idea that the free energy change after melting some of the ice remains constant ($\\Delta G = 0$, $K=1$), consider what would happen to the proportions of ice and water if you should mix together two different samples of ice and water, or simply add water, or ice, to a sample of ice and water, all at the melting point. Nothing! The proportions would remain constant. The point is that at equilibrium (at the melting point) there is no force driving formation of more ice *or* water.\n\n\nSuperheated ice1 (ice at a temperature above the melting point) is not an equilibrium state relative to water at the same T, because above the freezing point liquid water has a *lower* free energy than ice. You can in fact create superheated ice, but it will spontaneously melt to water if you remove the constraints that allowed you to create this state.\n\n\nIn the case of a beaker (a diathermal container, allowing heat transfer), there is no equilibrium because heat is constantly being transferred into the container, assuming the surroundings are above the melting point (although if heat is added slowly enough then the system is approximately at equilibrium at any given point). The way of thinking around this is to consider the Gibbs free energy at the initial and final *equilibrium* states of the system, before and after changes due to addition of heat.\n\n\n*References*\n\n\n1. Schubert, G. and Lingenfelter, R.E., Superheated Ice Formed by the Freezing of Superheated Water. Science **1970**, *168* (3930) pp. 469-470, DOI: 10.1126/science.168.3930.4.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/166420/why-does-column-chromatography-not-work-like-its-supposed-to
|
Why does column chromatography not work like it's supposed to?
|
I'm a few months into working in an organic synthesis lab and I consider myself to have a solid intuition for, well, how chemistry works. I have ran at least 50 columns at this point but no matter what experiment I run, there's always some concepts that seem off.
To start, I don't think my technique is the problem. I know for a fact I am packing the column well. I dry pack but run plenty of solvent cycles through it before loading. I try to load with as little solvent as possible, to the point where the top of the sand is still dry sometimes. I load with the same solvent I use for elution, so it is not too polar in comparison. I run the column slow (like abysmally slow, 2 drips a second) to not "flush" things out. I try to perfect technique and minimize other factors.
But I still run into problems. For example, everyone uses TLC plates to test different solvent systems for maximum separation. If my desired product dot is smack in the middle with Rf of 0.5 on the plate, I'm expecting it to come out within two cycle volumes on the column (1/Rf). But regardless, it seems like it always comes out right after the first cycle and it catches me by surprise. The compounds I'm isolating have very close Rfs, and it would be great if I could at least predict when it will come out so I can switch to smaller collection tubes. I realize that not using the recommended 0.2 Rf means I will need to be very precise with collection but I don't mind. What I DO mind is that the product doesn't even match the tested Rf.
Also, it seems like if I use any solvent system more polar than the recommended 0.2 Rf TLC, the product and all the impurities will wash out together with almost no separation at all. It's like all basic concepts of chromatography have gone out the window. The separation was incredible on the TLC plate, like I'm talking the product at an Rf of 0.5 and the impurities had Rf of 0.2. Then I try a column with 50 mL cycle volume and everything started coming out within 10mL of each other?
Also, collection tube TLC doesn't make much sense either. I take a TLC of a collection tube with the product, with the product showing Rf of, say, 0.5, but the tube also contains an impurity with an Rf of 0.05. How is that even possible? According to my cycle volume calculation (CV \* 1/Rf), an impurity with that Rf should come out many many collection tubes afterwards.
I just want to understand the actual reason why this is happening. I don't mind changing my procedures, but it is rather frustrating to not understand the science.
| 3
|
[] |
https://chemistry.stackexchange.com/questions/166414/question-on-thermodynamics-from-a-recently-conducted-examination
|
Question on Thermodynamics from a recently conducted Examination
|
This is a question from the recently conducted JEE Main exam (in India). I had solved it myself, but found the answer to be incorrect. Here's the question.[](https://i.stack.imgur.com/XY532.jpg)
Here is how I solved it:
$\Delta H = \pu{41.1 kJ mol-1}$
$R = \pu{8.314 J mol-1 K-1}$
$w=\pu{36 g}$
$$ \Delta n\_g=2$$
$$ \ce{\Delta H = \Delta U + \Delta n\_gRT}$$
$$ \ce{ \Delta U = (44.1 - (2×8.31×373)/1000) \pu{J K-1}}$$
$$\ce{\Delta U = \pu{(41.1 - 6.199) J K-1}}$$
$$\ce{\Delta U = \pu{34.9 J K-1}}$$
$$Ans. 35$$
However, the answer key states that the answer will be 38, instead of 35.
I feel that the answer can be 38 only if there is 1 mole of water instead of 2.
Kindly help, and point out my mistake, if any.
| 1
|
[
[
"\nYour mistake is that you took $\\Delta n\\_g=2$ when it should be $1$. Remember it is asking for the internal energy change for the *reaction* (this can be seen by the fact that kJ/mol is being asked, not kJ-remember dividing by moles makes this intensive so there is no dependency on number of moles taken). Thus, the moles of water taken is irrelevant and $\\Delta n\\_g$ only depends upon the stoichiometric coefficients. Since the reaction is $\\ce{H2O(l) -> H2O(g)}$, so $\\Delta n\\_g$=1.The mass of water given is just the examiner trying to trick(or rather, test) you.\n\n\nThat being said, I might add that the question is poorly framed and also somewhat ambiguous. This is because of 2 reasons:-\n\n\n1. It should be clearly mentioned that internal energy of vapourisation *of the reaction* is being asked and the only hint of this is kJ/mol which is easy to miss.\n2. Additionally, it should be made explicit that internal energy of vapourisation of the reaction $\\ce{H2O(l) -> H2O(g)}$ is being asked because it can vary depending on if it is $\\ce{H2O(l) -> H2O(g)}$ or $\\ce{2H2O(l) -> 2H2O(g)}$ or $\\ce{3H2O(l) -> 3H2O(g)}$ etc.(see addendum for more)\n\n\nHowever, some texts define enthalpy of vapourisation and internal energy of vapourisation as the $ \\Delta U$ and $\\Delta H$ of the reaction when **1 mol** of the compound is undergoing phase transition, and then the above 2 points don't hold because if moles are taken, then stoichiometric coefficient magnitudes are irrelevant and only their ratios are important.\n\n\nAddendum:\n\n\nJust in case you want some depth in this topic, remember that **in any calculation of $ \\Delta \\_rH$ and $ \\Delta \\_r U$ we\nassume the number of moles of the reactants/products to be same as that of the stoichiometric coefficients**. This is because by definition, $ \\Delta \\_rH$ and $ \\Delta \\_r U$ depend only on the reaction equation and not on particular reactions in which we can vary the number of moles being taken.If you calculate $ \\Delta H$ or $ \\Delta U$ of a particular reaction, then that cannot be defined as $ \\Delta \\_rH$ or $ \\Delta \\_r U$ unless the number of moles of every reactant and product matches the stoichiometric coefficients. Finally, note that stoichiometric coefficients need not be the simplest ratios i.e. $ \\Delta \\_rH$ will be different for each of $\\ce{H2O(l) -> H2O(g)}$, $\\ce{2H2O(l) -> 2H2O(g)}$ and $\\ce{3H2O(l) -> 3H2O(g)}$ because the stoichiometric coefficients are different.\n\n\n",
"6"
]
] |
https://chemistry.stackexchange.com/questions/166412/how-can-i-produce-red-iron-oxide-with-electrolysis
|
How can I produce Red Iron oxide with electrolysis
|
I am producing Iron Oxide with the intention of using it for pigmentation. My attempts so far have been to use electrolysis. My first attempt I used Borax as an electrolyte and a 12 volt adapter with 3-4ish amps. The result was a typical light orange rust color, however the process was rather slow, 24 hours resulted in about a tablespoon of rust. My next attempt I used table salt instead. This was far faster, it produced nearly a cup of black sludge in 10 hours. As per the tutorials I watched on youtube, I then filtered the matter through a coffee filter and a funnel, then heated the result in a pan. The result however was a rust color covered in a layer of black crust, as if I had burnt a sheet of cookies. Powdering the result gave me a powder of very dark brown color.The Residue left over on the coffee filter and in the container I used for electrolysis are the typical rust orange color I would expect. It is my understanding that heating the black matter that results from the electrolisis usually produces a more red color. So my question(s) are:
* how I can change my approach to produce red pigmentation using this process, and why may I have ended up with black residue instead of orange or red?
* What are the variables that change the pigmentation of rust?
+ Is it the speed at which it drys? the temperature in which it is heated? the electrolyte used in the process of electrolysis?
| 0
|
[
[
"\nRed iron oxide is Fe2 O3 , it forms in air at roughly 700F. In a water solution you will make \"rust\", brown/red. Rust is a complex mixture of hydroxides and hydrated oxides such as Fe3 O4 -H2O. At temperatures higher than 700 F you start getting black Fe3 O4 ( aka mill scale). At very high temperature you get black Fe O. Fretting corrosion will produce Fe2 O3 at room temperature but is not a practical source.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/166407/cfse-for-octahedral-complexes-with-more-than-one-type-of-ligands
|
CFSE for octahedral complexes with more than one type of ligands
|
How would be crystal field stabilization energy for the complex [Co(NH3)5Cl]²-? I guess that due to the presence of negative ligand Cl- 3 sets of d orbitals can be present with a dz²or dx²-y²orbital of a slightly little different energy.
| 0
|
[
[
"\nWhen more than one type of ligands are present with some being SFL and others being WFL then, the final effect of those ligands dominate which are greater in number. So, in the given compound, ammonia is more than chlorine so effect of SFL dominates and you get inner orbital complex\n\n\n",
"-2"
]
] |
https://chemistry.stackexchange.com/questions/166406/what-will-be-the-major-product-in-the-nitration-reaction-of-2-methyl-5-nitrophen
|
What will be the major product in the nitration reaction of 2-methyl-5-nitrophenol?
|
[](https://i.stack.imgur.com/iGOGY.png)
I know if the $\ce{-NO2}$ group be added at the ortho position with respect to the $\ce{-OH}$, hydrogen bonding between the *hydroxyl* and *nitro* groups would lead to stability.
But would that product be major considering steric factors?
| 1
|
[
[
"\nNitration of phenol with concentrated nitric acid gives 2,4,6-trinitrophenol . So, it is quite possible here that the corresponding substituted trinitrophenol may form in the reaction. But you have also added concentrated sulphuric acid so it may cause some other product to form.\n\n\nGenerally electrophilic aromatic substitution of phenol gives para product in slightly higher amount than ortho product due to inductive effect of oxygen. However in case of nitration, ortho product becomes higher than para product due to intramolecular hydrogen bond formation. So I think that it should be major product in your reaction also\n\n\n",
"-1"
]
] |
https://chemistry.stackexchange.com/questions/166402/can-proton-nmr-be-used-to-study-transition-metal-complexes
|
Can proton NMR be used to study transition metal complexes?
|
I prepared metal complexes with a Schiff base ligand L using cobalt(II) chloride hexahydrate, iron(III) chloride hexahydrate, and nickel(II) nitrate hexahydrate. I used 1H NMR to study the difference between them and maybe get some information about the structure of the formed complexes.
The results showed a slight upfield shift in almost every peak and the appearance of some new peaks (attached figure), which is a bit ambiguous for me, and I don't know how to interpret them:
[](https://i.stack.imgur.com/IzSAk.jpg)
| 3
|
[] |
https://chemistry.stackexchange.com/questions/166398/what-is-the-hybridization-of-oxygen-atoms-in-boric-acid
|
What is the hybridization of oxygen atoms in boric acid?
|
I was making the $\ce{H3BO3}$ structure and tried to think of hybridization of oxygen atoms and got confused between $\mathrm{sp^2}$ and $\mathrm{sp^3}$.
Boron has empty orbitals, so the lone pairs of oxygen can do backbonding with boron, resulting in $\mathrm{sp^2}$ hybridization. But at a moment only one out of three oxygen atoms would do so. What about other two oxygens?
| -1
|
[
[
"\nAll three oxygen atoms backdonate a π-electron pair and thus they are all $\\mathrm{sp^2}$. When you construct the molecular orbitals, you find that the π orbitals are as follows:\n\n\n* One orbital bonds the boron atoms to all three oxygen atoms in a double Y shape — one Y on each side of the molecular plane.\n* Two orbitals are nonbonding because they have electron density on the oxygen atoms only; the boron is a node. (This corresponds to two oxygen atoms having a second lone pair whenever you draw any one of the contributing valence-bond structures.)\n* Finally, an antibonding orbital, which has the same threefold rotational symmetry as the bonding one but the oxygen atoms contribute out of phase from the boron atoms.\n\n\nSince there are six π electrons (two from each oxygen atom but none from the electron-deficient boron), only the bonding and nonbonding orbitals are occupied and thus the molecule is bonded with the planar, $\\mathrm{sp^2}$-hybridized structure.\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/166390/searching-for-evidence-based-published-study-that-compares-disinfectant-residues
|
Searching for evidence-based published study that compares disinfectant residues after evaporation
|
I've been plowing through databases to find evidence-based research reporting on the residue remaining after evaporation of chemical disinfectants. I want to know if there is a study which states the amount of residue that disinfectants (which are commonly used for healthcare surfaces or cleanrooms) leave after evaporation in air-dry way (not via Residue on Evaporation RoE Test which includes heating the substance). These disinfectants are mostly common and include alcohol, chlorine and chlorine compounds, formaldehyde, glutaraldehyde (GTA), hydrogen peroxide (H2O2), iodophors, ortho-phthalaldehyde (OPA), peracetic acid, peracetic acid and hydrogen peroxide, phenolics, and quaternary ammonium compounds (quats).
I've found an article which states the amount of residue on some materials in healthcare equipment, but it doesn't say if the disinfectant has evaporated or not1 . There are some reports from ALS lab too, but the source is not clear enough(*see image below*) 2 .
1. C. Lerones, A. Mariscal, M. Carnero, A. García-Rodríguez, J. Fernández-Crehuet, Assessing the residual antibacterial activity of clinical materials disinfected with glutaraldehyde, o-phthalaldehyde, hydrogen peroxide or 2-bromo-2-nitro-1,3-propanediol by means of a bacterial toxicity assay,.*Clinical Microbiology and Infection* **2004,** *10*, 984–989. [DOI: 10.1080/00397918908050700.](https://doi.org/10.1111/j.1469-0691.2004.00967.x)
2. <https://www.contecinc.com/articles/disinfectant-residues-mitigation-and-management/>
[](https://i.stack.imgur.com/m2kcM.png)
Thanks community!
| 0
|
[] |
https://chemistry.stackexchange.com/questions/166389/purpose-of-1h-in-naming-of-substituted-heterocyclic-compouns
|
Bibenzimidazole nomenclature: Indicated H and locants
|
I am trying to understand how the locants work for the following molecule (locants added by me).
[](https://i.stack.imgur.com/GdlE2.jpg)
I added the locants following IUPAC Blue Book rules from 2013, P-14.4. Basically, first numbering heteroatoms in the ring, assigning the highest locant to the one with an indicated hydrogen, and then number clockwise around the ring (red locants).
I assumed the same would go for the second ring, so the name would be:
$1^1H,2^1H-1^2,2^7-\text{bibenzoimidazole}$
However, Chemdraw says the name is:
$1H,3'H-2,4'-\text{bibenzo[d]imidazole}$
I get the differences between the prime and composite locants (P-14.3.1), however it shows a numbering more like:
[](https://i.stack.imgur.com/Fz03T.jpg)
Is Chemdraw wrong or am I missing something here? If Chemdraw is right then I'd assume that for some reason the ring assembly has preference for a lower locant, but I can't find the rule that states this in the Blue Book.
| 3
|
[
[
"\nYour understanding of the numbering of the individual cyclic systems is correct. If we consider the following arbitrarily simplified separate compounds, the numbering is in accordance with your expectations; low locants are given first to the indicated hydrogen and then to the substituent according to the usual rule P-14.4.\n\n\n \n\n\nIn ring assemblies of two identical cyclic systems that are joined with a single bond, however, low locants are given first to the junction according to Rule P-28.2.1.\n\n\n\n> \n> **P-28.2.1** Ring assemblies with a single bond junction\n> \n> \n> (…)\n> \n> \n> Each cyclic system is numbered in the traditional way, one with unprimed locants, the other with primed locants. Lowest possible locants must be used to denote the positions of attachment. These locants must be cited in preferred IUPAC names; they can be omitted in general nomenclature when no ambiguity would result. \n> \n> (…)\n> \n> \n> \n\n\nAfter that, indicated hydrogen is named and numbered according to P-28.2.3.\n\n\nTherefore, the systematic name is **1*H*,3'*H*-2,4'-bibenzimidazole** and not 1*H*,1'*H*-2,7'-bibenzimidazole since the locants 2,4' for the junction are lower than 2,7'.\n\n\n\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/166387/how-to-synthesize-sodium-metabisulfite
|
How to synthesize sodium metabisulfite?
|
I am looking for ways to produce sodium bisulfite from $\ce{SO2}$ and $\ce{NaOH}$, but sources talking about $\ce{Na2S2O5}$ is quite scarce.
Do you have any recommendations of articles, books or any other type of source?
| -2
|
[
[
"\nWikipedia is a good start.\n\n\nWhen you treat a solution of suitable base, say sodium hydroxide or sodium bicarbonate with sulfur dioxide, you get sodium bisulfite.\n\n\n$$\\ce{SO2 + NaOH → NaHSO3}$$\n$$\\ce{SO2 + NaHCO3 → NaHSO3 + CO2}$$\n\n\nNow the same reaction can also lead to sodium metabisulfite. When the reaction is conducted in warm water, $\\ce{Na2SO3}$ initially precipitates as a yellow solid but when you apply more $\\ce{SO2}$, the solid dissolves and on cooling, it crystallizes to sodium metabisulfite.\n\n\n$$\\ce{SO2 + 2 NaOH → Na2SO3 + H2O}$$\n$$\\ce{SO2 + Na2SO3 → Na2S2O5}$$\n\n\nAlternatively, if you have heaps of sodium bisulfite, [you can simply heat it to get sodium metabisulfite.](https://chemistry.stackexchange.com/a/17324/17368)\n\n\nYou may refer to this book: Johnstone, H. F. (**1946**). \"*Sulfites and Pyrosulfites of the Alkali Metals*\". Inorganic Syntheses pt .2 pp. 162–167. doi:[10.1002/9780470132333.ch49](https://onlinelibrary.wiley.com/doi/10.1002/9780470132333.ch49)\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/166379/how-the-optical-fibre-ph-microsensors-work
|
How the optical fibre pH microsensors work?
|
Just browsed through the chemical sensor products of the company, called PreSens (<https://www.presens.de/products/ph/sensors>) in order to look for a precise method to measure pH change in a physiological solution (Dulbecco's Phosphate Buffer Saline 1M).
I just would like to know how the "Sensor Spot" and "Needle type" optical sensors work exactly. According to the brochure on the aforementioned page, the Sensor Spots seem to based on fluorescence quenching: basically a quencher (I believe the hydrogen ions or the hydronium ions? I'm not sure) reduce the detected fluorescence intensity of the fluorophore (some pH dependent dye incorporated into a membrane "spot").
I would like to know more about this mechanism, I leafed through numerous reviews, but I have not be able to find any detailed information about pH detection based on fluorescence quenching. What is the quencher here? Could you tell me what is the mechanism in this particular case?
Thank you for your time.
| 1
|
[] |
https://chemistry.stackexchange.com/questions/166375/do-you-always-need-to-balance-a-chemical-equation-to-find-the-percentage-yield-a
|
Do you always need to balance a chemical equation to find the percentage yield and limiting reagent?
|
[](https://i.stack.imgur.com/G46Vv.png)
I'm doing some problems through MIT open courseware and I've got one I'm absolutely stuck on. It details a three-step synthesis for hyperzine A, gives molecular weights and diagrams for individual reactants and the formula for the final result (huperzine A).
Then it asks you to find the limiting reagent for the first reaction. Every single source I've been able to find on this says that you need to balance the equation in order to do this, but the reactants seem so complicated I would have no idea how to begin doing this. It seems like we've got C10H16O(s) reacting with C8H5O4S2F6N(aq), but the result seems to be C11H15O3S1F3 (sorry, I have no idea how to format). They don't mention any other products, so my most obvious problem is that I have no idea what happens to the nitrogen, and even getting past that I have no idea how you would balance such a complicated reaction to work out mole ratios.
Question a asks about the limiting reagent, and you are only told about how much product is produced in question b, so I'm assuming to answer question a you don't need that information.
Furthermore, question b asks about percentage yield, and again, every source I've been able to find has stated you need to know the chemical equation (which I'm interpreting to mean 'balanced') to get this.
Since the only discussion of this problem I've seen online declares it 'easy' without any further explanation, I must be missing something big, but I've been bashing my head at this for many hours and have absolutely no idea where I'm going wrong. I've even accidentally spoiled myself on the answers at this point, but they haven't helped me to get any further with it.
| 0
|
[] |
https://chemistry.stackexchange.com/questions/166374/could-perchlorate-salts-allow-for-underwater-combustion
|
Could perchlorate salts allow for underwater combustion?
|
So apparently perchlorate salts can act as oxidizers for combustion (basically they can replace oxygen in the fire triangle, at least I think), and the reason why you can't have a fire underwater is because there's not enough oxygen gas in the water. So if one were to mix/dissolve perchlorate salts with a liquid fuel and made it hot, could the perchlorates allow it to combust?
| 0
|
[] |
https://chemistry.stackexchange.com/questions/166372/does-solid-sodium-hydroxide-react-with-carbon-dioxide-and-air-moisture
|
Does solid sodium hydroxide react with carbon dioxide and air moisture?
|
I heard a claim in the context of soapmaking that a stray grain of sodium hydroxide will decay into soda ash, i.e. sodium carbonate, on contact with air. This reaction happens in [an aqueous solution](https://chemistry.stackexchange.com/questions/64570/would-sodium-carbonate-from-carbon-dioxide-plus-hydroxide-be-formed-in-solution), but I'm wondering how fast it would happen with only the air moisture absorbed by the hydroxide grain.
More to the point, some of the sodium hydroxide grains I have for soapmaking seems to have stuck together into a solid lump by absorbing air moisture inside the closed container. Should I assume a nontrivial fraction of that has turned into sodium carbonate by now?
| 0
|
[] |
https://chemistry.stackexchange.com/questions/166371/use-of-boroniii-oxide-for-drying-acetone
|
Use of boron(III) oxide for drying acetone
|
Acetone undergoes [aldol self-condensation](https://en.wikipedia.org/wiki/Self-condensation) in the presence of acids or bases:
[](https://i.stack.imgur.com/mjUsC.png)
which makes drying agents such as 3A/4A molecular sieves, barium oxide, potassium hydroxide and carbonate (basic), copper(II) sulfate or aluminium(III) oxide (acidic) less effective and promotes formation of mesityl oxide to a significant degree.
Boron(III) oxide is considered as one of the most appropriate desiccants for acetone [[1, pp. 3967–3968](https://doi.org/10.1021/jo00414a038)]:
>
> As with $\ce{Me2SO},$ the root of the difficulty is the acidic α protons, which in this
> case compounds the drying problem not only by inflating apparent water content by exchange process but also by providing a pathway to self-condensation through enol intermediates. This facet of acetone chemistry makes the choice of a successful desiccant a delicate process. As Table IV shows, mild siccatives such as calcium sulfate are inept; more potent desiccants such as molecular sieves exhibit a short initial drying action but thereafter actually cause disastrous increases in water content by displacement of the condensation equilibrium. This interpretation was confirmed for molecular sieves and other basic desiccants such as barium oxide by gas chromatographic analysis which demonstrated the presence of mesityl oxide in the dried solvent (see Table IV).
>
>
> In summary, while both cupric sulfate and 3A molecular sieves are clearly at least useful preliminary desiccants, the agent *par excellence* for acetone is powdered boric anhydride.
>
>
> [](https://i.stack.imgur.com/oHGVn.png)
>
>
>
However, $\ce{B2O3}$ is a Lewis acid itself and reacts with water to produce boric acid, which should build up upon drying the solvent.
Why doesn't $\ce{B2O3}$ appear to be promoting aldol condensation, at least in the case of acetone?
### Reference
1. Burfield, D. R.; Smithers, R. H. Desiccant Efficiency in Solvent Drying. 3. Dipolar Aprotic Solvents. *J. Org. Chem.* **1978**, 43 (20), 3966–3968. DOI: [10.1021/jo00414a038](https://doi.org/10.1021/jo00414a038).
| 3
|
[] |
https://chemistry.stackexchange.com/questions/166369/boiling-and-evaporation
|
Boiling and Evaporation [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
---
**Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)
Closed last year.
[Improve this question](/posts/166369/edit)
A closed tank is filled with water completely and heated from outside. What will occur boiling or evaporation or both?
| -2
|
[
[
"\nIf you fill a tank completely with liquid water (no gas bubble) and start heating it, the liquid water will try to expand faster than the vessel since the coefficient of thermal expansion of water is larger than the corresponding value for typical vessel materials (e.g. steel). Thus the pressure will increase sharply so that no evaporation can occur.\n\n\nBecause of the strong pressure increase, the vessel will probably fail before reaching 100 °C. When the vessel fails, the pressure will quickly drop to ambient pressure. If the temperature is still below 100 °C, no boiling can occur; there will be just slow evaporation from the surface of the warm water.\n\n\nIf the temperature is already above 100 °C when the vessel fails, some flash evaporation will occur when the pressure quickly drops to ambient pressure. A part of the hot liquid water flashes to steam (which can be calculated using the enthalpy balance). The temperature drops to the boiling point (100 °C) at the new pressure. After that, there will be slow evaporation from the surface of the warm water.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/166362/what-does-it-mean-that-a-state-belongin-to-a-given-irrep-transforms-like-rx
|
What does it mean that a state belongin to a given irrep transforms like $Rx$, $Ry$ or $Rz$
|
The present question is related to [this other question](https://chemistry.stackexchange.com/q/166089/125277) I did few days ago.
Given a point group and the list of the irreps (see for example [here](https://www.webqc.org/symmetrypointgroup-oh.html)) the meaning of an irrep which transforms like $x$ or $x^2$ is clear to me.
Instead I do not understand what is meant when it is written that an irrep transforms like $Rx$.
Let me explain a bit better. Say that the group is the symmetry group o a crystal structure. It can be represented as a set of $3\times3$ matricies $M\_i$. These matrices can be used to transform the vector $v=(x,y,z)$. So I can clearly see how something which transforms like x would be affected: $v'=M\_iv$ and $v'=(x',y',z')$.
I can also assume that any function of $f(x,y,z)$, like $f(x,y,z)=x^2$ would be transformed like $f(x,y,z) \rightarrow f(x',y',z')$ (is this point true?). However I do not understand how $Rx$ would transform. It is not even a defined function, but a group of operations by itself.
| 1
|
[
[
"\n$R\\_x$ etc. are rotations about the indicated axis. See for example [this article in chem.libretexts](https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Symmetry_(Vallance)/01%3A_Chapters/1.14%3A_Character_Tables) or [this pdf](https://instruct.uwo.ca/chemistry/734b/Group%20Theory-Part%204%20Irreducible%20Representations%20and%20Character%20Tables.pdf)\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/166361/does-rubbing-alcohol-bleach-really-produce-chloroform
|
Does Rubbing Alcohol + Bleach really produce Chloroform?
|
A lot of sources on the internet claim that mixing rubbing alcohol with chlorinated bleach produces chloroform.
Rubbing alcohol is Isopropanol, and bleach is Sodium Hypochlorite. Neither of these are the reagents for the [Haloform reaction](https://en.wikipedia.org/wiki/Haloform_reaction). So I ask, is chloroform really produced, and if so, how?
| 1
|
[
[
"\nYes, rubbing alcohol and bleach could produce chloroform when mixed, if the alcohol is isopropyl-type rubbing alcohol and the bleach is hypochlorite-type bleach. Isopropyl alcohol is indeed a substrate for the haloform reaction, according to Fuson and Bull, *Chem. Rev.*, **15**, 275 (1934) and references therein.\n\n\nI can't recall the reference off the top of my head, but I believe in \"dilute\" solutions approaching \"bleach and alcohol\" the chlorination tends to continue and produces carbon tetrachloride as the main product.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/166359/does-the-boiling-point-of-ammonia-hydroxide-change-with-the-ratio-of-water-to-am
|
Does the boiling point of ammonia hydroxide change with the ratio of water to ammonia?
|
My kids and I are designing a science experiment and we could use help with the ratio of water to ammonia in an ammonia hydroxide mixture.
A little bit of background:
Initial goal: To convert the ammonia hydroxide into gas through solar / ambient heat with enough pressure to inflate a balloon.
The how: We have a metal tank that is painted black. The balloon is wrapped around the outlet of the tank at the top of the tank. We put the ammonia hydroxide in the tank when cool outside, then wait for the day to warm (we live in the high desert) and ammonia to turn into vapor.
Initial tests will fill the balloon but does not generate enough pressure to expand the rubber/plastic to properly inflate it.
Our idea, and the genesis of this question, is that if we use a higher concentration of ammonia we will a) boil faster at lower temperatures and b) might generate more vapor and create pressure to inflate the balloon.
Apologize in advance if this question is incomplete. Happy to provide more information. And thanks for your help!
| -1
|
[
[
"\nYour idea of increasing the ammonia concentration to increase the total vapor pressure over the solution is physically sound. For two-component mixtures, the total pressure will be a function of both solution composition and temperature. In the case of ammonia and water, since ammonia is the more volatile component (it is a gas at standard conditions) increasing its concentration will lead to a higher vapor pressure for the solution.\n\n\nAlthough the data are over 100 years old, a paper by Perman (*J. Chem. Soc.*, **83**, 1168 (1903)) gives some relevant information. Below is a table with the vapor pressures he measured for ammonia solutions at various temperatures and ammonia concentrations. You should be able to calculate the pressure ratios you're interested in from these data.\n\n\n[](https://i.stack.imgur.com/wRLua.png)\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/166357/how-does-the-stoichiometric-coefficient-affect-the-time-until-a-given-percentage
|
How does the stoichiometric coefficient affect the time until a given percentage of the reactant is consumed (if at all)?
|
### Problem from Serway's College Physics test bank:
---
The reaction $2NO\_2\rightarrow2NO+O\_2$ obeys the rate law:
$\large \frac{\Delta[O\_2]}{\Delta t} = (1.40\times10^{-2})[NO\_2]^2 \space at \space 500^\circ \space K $
If the initial concentration of $NO\_2$ is 1.00 M, how long will it take for the [$NO\_2$] to decrease to 35.8% of its initial value?
A) 45.9 s
B) 73 s
C) 128 s
D) 1.40 x 10-2 s
E) cannot be determined from this data
**ANSWER: C**
---
### My work
We are given the reaction rate for $O\_2$. Since for a given reaction $aA + bB \rightarrow cC + dD$, the rate of the reaction $r = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}$, I conclude that $\frac{\Delta[O\_2]}{\Delta t} = -\frac{1}{2}\frac{\Delta[NO\_2]}{\Delta t} \Rightarrow -\frac{\Delta[NO\_2]}{\Delta t} = 2\frac{\Delta[O\_2]}{\Delta t}$. Therefore:
$-\frac{\Delta[NO\_2]}{\Delta t} = 2 \times (1.40\times10^{-2})[NO\_2]^2 = (2.8\times10^{-2})[NO\_2]^2 \Rightarrow$
$\textbf k = 2.8\times10^{-2}$
Knowning that the reaction is second-order in $NO\_2$, I plug in $[NO\_2] = 0.358[NO\_2]\_0$ into the integrated rate law and solve for t as so:
$\frac{1}{0.358[NO\_2]\_0} = kt + \frac{1}{[NO\_2]\_0}$
$\frac{1}{0.358\times1M} - \frac{1}{1M} = 2.8\times10^{-2}\times t$
$t \approx 64 s$
---
### Answer given in the test bank
However, the textbook gives 128s as the answer, indicating that the reaction rate was NOT multiplied by 2 (from $\frac{\Delta[O\_2]}{\Delta t}$ to $-\frac{\Delta[NO\_2]}{\Delta t}$).
Is there a gap in my understanding of conversion between reaction rates according to the stoichiometric coefficient of the reactant/product being handled or did the test bank simply give an incorrect answer?
| -1
|
[
[
"\nYou should not multiply the rate constant by an stoichiometric coefficient.\n\n\nThe rate of this chemical reactionis given by\n\n\n$$ v = -\\frac{1}{2}\\frac{d|NO2\\_2|}{{dt}} = +\\frac{1}{2}\\frac{d|NO|}{{dt}} = +\\frac{1}{1}\\frac{d|O2\\_2|}{{dt}} = k |NO\\_2|^2 $$\n\n\nwhere k is the reaction rate constant. In your exercise, $k = 1,4\\times10^{-2} \\;mol^{-1}\\,dm^3\\,s^{-1}$.\n\n\nNotice that the stoichimetric coefficients appear in the reaction rate definition, thus the reaction rate does not depend on which reactant or product used to obtain it.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/166355/what-are-the-other-examples-of-compounds-undergoing-decarboxylation-in-the-peric
|
What are the other examples of compounds undergoing decarboxylation in the pericylic manner as shown by β-ketocarboxylic acids?
|
I know that β-ketocarboxylic acids undergo decarboxylation on light heating through an enol intermediate undergoing a pericyclic reaction. However, what other groups may be placed at the β position that may give the same decarboxylation reaction?
I haven't been able to find any reputable sources. However, my teacher gave us a question in which an imine at β position underwent decarboxylation following the same process as a β-ketocarboxylic acid would. Does this reaction happen and are there other groups that follow the exact decarboxylation mechanism when at the β position?
| 3
|
[] |
https://chemistry.stackexchange.com/questions/166348/can-polarizing-power-and-the-inductive-effect-both-explain-the-highly-covalent-b
|
Can Polarizing Power and the Inductive Effect both explain the highly covalent bonds of Mn2O7?
|
The Mn-O bonds of Mn2O7 have more covalent character than those of MnO2. This makes sense when imagining Mn2O7 to consist of Mn7+ ions and O2- ions; the high charge of the cations give them a high polarizing power, leading to the Mn-O bonds having high covalent character.
However, in nature, bonds are not "ionic with covalent character" or "covalent with ionic character," they are simply bonds. Therefore, I am curious if it is possible to explain the difference in the covalent character of Mn2O7 and MnO2 when viewing the bonds as covalent to start. In this case, is the explanation for the covalent bonds that the Mn atoms gain a high positive charge due to the highly electronegative oxygens, allowing the Mn to pull electrons back in? In other words, the oxygens inductively pull on each other's electrons by making the manganese more positively charged, leading to all of the electrons being more evenly shared between the manganese and oxygens?
The inductive effect and polarizing power explanations both seem to be describing the same phenomenon: the manganese atoms in Mn2O7 have a high positive charge (when viewing Mn2O7 as ionic, that charge is the +7 oxidation state; when viewing Mn2O7 as covalent, that charge is the partial positive charge due to the oxygens), causing them to pull electrons better, leading to highly covalent bonds with the oxygens. In general, can we say that highly positively charged atoms in a molecule pull electrons better (and therefore have a higher electronegativity), whether we are measuring that charge by the atom's oxidation state or by looking at the charge due to polar covalent bonds? It would make sense if this were true, because oxidation states represent the charges if all bonds are ionic, which is the extreme of polar bonds. Even if an atom's charge isn't as high as its oxidation state, I imagine that a higher oxidation state still corresponds to a higher actual charge.
On a side note, why do we only talk about polarizing power in the context of ionic bonding (or at least why have I only seen it used in the context of ionic bonding, such as in Fajan's Rules)? Can we talk about polarizing power of partial charges created due to electronegativity differences? For example, in the ketone RCOR', could we say that the carbonyl carbon has some polarizing power due to its partial positive charge, causing it to withdraw electrons from the R and R' groups? I think this is simply a way of describing the inductive effect. Maybe the reason we never talk about polarizing power in this kind of context because it is simpler to sum up the effect as "the carbonyl oxygen withdraws electrons from nearby atoms through the inductive effect?"
| -2
|
[
[
"\nYour understanding of the inductive effect and polarising power/polarisability is a bit inaccurate in some places.\n\n\nThe inductive effect is a result of the polarisation of sigma bonds (often as a result of differences in electronegativity.) This is first and foremost a *covalent effect*. For example, acyl chlorides are more reactive than aldehydes because the electronegative chlorine atoms pull electron density from the electrophilic carbonyl carbon, resulting in a stronger electrophile. The implicit fact that is not stated here is that this electron-withdrawing effect is through a sigma bond.\n\n\nOn the other hand, Fajan's Rule is a result of a cation with sufficiently high polarising power (high charge density) meeting an anion with sufficiently high polarisability (huge polarisable electron cloud often expressed as an anion of low charge density). Notice instead of dealing with the donation of electron density through a sigma bond where usually the discussion of electronegativity comes into play, Fajan's rule has little or no relation to electronegativity but rather the actual charge of the various species.\n\n\nSo, coming back to the question of Manganese (VII) Oxide and Manganese (IV) Oxide. From the view of inductive effects, the strength of the \"covalent\" bonds in these two species should inherently be the same due to the same difference in electronegativity. However, by applying Fajan's rule, it is obvious that because of the significantly higher polarising power of the Manganese (VII) species, Manganese (VII) Oxide has more covalent character.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/166347/when-does-the-angle-between-the-planes-in-tetrahedral-molecule-with-central-atom
|
When does the angle between the planes in tetrahedral molecule with central atom deviate from 90°?
|
Consider an atom X bonded to four other atoms A, B, C, D in a tetrahedral fashion with sp³ hybridisation. If A, B, C, D are the same, every bond angle is 109.47°, and the angle between the planes formed by joining nuclei of X, A, B and X, C, D is exactly 90° (treat nuclei as point sized since they are very small with respect to the atoms) if pairs A, B and C, D are identical elements (symmetrical molecules like methane or dichloromethane):
[](https://i.stack.imgur.com/TnkJW.png)
If A, B, C, D are different, then the bond angle will be different to minimise repulsion and maximize stability as per VSEPR theory. But what happens to the angle between the planes? My intuition says that it will always be 90° in any molecule and the reason is that *repulsions are minimised in that case*.
However, this reasoning is only a qualitative one, but I wish that this could be mathematically proved or in some other way using some very good logic. In the following figure you can see that due to decrease in angle between bonding planes, B and D come close to each other and also A and C come close to each other. So they will face steric repulsion from each other leading to instability:
[](https://i.stack.imgur.com/wGN7P.png)
In case the angle between the planes is not always 90°, what are these exceptions? In organic chemistry, many kind of exceptions frequently occur, so I was trying to think of some case where steric hindrance or torsional strain or angle strain or ring strain or some other reason may cause this angle to change.
The boat configuration of cyclohexane came into mind. I knew that there is flagpole interaction between two hydrogen atoms which get very close to each other. Maybe this repulsion may cause hydrogen atoms to deviate a little to the sides, causing *a little* change in angle between bonding planes in this example. This could be the first exception, but I am not very sure of it.
Also, such repulsion can happen in two ways: clockwise and counterclockwise rotation of the planes. For example, in the above figure a mirror image can be formed if B comes closer to C and D comes closer to A. Both of these compounds will be mirror image of each other and therefore may create optical activity. This optical activity if it happens can prove to be a practical confirmatory test to be sure weather this thinking is correct or not.
If I am correct, are there better examples where the angle between bonding planes is different from 90°?
| 0
|
[] |
https://chemistry.stackexchange.com/questions/166343/how-to-keep-kappa-carrageenan-lasting-at-room-temperature-for-weeks-or-months
|
how to keep kappa-carrageenan lasting at room temperature for weeks or months? (art project)
|
I'm helping an artist make an outdoor gelly structure (not for eating) and we want to keep it from molding. The temperatures are around 10-30 degrees celsius and it's in Copenhagen, Denmark in September, and hopefully, it can last for some weeks or more.
Right now we are experimenting with kappa carrageenan with a mix of CaCl2 and we are going to add locust bean gum, pectin, and sugar to see how it behaves.
But what would we add for conservation? Sodium benzoate? and is there a limit on how much to add? (we want it as transparent as possible)
We don't wanna use some too harmful chemical since we gonna have to get rid of it and work with it.
Thanks, any help is most appreciated. Have a nice day.
Resources I found so far:
In this paper, they say they can make it last 7 months at room temperature.
The paper mentions Potassium makes it last longer:
<https://idosi.org/wjdfs/wjdfs6(2)/18.pdf>
food exchange:
<https://cooking.stackexchange.com/questions/57752/how-much-citric-acid-to-use-to-preserve-vegetarian-jelly>
general booklet on caragenan:
[http://www.bisi.cz/cmsres.axd/get/cms$7CVwRhc3USVqgzxkKF96gI$2BChNrXcTq$2BOUdiEtz5TfYA$2Fg1ADRHMfXfdEjUsYQagqUs9N6byPOkok$3D](http://www.bisi.cz/cmsres.axd/get/cms%7Bb914bcb4-5f6b-4e51-9c8d-2da1ca475172%7D2BChNrXcTq%7B666e580c-4d58-4ef1-9a6d-cd3f5700b6fe%7D2Fg1ADRHMfXfdEjUsYQagqUs9N6byPOkok$3D)
| 1
|
[] |
https://chemistry.stackexchange.com/questions/166340/how-to-rationalize-independence-of-half-life-time-from-the-initial-concentration
|
How to rationalize independence of half-life time from the initial concentration for the first order reaction?
|
Using the rate expression for the first order kinetics and expressing the half-life time, it can be proven the half-life time $t\_\frac 1 2$ of the first order reaction is independent of its initial concentration:
$$t\_\frac 1 2 = \frac{\ln 2}{k}$$
Can we *intuitively* explain that the first order half-life time is independent of its starting concentration and remains constant *without* the aid of the derived expression for the half-life time said above?
For the zeroth order reaction, it can intuitively be explained, but for the first order reaction it seems a little hard.
| -1
|
[
[
"\nThe basic assumption is that the probability of reaction of a given molecule (or disintegration of a radioactive atom) in a given time interval is independent of past history, and depends only on the time interval. If it is assumed that the time interval we choose is sufficiently small the chance of reaction is *proportional* to this time interval and the exponential decay law can be derived from this. The constant of proportionality is called the rate constant $k$. These assumptions mean that we cannot predict exactly when a given molecule will react but because of the proportionality (the rate constant) we can after *many measurements* know the form of the decay. Of course we assume that all the molecules of a given type are identical and behave in the same way and independently of one another.\n\n\nWe can measure the average time our type of molecule takes to decay by making making many repeated measurements and this average is the reciprocal of the rate constant and is related to the half life as $\\ln(2)/k$.\n\n\n",
"3"
],
[
"\nA particle has some probability to decay (Radioactive decays are ideal illustrative examples.) or react during some time interval. With more particles, the count of particles is directly proportional to particle count that will do so during that interval.\n\n\nTherefore, the relative portion of particles reacting during that period does not depend on the amount nor concentration (assuming very high particle numbers converging to matter continuity). When the time interval leads to a portion being the half, it is the half life time.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/166337/is-the-glaser-coupling-reaction-possible-with-metal-alkyne-complexes
|
Is the Glaser coupling reaction possible with metal alkyne complexes?
|
As the Glaser coupling reaction is usually done with terminal alkynes with the hydrogen being straight, I wonder if the same reaction can occur in terminal alkynes connected to the backbone of a metal alkyne complexe.
As an example, in the trimerization reaction of acetylene, which happens with the formation of a metal alkyne complete; the hydrogens are bent in a v-shape to the carbon; not straight. [Alkyne Trimerisation](https://upload.wikimedia.org/wikipedia/commons/thumb/a/a6/SimplifiedMechTrimeriz.png/1024px-SimplifiedMechTrimeriz.png).
Would it be possible? If it weren’t the case, what would impede this reaction?
| 4
|
[
[
"\nThe hydrogen being straight corresponds to its sigma bond with carbon having more carbon $2s$ character and less carbon $2p$ character, which makes the covalent overlap somewhat less and the bond more polar. Thus the hydrogen is susceptible to deprotonation by the strong base that initiates the reaction.\n\n\nIf the bond angle becomes bent due to the electronic contributions from the metal, we have lost some $2s$ character in the carbon-hydrogen bond in favor of more $2p$, and the required deprotonation could be inhibited.\n\n\nNote that in the alkyne trimerization shown in the question, the first two acetylene molecules are effectively reduced before the third one displaces the metal. Such a reduction is not seen in the Glaser coupling.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/166336/what-is-the-necessary-and-sufficient-condition-for-a-mixture-to-be-ideal
|
What is the necessary and sufficient condition for a mixture to be ideal?
|
I am trying to understand some concepts from solution thermodynamics related to ideal mixtures and fugacity. **My main question is what is the necessary and sufficient condition for a mixture to be an ideal mixture?** Below is my attempt to answer this and some related follow up questions.
The fugacity of component $i$ in a mixture is typically defined by two requirements:
1. $f\_{i}(T,p,\{x\})=p\_{i}\exp\left(\frac{1}{RT}\left[\mu\_{i}(T,p,\{x\})-\mu\_{i}^{IGM}(T,p,\{x\})\right]\right)$
2. $\lim\_\limits{p \to 0}\left(\frac{f\_{i}(T,p,\{x\})}{p\_{i}}\right)=1$
Where IGM is an ideal gas mixture. Using the first part of the definition, we can rewrite the second condition as:
$$\lim\_\limits{p \to 0}\left( \mu\_{i}(T,p,\{x\})-\mu\_{i}^{IGM}(T,p,\{x\}) \right)=0$$
From this we can derive an expression for fugacity that we can calculate explicitly:
$$f\_{i}(T,p,\{x\})=p\_{i}\exp\left(\frac{1}{RT} \lim\_\limits{p’ \to 0} \int\_{p’}^{p} \left[ \bar{V}\_{i}(T,p,\{x\}) - \frac{RT}{p} \right] dp \right)$$
And from this we can show:
$$\frac{f\_{i}(T,p,\{x\})}{x\_{i} f\_{i}(T,p)}=\exp\left(\frac{1}{RT} \lim\_\limits{p’ \to 0} \int\_{p’}^{p} \left[ \bar{V}\_{i}(T,p,\{x\}) - \bar{V}\_{i}(T,p) \right] dp \right)$$
An ideal mixture is defined as a mixture where the chemical potential for any component $i$ is given by:
$$\mu\_{i}^{IM}(T,p,\{x\})=\mu\_{i}(T,p)+RT\ln x\_{i}$$
All other properties of an ideal mixture follow from this definition, since the chemical potential is equal to the partial molar Gibbs free energy. One key property is that $\bar{V}\_{i}(T,p,\{x\}) = \bar{V}\_{i}(T,p)$ at all conditions in an ideal mixture. That is, the volume change of mixing is zero. In turn this means the integrand above is always zero and the expression reduces to the Lewis-Randall rule:
$$f\_{i}^{IM}(T,p,\{x\})=x\_{i}f\_{i}(T,p)$$
From the first part of the definition of fugacity we can also express the chemical potential as:
$$\mu\_{i}(T,p,\{x\})=\mu\_{i}(T,p) +RT \ln \left( \frac{f\_{i}(T,p,\{x\})}{f\_{i}(T,p)} \right)$$
Then if we substitute the LR rule we obtain the definition of an ideal mixture:
$$\mu\_{i}^{IM}(T,p,\{x\})=\mu\_{i}(T,p) +RT \ln x\_{i}$$
In summary, the two requirements below seem to ensure that a mixture is ideal:
1. $\bar{V}\_{i}(T,p,\{x\}) = \bar{V}\_{i}(T,p)$
2. $\lim\_\limits{p \to 0}\left( \mu\_{i}(T,p,\{x\})-\mu\_{i}^{IGM}(T,p,\{x\}) \right)=0$
I have two follow up questions:
* Is this actually true? Does the volume condition always ensure the mixture is ideal? Or are there ideal mixtures where the volume of mixing is nonzero (i.e. $\bar{V}\_{i}(T,p,\{x\}) \ne \bar{V}\_{i}(T,p)$)?
* Is the logic of this argument backwards? I don’t understand the
mathematical reason why we are free to define fugacity with its
second condition. If we didn’t assume that then the volume condition
would not be enough to derive the the chemical potential definition
of an ideal mixture. Is the fugacity limit requirement actually some
kind of consistency condition due to the definition of an ideal
mixture?
| 1
|
[] |
https://chemistry.stackexchange.com/questions/166332/why-is-reported-effective-pore-size-smaller-that-the-one-declared-in-the-name-of
|
Why is reported effective pore size smaller that the one declared in the name of the molecular sieve type X?
|
The [molecular sieve](https://en.wikipedia.org/wiki/Molecular_sieve) type is often coded with `<number><letter>` combination, where `<number>` specifies the pore size in ångströms and `<letter>` refers to the Linde type of zeolite. For sieve types 3A, 4A and 5A the reported pore sizes ubiquitously match the code, i.e. being 3 Å, 4 Å and 5 Å, respectively, with virtually no discrepancy neither on the manufactures' websites nor in the literature.
However, the majority of the companies selling 13X type sieves across both EU and US ([ADCOA](https://www.adcoa.net/product/item-ms13x00-molec-sieves-type-13x-custom-packaged/), [VWR](https://uk.vwr.com/store/product/2338947/molecular-sieve-13x-1-0-nm-10-a-pellets-oe-3-2-mm-flukatm), [GeeJay Chemicals Ltd.](https://www.geejaychemicals.co.uk/molecular-sieves), [Lawrence Industries](https://www.l-i.co.uk/products/molecular-sieve-air-gas-drying-zeolite), [MTE Group](https://www.mte-process.com/mte-catalyst-support/molecular-sieves/), …) report nominal pore size 10 Å instead of 13 Å. Analogously, the spec sheets for the mol sieves type 10X report the pore diameter 8 Å instead of expected 10 Å.
Surprisingly, there wasn't any mismatch in older literature and patents [1, 2], and the 13X sieves were claimed to have 13 Å pores, which seems logical. An editorial remark of a similar nature has been made for the 2006 paper [[3](https://doi.org/10.1063/1.2272835)] pointing out the inconsistency between expected pore size for 13X zeolite and the pore size reported by Sigma-Aldrich which authors used in their work. Based on discrepancy of the data presented in the literature at that time, the authors suggested that
>
> microscopic structure of commercially produced 13X may not be fully independent of the sample preparation procedure
>
>
>
and from the rest I draw a conclusion that they infer that commercially available sieves of 10X and 13X types could be [faujasite](https://en.wikipedia.org/wiki/Faujasite), zeolite of Linde type X and synthetic zeolites of *similar* composition, which could deviate from one manufacturer to another. I am not finding this completely convincing since the reported pore size doesn't fluctuate around one number, it's consistently understated by 2 Å and 3 Å for 10X and 13X sieves, respectively.
What exactly is going on here ~~and where can I get a refund for my missing 3 ångströms~~? Were the initial structures of the X type zeolites and, accordingly, the pore dimensions, incorrectly determined? Or did the manufacturers changed their product and continued to ship it using the former name?
### References
1. Meyer, J. R. *Method for Producing Exceptionally Pure Hydrogen*. [US3011589A](https://patents.google.com/patent/US3011589A/en), December 5, **1961**.
2. Read, P. L. *Sorption Vacuum Pump*. [US3264803A](https://patents.google.com/patent/US3264803A/en), August 9, **1966**.
3. Swenson, J.; Jansson, H.; Howells, W. S.; Longeville, S. *Reply to “Comment on ‘Dynamics of Water in a Molecular Sieve by Quasielastic Neutron Scattering’ ” [J. Chem. Phys. 125, 077101 (2006)]*. *J. Chem. Phys.* **2006**, 125 (7), 077102. DOI: [10.1063/1.2272835](https://doi.org/10.1063/1.2272835).
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https://chemistry.stackexchange.com/questions/166331/is-bents-rule-an-alternative-explanation-for-the-inductive-effect
|
Is Bent's Rule an alternative explanation for the Inductive Effect?
|
Today I read that Bent's Rule can be used to explain the inductive effect (I was reading the Consequences section [of this Wikipedia article](https://en.wikipedia.org/wiki/Bent%27s_rule)). If I understand it correctly, for a molecule CL3R, the more electronegative the substituents (denoted L) are, the more s character can be saved for the hybrid orbital that the C uses for the C-R bond. Higher s character for the hybrid orbital means lower energy, so when the C-R bond forms, the resulting molecular orbital has more electron density around the carbon. This means that the carbon withdraws electrons better, so the electron-withdrawing property of the substituents has been transmitted to the carbon.
However, when I originally learned about the inductive effect, I understood it through a Coulombic argument: the three electronegative substituents withdraw electrons from the carbon, creating a partial positive charge on the carbon, which gives it a better pull on the electrons in the C-R bond (in other words a higher electronegativity).
Are Bent's Rule and Coulomb's Law simply two separate ways of justifying the inductive effect, or should they be seen as working together to produce the inductive effect?
And of course, are these both just ways of simplifying the molecular orbital explanation of the inductive effect?
| 3
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https://chemistry.stackexchange.com/questions/166329/vinylic-nucleophilic-substitution
|
Vinylic Nucleophilic Substitution [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/166329/edit).
Closed last year.
[Improve this question](/posts/166329/edit)
While reading "Francis A. Carey, Richard J. Sundberg - Advanced Organic Chemistry Part A. Structure and Mechanisms-Springer (2007)", I came across the following schematic for addition of electrophiles to vinyl ethers:
[](https://i.stack.imgur.com/WwNEW.png)
I understand that such a reaction has application in polymer chemistry. For example, under the right acidic conditions, methylvinylether will form polyvinylmethylether:
[](https://i.stack.imgur.com/9mTYY.png)
Under what conditions will the electrophile react with the lone-pair of oxygen instead of the $\beta$-Carbon?
In another section, I found that hard-hard and soft-soft interactions are preferred over hard-soft interactions:
[](https://i.stack.imgur.com/CpqMK.png)
Many questions arise from reading this. Is the lone pair on the oxygen atom hard or soft? And accordingly, what kind of electrophile should be used? Do stearic factors play a role?
It has been pointed out in the comments that addition of electrophile to oxygen is a reversible and non-productive. What I was thinking was something like this:
[](https://i.stack.imgur.com/XMutB.png)
I guess the reaction could be called a **vinylic nucleophilic substitution**.
| 1
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[] |
https://chemistry.stackexchange.com/questions/166327/does-no-optical-rotation-always-implies-optical-inactivity
|
Does no optical rotation always implies optical inactivity? [duplicate]
|
**This question already has an answer here**:
[Are there chiral compounds that don't rotate plane-polarized light?](/questions/103048/are-there-chiral-compounds-that-dont-rotate-plane-polarized-light)
(1 answer)
Closed last year.
This question popped up in my mind in reference to this question,
[Is a compound optically active if plane polarised light is deflected by an angle of n\*(2π) angles?](https://chemistry.stackexchange.com/q/166324/124759)
Suppose that I give a chemistry exam in future in which a question is asked like this,
>
> A pure organic compound(not a racemic mixture) was analysed once and it was found that that there was absolutely zero rotation of PPL when passed through it(not even slight rotation as in cryptochirality). This means that it must be an optically active compound.
>
>
>
Now there are two options whether the above statement is true or false. What will be the answer? Assume that the organic compound is pure ie it is not a mixture (not even a racemic mixture). Also assume that the enantiomers (if possible) are separable which means that there is nothing like amine inversion or any other similar mechanism through which enantiomers can interconvert into each other.
Though it might be quite intuitive that the answer should be false but think about the person who set this question. (It has quite often happened with me that even after having the correct logic, my answer is consider wrong by majority.)
| -1
|
[
[
"\n\n> \n> Now there are two options whether the above statement is true or false. What will be the answer?\n> \n> \n> \n\n\nI choose to *false*. We have two cases for no rotation of plane polarized light. One of the case is that the compound is optically inactive. And the other case is when the analyte is a [racemic mixture](https://en.m.wikipedia.org/wiki/Racemic_mixture).\n\n\nWikipedia states that,\n\n\n\n> \n> A racemic mixture, or racemate, is one that has equal amounts of left- and right-handed enantiomers of a chiral molecule.\n> \n> \n> \n\n\nSo the rotation of plane polarized light is a combined effect of nature of compound and nature of analyte. So no rotation *doesn't implies that the compound is optically active*.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/166324/is-a-compound-optically-active-if-plane-polarised-light-is-deflected-by-an-angle
|
Is a compound optically active if plane polarised light is deflected by an angle of n*(2π) angles? [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
---
**Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)
Closed last year.
[Improve this question](/posts/166324/edit)
Is a compound optically active if plane polarised light is deflected by an angle of n\*(2π) angles(like 360°); as there won't be any deflection in the analyser of the polarimeter?
| -1
|
[
[
"\n**Yes**\n\n\nThe actual rotation you observe in a polarimeter depends on several factors: the path length, the concentration (if in solution) and the specific amount of optical rotation caused by the compound.\n\n\nIf you observe the actual rotation at a single concentration you might observe no *apparent* effect because the rotation was 360°. But at a different concentration or a different path length with the same substance you *would* observe a rotation.\n\n\nSo it is pretty easy to avoid the mistake of assuming no rotation by simply taking more than one measurement while varying the other factors.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/166321/how-to-rewrite-kinetic-mass-balance-to-get-to-proper-si-units
|
How to rewrite kinetic mass balance to get to proper SI units?
|
Suppose from kinetic mass balance I get the following differential equation of the molarity substances $A$,$B$, and $C$ in a reactor, with molar in- or outflow rate $\phi$, and rate constant $k$:
$$
\frac{\mathrm{d}}{\mathrm{d}t}[C]
= k[A]^2[B]^3 + \phi{C\_{\mathrm{in}}} - \phi{C\_{\mathrm{out}}}
$$
Where the dimensions convert to:
$$
[\pu{mol L-1 s-1}]
= [\pu{s-1}] [\pu{mol5 L-5}] + [\pu{mol s-1}] - [\pu{mol s-1}]
$$
Which can be simplified by adding the flow rate terms to:
$$
[\pu{mol L-1 s-1}]
= [\pu{s-1}] [\pu{mol5 L-5}] \pm [\pu{mol s-1}]
$$
By cancelling similar terms mole,seconds on both sides to:
$$[\pu{L-1}] = [\pu{mol4 L-5}] \pm 1$$
This seems that something is off about the rate constant, but also that there is no $\pu{L-1}$ term in the molar flow rate.
I don't expect exact answers to this, but could someone point me the right way, or send some clear explanation on either reaction rate constants or on kinetic mass balances? It seems I cannot find clear explanations, not even in textbooks.
| 0
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[] |
https://chemistry.stackexchange.com/questions/166319/does-e1-mechanism-always-imply-first-order-reaction
|
Does E1 mechanism always imply first order reaction?
|
[Oxford University Press, Okuyama & Maskill: Organic Chemistry — Chapter 13: Multiple Choice Questions, Question 1](https://global.oup.com/uk/orc/chemistry/okuyama/student/mcqs/ch13/):
>
> Which of the following statements regarding the E1 mechanism is wrong?
>
>
> **a**) Reactions by the E1 mechanism are unimolecular in the rate-determining step.
>
> **b**) Reactions by the E1 mechanism are generally first order.
>
> **c**) Reactions by the E1 mechanism usually occur in one step.
>
> **d**) Reactions by the E1 mechanism are multi-step reactions.
>
>
>
Option **c** is the correct answer, but I have a few questions regarding option **b** being true:
1. Is the reaction which follows E1 mechanism *always* a first order reaction? Are there any examples with the second or higher orders?
2. Dehydration of secondary alcohol in the presence of concentrated sulfuric acid follows E1. What is the order for this reaction?
3. Does the number one in E1 represent molecularity or order of the reaction? According to me, it is usually molecularity of RDS, not the order, because option **b** says E1 reactions are *generally* 1st order, not *always*.
| -2
|
[
[
"\nI think the problem lies in definition of terminologies. There are various kinds of reaction which occur in multiple steps, but just because one of the elementary step reaction follows SN1 or E1, it does not mean the whole reaction will have the first order.\n\n\nSame thing here. The cation formed when alcohol attacks on $\\ce{H+}$ undergoes SN1 and the rate will be only proportional to concentration of this cation. Further, the concentration of this cation is proportional to both hydrogen ion concentration and alcohol concentration.\n\n\nIn conclusion, E1 is always the first order reaction, but the dehydration of alcohols is the second order reaction (or pseudo first order if proton is abundant) because water expulsion is just one of the elementary steps happening in the reaction.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/166315/are-there-any-chemistry-research-classification-systems
|
Are there any chemistry research classification systems?
|
There are research classification systems for mathematics ([MSC](https://en.wikipedia.org/wiki/Mathematics_Subject_Classification)) and physics ([PACS](https://en.wikipedia.org/wiki/Physics_and_Astronomy_Classification_Scheme) and subsequently [PhySH](https://en.wikipedia.org/wiki/PhySH)) which help the researchers to navigate through codes composed by letters/numbers made by associations that specialise in those pieces of human knowledge. Are there any classification systems of research in chemistry?
| 1
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[] |
https://chemistry.stackexchange.com/questions/166313/what-does-it-mean-that-the-standard-molar-entropy-value-is-the-amount-of-energy
|
What does it mean that the standard molar entropy value is the amount of energy that a substance must have to exist at a certain temperature?
|
What does it mean that the standard molar entropy value is the amount of energy that a substance must have to exist at a certain temperature?
According to this reference: <http://khimiya.org/volume15/Entropia.pdf> the meaning of standard molar entropy is the relative amount of energy that the substance must have to exist at for example 298 Kelvin.
An example is given of diamond and graphite, which has 2.4 J/K and 5.7 J/K respectively. I see that graphite has less strong bonds than diamond, i.e. diamond has higher potential energy stored inside its bonds. But I don't understand how that translates into a lower amount of energy needed to exist at a certain temperature? Is this the energy that needs to be supplied to the diamoned to raise its temperature to 298 kelvin? But isn't it than just a kind of heat capacity? What happens if you did not transfer enough energy to make it exist stably at that temperature?
>
> 1. The meaning of standard molar entropy values
> When seen in their relation to the energy content of a substance, standard molar entropy
> values, $S^0\_{298}$, give useful insight that is absent when - as is usual -
> those J/K values are treated as merely abstract numbers to be added
> or subtracted in determining a $dS\_0$ reaction . A $S^0\_{298}$ value for a
> substance is the number of joules of energy/T transferred
> incrementally (reversibly, from the surroundings at each T) to a mole
> of substance from 0 K to 298 K. Thus, this number is a rough indicator
> or approximate index (not the joules dispersed at 298 K, nor the total
> joules dispersed from 0 K to 298 K!) of the relative amount of energy
> that the substance must have to be exist stably at 298 K. This is why
> $S^0\_{298}$ values illuminate e.g., the difference in rigidity of bonding:
> the more rigid bonds of diamond (2.4 J/K) vs. the looser interatomic
> bonds in graphite (5.7 J/K) ,
>
>
>
| 0
|
[
[
"\nThe wording of the cited paragraph is confusing, but it does not say that \"the standard molar entropy value is the amount of energy that a substance must have to exist at a certain temperature\" (as claimed in the question). In fact the author says specifically that this is not the case. Rather, it is \"is a rough indicator or approximate index\" of the amount of energy.\n\n\nMathematically, the change in internal energy of a substance as it is heated (at constant volume to avoid PV work) from absolute zero to a temperature T would be $$\\Delta U = \\int\\_0^T C\\_V dT,$$ where $C\\_V$ is the heat capacity at constant volume and is a function of *T*. For one mole of substance, we would use the molar heat capacity $\\overline{C}\\_V$.\n\n\nThis is similar to but not the same as the expression for absolute entropy, which is $$S^\\circ\\_T = \\int\\_0^T \\dfrac{C\\_P}{T} dT,$$\nwith the crucial difference being the *T* in the denominator. (The difference between $C\\_P$ and $C\\_V$ is of secondary importance.)\n\n\nAs long as the change in the heat capacity as a function of temperature is similar between two substances, then the relative ranking based on absolute entropy will be similar to a ranking of the internal energy, which is what the author means when he says that the former can be used as \"a rough indicator or approximate index\" of the latter. But clearly they are by no means the same and I would argue that it's somewhat misleading to suggest that the entropy should be used in this way.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/166305/iupac-name-for-dicarboxylic-acid-with-aldehyde-and-amine-groups
|
IUPAC name for dicarboxylic acid with aldehyde and amine groups
|
>
> The IUPAC name of the structure is:
>
>
> [](https://i.stack.imgur.com/sQvO5.png)
>
>
> (**A**) 3-amino-2-formyl butane-1, 4-dioic. Acid
>
> (**B**) 3-amino-2, 3-dicarboxy propanal
>
> (**C**) 2-amino-3-formyl butane-1, 4-dioic acid.
>
> (**D**) 1-amino-2-formyl succinic acid
>
>
>
Why is the correct answer option **C** and not option **A**? Can't we also name this compound by IUPAC nomenclature as 2-formyl-3-amino butane-1, 4 dioic acid? If we take the right carboxylic acid functional group COOH as 1, then number the carbon atoms, then CHO takes 2, so formyl is 2, right? Also, doesn't formyl have greater seniority over amino?
| 1
|
[
[
"\nThe preferred name of the parent structure is **butanedioic acid**. The name ‘succinic acid’ is retained for general nomenclature with functionalization but no substitution is allowed (see Rule P-65.1.1.2.2); thus, it cannot be used in this case.\n\n\nThe terminal locants ‘1,4’ of butanedioic acid are not cited.\n\n\n\n> \n> **P-14.3.4.1** Terminal locants are not cited in names for mono- and dicarboxylic acids derived from acyclic hydrocarbons and their corresponding acyl halides, amides, hydrazides, nitriles, aldehydes, amidines, amidrazones, hydrazidines, and amidoximes, when unsubstituted or substituted on carbon atoms.\n> \n> \n> \n\n\nSimple substituent groups that are named by means of prefixes (such as ‘amino’ and ‘formyl’) are arranged alphabetically.\n\n\n\n> \n> **P-14.5** ALPHANUMERICAL ORDER\n> \n> \n> Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved\n> \n> \n> Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.\n> \n> \n> (…)\n> \n> \n> **P-14.5.1** Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.\n> \n> \n> \n\n\nTherefore, the correct alphanumerical order for the compound given in the question is ***x*-amino-*y*-formylbutanedioic acid** (not ‘*x*-formyl-*y*-aminobutanedioic acid’).\n\n\nThe locants *x* and *y* are used to indicate positions of the parent structure at which modifications represented by suffixes occur. The relevant rules concerning the numbering of locants for substituent prefixes are:\n\n\n\n> \n> **P-14.3.5** Lowest set of locants\n> \n> \n> The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; (…)\n> \n> \n> \n\n\nand\n\n\n\n> \n> **P-14.4** NUMBERING\n> \n> \n> When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:\n> \n> \n> (…)\n> \n> \n> (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;\n> \n> \n> (g) lowest locants for the substituent cited first as a prefix in the name;\n> \n> \n> (…)\n> \n> \n> \n\n\nIn accordance with Rule (f), the compound given in the question could be named as ‘2-amino-3-formylbutanedioic acid’ acid as well as ‘3-amino-2-formylbutanedioic acid’ since the locant set ‘2,3’ is the same in both cases. However, according to Rule (g), this example is named as **2-amino-3-formylbutanedioic acid** rather than ‘3-amino-2-formylbutanedioic acid’ since amino is cited first as a prefix in the name.\n\n\n\n",
"8"
]
] |
https://chemistry.stackexchange.com/questions/166301/can-glycerin-crack-a-glass-when-heated
|
Can glycerin crack a glass when heated? [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/166301/edit).
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[Improve this question](/posts/166301/edit)
Is it safe to put glycerin in double wall glass candle ( between the walls ) for decor purposes. I know they put it in vapes and double glass wall coffee tumblers. Is there a possibility of it to crack the either inner wall( the one that will be heated by the wax first ) or outside one if heated by candle burning? In condition if chamber with glycerin is closed .Thank you !
| -4
|
[
[
"\nNot chemically, but from usual, thermal or mechanical reasons.\n\n\nDirect flame contact or frequent or regular thermal shocks can crack glass. Heating resistant glass significantly decreases the risk, but does not eliminate it. The prolonged history of heating, together with tiny surface damages, increases the risk.\n\n\nHeated liquids can crack glass, if kept in closed container. Either directly by thermal dilation, either indirectly by vapour pressure. Minor water content can cause significant pressure if heated enough.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/166300/confusing-statement-about-the-pauli-exclusion-principle
|
Confusing statement about the Pauli exclusion principle
|
While reading "Francis A. Carey, Richard J. Sundberg - Advanced Organic Chemistry Part A. Structure and Mechanisms-Springer (2007)", I came across the following:
[](https://i.stack.imgur.com/J8Cq9.png)
The paragraph says that there are two major reasons for electron pairs repelling each other.
1. Electrostatic repulsion
2. Pauli exclusion principle
Now, the statement says that, in accordance with the Pauli exclusion principle, only two electron can occupy the same point in space and that those two electrons must have opposite spins. However, as far as I know, that is not the statement of Pauli exclusion principle, or is it equivalent to the following statement?
>
> No two electrons can have the same four quantum numbers.
>
>
>
This is a tricky question, since we really shouldn't be talking about electrons occupying a single point in space (given the uncertainty principle). Also, wouldn't electrostatic repulsion be enough to assert that no two electrons can occupy the same point in space regardless of the spin?
| 0
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[
[
"\nAlthough this is not a common way to represent the Pauli exclusion principle, it is indeed one consequence of it. Specifically, we start with the statement that a system of identical electrons must be antisymmetric with respect to any exchange of the electrons, which is the actual postulate that Pauli proposed.[The derivation of this bit is quite complicated and comes from relativistic quantum field theory.] That is, $$\\psi(q\\_1, q\\_2, . . . q\\_n)=-\\psi(q\\_2, q\\_1,. . . . q\\_n)$$ where each \"$q\\_i$\" is an electron described by a given set of quantum numbers.\n\n\nAlthough we don't consider electrons to be localized at specific points in space, we can consider the value of their wavefunctions as a function of three spatial coordinates and a spin coordinate, i.e. $x,y,z$ and $m\\_s$ or $\\theta, \\phi, r$ and $m\\_s$. If these are equal for $q\\_1$ and $q\\_2$ (ie two particles have the same spatial location and spin), we would have that $$\\psi(q\\_1, q\\_1, . . . q\\_n)=-\\psi(q\\_1, q\\_1,. . . . q\\_n)$$ which is only true if $\\psi =0$, which means that the wave function associated with this set of electrons has zero probability of being in a state where two electrons of the same spin also have the same spatial coordinates.\n\n\nThere's a more complete description of this in Levine's Quantum chemistry book in chapter 10.\n\n\nUPDATE based on comments: Since $\\psi$ is a continous function, the fact that it has zero amplitude when two electrons have the same spatial and spin values also means that the amplitude approaches zero as two electrons get closer together, which is a way of saying that the probability (amplitude squared) of the system existing in a state where two electrons are close together (but not in the same position) is low and approaches zero as the electrons get closer together (because it necessarily equals zero for the state where they are at the same point). We describe this result as \"repulsion\" of the electrons from one another.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/166298/isolation-is-a-consequence-of-definition
|
Isolation is a Consequence of Definition [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
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Closed last year.
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Let's consider a ballon, such that it has an adiabatic wall, and there is no exchange of heat with the surroundings. Now, let the balloon wall be completely impermeable to the movement of matter across the boundary. If so, then the ballon is theoretically an isolated system. However, if we stretch the balloon, to increase its volume, then the temperature of the air inside the balloon will decrease. Will that mean work is done by the system on the surroundings, despite it being isolated?
Here I thus feel that the concept of isolation is actually a consequence and highly bound by definition. The balloon continues to remain isolated unless, some work is done on it, leading to it losing its property of isolation if some work is done on it.
Is my argument correct?
| -2
|
[
[
"\nFor a nice definition of an isolated system in thermodynamics see [the Wikipedia article on isolated systems](https://en.wikipedia.org/wiki/Isolated_system). The headline is:\n\n\n\n> \n> In physical science, an isolated system is either of the following:\n> \n> \n> * a physical system so far removed from other systems that it does not interact with them.\n> * a thermodynamic system enclosed by rigid immovable walls through which neither mass nor energy can pass.\n> \n> \n> \n\n\nYour balloon is not an *isolated system* if it can do work on its surrounding or have work done on it by its surroundings.\n\n\n",
"6"
]
] |
https://chemistry.stackexchange.com/questions/166296/calculation-of-canonical-orthogonalized-basis-from-overlap-matrix
|
Calculation of canonical orthogonalized basis from overlap matrix?
|
From Equation [3.169] in David Cook's [Book](https://xn--webducation-dbb.com/wp-content/uploads/2020/11/Oxford-Science-Publications-David-B.-Cook-Handbook-of-computational-quantum-chemistry-Oxford-University-Press-USA-1998.pdf) , I understand that we use eigenvalues and eigenvectors of the overlap matrix to calculate an orthonormal matrix using the formula $$X=Us^{-1/2}$$, where $U$ is the matrix of eigenvectors and $s$ is the matrix of eigenvalues for the overlap matrix. But, when I tried to calculate the dot product between two vectors from the new matrix $X$, I do not get $\phi\_i\phi\_j=1$ or $0$ depending on if $i=j$ or $i\neq j$. Does this mean that I have to normalize $X$?
| -1
|
[
[
"\nIt was not a requirement that *X* had to be orthogonal. The only requirement was that $X^{\\dagger}SX$ must be orthonormal.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/166295/is-there-an-equivalent-term-for-favourable-and-non-favourable-entropy
|
Is there an equivalent term for favourable and non-favourable entropy?
|
For Gibbs energy, we have 'exergonic' or 'endergonic'
For Enthalpy, we have 'exothermic' or 'endothermic'
It seems logical there should be an equivalent pair of terms for entropy, but I can't seem to find them. Any help would be appreciated; if I'm just blind and haven't seen the terms apologies!
| 1
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[] |
https://chemistry.stackexchange.com/questions/166294/confusing-concentration-calculation-about-vivianite-solubility
|
Confusing concentration calculation about vivianite solubility
|
I'm confused about the results of [PO4] and [Fe(II)] calculated in [a paper](https://aslopubs.onlinelibrary.wiley.com/doi/pdf/10.4319/lo.2003.48.2.0929). The example they posted is shown in the figure below:
[](https://i.stack.imgur.com/BbcLF.png)
I tried to calculate the phosphate and Fe(II) concentration. The calculation process is below:
$\ce{ Fe3(PO4)2 8H2O <=> 3Fe^2+ + 2PO4^3- + 8H2O}$
Let's set the concentration of $\ce{Fe^2+}$ as 3x mol/L, concentration of $\ce{PO4^3-}$ as 2x mol/L, then:
(3x)^3\*(2x)^2 = 10^-36, which makes x = 2.47$\*$10^-8
So concentration of $\ce{Fe^2+}$ is 3x = 7.42$\*$10^-8 mol/L, concentration of $\ce{PO4^3-}$ is 2x = 4.95$\*$10^-8 mol/L.
However, the paper said that Fe(II) and P concentration is 1.53$\*$10^-5 mol/L and 1.02$\*$10^-5 mol/L, respectively.
Here are my two questions:
(1) Why my calculation is not the same (even in the magnitude) as the paper's results?
(2) Why do they mention "pH = 7", I couldn't find where is this "pH" take part in the solubiity calcculation.
| 1
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[] |
https://chemistry.stackexchange.com/questions/166287/is-there-a-liquid-with-refractive-index-1-5-to-1-6-that-is-substitute-to-bromofo
|
Is there a liquid with refractive index 1.5 to 1.6 that is substitute to bromoform?
|
So I am tasked with making my countries standards sustainable, wherever possible and trying to address any of the 17 SDGs. I came across a test in a certain material specification code that uses microscope and bromoform of AR or GR grade for measuring glass content in the material. USEPA classifies bromoform as possible human carcinogen and I was wondering if it is just used for its' refractive index properties, we could substitute it with safer materials. I found that correct proportions of glucose or sucrose can give up to 1.4 to 1.45 refractive indices. I was wondering if there was a liquid with 1.6 to 1.7 which could be a substitute to bromoform?
Rest physical conditions remain normal as in Temperature is around 27C and material should be transparent and operate in Visible light spectrum etc.
Thanks in advance for all the efforts and help!
| 2
|
[
[
"\nIf you can make do with a a refractive index of ~1.5, look at [immersion oils](https://en.wikipedia.org/wiki/Oil_immersion) used for microscopy. Since they are widely used in laboratories without restrictions such as a fume hood, they're likely nontoxic, though consult the chemical safety data sheets (SDS) for the particular ones of interest.\n\n\nVarious companies offer higher nD liquids. For example, Cargille has sets of liquids ranging from 1.3 to **1.8**. The [SDS for Cargille Refractive Index Liquid Series A](https://cargille.com/wp-content/uploads/2018/06/Cargille-Refractive-Index-Liquid-Series-A-nD1-571-1-640-USA.pdf), which includes 1.6 nD, states, \"IARC (International Agency for Research on Cancer): None of the ingredients are listed.\" That said, if you have questions on toxicity, contact the manufacturer for more specific information.\n\n\n",
"7"
]
] |
https://chemistry.stackexchange.com/questions/166286/can-we-make-chemical-bonds-using-light-instead-of-breaking-them
|
Can we make chemical bonds using light, instead of breaking them?
|
I am trying to bridge a conceptual understanding of how light affects both breaking and forming bonds. When bonds are broken, energy is absorbed (endothermic process). Conversely, when bonds are made, energy is released (exothermic process).
We can calculate the necessary wavelength that is required to break bonds using the equation $\lambda=\hbar c/E.$ For example, light with a wavelength of about $\pu{241 nm}$ is needed to break the bonds of a $\ce{O2}$ molecule (assuming you know the bond dissociation energy).
Is it possible to use a specific wavelength to get the atoms to bond again? If so, is there an equation to calculate this for the atoms involved? I’ve been reading that lasers are used to cool atoms which leads me to think that since atoms are absorbing and reemitting the energy from the lasers to lower its kinetic energy, bonds could be formed when the it's low enough.
All I have been finding is how to break bonds using light. Is there light of a specific frequency that could be used to form a bond between given atoms in proximity?
| 3
|
[
[
"\nSurely, if light can break bonds, it can make bonds as well.\n\n\nWe find some simple examples in high school textbooks as well, such as the chlorination of olefin compounds under the presence of UV light. This is an example of a photo-initiation reaction, where the first step is dependent on light. You can find more about this in this article1. Here is a schematic from the same article:\n\n\n[](https://i.stack.imgur.com/jIywg.png)\n\n\nDo note that halogenation often occurs as a substitution reaction $\\alpha$ to the double-bond. Under special circumstances (e.g., absence of $\\alpha$-$\\ce{H}$), halogenation can be made to occur as an addition reaction at the double-bond as well.\n\n\nWe can also consider some more interesting examples. Following reaction is from this study2\n\n\n[](https://i.stack.imgur.com/VEEjI.png)\n\n\nAnother schematic from the same article:\n\n\n[](https://i.stack.imgur.com/QaWye.png)\n\n\nThis concept is called photo-dimerization3:\n\n\n\n> \n> The photodimerization is an example of a direct photoreaction where every step for polymer build-up is initiated by an absorbed photon, thus every single reaction step is dependent on the quantum yield of the photoreaction (generally very much smaller than one).\n> \n> \n> \n\n\n### References\n\n\n1. Schönberg, A. (**1968**). Photohalogenation. In: Preparative Organic Photochemistry. Springer, Berlin, Heidelberg. DOI: [10.1007/978-3-642-87918-0\\_37](https://link.springer.com/chapter/10.1007/978-3-642-87918-0_37)\n2. \"Light- and heat-triggered reversible luminescent\nmaterials based on polysiloxanes with anthracene groups\" by Han et.al. RSC Adv., **2017,7**, 56489-56495, DOI: [10.1039/c7ra12201b](https://pubs.rsc.org/en/content/articlehtml/2017/ra/c7ra12201b)\n3. *The Encyclopedia of Materials : Science and Technology.* by K.H.J Buschow, **2001**, Pages 6946-6951\n\n\n",
"5"
],
[
"\n\n> \n> Is it possible to use a specific wavelength to get the atoms to bond again?\n> \n> \n> \n\n\nNot directly. The crucial step in photochemistry is the absorption of a photon. Usually, this results in a molecule in an excited electronic state. Depending on the molecule, this will make a bond weaker, or break a bond.\n\n\nAn example of a simple (the simplest) molecule is the hydrogen molecular ion, $\\ce{H2+}$. It has a single electron, making things so simple you can actually calculate the electronic states from scratch. If you excite the electron, the molecule will fall apart. You can read more about it on [this page](http://www2.mpq.mpg.de/%7Ehaensch/h2+/introduction.htm).\n\n\n\n> \n> I am trying to bridge a conceptual understanding of how light affects both breaking and forming bonds. When bonds are broken, energy is absorbed (endothermic process). Conversely, when bonds are made, energy is released (exothermic process).\n> \n> \n> \n\n\nAs Poutnik states in the comments to the question, making a bond is exothermic, so adding energy through photons is not productive for bond formation. However, for more complex photoreactions, absorbance of a photon is often the first step.\n\n\nThe most famous examples are cyclo-additions, or [electrocyclic reactions](https://en.wikipedia.org/wiki/Electrocyclic_reaction). Some work without light (often indicated by \"$\\Delta$\" for just heating up) while others require photoactivation (often indicated by \"ℎ$ \\nu$\" for light with appropriate wavelength).\n\n\n[](https://i.stack.imgur.com/FyGXw.png)\n\n\nIn the photochemical reaction sketched above, the absorbed photon results in a electronic excitation that does not break the molecule, but changes in electronic symmetry, allowing the reaction to happen. The product in the example above also is in an excited electronic state.\n\n\n\n> \n> Is there light of a specific frequency that could be used to form a bond between given atoms in proximity?\n> \n> \n> \n\n\nYes, but not in a single-step reaction where the bond \"is made by the light\", like welding two metal pieces together. Instead, using the welding analogy, you partially melt one work piece, and then attach it to the second work piece in a separate step that would not happen without the first step. Like all analogies, this does not capture the whole story, of course.\n\n\n\n> \n> Is it possible to use a specific wavelength to get the atoms to bond again?\n> \n> \n> \n\n\nIf the bond-breaking reaction is photochemical and there is an intermediate where the product is electronically excited before you get the ground state product, you can try to excite it to that state again to get the bond-forming reaction. This is non-equilibrium chemistry, so you can choose the wavelength (either exciting the reactant or the product) to manipulate the direction of the reaction away from the equilibrium concentrations, like the example in the answer by ananta.\n\n\n",
"5"
],
[
"\nSuch a reaction would likely be indirect, with the bond formed by an intermediate that the electromagnetic radiation produces.\n\n\nOne of the simplest examples of such a reaction involves the noble gas helium. When helium gas is exposed to an electrical discharge, the electromagnetic radiation in this discharge ionizes some of the atoms, which then combine with still-neutral helium atoms to form the $\\ce{He\\_2^+}$ cation (see [Wikipedia](https://en.wikipedia.org/wiki/Helium_dimer)). This is a true covalently bonded species, with bond order 1/2, because two of the valence electrons are in the bonding orbital and only one is in an antibonding orbital.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/166285/how-does-one-derive-a-kd-from-an-equilibrium-titration-experiment-i-am-definite
|
How does one derive a KD from an equilibrium titration experiment? I am definitely making a mistake somewhere
|
Any help would be **greatly** appreciated. Thank you in advance.
If I have an antibody A and a target B, and experimentally titrate the antibody against a single concentration of B, and then measure the % of B that is bound after the solutions reach equilibrium, I should be able to determine the KD of the interaction.
Source below (and others) shows that determining KD is as simple as determining the EC50 of such a curve. from:<https://www.sciencedirect.com/topics/nursing-and-health-professions/maximum-binding-capacity> 50% occupancy
[](https://i.stack.imgur.com/OBvmg.png)
How does that square with the following equation for KD?
KD = ([A]\*[B]) / [AB] where A is unbound antibody, B is unbound antigen, and AB is the bound complex
Let's say I have 10 pM of target antigen, 50% of which is bound when incubated to equilibrium with 10 pM of antibody. That would mean the antibody and antigen are each 50% bound and 50% unbound.
KD = 5pM \* 5pM / 5pM KD = 5 pM
So I have calculated a 5 pM KD, not the 10 pM that I would think would be similar to the graph below (assuming they had pM units on the X axis) I'm assuming these types of example graphs refer to the input concentration, right?
Even more confusing, what if the antibody is a really tight binder, such that 10pM of antibody is able to essentially bind 100%, such that 10 pM antibody is able to bind 50% of 20pM of antigen. (1:1 binding assumed)
KD = ~0.001pM \* ~10pM / ~10pM
Now you have a KD of 0.001pM, but the graph would suggest that 10 pM of antibody is necessary to bind to 50% of the target antigen.
I'm sure I'm missing something. If anyone could help it would be greatly appreciated.
| 1
|
[
[
"\n\n> \n> So I have calculated a 5 pM KD, not the 10 pM that I would think would be similar to the graph below (assuming they had pM units on the X axis) I'm assuming these types of example graphs refer to the input concentration, right?\n> \n> \n> \n\n\nYour calculation is correct. The concentration in the graph refers to the free concentration of ligand of the receptor. The receptor will be present at very low concentration, and the ligand is often cheap (if it is an approved drug), so it is fine to use it in large excess. In those cases, free ligand concentration is similar to total ligand concentration. However, the graph could be labelled more accurately.\n\n\n\n> \n> Let's say I have 10 pM of target antigen, 50% of which is bound when incubated to equilibrium with 10 pM of antibody. That would mean the antibody and antigen are each 50% bound and 50% unbound.\n> \n> \n> \n\n\nIn the case of antibody and antigen, both binding partners might be expensive, so using them at a 1:1 concentration ratio might make sense. In this case, you would calculate the free concentration like you did. So when you have equal concentrations of complex and free antibody, the concentration of free antigen is equal to the dissociation constant. Or when you have equal concentrations of complex and free antigen, the concentration of free antibody is equal to the dissociation constant. In your example, that would be 5 pM.\n\n\n### The graphs\n\n\nThe graph has a couple of issues. There should be a unit on the concentration axis (the y-axis is a percentage of full binding, so that is correct without units, but should have a percentage sign). The concentration axis should be labeled as \"free ligand concentration\". Often, as in the graph below, the free ligand concentration is simply written as $[\\mathrm{L}]$. If you want to specify the initial (or total) concentration, the symbol you would use is $[\\mathrm{L}]\\_\\mathrm{total}$.\n\n\n\n> \n> [](https://i.stack.imgur.com/Vd64a.png)\n> \n> \n> \n\n\nThe shape of the graph looks different because the concentration axis is linear, not logarithmic as in the graph posted in the question. The source for the graph and explanation is [http://people.reed.edu/~glasfeld/Chem391/notes/ChLigand\\_2015.pdf](http://people.reed.edu/%7Eglasfeld/Chem391/notes/ChLigand_2015.pdf).\n\n\nIf you google \"graph occupancy ligand concentration\", you will find that many graphs do not carefully distinguish between free ligand concentration and total ligand concentration. This is probably because these two concentration are quite similar when the ligand is in large excess over the binding protein. There is also the occasional confusion between dissociation and association constant, so you have to use your critical thinking skills when coming across figures in the wild (and elsewhere).\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/166280/reaction-of-alkane-with-alkene-in-acidic-medium
|
Reaction of alkane with alkene in acidic medium
|
[](https://i.stack.imgur.com/8BUdu.jpg)
This question was asked in jee mains 2022 june 25 shift 1. I have never seen such a kind of reaction before and also could not find it anywhere on internet. But I guessed that the proton will add to primary carbon of the double bond and form tert-butyl carbocation while isobutane will release proton to form isobutyl carbanion. They will then add to each other to form the product D. This is why I marked it.
However the released answer key shows that option B is correct. One possible thought that came in my mind was that bulky bases like tertbutoxide ion make Hoffman product major because they can't attack middle carbon due to steric hindrance. Same case happen here also with nucleophile. So, instead primary carbocation would form and get attacked to form that product in option B.
However I am not sure of myself. My query is not regarding this specific question but asking in general wether such kind of reactions with alkane and alkene actually happen? What is their mechanism? And most importantly how to decide wether a given base or nucleophile is bulky or not?
| 5
|
[
[
"\nThis reaction is actually called as\n\"alkylation of alkenes\". If I had known *this* term, I could have directly searched it on internet but won't have asked it on SE ;)\n\n\n<https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/10%3A_Alkenes_and_Alkynes_I_-_Ionic_and_Radical_Addition_Reactions/10.10%3A_Alkylation_of_Alkenes>\n\n\nThis website shows that according to the scientist P.D. Bartlett, a carbocation can react with a hydrocarbon with tertiary carbon by transferring hydride (H-). This is reminiscent of the familiar cannizaro reaction where also this hydride transfer takes place. A chain reaction occurs leading to formation of tertbutyl cation adding with isobutene to form a carbocation which then extracts hydride from isobutane, and then the cycle continues.\n\n\nI have found similar explanation by coaching channels on YouTube, like this one.(watch from 17:33)\n<https://youtu.be/T6TUROPO4eM>\nThough it is in hindi,but you can still see the structures\n\n\nThe catalyst required could be any acid like sulphuric or hydrofluoric acid etc but this website says that this particular reaction can also occur without catalyst at high temperatures (and is just lesser feasible economically). [https://www.sciencedirect.com/topics/physics-and-astronomy/alkylation#:~:text=Conventional%20paraffin%20(alkane)%E2%80%93olefin,yielding%202%2C3%2Ddimethylbutane](https://www.sciencedirect.com/topics/physics-and-astronomy/alkylation#:%7E:text=Conventional%20paraffin%20(alkane)%E2%80%93olefin,yielding%202%2C3%2Ddimethylbutane).\n\n\nAs expected there is no carbanion formation. The reaction mechanism is therefore quite similar to free radical halogenation in the sense that there is an initiation step and then propagation step although there is no radical involved.\n\n\n",
"7"
]
] |
https://chemistry.stackexchange.com/questions/166277/effect-of-beta-branching-and-carbocation-rearrangement-on-sn1-reaction-rates
|
Effect of beta branching and carbocation rearrangement on SN1 reaction rates
|
My book (NCERT class 12 Chemistry) asks
>
> Predict the order of reactivity of the four isomeric bromobutanes in SN1 and SN2 reactions.
>
>
>
The answer given for SN1 is
>
> 1-bromobutane < 1-bromo-2-methylpropane < 2-bromobutane < 2-bromo-2-methylpropane
>
>
>
My confusion is between (I) 1-bromobutane and (II) 1-bromo-2-methylpropane. The book explains
>
> the carbocation derivative derived from (II) is more stable than derived from (I) because of a **greater electron donating inductive effect** of (CH3)2CH- group.
>
>
>
However, I have been taught that **hyperconjugation** is (usually) a stronger effect than induction. Since the carbocation from (I) would have a greater number of alpha hydrogens, that should be more stable.
However, the carbocation in (II) can rearrange to form a tertiary carbocation, whereas (I) can only rearrange to secondary. How significant will rearrangement be when considering the reaction rates? As an additional example, how will reaction rates of 2-bromopentane and 1-bromo-3-methylbutane (where tertiary carbocation is formed after two rearrangements) compare?
| 1
|
[] |
https://chemistry.stackexchange.com/questions/166273/full-list-of-cas-numbers
|
Full list of CAS numbers?
|
I'm trying to find a full list of CAS Registry Numbers, but it looks like it's not in open access.
Is it a proprietary database that can't be downloaded easily?
| 1
|
[] |
https://chemistry.stackexchange.com/questions/166272/is-elongation-a-measure-of-plastic-deformation
|
Is elongation a measure of plastic deformation?
|
I did a tensile test of a metallic alloy, and I want to calculate it's elongation. Would it simply be the (failure strain - yield strain)?
For reference, I got the following values from the test:
Yield strain: 0.015
Yield stress: 404 MPa
Failure strain: 0.22
Failure stress: 592 MPa
Would the elongation simply be 0.22-0.015 = 0.205
0.205 \* 100 = 20.5%? I.e. simply the amount of strain that was done plastically?
Thanks in advance!
| -1
|
[
[
"\nI measured elongation and reduction of area on many bars, shapes, materials. Never calculated it, never heard on anyone calculating elongation. The elongation is the primary measure of ductility. The tensile machine graph puts out a strain which is interesting , but we always measured the bars to be sure. That also give you a look at the fracture surface which can also be informative. What are the units of your 0.015 strain; % ?, mm ? ( inches) of what gage length? Also various specifications have different definitions of yield.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/166266/how-to-neutralize-bleach-in-a-washing-machine
|
How to neutralize bleach in a washing machine
|
i’ve seen a couple threads on this website about neutralizing bleach but none of them really say how to do it.
for some context, someone put 7 cups of bleach in our washing machine along with the water without any clothes inside because they wanted to disinfect the washing machine from a specific pathogen. They talked to Clorox on the phone and they said it was okay to use more bleach but now they’re double backing and saying that you weren’t supposed to use more than half a cup.
i’ve done multiple rinse cycles sense in order to wash away any residue but were nervous that because it was so much bleach that the cycles aren’t effective or that we need to do more and will never know exactly when it’s OK to use again.
So I just want to nip it in the bud and neutralize whatever residue that’s in there. I’ve seen people say to use Hydrogen peroxide but I obviously don’t want to cause any accidents if there are residue in there. I’ve seen a different product but I don’t know if it’s safe to use in the washing machine or how much to use to neutralize 7 cups of bleach.
This experience has been extremely stressful, I don’t want clothing to have residue bleach on it and possibly put pets and people in danger.
\*\*\* addition from og post\*\*\* I’ve read online that vinegar is not a good one i don’t know if that because of the possible reaction or because it’s not a neutralizer. Also does any one know of a way to test for residue?
Any advice would greatly be appreciated
| 0
|
[] |
https://chemistry.stackexchange.com/questions/166258/equivalence-point-in-relation-with-law-of-equivalence
|
Equivalence point in relation with law of equivalence
|
Why is that we apply law of equivalence only at equivalence point during a titration. Isn't the law valid always? In the sense, cant we use law equivalence even before we reach the equivalence point?
| 0
|
[
[
"\nWe can, but properly.\n\n\n* During titration, the molar amount of the used titrant is equivalent to the respective molar amount of the analyte **it has already reacted with**.\n* At the equivalence point, the molar amount of the used titrant is equivalent to the respective total molar amount of the analyte **present in the sample**.\n* After the equivalence point, the molar amount of the titrant **that reacted** is equivalent to the respective total molar amount of the analyte present in the sample.\n\n\nSo the amount of titrant that reacted is always equivalent to the amount of analyte that reacted. At the equivalence point, this is equal to the total amount of analyte and the amount of titrant added. At other stages, there will be excess of analyte or titrant, so the law is valid for the amount that reacted.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/166253/regioselectivity-of-sn2-reaction-on-adenine
|
Regioselectivity of SN2 reaction on adenine
|
I found the following question in my examination:
[](https://i.stack.imgur.com/fYLEj.jpg)
---
I was able to make out from the options that the usage of a base is to remove the supposedly acidic hydrogen from either the Nitrogen $A$ or $B$. Then the Nitrogen with a negative charge would attack $\ce{CH3I}$ through an $\ce{SN2}$ path and hence overall, the Hydrogen is replaced by methyl.
Now I got stuck on deciding which Nitrogen would be involved here? Both A and B have reduced electron density as their lone pairs are delocalised.
A point of difference maybe that $A$ isn't directly responsible for aromaticity in the 6 membered ring while $B$ is directly involved in aromaticity of the 5 membered ring.
But I don't know if that creates a difference. It would be great if someone could highlight where I'm going wrong and how was I supposed to judge in the examination as I did not have access to $\ce{pK\_a}$ values and other experimental data.
---
Some papers online suggest existence of different paths than this, but considering the level of the exam (JEE-main) it came in, real experimental reaction isn't much considered and more of theoretical approach is considered. I'll reframe my question.
Say that attack had to take place by either Nitrogen 'A' or, 'B' (Which is what the question hints us by using a base and so are the options framed in such a way to confuse)
Which Nitrogen would be preferred?
The correct answer given is (4) that is, Nitrogen $B$ is considered. How can we justify and explain this ?
| 2
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[
[
"\nIf a strong base (not just the methyl iodide and the solvent) is present, then it will tend to deprotonate the $\\ce{N-H}$ nitrogen on the five-membered ring, as this proton is about as acidic as one on an alcohol hydroxyl group; compare the $\\mathrm{pK\\_a}$ values given for [indole](https://en.wikipedia.org/wiki/Indole) and [ethanol](https://en.wikipedia.org/wiki/Ethanol). This $\\ce{N-H}$ proton is exceptionally acidic because the negative charge left behind by removing the proton is effectively delocalized around the aromatic five-membered ring.\n\n\nSo the base removes that proton, and the deprotonated nitrogen then replaces the proton with the electrophile forming product (4).\n\n\nIncidentally, once the five-membered ring is deprotonated nothing really prevents the electrophile from (alternatively) attacking the *other* nitrogen atom on that ring. That would give a different product from any of the choices given.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/166239/why-is-orbital-energy-not-the-mean-between-ionization-energy-and-electron-affini
|
Why is orbital energy not the mean between ionization energy and electron affinity when the orbital has two electrons?
|
In [this question](https://chemistry.stackexchange.com/questions/166206/ferrocenes-band-gap-is-supposed-to-make-it-colorless-but-it-isnt-why-is-it-so) it is asked why ferrocene is colored when the HOMO-LUMO gap seems to be beyond the visible light range. I tried to answer that orbital energies change with electronic transitions and therefore the transition energy need not match the difference in orbital energies. However, my answer crumbled under unexpected pressure and I wound up deleting it.
I tried to use a hydrogen atom as an example of changing orbital energy levels. When an electron is added to make a hydride ion, the combined energy of both electrons in the 1s orbital should be -13.6 eV (from the ionization energy of the neutral atom) plus -0.8 eV (from the affinity for the additional electron). With this total energy distributed between two electrons in the orbital the 1s orbital energy seemingly should br -7.2 eV (per electron) in the anion, versus -13.6 eV for the neutral atom when there is only one electron in the orbital. But I was barraged with comments that this analysis is wrong; the orbital energy in the hydride ion is only -0.8 eV from just the electron affinity. I do not understand why, and am baffled because this rendering seems inconsistent with conservation of energy.
* Why does only the electron affinity figure into the orbital energy?
* How does such a rendering square with conservation of energy?
Please don't tell me I am wrong, direct me to getting right.
| 2
|
[
[
"\nI think the confusion stems from a misunderstanding of the meaning of the orbital energies.\n\n\nConsider the hydrogen atom and its orbitals. You can obtain the energy of the electron in its groudn state using the Schrödinger equation\n$$ \\hat{H} \\Psi = E \\Psi $$\nFor this atom you can write the hamitltonian (in atomic units) as\n$$ \\hat{H} = -\\frac{1}{2}\\nabla^2 -\\frac{1}{R}$$\nwhere $-\\frac{1}{2}\\nabla^2$ represents the electron kinetic energy (operator) and $\\frac{1}{R}$ represents the Coulomb atraction between the electron and the hydrogen nucleus. Thus, the energy of the electron calculated corresponds to the binding energy of the electron to the nucleus or, if you prefer, the energy needed to ionize the atom.\n\n\nConsider now the H$^-$ anion, the atomic system is different, you have an hydrogen nucleus and two electrons that interact amongst themselves. Thus, the hamiltonian is different too, and, accordingly, the orbitals and its energies are different from those of hydrogen.\n\n\nThe hamiltonian for the anion has the form: \n\n$$ \\hat{H} = -\\frac{1}{2}\\nabla^2(1) -\\frac{1}{2}\\nabla^2(1) -\\frac{1}{R\\_1} -\\frac{1}{R\\_2} + \\frac{1}{r\\_{12}} $$\nwhere there appears a new term in the hamiltonian $+\\frac{1}{r\\_{12}}$ that represents the repulsion of the electrons.\n\n\nAs @Infinite has mentioned in his answer, you can describe the properties of the anion ignoring the electron-electron interaction.\n\n\n",
"4"
],
[
"\nThe orbital energy simply refers to the ionisation energy. So the orbital energy of the hydride ion should be numerically equal to the electron affinity value of the hydrogen atom.\n\n\nAnd there is no violation of law of conversation of energy. For removing the first electron we should supply 0.8 eV and for removing second electron we should supply 13.6 eV for an atom. And I see that there is no loss of energy ( $\\because$ the total energy of hydride ion is (13.6+0.8) eV in magnitude).\n\n\nAn explanation to the increased orbital energy from -13.6 eV to -0.8 eV is due to the electron-electron repulsions.\n\n\n\n\n---\n\n\nPlease correct me if I did anything wrong because I don't have a good background in quantum mechanics.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/166237/amorphous-silica-diatomaceous-earth-dissolution-and-precipitation-using-househ
|
amorphous silica (diatomaceous earth) dissolution and precipitation using household bleach or ammonia [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
---
**Personal medical questions** are off-topic on Chemistry. We can not safely answer questions for your specific situation and **you should always consult a doctor for medical advice**.
Closed last year.
[Improve this question](/posts/166237/edit)
I'm a layman not a chemist. My question is whether it's possible to dissolve amorphous silica in room temperature water mixed with a commonly available alkaline household product and then have it precipitate naturally out of solution over a few days or weeks. To dissolve a kilo of amorphous silica, what would be the right amount of over-the-counter laundry bleach or household ammonia and room temperature water?
I'm looking for a way to prevent nests of tiny stinging red fire ants under patio pavers that will be laid on sand and jointed with sand. Diatomaceous earth (amorphous silica) is said to be fatal to ants when they come into contact with it, so I wondered if spraying the screeded sand underlayment *in situ* and also the sand in the joints until it was quite wet with the diatomaceous earth solution might do the trick.
I also thought it would pose [a lower health risk](https://pubmed.ncbi.nlm.nih.gov/11876495/) and involve much less discomfort to mix up a batch while wearing a respirator and gloves than to work with the diatomaceous earth in powder form, mixing it dry with the sand, over the several weeks it will take me to lay the pavers.
| 0
|
[
[
"\nOnce it's been [wet, diatomaceous earth is less effective](https://gardenerspath.com/how-to/disease-and-pests/diatomaceous-earth/), so putting it into water, even as a suspension, defeats it's purpose. For that matter, it needs to be reapplied if wet by rain or damp earth. According to [BugTech](https://www.bugtech.com/diatomaceous-earth-benefits-as-a-natural-pesticide/), \"sharp edges make tiny cuts in an insect’s exoskeleton upon contact, and the diatomaceous earth then absorbs the oils and fats from its outer layer.\" If the silica is wet, it will hydrophilic, not oleophilic, and would not absorb oils.\n\n\nNon-chemistry suggestion: disperse it with air. Use an old container from \"talcum\" (now starch, for safety) powder, or a container for boric acid insecticide. These squeeze containers are designed to spray out the powder, propelled by air. Diatomaceous earth is [sold in air-spray bottles](https://rads.stackoverflow.com/amzn/click/com/B06X952WC5), too. It *might* be feasible to use a sandblasting gun on large areas, with proper respirator -- though at the risk of silicosis.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/166226/why-does-symmetrical-distribution-of-electrons-in-d-and-f-orbital-lead-to-greate
|
Why does symmetrical distribution of electrons in d and f orbital lead to greate stability of these or orbitals in comparison to s orbital? [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/166226/edit).
Closed last year.
[Improve this question](/posts/166226/edit)
My high school textbook mentions the following as on of the reason for why Cr and Cu have thier 3d orbitals filled completely before their 4s orbital.
>
> The extra stability of d and f orbital is due to:
> Symmetrical distribution of electron :It is well known that symmetry leads to stability . The completely filled or half - filled subshell have symmetrical distribution of electron in them and are therefore more stable . This effect is more dominant in d and f - orbitals . This means three or six electrons in p - subshell , 5 or 10 electrons in d subshell and 7 or 14 in f - subshell forms a stable arrangement
>
>
>
I don't understand why symmetry should lead to greater stability. So I asked my teacher but I was unsatisfied by the explanation and looked over some books and websites but I couldn't understand why symmetrical distribution of electrons should lead to greater stability.
The completely filled s orbital also has Symmetrical Distribution of electrons(at least According to me ,correct me if am wrong) so why are the d orbitals filled completely before the s orbital?How does symmetry leads to greater stability?
| -2
|
[
[
"\nI tried to add this as a comment, but I don't have 50 reputation points so I'm replying here.\n\n\nI don't think symmetry of half-filled orbitals plays much of a role in the stability, look at the electronic configurations of Nb, Tc, Ru, Pt for example. They too have exceptional electronic configurations where the electron enters d orbital instead of s orbital, but in their cases it is not due to partial or full filling of the d orbital. Also, W (tungsten) has a configuration of d4s2, and not d5s1 as would be expected had symmetry of orbitals been the sole reason for stability.\n\n\nI can't prove this, but it might be because s orbital being small and spherical in shape has less distribution of electrons, and consequently has greater electron-electron repulsion as compared to the d orbital, which might be resulting in these particular electron configurations with an electron in the d orbital instead of the s orbital to be more energetically favourable by having less e--e- repulsions.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/166224/why-are-nitriles-less-basic-than-amides
|
Why are nitriles less basic than amides?
|
Like nitriles have a localised lone pair which amides do not have, as their lone pair is delocalised and still amides are more basic. Why?
I could not find any satisfactory explanation so please explain in detail.
| 0
|
[] |
https://chemistry.stackexchange.com/questions/166220/how-can-we-define-molar-conductivity-as-the-conductance-of-1-ml-of-electrolytic
|
How can we define molar conductivity as the conductance of 1 ml of electrolytic solution?
|
I've been taught that conductance ($G$) of an electrolytic solution is the inverse of the resistance of the solution, and conductivity ($\kappa$) is the inverse of the resistivity. From this, we have
$$\frac 1 G = \frac 1 \kappa \frac l A$$
where $l$ is the distance between electrodes and $A$ is the area of the electrodes.
$$\kappa = G \frac l A$$
Therefore, when $l=1$ cm and $A=1$ cm$^2$, $\kappa = G$, i.e.
>
> $\kappa$ can be defined as the conductance of $1$ ml of the electrolytic solution.
>
>
>
I don't think the above definition is a good definition. When $l=2$ cm and $A=0.5$ cm$^2$, the volume between the electrodes is still $1$ ml, but the conductivity is
$$\kappa=G \frac 2 {0.5} = 4G \neq G$$
However, in the class I attended, they proceeded to use this definition to derive the formulae for molar and equivalent conductivity by the unitary method (if the conductivity of $1$ ml is $\kappa$, then the conductivity of $1000$ ml is $1000 \kappa$ ...)
Is this a correct definition? Am I missing something?
| 0
|
[
[
"\nSolution conductivity depends only on properties of solution itself (intensive property).\n\n\nSolution conductance depends on solution conductivity AND geometry of the measurement.\n\n\nConductivity has\n\n\n* the integral definition $G = \\kappa \\cdot \\frac Al$\n* the differential definition $\\vec J = \\kappa \\cdot \\vec E$, defined by the current density and potential gradient.(\\*)\n\n\nConductivity (in $\\pu{S cm-1}$) is *numerically* equal to conductance of 1 mL *if and only if* the volume has the shape of a cube with 1 cm long edge.\n\n\nConductance for 10 cm cube is 10 times greater, but conductivity is the same, depending just on the solution itself.\n\n\nMolar conductivity is not defined by conductance, but by conductivity, divided by molar concentration.\n\n\n\n\n---\n\n\n(\\*) For solid phases, conductivity is generally a tensor and vectors of potential gradient and current density need not to be parallel for anisotropic materials.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/166209/why-is-oxygen-an-anomaly-in-this-context
|
Why is oxygen an anomaly in this context?
|
In *A Very Short Introduction: Earth System Science* by Tim Lenton, the author discusses the composition of atmosphere:
>
> the first observations from land-based telescopes showed that Mars had an atmosphere dominated by carbon dioxide, just as would be expected in the absence of life. So too does Venus. But the Earth has a remarkable atmosphere, containing a chemical cocktail of highly reactive gases, sustained by life.
>
>
> **Oxygen is the prime anomaly**—at just over a fifth of Earth’s atmosphere it is essential for our existence as mobile, thinking animals, but without photosynthesis to create it, oxygen would be a very rare trace gas. Mixed in with oxygen are gases like methane that react eagerly with it—so much so that they are on the verge of combusting together.
>
>
>
Why is oxygen an anomaly in this context?
| 6
|
[
[
"\n**Because oxygen is so reactive it won't be present in a planetary atmosphere unless there is a source continuously producing it**\n\n\nIn the context of things found in planetary atmospheres, oxygen is a major anomaly (or, to put this another way, the earth has a very unusual atmosphere).\n\n\nThe reason is simple. Oxygen is so reactive that an atmosphere can only contain a significant amount if there is something continuously producing large amounts of it. On earth, that is life or more specifically living things that photosynthesise.\n\n\nIf photosynthesis stopped on earth it is estimated that there would be about enough oxygen to last only 20,000 years (based on the estimated rate of photosynthesis which is broadly in balance with the rate of consumption). Even if all the air-breathing animals died the oxygen would get used up pretty quickly by other processes like burning, rusting and geological weathering.\n\n\nThe atmospheres of planets typically depend on geological processes. But though some of these can produce some oxygen, there are many, faster, processes that use it up. So we don't expect to see free oxygen in a planetary atmosphere where the planet has no life.\n\n\n",
"10"
],
[
"\nFor the reasons mentioned in the quotation, oxygen -- unlike other major atmospheric components like nitrogen, argon or carbon dioxide -- cannot survive from primordial times and must be created by some active, ongoing process.\n\n\nOn Earth that process is, of course, life. But on some other worlds there are alternatives, namely spalling of oxygen atoms off rock or ice by ultraviolet radiation. Jupiter's moon [Ganymede](https://en.wikipedia.org/wiki/Ganymede_(moon)), with its water-ice surface, has a thin oxygen-bearing atmosphere.\n\n\n",
"6"
],
[
"\nOxygen is not an anomaly when talking about \"reactive gases\" in the atmosphere as the author clearly mentions that oxygen is also highly reactive. In fact there are less reactive gases (nitrogen) present. I think the author means to say that oxygen is the anomaly when comparing atmospheric composition of planets. Other planets will contain all kinds of gases (but not oxygen). [Here](https://enigmar.net/astronomy-universe-centre-science-sun-stars-planets-milky-way/the-planets-mercury-venus-mars-jupiter-saturn-uranus-neptune-orbits-sidereal-solar-system/atmospheres-of-planets-venus-mars-earth-jupiter-saturn-neptune-uranus/#:%7E:text=The%20atmosphere%20of%20the%20planets%2C%20from%20venus%20to,a%20surface%20temperature%20hot%20enough%20to%20melt%20lead.), it says that planetary atmospheres consist of many gases including carbon dioxide, methane, ammonia, and hydrogen and helium on distant planets. The Earth has a unique atmosphere of oxygen which helps sustain life.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/166206/ferrocenes-band-gap-is-supposed-to-make-it-colorless-but-it-isnt-why-is-it-so
|
Ferrocene's band gap is supposed to make it colorless but it isn't. Why is it so?
|
Wang and Islam [[1](https://doi.org/10.48550/ARXIV.1509.05122)] state that the ferrocene molecule's band gap, i.e. the HOMO-LUMO gap at the Kohn-Sham level, is 5.03 eV, which corresponds to an absorption wavelength of at most 240 nm.
So, ferrocene should be colorless — except that it isn't. I found out that liquid water, which has an even higher band gap (7 eV) than ferrocene, is still colored [because of vibronic, and not electronic, excitations](https://en.wikipedia.org/wiki/Color_of_water).
Does the same hold for ferrocene as well, and, if it does, do the expected effects to the HOMO-LUMO gap of e.g. π-electron withdrawing/donating substituents (which would change the color of the parent molecule, were that color be dependent on the gap — again, except that it isn't) “*change*” the color in the “*expected*” ways?
P.S. Please read my long string of comments, involving explicit calculations done for the ionisation potential/negated Kohn-Sham HOMO energy of the hydride anion, on Oscar Lanzi's answer before adding more answers.
### Reference
1. Wang, F.; Islam, S. Impact of ionization of ferrocene: EOES of α- and β-electrons and the fingerprint orbital 8a₁’ of ferrocenium. **2015**. DOI: [10.48550/ARXIV.1509.05122](https://doi.org/10.48550/ARXIV.1509.05122).
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https://chemistry.stackexchange.com/questions/166205/zwitterionic-form-in-amino-acid
|
Zwitterionic Form in Amino Acid
|
In my laboratory workbook, it has been mentioned that the zwitterionic form of the amino acid is the dominant form of the amino acid at pI and no movement of the molecules takes place under an external electric field. I am a bit confused by this statement.
It is clearly not possible for all the molecues to exist in the zwitterionic form. Some of them must exist in any of the other charged forms, even at pI. This should lead to these few charged molecules, moving in an electric field.
My question is:
* Is it true that all molecules are not zwitterions at pI?
* When we say no migration of ions takes place at pI, do we actually mean no "NET" migration of ions occurs i.e. whatever number of ions move towards one side, causes an equal number of ions to move towards the other?
| 0
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[] |
https://chemistry.stackexchange.com/questions/166203/how-do-i-know-the-order-of-the-reaction
|
How do i know the order of the reaction? [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/166203/edit).
Closed last year.
[Improve this question](/posts/166203/edit)
$$\ce{2N2O5 ↔ 2N2O4 + O2}$$ is this reaction first order or second order ? And why ?
I know first order is when duplicating the concetration the velocity duplicates too, and secod order reaction is of order n if, when doubling the concentration, the speed increase by $\pu{2^n}$. For this cases i need the velocitys and concentrations but i only have the initial concetration and 5 hal life times
[](https://i.stack.imgur.com/i4eSi.png)
$$\ce{N2O5}= 2,4 mols\_0$$
I dont know what to do, cuz to know, would have to have this data . Am i wrong ?
| 0
|
[
[
"\nOrder is an experimental data. It is never possible to derive the order of a reaction from its chemical equation. The order has to be determined by measuring the rate of a chemical reaction (in $\\pu{mol/s}$, or in $\\pu{mol L^{-1} s^{-1}}$) , and plotting a function of this rate vs. time. If the logarithm of the rate decreases linearly with time, the order is $1$. If the reciprocal of the rate increases linearly with time, the order is $2$. This is not dependent on the chemical equation of the reaction. In both cases, the slope of the line is the rate constant, $\\pu{k\\_{1}}$ or $\\pu{k\\_{2}}$.\n\n\nThis is what you should have to do with your experimental results about the decomposition of $\\ce{N2O5}$. But here all you know about this reaction is its half-life. And this parameter (half-life) is a characteristics and a constant of the first-order reactions. So from this very property, you can state that the reaction is first order, without doing any calculations.\n\n\nAs the half-life is known at different temperatures, it probably means that, in a second time, you would have to calculate the activation energy of the reaction.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/166201/should-change-in-enthalpy-be-zero-in-galvanic-cells
|
Should change in enthalpy be zero in galvanic cells?
|
When an exothermic redox reaction happens by just mixing the reagents, the change in enthalpy at constant pressure equals heat which is consistent with the fact that:
>
> For a closed system at constant pressure, change in enthalpy equals heat.
>
>
>
This can be derived from the definition of enthalpy and the first law of Thermodynamics.
When the same reaction takes place in a galvanic cell, there is no heat involved so the change in enthalpy must be zero. But as we know, the energy in galvanic cells is released in the form of electrical work. So $ΔH$ must be translated into electrical work which is inconsistent with the fact that $ΔH = Q\_p$ (with $Q\_p$ being zero in our case).
I want to understand where is the flaw in the above reasoning and how can we relate the change in enthalpy with the electrical work done by the galvanic cell.
I have looked here [Link1](https://chemistry.stackexchange.com/questions/23684/heat-given-off-from-an-electrochemical-cell-compared-to-mixing-reactants?rq=1) but the accepted answer is in contrast with the statement in the quotes (see also [Link2](https://physics.stackexchange.com/questions/600131/why-does-heat-of-reaction-equal-the-change-in-enthalpy-for-chemical-reactions) for the quoted statement).
| 1
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https://chemistry.stackexchange.com/questions/166199/what-is-the-standard-method-for-activating-molecular-sieves-with-microwave-radia
|
What is the standard method for activating molecular sieves with microwave radiation?
|
Typically, molecular sieves are dried under vacuum at elevated temperatures (about 200 to 300 °C depending on pore size) for several hours.
Alternatively, microwave heating has been used for preparation of mol sieves at least since 1980s and takes minutes instead of hours, but a standard protocol for this procedure appears to be something left to be desired.
Early technical report [Microwave regeneration of molecular sieves](https://inis.iaea.org/search/search.aspx?orig_q=RN:19034229) [1] illustrates the efficiency of the method: over 90 % of water was removed under 10 min mark.
However, as convenient as it sounds, there is an issue of the microwave radiation being equally readily absorbed by both water and the sieves.
This causes overheating the sieves (up to 500 °C and higher) and, arguably more importantly, temperature non-uniformity causing sintering, change in porosity and cracking of the sieves.
The general recommendation was to use short regeneration cycles at full power and apply purge gas (air) flow with temperature control.
The report, however, only relies on the data obtained for microwave source operating at 2450 MHz and one of the recommendation for future studies was to test other frequencies.
Kappe's work from 2009 [[1](https://doi.org/10.1071/CH08450)] draws a similar conclusion and calls for further optimization:
>
> However, the use of these materials in conjunction with microwave heating is further complicated because molecular sieves are known to be very strong microwave absorbers.[10]
> In fact, one of the best ways to dry/activate molecular sieves is to irradiate them with microwaves.[11]
>
>
> […]
>
>
> [11] Although many chemists apparently use this method in the laboratory, we were not able to find an appropriate reference in the chemical literature.
> According to our experience, the procedure given in the Experimental Section provides reproducible results but no efforts were made to quantitatively assess the water content of the sieves obtained using this protocol.
> We believe that main reason to use microwaves is convenience due to rapid heating.
> However, care must be taken not to overheat the sieves by prolonged microwave irradiation.
> This will result in material with decreased efficiency for water adsorbance.
>
>
>
There are numerous websites with various recommendations for using microwave radiation for mol sieves activation ([Reddit](https://www.reddit.com/r/chemistry/comments/7k2joi/drying_molecular_sieves_in_the_microwave/), [Chemtips](https://chemtips.wordpress.com/2014/12/01/how-to-activate-molecular-sieves/), [Sciencemadness](http://www.sciencemadness.org/talk/viewthread.php?tid=156132)) and papers targeting narrowed-down applications [3], but it looks like a standardized procedure with recommended microwave frequency range and heating time strategy from an authoritative source is missing, and the recommendations are sometimes contradicting.
If the answer turns out to be dependent on morphology of the sieves, for the sake of simplicity it probably would make sense to post one for the most commonly used ones in the lab setting (3 Å beads, 8–12 mesh, I presume).
### References
1. Singh, V. P. *Microwave Regeneration of Molecular Sieves*; F84023; Canada, **1984**. ([PDF](https://inis.iaea.org/collection/NCLCollectionStore/_Public/19/034/19034229.pdf))
2. Baghbanzadeh, M.; Kappe, C. O. Can Molecular Sieves Be Used as Water Scavengers in Microwave Chemistry? *Aust. J. Chem.* **2009**, 62 (3), 244. DOI: [10.1071/CH08450](https://doi.org/10.1071/CH08450).
3. Sârbu, E.; Călinescu, I. Microwave Assisted Regeneration of 3Å Molecular Sieves Used for Ethanol Dehydration. *U.P.B. Sci. Bull., Series B* **2014**, 76 (1).
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[] |
https://chemistry.stackexchange.com/questions/166196/can-thiamine-pyrophosphate-be-prepared-from-thiamine-hydrochloride-by-chemical-m
|
Can Thiamine Pyrophosphate be Prepared from Thiamine Hydrochloride by chemical means?
|
Is there a quick and easy way to prepare Thiamine Pyrophosphate in the lab (i.e. without using micro-organisms) starting by Thiamine Hydrochloride (by a simple displacement reaction maybe)?
| 0
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[] |
https://chemistry.stackexchange.com/questions/166191/what-does-the-word-dynamics-mean-in-the-context-of-proteins-or-biomolecules-in
|
What does the word "dynamics" mean in the context of proteins or biomolecules in general?
|
What does it mean by “*dynamics*” when we say “*dynamics of protein*” or “*dynamics of biomolecules*”? For instance, McCammon's paper has the title "*Protein Dynamics*" [[1](https://doi.org/10.1088/0034-4885/47/1/001)]. What does the word “*dynamics*” mean in this case?
Please, supply me with at least one reference.
### Reference
1. McCammon, J. A. Protein Dynamics. *Rep. Prog. Phys.* **1984**, 47 (1), 1–46. DOI: [10.1088/0034-4885/47/1/001](https://doi.org/10.1088/0034-4885/47/1/001).
| 0
|
[
[
"\nFrom Lottspeich' *Bioanalytics*, subsection *Determination of Protein Dynamics* [1, p. 463] (strong emphasis mine):\n\n\n\n> \n> ### Determination of Protein Dynamics\n> \n> \n> Proteins are not rigid, static entities, but rather exist as ensembles of conformations.\n> **The internal motions, which give rise to transitions between different structural states, are collectively referred to as protein dynamics** and occur on a wide\n> range of time scales (ns–s).\n> Protein dynamics play crucial roles in protein functions particularly for the interaction with binding partners, in enzyme catalysis, and in allosteric regulation.\n> Luckily, many NMR spectroscopic parameters depend both on the mobility of the whole molecule (translational or rotational motions) and on internal motions (transitions between different conformations or bond rotations), providing insight into a wide variety of dynamic processes ranging from fast fluctuations (lasting picoseconds) to slower conformational changes (lasting microseconds or more).\n> \n> \n> \n\n\nProtein dynamics studies have been used in biochemistry for studying genomics [[2](https://doi.org/10.1126/science.291.5505.843)], enzymology [[3](https://doi.org/10.1146/annurev-biochem-051710-133623)] and many other areas, including [GPCRs activity](https://en.wikipedia.org/wiki/G_protein-coupled_receptor) research which lead to the [2012 Nobel Prize in chemistry](https://www.nobelprize.org/prizes/chemistry/2012/summary/).\n\n\n### References\n\n\n1. *Bioanalytics: Analytical Methods and Concepts in Biochemistry and Molecular Biology*, 1st ed.; Lottspeich, F., Engels, J. W., Eds.; Wiley: Weinheim, **2018**. ISBN 978-3-527-33919-8.\n2. Misteli, T. Protein Dynamics: Implications for Nuclear Architecture and Gene Expression. *Science* **2001**, 291 (5505), 843–847. DOI: [10.1126/science.291.5505.843](https://doi.org/10.1126/science.291.5505.843).\n3. Klinman, J. P.; Kohen, A. Hydrogen Tunneling Links Protein Dynamics to Enzyme Catalysis. *Annu Rev Biochem* **2013**, 82, 471–496. DOI: [10.1146/annurev-biochem-051710-133623](https://doi.org/10.1146/annurev-biochem-051710-133623).\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/166185/can-hydrogen-fluoride-form-three-or-more-hydrogen-bonds
|
Can hydrogen fluoride form three or more hydrogen bonds? [duplicate]
|
**This question already has an answer here**:
[How many hydrogen bonds are formed by water and by HF?](/questions/69370/how-many-hydrogen-bonds-are-formed-by-water-and-by-hf)
(1 answer)
Closed last year.
As far as I know, each $\ce{HF}$ molecule has two hydrogen bonds, one formed by its hydrogen atom and one which the $\ce{F}$-atom forms with hydrogen atom of a third $\ce{HF}$ molecule.
In other molecules, like $\ce{H2O}$, where both of oxygen's lone pairs are used in hydrogen bonding each water molecule forms as many as four hydrogen bonds. Why does fluorine, which has three lone pairs, not use all its lone pairs up to form three hydrogen bonds instead?
| 1
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[
[
"\nHydrogen fluoride has only one hydrogen atom per fluorine atom, making a branched hydrogen-bond network difficult to form. (This is one reason hydrogen fluoride also forms hydrogen-bonded structures in the gas phase, unlike water where hydrogen-bond branching is more likely to occur and drive condensation.)\n\n\n[Ammonium fluoride, $\\ce{NH4F}$](https://en.wikipedia.org/wiki/Ammonium_fluoride), is not so constrained. In this compound each fluorine can use all four of electron pairs to form bonds that are apparently both hydrogen bonds and ionic bonds:\n\n\n\n> \n> Ammonium fluoride adopts the wurtzite crystal structure, in which both the ammonium cations and the fluoride anions are stacked in ABABAB... layers, each being tetrahedrally surrounded by four of the other. There are N−H···F hydrogen bonds between the anions and cations.[1] This structure is very similar to ice, and ammonium fluoride is the only substance which can form mixed crystals with water.[[2]](https://doi.org/10.1038/173316a0)\n> \n> \n> \n\n\n**Cited References**\n\n\n1. A. F. Wells, Structural Inorganic Chemistry, 5th ed., Oxford University Press, Oxford, UK, 1984.\n2. Brill, R.; Zaromb, S. \"Mixed Crystals of Ice and Ammonium Fluoride\". *Nature.* **173** (4398): 316–317. <https://doi.org/10.1038/173316a0>.\n\n\n",
"4"
],
[
"\nIn a true intermolecular hydrogen bond, the X-H---Y bond angle is approximately 180$^\\circ$. Thus, each H atom can only participate in one H-bond.\n\n\nIn pure HF, the ratio of H:F is necessarily 1:1. Since each H-bond requires one H and one F and each H can only participate in one H-bond, the F atoms necessarily participate (on average) in only one H-bond each as well.\n\n\n",
"3"
],
[
"\n$\\ce{HF}$ can not form two hydrogen bonds since the two positively charged $\\ce{H}$ atoms would repel each other.\n\n\n[](https://i.stack.imgur.com/iyfU2.png)\n\n\nAlso, unlike it has been said in the comments, $\\ce{H}$-bonds do involve orbital overlap and are not a consequence of electrostatic attraction alone. Although the overlap does not result in very strong bonds, as is the case with overlapping in covalent bonds, $\\ce{H}$-bonds can be surprisingly stable ($\\Delta G \\approx 3\\text{kJ/mol}$).\n\n\nNever thought I would find a possible explanation while comparing neighbor interactions. If we assume that the the zig-zag structure for $\\ce{HF}$ is correct, we can not rule out the repulsion caused by the lone pairs of fluorine.\n[](https://i.stack.imgur.com/wc17C.png)\n\n\n",
"-3"
]
] |
https://chemistry.stackexchange.com/questions/166184/stability-comparison-of-europiumii-and-ytterbiumii-ions
|
Stability comparison of europium(II) and ytterbium(II) ions
|
“Which one of the lanthanoids given below are most stable in divalent form?” This question was asked in JEE Main 2022 (Indian entrance test for engineering). Now the question had both $\ce{Eu^2+}$ and $\ce{Yb^2+}$ as the options.
My thought was that since $\ce{Yb^2+}$ has $f^{14}$ configuration it should be somewhat more stable than the half filled $\ce{Eu^2+}$. The answer key however says that $\ce{Eu^2+}$ is more stable. It would be helpful if someone could explain why that could be true. Relevant data of Reduction potentials would be helpful as well ( I tried to find it but without much success).
| 0
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[] |
https://chemistry.stackexchange.com/questions/166180/crystal-field-splitting-in-octahedral-complex
|
Crystal Field Splitting in Octahedral Complex
|
So i was going through my 1st year engineering chemistry textbook as a i have test in about a few weeks. So my textbooks says:
"Repulsion by six ligands in an octahedral complex splits the d orbitals
on the central metal into t2g and eg levels. It follows that there is a corresponding repulsion between the d electrons and the ligands. If the d
electrons are symmetrically arranged, they will repel all six ligands equally.
Thus the structure will be a completely regular octahedron."
So i understood the first 2 lines. Basically what my understanding says is that in an octahedral complex the ligands usually approach the central metal along the x, y and z axis. So obviously the orbitals which lie along the axis (dx2-y2 and dz2) will rise in energy which is your eg level than compared to the orbitals which lie in between the axis ( dxy, dyz, dxz) which your t2g levels. What i am not able to understand is the part where they say "If the d electrons are symmetrically arranged, they will repel all six ligands equally". The electrons in eg level rise in energy because they face more repulsion from the ligands right? so how is it possible that the d electrons will repel all ligands equally? Shouldn't the electrons in the eg level repel them more than the electron in the t2g level? I even tried to make my self understand by taking an example of a d5 configuration(which is symmetrically arranged) in the presence of a weak field ligand where two of your electrons would be in eg and the other 3 electrons would be in t2g level, but still I wasn't able to understand how is it that the d electrons are repelling the ligands equally?
| 1
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[] |
https://chemistry.stackexchange.com/questions/166178/why-isnt-formyl-chloride-stable
|
Why is formyl chloride unstable while higher acyl chlorides are stable?
|
From Bruice's *Organic Chemistry*, section *18.6 Friedel–Crafts acylation of benzene* [1, pp. 876–877]:
>
> *Friedel–Crafts acylation* places an acyl group on a benzene ring, and *Friedel–Crafts alkylation* places an alkyl group on a benzene ring.
> An acyl chloride or an acid anhydride is used as the source of the acyl group needed for a Friedel–Crafts acylation.
>
>
> [](https://i.stack.imgur.com/fK3zY.png)
>
>
> […]
>
>
> The synthesis of benzaldehyde from benzene poses a problem because formyl chloride, the acyl halide required for the reaction, is unstable and must be generated in situ.
> The **Gatterman–Koch reaction** uses a high-pressure mixture of carbon monoxide and HCl to generate formyl chloride and uses an aluminum chloride–cuprous chloride catalyst for the acylation.
>
>
> [](https://i.stack.imgur.com/lS0qt.png)
>
>
>
From this excerpt we infer that formyl chloride is unstable.
Is it just a remark, or does it affect the reaction?
If *formyl* chloride is unstable, why is *acyl* chloride stable?
The answer by TheSimpliFire to [Stability of formyl chloride](https://chemistry.stackexchange.com/questions/107557/stability-of-formyl-chloride) suggests vigorous decomposition of formyl chloride by water, but the Gatterman–Koch reaction proceeds under anhydrous conditions. So, why is formyl chloride unstable even in water-free environment?
### Reference
1. Bruice, P. Y. *Organic Chemistry*, 8th ed.; Pearson: Upper Saddle River, NJ, **2016**. ISBN 978-0-13-404228-2.
| 9
|
[
[
"\n\n> \n> The simplest stable acyl chloride is ethanoyl chloride or acetyl chloride; methanoyl chloride (formyl chloride) is not stable at room temperature, although it can be prepared at –60 °C or below. ([Wikipedia](https://en.wikipedia.org/wiki/Acyl_chloride?wprov=sfla1))\n> \n> \n> \n\n\nThe instability of $\\ce{HCOCl}$ is caused by ease of elimination of HCl from its molecules. Cl is decent leaving group and after it's gone, remaining acylium cation has a very acidic hydrogen instead of alkyl present in other acyl halides. Thus easier breaking of C-H bond vs C-C bond causes drastic difference in thermal stability.\n\n\nNote also the formyl chloride can be considered not only the derivate of formic acid, but also a derivate of formaldehyde.\n\n\n",
"9"
],
[
"\n**More stable formyl compounds**\n\n\nWhile formyl chloride is unstable, [formyl *fluoride*](https://en.wikipedia.org/wiki/Formyl_fluoride), with a stronger bond from the carbonyl carbon to the halogen, is marginally more stable. It can be made by a double displacement reaction:\n\n\n\n> \n> $\\ce{HCOONa + C6H5C(O)F → FC(O)H + C6H5COONa}$[[1]](https://doi.org/10.1021%2Fcr00080a001)\n> \n> \n> \n\n\nThe compound is still unstable enough at ambient temperature to require special procedures to prepare and store, as it decomposes autocatalytically. Low temperature preparation and storage over alkali metal fluorides (which absorb the hydrogen fluoride catalyst) are employed to stabilize the product.\n\n\n[Formyl cyanide](https://en.wikipedia.org/wiki/Formyl_cyanide), which does not have a good proton-acceptor atom, is stable enough to be made at high temperature. Various synthetic methods are available [[2]](https://doi.org/10.1021/ja00032a001)[[3]](https://doi.org/10.1016/0009-2614(88)87435-9)[[4]](https://doi.org/10.1006/jmsp.1995.1183).\n\n\n**Cited Reference**\n\n\n1. Olah, G. A.; Ohannesian, L.; Arvanaghi, M. (1987). \"Formylating Agents\". *Chemical Reviews*. **87** (4): 671–686. <https://doi.org/10.1021%2Fcr00080a001>.\n2. Lewis-Bevan, Wyn; Gaston, Rick D.; Tyrrell, James; Stork, Wilmer D.; Salmon, Gary L. (March 1992). \"Formyl cyanide: a stable species. Experimental and theoretical studies\". *Journal of the American Chemical Society*. **114** (6): 1933–1938. <https://doi.org/10.1021/ja00032a001>.\n3. Bogey, M.; Destombes, J.L.; Vallee, Y.; Ripoll, J.L. (May 1988). \"Formyl cyanide: Efficient production from allyloxyacetonitrile and its millimeter-wave spectrum\". *Chemical Physics Letters*. **146** (3–4): 227–229. Bibcode:1988CPL...146..227B. <https://doi.org/10.1016/0009-2614(88)87435-9>.\n4. Bogey, M.; Demuynck, C.; Destombes, J.L.; Vallee, Y. (August 1995). \"Millimeter-Wave Spectrum of Formyl Cyanide, HCOCN: Centrifugal Distortion and Hyperfine Structure Analysis\". *Journal of Molecular Spectroscopy*. **172** (2): 344–351. Bibcode:1995JMoSp.172..344B. <https://doi.org/10.1006/jmsp.1995.1183>\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/166177/which-is-correct-molybdate-or-molybdenate
|
Which is correct - molybdate or molybdenate?
|
Most websites (including Wikipedia) say that the anion MoO42- is called a molybdate ion, although some websites use molybdenate instead. Which one is correct, and if one of them is correct, is the other term still allowed?
| 5
|
[
[
"\nAccording to Table X of the Red Book (‘Nomenclature of Inorganic Chemistry – IUPAC recommendations 2005’), the anion name for molybdenum is **molybdate.**\n\n\n",
"9"
],
[
"\nIf you do a Google ngram of the terms 'molybdate' and 'molybdenate', you will find 'molybdate' standing out.\n\n\n[](https://i.stack.imgur.com/mcvBz.png)\n\n\nNo one known why the 'en' was kept when you can simply shorten it to molybd-*ate* instead of molybden-*ate* (like phosph-ate and not phosphor-ate). However, the latter didn't got totally eliminated. Some research papers([here](https://www.researchgate.net/publication/292583045_Inhibitive_effects_of_molybdenate_on_iron_artifacts_during_desalination_process) is a recent example) and textbooks still uses 'molybdenate'. The following image is from a physics textbook published in 2020.\n\n\n[](https://i.stack.imgur.com/BsEFp.png)\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/166176/does-the-charge-on-an-atom-impact-the-polarization-of-covalent-bonds
|
Does the charge on an atom impact the polarization of covalent bonds?
|
So far, I have generally seen polarization power only discussed in terms of ionic bonds: small, highly charged cations have more polarization power, and large anions are more polarizable. The more polarization power and polarizability, the greater the covalent character of an ionic bond.
Generally, one can estimate the polarity of a covalent bond by looking at the differences in the electronegativities of the two atoms involved in the bond. However, I am wondering whether one can apply the ideas of polarizing power to atoms in covalent bonds, specifically the idea that net charge impacts the polarity of the bond. Does a higher positive net charge on one atom cause electrons in covalent bonds to be pulled towards that atom? This is what I would conclude from a polarization power argument, because an atom with a positive net charge has more polarization power. Similarly, would an atom with a negative charge pull less electrons? These ideas also seem to coincide with a Coulomb's Law argument that a more positive charge would attract electrons better, and a more negative charge would not attract them well.
One piece of evidence supporting this idea is the inductive effect; polarization propagates down a chain of sigma bonds because partially positive carbons are able to polarize carbon-carbon bonds. This polarization occurs in spite of the fact that the difference in electronegativity between the two carbon atoms is 0. This seems to imply that one has to consider both electronegativity and the charge on the atoms when determining the polarity of bonds. [](https://i.stack.imgur.com/GESmV.png)
There are two particular cases that I am curious about: metals in coordination complexes and formal charges. Consider two complexes: one with a Mn (II) center and one with a Mn (VII) center. Will the Mn (VII) center withdraw the electrons of the M-L dative bonds better than an Mn (II) center (in turn inducing an electron withdrawal throughout the rest of the ligands)? For example, would an NH2 ligand be less basic when bonded with an Mn (VII) metal center than when bonded to an Mn (II) metal center, because the Mn (VII) center is able to withdraw electrons better (even though an Mn-N bond, with the same electronegativity difference between Mn and N, exists in both cases)?
As for formal charges, would an atom with a positive formal charge (such as the nitrogen in NH4+) withdraw electrons better than one with a negative formal charge (such as the nitrogen in NH (2-))? I know that formal charge does not always represent the true charge on an atom, but I am thinking that it could be used to compare the relative amounts of charge on related species - for example, I know most electron density is concentrated on the nitrogen in both NH3 and NH4+, but I assume that there would be a bit less electron density on the nitrogen in the NH4+ because the lone pair electrons of NH3 are now shared with a hydrogen.
I am also aware that there are other factors to consider when looking at the electron distributions of complexes and formal charge, such as pi donors/acceptor ligands and favored resonance states. However, I am wondering about the effects of charge on the polarization of bonds. Thank you for your help.
| 0
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[
[
"\nEverything is correct here, including the explicit examples- the ammonium group is a very strongly inductively electron withdrawing group in electrophilic aromatic subtitution, far negating any hyperconjugative mesomeric electron donation effects, and \"permanganamide\" is likely to be strongly acidic, with the Mn-O and Mn-N bonds being highly covalent.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/166167/searching-for-electronic-spectroscopy-reference-book
|
Searching for electronic spectroscopy reference book
|
I'm looking for a reference book that will help me describe UV-VIS spectra of organic compounds in terms of Q band, MLCT band, Soret band, and so on. Is something like that available?
Thanks
| 0
|
[] |
https://chemistry.stackexchange.com/questions/166161/why-is-hybridization-different-in-a-ring
|
Why is hybridization different in a ring? [duplicate]
|
**This question already has answers here**:
[Why is the lone pair of pyridine's nitrogen atom not part of the aromatic ring?](/questions/50266/why-is-the-lone-pair-of-pyridines-nitrogen-atom-not-part-of-the-aromatic-ring)
(2 answers)
[sp² hybridized orbital](/questions/27309/sp%c2%b2-hybridized-orbital)
(1 answer)
Closed last year.
Why is nitrogen $sp^3$ hybridized in a compound such as NH3 but $sp^2$ hybridized in something like pyrrole, which has a ring structure? In both cases there are three bonds and one lone pair.
| -4
|
[
[
"\nOne might argue that the nitrogen in pyrrole does not really have a lone pair we can identify like that in ammonia or for that matter pyrrolidene ($\\ce{C5H9N}$, adding enough hydrogen atoms to saturated the five-membered ring in pyrrole). In the pyrrole molecule, what \"should be\" the lone pair on nitrogen is instead coupled with the other pi electrons to form a continuous ring of conjugated pi bonding. Such a coupling is commonplace when it forms an aromatic ring, meaning in the most common cases $4n+2$ electrons in this ring (like the six pi electrons in pyrrole). There are also cases where what seems to be an *empty* orbital really isn't empty because of a similar coupling (like the [cyclopropenyl cation](https://en.wikipedia.org/wiki/Cyclopropenium_ion)). You should see more examples of this effect as you study aromatic compounds.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/166153/why-do-atoms-make-bonds
|
Why do atoms make bonds? [duplicate]
|
**This question already has answers here**:
[Fundamental forces behind covalent bonding](/questions/710/fundamental-forces-behind-covalent-bonding)
(5 answers)
Closed last year.
I have been studying chemistry for a long time in school till now, but, what I recently realised I do not understand is why do atoms really make bonds, why do they want to gain or lose electrons or why do they want to attain noble gas configurations when it results in putting a charge on them and comparing to the original condition, let's say oxygen has 8 electrons and 8 protons, why isn't this stable?
This question came to me when I was trying to understand why the ionisation enthalpy can never be negative and how an atom like Na only exhibits it's electropositivity when provided with some energy and is stable until then.
Thank you in advance for your answer
PS: I have realised in the past few days that I really understand very little chemistry:(
| -2
|
[
[
"\nI'd say for the same reason why an apple fells to the ground. Because there is an energy gain. Atoms and electrons follow the electromagnetic forces through which they interact, in the same way the apple and our planet do. Just in the case of the apple it is a two body problem with classical mechanics. In the case of bonds it is a few (many) body problem. This makes it much more difficult to visualize it and/or to have an intuitive picture. Luckily we have math and numerical simulations to help us with that.\n\n\n",
"3"
],
[
"\n\n> \n> why do atoms really make bonds, [...]\n> \n> \n> \n\n\nWe never know why, but we can study when they do and when they don't make bonds. And mostly (on earth) they do make bonds.\n\n\n\n> \n> [...] why do they want to gain or lose electrons or why do they want to attain noble gas configurations when it results in putting a charge on them and comparing to the original condition, let's say oxygen has 8 electrons and 8 protons, why isn't this stable?\n> \n> \n> \n\n\nA single (isolated) oxygen atom will keep is electrons, as will a single sodium atom. Making ionic compounds from isolated atoms (a thought experiment), there is usually a net cost of transferring electrons to make the anions and cations, and a large gain when these cations and anions combine to form ionic compounds. The cost is also offset when cations and anions are solvated in water. Mostly, the atoms in elements are already bonded in some way (metallic bonds for metals, covalent bonds for non-metals), so when compounds form from elements, you also have to consider breaking those bonds. Some elements react with many other elements, while others (like noble metals such as gold or platinum) don't.\n\n\n",
"2"
],
[
"\n**Because some molecules have lower energy that the isolated atoms they are made of**\n\n\nIsolated atoms are usually stable if they don't interact with anything else. But, when things interact, you need to start thinking about what can happen and one way to do this is to understand the energy levels of the isolated atoms compared to the molecule.\n\n\nIn some cases when isolated atoms interact, nothing happens. Helium atoms, for example, just bounce off each other and no bonds are formed.\n\n\nBut two isolated hydrogen atoms, if they meet, will often produce a stable hydrogen molecule. When this happens, energy is released and we can measure how much. The hydrogen molecule, we say, has *lower energy* than two isolated hydrogen atoms. This we call a chemical bond.\n\n\nBut why is a molecule lower in energy than two isolated atoms? It is partially about electromagnetic forces (two positive protons attract two negative electrons in a hydrogen molecule). But knowing whether the *net* forces lead to a bond involves a lot of detailed calculations and you need a fair amount of quantum mechanics to even approximate the answer. The summary of the answer is that the amount of energy involved in the electromagnetic field of two isolated hydrogen atoms is a lot more than the energy involved in a hydrogen molecule. This isn't true for two helium atoms, which is why they keep to themselves and don't form bonds. But the details of those calculations are advanced.\n\n\nA great deal of chemistry is about observing when and how bonds form and deriving general rules from observation that explain what happens without having to do those calculations (which are, in general, extremely hard). But the general rule is all about whether the amount of energy involved is lower with a bond than it is with isolated atoms.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/166145/does-iron-rust-because-of-impurities-found-in-it-ex-carbon
|
Does iron rust because of impurities found in it (ex. carbon)?
|
If we, theoretically, get a piece of ideally 100% pure iron and it is left in moist air? Will it rust? My understanding of iron rust is that the Iron itself becomes the ANODE and carbon impurities (as an example of impurities) are CATHODE, so if this cathode is not there will iron itself become anode and cathode and rust?
| 0
|
[
[
"\n**Pure iron will rust but far, far more slowly than iron with impurities**\n\n\nThe key point here is not whether iron will rust or not: it is about the speed of rusting.\n\n\nVery pure iron will rust but very, very slowly. But even tiny impurities will promote much faster rusting. Many iron alloys (which means most iron you will ever see as very pure iron is rare) contain small amounts of carbon or other substances. On the surface of the iron these can promote tiny electrical potential that greatly speed the process of corrosion. The opposite is also true: some \"stainless\" alloys contain deliberately added ingredients that promote the formation of a strong, inert oxide layer on the surface that inhibits such reactions (making the substance a little like aluminium whose surface oxide layer is strong and inert, protecting a very reactive metal from oxidation reactions that would otherwise consume it quickly).\n\n\nThe issue with iron in general is that iron oxide is not strong and does not form a protective layer because it is flaky and friable. So, once started, rusting tends to get worse and happen faster over time.\n\n\nEven pure iron will rust quickly in the wrong environment. Surface contamination will create impurity sites that can promote the electrochemical reactions that accelerate rusting. But those reactions need water, so keeping water away from the surface can slow any reactions very substantially. But, obviously, many varieties of impure iron can rust quickly when any water is present as the contaminants that promote the reactions are already present in the surface of the iron.\n\n\nIn summary, pure iron will rust. But slowly even when water is present. Impure iron will rust faster. All forms of iron will rust quickly if the environment adds other surface contaminants that can promote the electrochemical reactions that make rusting happen.\n\n\n",
"2"
],
[
"\nIt depends; sulfide inclusions will enhance oxidation. Chloride surface contamination will promote oxidation. Phosphorus inclusions will slow oxidation . Copper and chrome \"impurities\" will also slow oxidation. The latter three elements are the basis of \"weathering\" carbon steels like Corten. The carbon affect you describe causes \"graphitic corrosion\" of flake type cast irons ; oxidizing the iron ,leaving carbon flakes.\n\n\n",
"-1"
],
[
"\nThree substances are required to transform iron into rust : air (oxygen), water, and a impurity at the surface of the metal. If one of these substances is missing, iron will not rust. The whole operation occurs at the contact of the impurity and iron and it has an electric origin. The reduction occurs at the impurity working as a cathode. It is well described in N. N. Greenwood and E. Earnshaw, Chemistry of the Elements, Pergamon 1983, p.1250\n\n\n",
"-6"
]
] |
https://chemistry.stackexchange.com/questions/166143/probability-density-and-radial-distribution-function-of-finding-the-most-probabl
|
Probability density and radial distribution function of finding the most probable distance of electron in 2p orbital in hydrogen atom
|
Referring to the answer by DSVA ([Most probable point for finding an electron in the 1s orbital of a Hydrogen atom](https://chemistry.stackexchange.com/q/84726/89782))
>
> There's a maximum of finding the electron at a certain distance away from the core (but not a single point at that distance)
>
>
>
I face a problem in solving the maximum probability of finding electron in a 2p orbital. $$\psi=R\_{2,1}Y\_{1,0}=\sqrt\frac{1}{32\pi a\_o^3} \left(\frac{r}{a\_o}\right) \exp\left(\frac{-r}{2a\_o}\right) \cos\theta $$
Using probability density function, differentiate and equate to zero$$ \frac{d\psi^2}{dr}=constants\left(2r-\frac{r^2}{a\_o}\right)\exp\left(\frac{-r}{a\_o}\right)=0$$
I obtain$$r=2a\_o$$
Using radial probability distribution, differentiate and equate to zero$$P(r)=r^2|R(r)|^2 $$
Referring Atkins' Physical Chemistry (pg. 312), it is stated that spherical harmonics is normalised to 1.$$\frac{dP}{dr}=constants\left(4r^3-\frac{r^4}{a\_o}\right)\exp\left(\frac{-r}{a\_o}\right)=0 $$
I obtain$$r=4a\_o$$
My question: When should we use radial probability or probability density to find maximum probability of finding electron and its most probable distance? What does the difference of values mean?
| 3
|
[
[
"\nWhat you are interested in is the maximum probability of finding the electron at a given position.\n\n\nIn general, the probability of finding an electron in a volume $\\pmb{V}$ (not necessarily infinitesimal) is\n\n\n$$ \\int\\_{\\pmb{V}} |\\Psi(\\pmb{r})|^2 d^3 \\pmb{r}$$\n\n\nThis integral is, so far, abstract, because we haven't defined a coordinate system yet. If you want to use spherical coordinates - which is a common thing to do -, then you can translate this integral to\n\n\n$$ \\int\\_\\pmb{V} |\\Psi(\\pmb{r})|^2 r^2 \\sin \\theta dr d\\theta d\\phi$$\n\n\nSo if you want the maximum probability, then what you are interested is in maximizing the integrand with respect to $r$\n\n\n$$ \\max\\_r |\\Psi(\\pmb{r})|^2 r^2 \\sin \\theta$$\n\n\nIf you then take the derivative of this quantity with respect to $r$ and set it to zero, you end up with\n\n\n$$ -\\frac{r^3 e^{-\\frac{2 r}{a\\_0}} \\left(r-2 a\\_0\\right) \\sin (\\theta ) \\cos ^2(\\theta )}{16 \\pi a\\_0^6} =0$$\n\n\nwhich has a solution indeed $r=2 a\\_0$.\n\n\nA common mistake is finding the maximum of the wavefunction itself is the fact that it only takes into account the decay of the wavefunction with the distance from the nucleus, but it doesn't account for the volume of the infinitesimal disc at that range. Imagine if the wavefunction would be constant up to a certain critical radius $r\\_c$, and zero outside(I know it breaks quantum mechanics but don't think too hard about that for a second): the maximum probability of finding it would be right at $r\\_c$, because that radius has the largest geometrical probability to find the constant electron density in.\n\n\nThis is actually quite often misunderstood in the context of the definition of the Bohr radius: it is not the maximum of the electron density of the hydrogen atom, but of the probability density. I hope the distinction is clearer now.\n\n\n(Two minor mistakes: I'm almost certain you need to include an angular factor in the probability density which you missed, and you shouldn't take the second derivative in the first case - which is irrelevant, anyways).\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/166138/are-carboxylic-anhydrides-in-graphene-oxides-considered-adsorption-sites-for-met
|
Are carboxylic anhydrides in graphene oxides considered adsorption sites for metals? If so, are they more or less reactive than carboxylic acids?
|
The literature abundantly addresses the fact that carboxylic acids present in graphene oxides act as efficient adsorption sites for metals in aqueous solutions (e.g., HE et al., 2021; ZHAO et al., 2019; NOVÁCEK et al., 2017; WANG et al., 2018). Wang et al. (2018), for example, showed through DFT measurements that the adsorption energies of the complexes (–COOH)/Co, GO(–OH)/Co, and NGO/Co, where N corresponds to nitrogen-containing functional groups introduced into graphene oxide and Co to cobalt ions, were 44.05 kcal/mol, 16.44 kcal/mol, and 6.33 kcal/mol, in that order. This result was corroborated by the performance of the adsorbents in removing Co(II) from the water (GO (0.74 mmol/g) > NGO (0.46 mmol/g)). However, I have not found anything regarding whether carboxylic anhydrides, which can be identified via potentiometric titration, also capture metals through their complexation and/or via ion exchange. I would like to know if they can and if their reactivity is lower than, equal to, or higher than that of carboxylic acids in the water, present in graphene oxides, for the adsorption of metals.
Thanks in advance!
HEA, Lei; WANG, Lei; ZHU, Haomiao; WANG, Zhe; ZHANG, Luxia; YANG, Lutao; DAI, Yong; MO, Hong; ZHANG, Jun; SHEN, Jian. A reusable Fe3O4/GO-COOH nanoadsorbent for Ca2+ and Cu2+ removal from oilfield wastewater. Chemical Engineering Research and Design, v. 166, n. 171, p. 248-258, jan. 2021.
ZHAO, L.; CHEN, J.; XIONG, N.; BAI, Y.; YILIHAMU, A.; MA, Q.; YANG, S.; WU, D.; YANG, S. Carboxylation as an effective approach to improve the adsorption performance of graphene materials for Cu2+ removal. Science of the Total Environment, v. 682, p. 591-600, set. 2019. Elsevier BV. <http://dx.doi.org/10.1016/j.scitotenv.2019.05.190>.
NOVÁČEK, M.; JANKOVSKÝ, O.; LUXA, J.; SEDMIDUBSKÝ, D.; PUMERA, M.; FILA, V.; LHOTKA, M.; KLÍMOVÁ, K.; MATĚJKOVÁ, S.; SOFER, Z. Tuning of graphene oxide composition by multiple oxidations for carbon dioxide storage and capture of toxic metals. Journal of Materials Chemistry A, v. 5, n. 6, p. 2739-2748, 2017. Royal Society of Chemistry (RSC). <http://dx.doi.org/10.1039/c6ta03631g>.
WANG, X.; LIU, Y.; PANG, H.; YU, S.; AI, Y.; MA, X.; SONG, G.; HAYAT, T.; ALSAEDI, A.; WANG, X. Effect of graphene oxide surface modification on the elimination of Co(II) from aqueous solutions. Chemical Engineering Journal, v. 344, p. 380-390, jul. 2018. Elsevier BV. <http://dx.doi.org/10.1016/j.cej.2018.03.107>.
| 2
|
[
[
"\nThere are no surprises that the metal adsorption literature is silent about carboxylic anhydride on carbon surfaces. These groups do exist on carbon surfaces but as you can guess from the name \"anhydride\" and \"water\" do not get along very well. This is why I had asked you to clarify your medium? These anhydride groups will enventually hydrolyze to carboxylic acid groups. There is a very nice paper \"Formation and chemistry of carboxylic anhydrides at the graphene edge\" [RSC Adv., 2015,5, 104198-104202, DOI https://doi.org/10.1039/C5RA23209K], that shows the hydrolysis of carboxylic anhydride in water and in humid air. This process is relatively fast as show in the Supporting File.\n\n\n[](https://i.stack.imgur.com/KXR4n.png)\n\n\nIn the field of carbons and surfaces, I would advise to read at 12 papers before arriving at a conclusion. Surfaces are very very difficult to analyze chemically and a lot of speculation exists.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/166125/what-exactly-happens-in-a-chemical-potential-energy-graph
|
What exactly happens in a Chemical Potential Energy Graph? [duplicate]
|
**This question already has an answer here**:
[Why is activation energy drawn in a potential energy diagram in reactions?](/questions/119255/why-is-activation-energy-drawn-in-a-potential-energy-diagram-in-reactions)
(1 answer)
Closed last year.
Along the x-axis, the progress of the reaction is observed, while along the y-axis, the potential energy of the reactants/products are observed. They keep rising until they attain the required activation energy for breaking of older bonds, and formation of new ones. But what exactly is this activation energy? Is it a form of potential energy (stored within the chemical bonds), or is it a form of kinetic energy (vibrations of the particles)? I don't really understand why, if the activation energy is just potential energy, it results in collisions (without any motion) and thus forms new compounds, given that, we all know, potential energy is the energy of a body at rest.
If it is not so, then what is the role of chemical potential energy into all of this, at all?
I am still a student, stuck within concepts, so please, keep the answer as brief as possible...
| -1
|
[
[
"\nIt is just potential energy: the potential energy needed for the reaction to happen.\n\n\nThis potential energy obviously needs to come from somewhere, and that's where kinetic energy comes in. If the molecules have high enough kinetic energy at the point of the collision for that reaction to happen, then part of that kinetic energy is converted to potential energy, and the reaction proceeds.\n\n\n[](https://i.stack.imgur.com/zeoMW.gif)\n\n\nThis is also why most reactions get faster as you increase the temperature. Higher temperature means more kinetic energy, meaning more successful reactive collisions.\n\n\n**Edit:** I should add that there seems to be some sort of misconception in your question. There's nothing along a potential energy graph that would result in a collision.\n\n\n",
"1"
],
[
"\nThe potential energy is the energy due to the positions of the atoms and their electrons have in a molecule that is a sort of combination of reactant and product, i.e. a transient structure from the path from reactants to products, some times called a 'transition state'. This barrier can be overcome by collisional (kinetic) energy between the reactant and product but also by thermal energy since by the Boltzmann distribution there is a small chance that a molecule may have, for a short period of time, many times the average energy expected at that temperature. Thus reactants may collide many millions of times before a reaction occurs, if the barrier is high, or only a few times if the barrier is low relative to average thermal energy. Consequently we observe rate constants in different types of reaction that vary from $10^{12}$/sec to less that 1 per hour.\n\n\n",
"0"
],
[
"\nMaybe this additional perspective will help. Reactions involve bond breaking and bond making. Breaking bonds results in an increase in potential energy (there is less attraction between atoms; you must add energy to break bonds, therefore, increase in PE). Bond Making results in a decrease in potential energy (there is more attraction between atoms; less obvious, energy is released when bonds form). So, in the first part of the reaction bond breaking dominates and the potential energy is increasing. In the latter stages of the reaction bond making dominates and potential energy is decreasing. Thus, the graph goes up and then it goes down. This is a very simplistic view, but it will give you a better perspective.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/165957/calculating-the-amount-of-water-condensed-when-cooling-below-the-dew-point
|
Calculating the amount of water condensed when cooling below the dew point
|
I've been given the following question:
>
> Calculate the dew point temperature of a 60% humid air at $30^oC$. How much water mass wil condensate out of 1 $m^3$ of such air when the temperature is lowered to $5^oC$?
>
>
>
I tried to find the dew point temperature corresponding to those conditions, which is $T\_d=21.55^oC$ and then I took the density of the partial water at that temperature and at $5^oC$ from thermodynamic stables. That would be
$$
\rho=0.01943[ \frac{kg}{m^3}] ~@~ 22^oC
, \rho=0.006797[ \frac{kg}{m^3}] ~@~ 5^oC
$$
And I calculated the mass condensated by $m=(\rho\_{22} - \rho\_5)\*[m^3]=0.01264 [kg]$
But according to our course professor, he says that all of the water vapor will condense. So
$
m=\rho\_{22}\*[m^3]=0.01943[kg]
$
Why is that? Does always all the water condense when cooling in constant pressure below the dew point temperature? Will it ever depend on the specific temperature I'm cooling to (in our case, $5^oC$)? Are there scenarios where my method would be correct?
Edit: It wasn't given in the question, but i think he assumed that we are cooling the air in constant pressure. First I calculated the mass of water vapor at $30^oC$ and I used the ideal gas equation with the pressure of $P(30,\nu)=RH\*P\_{sat}(30^oC)$, where RH stands for relative humidity.
If I look at the following $T-\nu$ diagram,
[](https://i.stack.imgur.com/YnJ3i.png)
the pressure line of the dew point is $P\_{sat}(22^oC)$, and if the temperature is further lowered to $5^oC$ I get
$P(5^oC,v)=Psat(22^oC)=> 5^oC$
which is far below the saturation line. I know that cooling below the dew point will condensate the excess water vapour. In that case, will all of the mass that I calculated in the start of the process condence (0.0126 kg)?
| 2
|
[
[
"\nThe maximum vapor pressure of water is $\\pu{4243 Pa}$ at $\\pu{30°C}$, and $\\pu{872 Pa}$ at $\\pu{5°C}$. If the original humidity is $\\pu{60}$%, it means that the initial vapor pressure is $\\pu{0.600 · 4243 Pa = 2546 Pa}$.\n\n\nThe initial amount of water in $\\pu{1 m3}$ humid ($\\pu{60}$%) air at $\\ce{30°C}$ is : $\\ce{n\\_{H2O} = pV/RT = \\frac{2546 Pa·1 m^3}{8.314 J K^{-1}mol^{-1}·303 K} = 1.0106 mol. This amount of water weighs 18.19 g}$.\n\n\nIf cooled down to $\\pu{5°C}$, $\\pu{1 m^3}$ air has a volume equal to $\\pu{\\frac{278}{303} 1 m^3 = 0.917 m^3}$.\nAt $\\pu{5°C}$, the amount of water remaining in $\\pu{0.917 m3}$ air (saturated in water) is :\n\n\n$\\ce{n'\\_{H2O} = pV/RT = \\frac{872 Pa · 0.917 m^3}{8.314 J K^{-1} mol^{-1} 278 K} = 0.345 mol}$. This amount of water weighs $\\pu{6.22 g}$\n\n\nThe water condensed during the cooling operation weighs : $\\pu{18.19 g - 6.22 g = 11.97 g}$.\n\n\nSo the proportion of water condensed during the cooling to $\\pu{5°C}$ of $\\pu{1 m^3}$ air taken at $\\pu{30°C}$ with $60$% humidity is $\\pu{\\frac{11.97 g}{18.19 g} = 65.80}$%. It is much smaller than $\\pu{100}$%. This result is far from your teacher's statement.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/165950/which-quantum-numbers-does-the-orbital-angular-momentum-depend-on
|
Which quantum numbers does the orbital angular momentum depend on? [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
---
**Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)
Closed last year.
[Improve this question](/posts/165950/edit)
I learned from a question that I recently solved that the orbital angular momentum depends on both the azimuthal and magnetic quantum numbers.
I did not expect this because the formula for the orbital angular momentum is $ \sqrt{l(l+1)} \frac {h}{2π} $ and so I expected it to only depend on the azimuthal quantum number.
So, why does orbital angular momentum depend on the magnetic quantum number? A detailed explanation will be appreciated.
| -3
|
[
[
"\n*Orbital angular momentum*, like any other form of angular momentum, is a vector quantity.\\* Thus, it has both a magnitude as well as a direction.\n\n\nThe *magnitude* of the orbital angular momentum is given by the formula you wrote, and is thus only dependent on the quantum number $l$:\n\n\n$$|L| = \\hbar\\sqrt{l(l+1)}.$$\n\n\nHowever, the *orientation* of the angular momentum vector is partly described by the quantum number $m$. In particular, the $z$-component of the orbital angular momentum is given by\n\n\n$$L\\_z = m\\hbar.$$\n\n\nIt's a quirk of quantum mechanics, however, that you can't simultaneously know all three components $(L\\_x, L\\_y, L\\_z)$, so this is the best we can really do. Ultimately, this is because the operators $L\\_x, L\\_y, L\\_z$ don't commute with one another.\n\n\nSo, given that we know the magnitude $|L|$ as well as the $z$-component $L\\_z$, we can say that the orbital angular momentum lies somewhere along a *cone*, as the following picture from Atkins' *Molecular Quantum Mechanics* (4th ed.) shows. Atkins uses the general symbol $I$ to refer to some kind of angular momentum; the fact that it's orbital angular momentum doesn't change anything. Indeed, the [Wikipedia page on magnetic quantum number](https://en.wikipedia.org/wiki/Magnetic_quantum_number) has a very similar picture, as pointed out in the comments.\n\n\n[](https://i.stack.imgur.com/wgcme.png)\n\n\n\n\n---\n\n\n\\* Pseudovector, for the picky.\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/165936/reaction-between-1-2-bisbuta-1-3-dien-1-ylcyclohexan-1-ol-and-potassium-hydrid
|
Reaction between 1,2-bis(buta-1,3-dien-1-yl)cyclohexan-1-ol and potassium hydride in THF
|
>
> Which product is formed in the following reaction?
>
>
> [](https://i.stack.imgur.com/Qcenu.png)
>
>
>
I am quite certain that we deal with an [oxy-Cope rearrangement](https://en.wikipedia.org/wiki/Oxy-Cope_rearrangement) that works at room temperature, but I don’t know how to draw the product given the symmetry.
| 7
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[
[
"\nYou are correct, it is indeed an oxy-Cope rearrangement and the macroexpansion gives the 14-membered ketone. From Paquette [[1, p. 13979](https://doi.org/10.1016/S0040-4020(97)00679-0)]:\n\n\n\n> \n> In like fashion, the macroexpansion of **38** and related compounds proceeds in a stepwise manner via enolates **39a** and **39b**. Since **39a** does not accumulate as the reaction proceeds, the second purely Cope rearrangement must be accelerated relative to the first. Trienone **40** has served as a precursor to (+)-muscone and (−)-(3*Z*)-cembrene A.\n> \n> \n> [](https://i.stack.imgur.com/MrY12.png)\n> \n> \n> \n\n\nThe reaction was reported by Wender et al. [[2, p. 3349](https://doi.org/10.1021/ja00348a072)]:\n\n\n\n> \n> When treated with KH in THF at room temperature, both *trans*- and *cis*-**10** gave a mixture of two *E*/*Z* isomers of the 15-membered tetraeneone **11** (61% and 40%, respectively). Catalytic hydrogenation of this mixture gave a 97% yield of muscone (**12**).\n> \n> \n> [](https://i.stack.imgur.com/lVz05.png)\n> \n> \n> \n\n\nStructural assignment based on IR and NMR studies suggest the resulting product is the *E*,*E*,*E*-trienone [[3](https://doi.org/10.1016/S0040-4020(01)93271-5)]. The process of ring expansion via Cope rearrangement appears to be initially investigated by Cookson and Singh in late 1960s in the context of obtaining macrocyclic musk ketone [[4](https://doi.org/10.1039/j39710001477)]:\n\n\n\n> \n> Reaction of the keto-allene (**III**) with vinylmagnesium bromide in tetrahydrofuran gave the vinyl alcohol (**VII**) in good yield.\n> \n> \n> …compound (**VII**) undergoes Cope rearrangement to the enol of (**VIII**) before it fragments. Complete rearrangement through to the ketone (**VIII**) before fragmentation cannot have occurred, for there\n> were distinct differences in the mass spectra of the alcohol (**VII**) and ketone (**VIII**)…\n> \n> \n> The compound (**VII**) though fairly stable, showed the appearance of a carbonyl band, evidently by rearrangement to (**VIII**), when left for a prolonged period.\n> \n> \n> [](https://i.stack.imgur.com/X3ETt.png)\n> \n> \n> \n\n\n### References\n\n\n1. Paquette, L. A. Recent Applications of Anionic Oxy-Cope Rearrangements. *Tetrahedron* **1997**, 53 (41), 13971–14020. DOI: [10.1016/S0040-4020(97)00679-0](https://doi.org/10.1016/S0040-4020(97)00679-0).\n2. Wender, P. A.; Holt, D. A.; Sieburth, S. M. Practical Method for α-Alkenyl Ketone Synthesis Based on a Facile Reductive Rearrangement of Alkynyl Halohydrins. *J. Am. Chem. Soc.* **1983**, 105 (10), 3348–3350. DOI: [10.1021/ja00348a072](https://doi.org/10.1021/ja00348a072).\n3. Wender, P. A.; Sieburth, S. McN.; Petraitis, J. J.; Singh, S. K. Macroexpansion Methodology. *Tetrahedron* **1981**, 37 (23), 3967–3975. DOI: [10.1016/S0040-4020(01)93271-5](https://doi.org/10.1016/S0040-4020(01)93271-5).\n4. Cookson, R. C.; Singh, P. Expansion of the Ring of Cyclododecanone by Two and Four Carbon Atoms. *J. Chem. Soc., C* **1971**, 1477. DOI: [10.1039/j39710001477](https://doi.org/10.1039/j39710001477).\n\n\n",
"8"
]
] |
https://chemistry.stackexchange.com/questions/165935/when-a-compound-is-soluble-in-a-dilute-mineral-acid-is-it-necessary-that-it-is
|
When a compound is soluble in a dilute mineral acid, is it necessary that it is also soluble in that concentrated acid?
|
For example, Barium Phosphate is soluble in dilute mineral acids. Does that also mean that it is also soluble in concentrated form of those acids?
| -2
|
[
[
"\nWell. Barium phosphate $\\ce{Ba3(PO4)2}$ is not soluble at all in sulfuric acid $\\ce{H2SO4}$ solutions, whatever their concentrations, because it forms barium sulfate $\\ce{BaSO4}$ which is an insoluble compound. On the other hand it is soluble in diluted solutions of hydrochloric acid, according to $$\\ce{Ba3(PO4)2 + 6 HCl -> 3 BaCl2 + 2 H3PO4}$$ Unfortunately it is not soluble in concentrated $\\ce{HCl}$ solutions, because it produces some barium chloride $\\ce{BaCl2}$ which has the strange property of being insoluble in concentrated $\\ce{HCl}$ solutions, as the solubility product of $\\ce{BaCl2}$ is exceeded.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/165933/shouldnt-the-cl-be-placed-beside-the-charged-n-ion-when-writing-the-structura
|
Shouldn't the Cl- be placed beside the charged N+ ion when writing the structural formula of Benzenediazonium chloride?
|
I found the structural formula of Benzenediazonium chloride written on a website:
[](https://i.stack.imgur.com/RXYAr.png)
I believe that the Cl- ion is stabilizing the N+ ion.
[](https://i.stack.imgur.com/YnRpw.png)
So, shouldn't the structural formula be written like the above instead?
| 0
|
[
[
"\nWith respect to the point of your question, both ion placements are acceptable for the specified form of the salt's structural formula.\nFrom section *GR-7.2 Positioning of components* [[1, p. 383](https://doi.org/10.1351/pac200880020277)] (emphasis mine):\n\n\n\n> \n> The various components of a salt may be positioned relative to each other in one of two ways. If one of the components consists of a single atom, it may be “paired up” with an oppositely charged atom in a larger component, with the positioning of the larger fragment determined by other considerations as discussed in GR-3. The fragments should be positioned as if there were a single bond between the cationic\n> and anionic centers, even if this requires that the cation be drawn on the right side of the anion. The positioning of the smaller fragment will then usually be determined according to other recommendations (that straight chains should not have arbitrary bends in them, for example).\n> \n> \n> [](https://i.stack.imgur.com/dxbjk.png)\n> \n> \n> *Alternatively, the components of a salt may be depicted next to each other. This positioning is acceptable for single-atom components, and preferred for larger ones.* If the components are of similar size, positively charged components should preferably be positioned to the left of negatively charged ones; if they are significantly different in size, the larger components should be positioned to the left of the smaller ones.\n> \n> \n> [](https://i.stack.imgur.com/god0t.png)\n> \n> \n> \n\n\n### Notes\n\n\n1. Negative charge should be depicted with superscripted minus $(\\ce{Cl-}),$ not en dash $(\\ce{Cl^\\text{-}}).$\n2. Although being acceptable, a circle is rarely used for the depiction of aromatic systems in publications beyond introductory level chemistry.\n3. Chemical names are common nouns and inherit the capitalization rules: do not capitalize chemical names in the middle of the sentence.\n4. An inline semistructural formula of benzenediazonium chloride could be $\\ce{Ph-N2+ Cl-}.$\n5. Even in the crystal structure ([[2](https://doi.org/10.3891/acta.chem.scand.17-1444)]; [CSD Entry: BZDIZC](https://www.ccdc.cam.ac.uk/structures/Search?Ccdcid=BZDIZC&DatabaseToSearch=Published)) $\\ce{Cl-}$ anion is equally distanced from nitrogen atoms in diazonium (an isosceles triangle with the legs of ~3.55 Å or ~3.23 Å):\n[](https://i.stack.imgur.com/iGlor.png)\n\n\n### References\n\n\n1. Brecher, J. Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008). *Pure Appl. Chem.* **2009**, 80 (2), 277–410. DOI: [10.1351/pac200880020277](https://doi.org/10.1351/pac200880020277). (Free Access)\n2. Rømming, Chr.; Karvonen, P.; Holm, A.; Nielsen, P. H.; Munch-Petersen, J. The Structure of Benzene Diazonium Chloride. *Acta Chem. Scand.* **1963**, 17, 1444–1454. DOI: [10.3891/acta.chem.scand.17-1444](https://doi.org/10.3891/acta.chem.scand.17-1444). (Free Access)\n\n\n",
"9"
],
[
"\nYou are right. But it is not always possible to print this formula with $\\ce{Cl-}$ ion **UNDER** the diazo line. It depends on the programs available on your computer, specially if you replace the benzene ring by the formula $\\ce{C6H5}$, and if you want to type it on one line in a text.\n\n\n",
"-3"
]
] |
https://chemistry.stackexchange.com/questions/165932/if-sulfur-is-directly-below-oxygen-on-the-periodic-table-why-isnt-so-the-most
|
If sulfur is directly below oxygen on the periodic table, why isn't SO the most common sulfur oxide? Instead of third, behind SO2 and SO3? [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/165932/edit).
Closed last year.
[Improve this question](/posts/165932/edit)
Chalcogen like oxygen and sulfur want to 'grab' (or share) two electrons to get to eight, if possible, or maybe 'give up' (or share) six, if necessary, to again get to eight... Right?
So, why is sulfur dioxide, where the sulfur shares four electrons (usually), more common than sulfur monoxide or sulfur trioxide?
Isn't sulfur dioxide a higher energy, less stable molecule?
| -4
|
[
[
"\nNobody really knows why $\\ce{SO}$ is such a rare and unstable substance. It is an orange gas at room temperature. The only way to synthesize $\\ce{SO}$ is to reduce thionyl chloride $\\ce{SOCl2}$ with some reagent eager to react with the chlorine atom, like sodium, silver or tin : $$\\ce{SOCl2 + 2 Na -> SO + 2 NaCl}$$ The strangest property of this gas is its reaction with water. It is decomposed with the smallest amount of water vapor according to the disproportion equation : $$\\ce{3SO + H2O -> H2S + 2 SO2}$$ It is an extraordinary reaction, as it requires $3$ molecules $\\ce{SO}$. Another oxide is known to disproportionate with water : $\\ce{NO2}$. But there is a good reason for this behavior : $\\ce{NO2}$ has an uneven number of electrons. It is not the case for $\\ce{SO}$.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/165919/nylon-6-6-synthesis-yield
|
Nylon 6,6 synthesis yield
|
This is the nylon 6,6 reaction
[](https://i.stack.imgur.com/nMgi2.jpg)
Having these data available, I would need to know the theoretical quantities, in grams, of the monomers
[](https://i.stack.imgur.com/KDaPN.jpg)
My attempt:
Adipoyl chloride is our limiting reactant so theoretical mass is the weighted mass.
Hexamethylenediamine and adipoyl chloride are in 1:1 ratio, so amine theoretical mol are the same of adipoyl chloride: if the calculations are correct, $0.0022\;\text{mol}$.
From which, the theoretical mass of amine will be
$$ 0.0022\;\text{mol} \times 116.21\;\text{g/mol} = 0.2557\;\text{g} $$
Any advice?
| 2
|
[] |
https://chemistry.stackexchange.com/questions/165918/how-to-define-a-chemical-bond-computationally
|
How to define a chemical bond computationally?
|
I'm working with an amorphous system. With oxygen, my system has both covalent and ionic bond forming cations. I utilise the [Wannier centre](https://en.wikipedia.org/wiki/Wannier_function) to define a covalent bond, and my theory is that if the Wannier centre is near to the line connecting the cation and anion, it is a covalent bond.
However, this method fails to identify an ionic bond as there is no region of high electronic charge density which can be localized to obtain Wannier center. How can an ionic bond be defined using a computational technique?
| 0
|
[] |
https://chemistry.stackexchange.com/questions/165917/is-calcium-hypochlorite-as-effective-as-sodium-hypochlorite-in-sanitizing-a-swim
|
Is calcium hypochlorite as effective as sodium hypochlorite in sanitizing a swimming pool from algae?
|
**Question**
From this [question](https://chemistry.stackexchange.com/q/165903/101416) we know that what swimming pool stores refer to as "liquid chlorine" (sold in 10 gallon jugs) is most likely [sodium hypochlorite ($\ce{NaOCl}$)](https://chemistry.stackexchange.com/a/165904/101416). In the spirit of [Least Publishable Units (LPUs)](https://en.wikipedia.org/wiki/Least_publishable_unit) I'll ask a brief sequel separate from the first question.
The worst is when microorganisms (algae) overrun the pool. In the presence of summer heat, ample sunlight, and perhaps some nutrients (just phosphates?), the green population can multiply rapidly. *Is calcium hypochlorite ($\ce{Ca(ClO)2}$) (sold in crystal form) as effective as sodium hypochlorite ($\ce{NaOCl}$) in sanitizing a swimming pool from algae?*
**Context**
A certain talented [fictional](https://www.imdb.com/title/tt0903747/) chemistry teacher whose fame rests on illicit activities apparently had little difficulty maintaining good water quality in his swimming pool. Yet anyone who has tried will attest that maintaining immaculate water clarity is sometimes a nontrivial objective.
This is the second in what may be a series of questions trying to understand the basics of maintaining a swimming pool. A quick search online will show an *enormous* number of results—all amateurish and none technical, hence the need for learned opinion.
| 3
|
[
[
"\n[Yim et al. [1]](https://doi.org/10.4142/jvs.2021.22.e11) studied the effectiveness of various hypochlorite salt treatments against a particular strain of bacterial cells and Speers. They found that calcium hypochlorite was much more effective than sodium hypochlorite... but it should he used with care because if its high corrosivity:\n\n\n\n> \n> Although HTH [calcium hypochlorite] is effective for decontaminating these B. anthracis strains, it is considered extremely corrosive to metals, skin and mucous membranes, eyes, and respiratory and gastrointestinal tract [2]. Therefore, to avoid material corrosion or toxicity, alternative disinfectants with high efficacy are needed.\n> \n> \n> \n\n\n**References**\n\n\n1. Jin-Hyeok Yim, Kwang-Young Song, Hyunsook Kim, Dongryeoul Bae, Jung-Whan Kun-Ho Seo (2021). \"Effectiveness of calcium hypochlorite, quaternary ammonium compounds, and sodium hypochlorite in eliminating vegetative cells and spores of Bacillus anthracis surrogate\". *J Vet Sci.* 2021 Jan; **22**(1): e11.\nPublished online 2021 Jan 18. <https://doi.org/10.4142/jvs.2021.22.e11>\nPMCID: PMC7850788\nPMID: 33522163\n2. Rogers JV, Ducatte GR, Choi YW, Early PC (2006). A preliminary assessment of Bacillus anthracis spore inactivation using an electrochemically activated solution (ECASOL) *Lett Appl Microbiol.* 2006; **43**:482–488.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/165910/why-do-we-have-a-negative-current-in-cyclic-voltammetry-curves
|
Why do we have a negative current in cyclic voltammetry curves?
|
Why do we have a negative current in cyclic voltammetry curves?
here are my thoughts. We have two conventions in reporting data of CV curves: the IUPAC convention and the US convention. In the IUPAC convention, the potential scan direction is from lower potential to higher and the US convention is reversed of the IUPAC convention (from higher potential to lower potential). So I think it's all about the direction of current in an electrochemical cell. because anodic current will be positive in the IUPAC convention and negative in the US convention. Therefore, maybe It does not relate to cathodic or anodic current. Can anyone explain more? what is the concept of the negative current?
| 0
|
[] |
https://chemistry.stackexchange.com/questions/165905/what-are-oxidation-numbers-of-each-element-in-cuh
|
What are oxidation numbers of each element in CuH?
|
My chemistry teacher said hydrogen when bonded with metals has oxidation number (O.N.) of −1 *except* CuH: in CuH the hydrogen has O.N. +1 and the copper has O.N. −1.
But on the internet it is ubiquitously told that the O.N. of hydrogen is −1 in CuH too, and the compound's name is [copper(I) hydride](https://en.wikipedia.org/wiki/Copper_hydride). I am confused which one is correct. If hydrogen has O.N. of +1 in CuH, why is it so?
| 2
|
[
[
"\nKorzhavi and Johanssen [1](https://www.researchgate.net/publication/267401061_Svensk_Karnbranslehantering_AB_Thermodynamic_properties_of_copper_compounds_with_oxygen_and_hydrogen_from_first_principles) performed first-principles calculations on the electronic energy levels of several compounds of copper, hydrogen and oxygen. They find good agreement between derived and experimental properties of stable compounds such as $\\ce{Cu2O}$ and correctly predict the instability of other candidate compounds such as $\\ce{CuOH}$ and $\\ce{CuH2}$.\n\n\nFor $\\ce{CuH}$ the calculated density of electron states shows the hydrogen-based 1s states filled, and lower in energy than the copper-based 3d states. This is similar to the oxygen- and hydroxyl-based states in the oxide and hydroxide. The copper hydride result thereby identifies that compound as a metal hydride with positive metal oxidation state.\n\n\n[](https://i.stack.imgur.com/9jTLd.png)\n\n\n**Reference**\n\n\n1. Korzhavyi, Pavel & Johansson, Borje. \"Thermodynamic properties of copper compounds with oxygen and hydrogen from first principles.\" Report to Svensk Kärnbränslehantering AB, Texhnical Report TR-10-30, February 2010. Retrieved 19 June 2022.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/165903/are-liquid-chlorine-and-super-shock-chlorine-the-same
|
Are liquid chlorine and super-shock chlorine the same?
|
To shock a swimming pool one uses either calcium hypochlorite ($\ce{Ca(ClO)2}$), commonly sold from at least one major brand in pouches,
[](https://i.stack.imgur.com/tkQQBm.png)
or the more obscure "liquid chlorine".
[](https://i.stack.imgur.com/rrkSHm.png)
I'm saying "obscure" because labels for liquid chlorine never list the compound.
But unlike the result of dissolving $\ce{Ca(ClO)2}$ in water, which results in, by definition, a bleached (white) color, liquid chlorine is, somewhat surprisingly, yellow.
Are liquid chlorine and super-shock chlorine the same?
| 0
|
[
[
"\nLiquid chlorine is probably bleach, which is a solution of sodium hypochlorite $\\ce{NaOCl}$. This substance is obtained by dissolving chlorine $\\ce{Cl2}$ in a solution of sodium hydroxide $\\ce{NaOH}$, according to $$\\ce{Cl2 + 2 NaOH -> NaOCl + NaCl + H2O}$$ Chlorine is produced by adding an acid to this solution : the acid destroys $\\ce{NaOH}$ and produced free chlorine $\\ce{Cl2}$ by reversing the previous equation (which should be written as an equilibrium). Even weak acids like carbonic acid (and $\\ce{CO2}$) can do the job. This is why the solution deserves to be called \"liquid chlorine\".\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/165902/evaporation-of-zinc-oxide
|
Evaporation of zinc oxide
|
My question involves the evaporation of silicate material (e.g., rocks, minerals, geologic material), specifically I am trying to work the number of electrons (*n*) that are involved in the evaporation reactions described by [Sossi et al. (2019)](https://www.sciencedirect.com/science/article/abs/pii/S0016703719303606) and [Sossi and Fegley (2018)](https://pubs.geoscienceworld.org/msa/rimg/article-abstract/84/1/393/566363/Thermodynamics-of-Element-Volatility-and-its?redirectedFrom=fulltext).
As an example, I will discuss the evaporation of zinc oxide (ZnO), which is shown in multiple sources (see footnotes) as:
[](https://i.stack.imgur.com/x4VmQ.png)
Sossi et al. (2019) state that this reaction involves two electrons (*n*=+2), however they do not indicate how they came to this value, so I have been trying to work it out myself. I should also note that they describe these reactions as **producing a metal bearing gas**.
I have two lines of thinking for determining *n* for this reaction:
***First line of reasoning***
ZnO(l) has two components with charges Zn = +2 and O = -2
Zn(g) evaporates as a metal, so the charge is Zn = 0
and
1/2O2 is just oxygen gas with charge 1/2O2 = 0.5\*(-2\*2) = -2
This line of reasoning implies that there was a change in the charge on Zn of -2. In other words, two electrons were lost, so two electrons were involved in the reaction. It involves the production of gaseous zinc metal, which fits nicely into the description of a metal bearing gas, at least in my mind. **However, this leaves the reaction unbalanced because the products have a -2 charge, which leads me to my second line of reasoning.**
***Second line of reasoning***
ZnO(l) his a stable compound, so ZnO = 0 charge
Zn(g) evaporates as a cation, so the charge is Zn = +2
and
1/2O2 is just oxygen gas with charge 1/2O2 = 0.5\*(-2\*2) = -2
Now the reaction is charge balanced, **but I am not clear on how two electrons were involved in this case, unless I can consider the two electrons that "went with" Zn after the evaporation of ZnO** as the two electrons that are involved in the reaction. However, I don't know if Zn++(g) counts as a metal bearing gas, which the authors say is present.
I have been tying myself in knots trying to work this out, so any guidance in the right direction would be helpful.
*Footnotes*
Other sources that discuss the evaporation of ZnO include [Anthrop and Searcy (1964)](https://pubs.acs.org/doi/pdf/10.1021/j100790a052?casa_token=1KAaaACIL_kAAAAA:iagmpmVhquDMTO8Imm7jQpA4PblgJYkPoGqc6wni_s26NBerBCdIwkjxMbX-sfgcE51FpaU5rU49e0PURw) and [Kodera et al. (1968)](https://www.journal.csj.jp/doi/abs/10.1246/bcsj.41.1039)
| 1
|
[
[
"\nYour following sentence is wrong (I copy it in italics without change)\n\n\n*1/2O2 is just oxygen gas with charge 1/2O2 = 0.5*(-2*2) = -2*\n\n\nNo! There is no charge $-2$ on the oxygen atom in the molecule $\\ce{O2}$, and no charge at all in the molecule $\\ce{O2}$ There is a charge $-2$ on the oxygen atom in the initial oxide $\\ce{ZnO}$. But during the decomposition of $\\ce{ZnO}$, the two electrons available on the Oxygen ion are transferred to the zinc ion $\\ce{Zn^{2+}}$ which becomes a non-charged atom $\\ce{Zn}$.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/165896/does-e1cb-take-place-over-here
|
Does E1cB take place over here?
|
[](https://i.stack.imgur.com/ZRsLbm.png)
I formed this product (reason: alkene has 6 alpha hydrogens)
---->
[](https://i.stack.imgur.com/dGdTfm.png)
However, it turns out, this is the answer:
[](https://i.stack.imgur.com/vEIWrm.png)
The reason why I suspect this is e1cb is because NO2 could stabilize the carboanion (conjugate base) formed after reaction with NaOCH3
I would also like to know if a substrate has two sites for E2 elimination to occur, how do I decide site which would be preferred. Is it the one with the most acidic hydrogen or the one which would lead to a more stable alkene?
| 1
|
[] |
https://chemistry.stackexchange.com/questions/165890/does-manganese-reacts-with-water-at-not-standard-conditions
|
Does manganese reacts with water at "not" standard conditions?
|
In [my answer](https://chemistry.stackexchange.com/a/165867/17368), I stated that manganese is said to not react with water *under normal conditions* although some sources say it reacts with water to liberate hydrogen gas. Does it implies that it reacts with water at harsh conditions?
[WebElements](https://webelements.com/manganese/chemistry.html) and [ChemLibreTexts](https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/3_d-Block_Elements/Group_07%3A_Transition_Metals/Chemistry_of_Manganese) states that "Manganese does not react with water *under normal conditions*" whereas [Lenntech](https://www.lenntech.com/periodic/elements/mn.htm) says "it reacts with water (**it rusts like iron**) and dissolves in dilute acids." What does rusting here mean? Does manganese rust like iron?
In [some sources](http://manganese.atomistry.com/chemical_properties.html) and *Concise Inorganic Chemistry* by J.D.LEE states "At room temperature very slowly reacts with water"
$$\ce{Mn + 2H2O -> Mn(OH)2 + H2}$$
and [some says that it reacts and oxidizes all the way to permanganate.](https://pilgaardelements.com/Manganese/Reactions.htm) The results are inconsistent and sometimes contradictory.
**Question**: Does manganese reacts with water? If so, in what condition (standard or harsh)?
| 4
|
[
[
"\nThe Lenntech phrasing \"reacts with water (it rusts like iron)\" is incomplete. [Wikipedia](https://en.wikipedia.org/wiki/Manganese) states that the reaction occurs with \"water containing dissolved oxygen\". This would actually be like iron, which also requires dissolved oxygen to react in water. With either metal the rusting reaction involves the metal being oxidized to ions while water and oxygen are reduced to hydroxide ions, leading to the formation of $\\ce{Mn(OH)2}$ or $\\ce{Fe(OH)2}$; but these hydroxides are prone to additional oxidation by oxygen or air to form oxides or oxide-hydroxides with the metal in higher oxidation states.\n\n\nThe above mechanism requires dissolved oxygen as one of the reactants to form the hydroxide and oxide ions; neither metal appreciably displaces hydrogen from otherwise pure water at room temperature. However manganese, again like iron, will react with steam to form oxide(s) plus hydrogen.\n\n\n",
"2"
],
[
"\nthank you for the excellent question. It was really thought provoking indeed. Here's the answer :\n\n\n1. Yes, it (Manganese) reacts with water.\n2. As Concise Inorganic Chemistry by J.D.LEE states it reacts with water in STANDARD conditions (if given enough time; which basically means the rate of the reaction is super slow)\n3. As it will react in standard conditions, it might also react in harsh conditions (ie. 100°C or 50°C). \"Harsh\" conditions weren't properly defined in the question. So, I am unsure about what you meant there.\n4. \"Rusts like iron\" means it oxidizes like iron. However, from the research done by Mingfei Zhou, Luning Zhang, Limin Shao, Wenning Wang, Kangnian Fan, Qizong Qin, it seems like Manganese Hydroxide is created upon reaction and not Manganese oxide as suggested by \"rusts like iron\".\n\n\n",
"1"
],
[
"\n**Reaction of manganese with water**\n\n\nThe interaction of manganese with water can be described by the following reaction equation: $$\\ce{Mn(s) + 2H2O(l) -> Mn(OH)2 (s) + H2 (g)}$$\nMost likely, the reaction is slowed down due to the poor solubility of the resulting [manganese (II) hydroxide](https://en.wikipedia.org/wiki/Manganese(II)_hydroxide) which coats the surface of the reacting manganese and prevents the metal from reacting with water. It is reasonable to say that manganese will react with warm water faster as stated in [4].\n\n\n**References**\n\n\nHope my answer will bring a bit of clarity to the problem. It is also supplemented with some reference links concerning the interaction of manganese with water:\n\n\n1. Ronald L. Rich. *Inorganic Reactions in Water (1st ed.)*: \"Water\", p. 153.\n2. C.E. Housecroft, A.G. Sharpe. *Inorganic Chemistry*: \"21.8 Group 7: manganese\", p. 738.\n3. N.N. Greenwood, A. Earnshaw. *Chemistry of the Elements (2nd ed.)*: p. 1044.\n4. P. Patnaik. *Handbook of Inorganic Chemicals*: p. 542.\n\n\n",
"1"
]
] |
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