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Cell voltage and Gibbs free energy .txt | So let's see what the relationship is. So, let's finally examine and see what the relationship is between cell voltage and changing gear free energy. Well, we see that changing gear's free energy is equal to negative of electrical work done. Well, why is that? Well, this guy is simply the amount of free energy available to do useful work, while with this guy, the electrical work is the amount of energy transformed from chemical energy to electrical energy in the form of moving electrons. And this is also energy used to do useful work. |
Cell voltage and Gibbs free energy .txt | Well, why is that? Well, this guy is simply the amount of free energy available to do useful work, while with this guy, the electrical work is the amount of energy transformed from chemical energy to electrical energy in the form of moving electrons. And this is also energy used to do useful work. So that's why the magnitude of these guys is the same, but the sign is different. So, why is it that we have a negative sign in front of the electrical work? Well, let's see why. |
Cell voltage and Gibbs free energy .txt | So that's why the magnitude of these guys is the same, but the sign is different. So, why is it that we have a negative sign in front of the electrical work? Well, let's see why. The negative sign accounts for the fact that for a product favorite reaction, changing gibsi energy is negative while cell voltage is positive. So that's why we need to add that negative sign in. So, from before, we saw that electrical work is equal to number of moles times starbase constant times our cell voltage. |
Cell voltage and Gibbs free energy .txt | The negative sign accounts for the fact that for a product favorite reaction, changing gibsi energy is negative while cell voltage is positive. So that's why we need to add that negative sign in. So, from before, we saw that electrical work is equal to number of moles times starbase constant times our cell voltage. So change and gives the energy is equal to negative m times f times cell voltage. So that's our reaction. That's our equation. |
Cell voltage and Gibbs free energy .txt | So change and gives the energy is equal to negative m times f times cell voltage. So that's our reaction. That's our equation. Now, suppose we have an electrochemical cell able to take cell that's composed of the following redox reaction. So it has an oxidation reaction and a reduction reaction. Our zinc solid is oxidized into zinc ion, while our copper ion is reduced to copper solid. |
Cell voltage and Gibbs free energy .txt | Now, suppose we have an electrochemical cell able to take cell that's composed of the following redox reaction. So it has an oxidation reaction and a reduction reaction. Our zinc solid is oxidized into zinc ion, while our copper ion is reduced to copper solid. So, this guy releases two electrons, while this guy takes those two electrons, gains two electrons forming our copper solid. Now, these electrons in the end, end up crossing out. Now, our cell voltage is 1.1 volt. |
Cell voltage and Gibbs free energy .txt | So, this guy releases two electrons, while this guy takes those two electrons, gains two electrons forming our copper solid. Now, these electrons in the end, end up crossing out. Now, our cell voltage is 1.1 volt. So let's use this formula to approximate how much energy this battery or this electrochemical cell can produce with the charge that it has. So, the change in gates free energy is equal to negative of the number of moles of electrons. Well, in this case, I specifically left these guys in to show you how many moles of electrons we have. |
Cell voltage and Gibbs free energy .txt | So let's use this formula to approximate how much energy this battery or this electrochemical cell can produce with the charge that it has. So, the change in gates free energy is equal to negative of the number of moles of electrons. Well, in this case, I specifically left these guys in to show you how many moles of electrons we have. Well, zinc loses two moles of electrons, and this guy gains two moles of electrons. That means we're dealing with two moles of electrons times our Faradays constant 96,500 approximately Coulombs per mole of electron times our cell voltage or electromotive force 1.1 volt. We see that these guys cancel out and our units are left Coulombs times voltage, which is simply joules. |
Cell voltage and Gibbs free energy .txt | Well, zinc loses two moles of electrons, and this guy gains two moles of electrons. That means we're dealing with two moles of electrons times our Faradays constant 96,500 approximately Coulombs per mole of electron times our cell voltage or electromotive force 1.1 volt. We see that these guys cancel out and our units are left Coulombs times voltage, which is simply joules. So in the end we get 212,300 joules of work. It's produced is released in the form of moving electrons or moving charge from the anode to the capital from this battery. And this amount of energy can be used to do useful work. |
Cell voltage and Gibbs free energy .txt | So in the end we get 212,300 joules of work. It's produced is released in the form of moving electrons or moving charge from the anode to the capital from this battery. And this amount of energy can be used to do useful work. For example, run a motor or power a light bulb, for example. So this becomes very useful and we'll do many more examples using this in the future. So I already spoke about this briefly in another lecture, but I want to talk about it once more. |
Cell voltage and Gibbs free energy .txt | For example, run a motor or power a light bulb, for example. So this becomes very useful and we'll do many more examples using this in the future. So I already spoke about this briefly in another lecture, but I want to talk about it once more. Now, the reason that a D battery has more energy than AAA battery is not because a difference of cell voltage. In fact, the cell voltage for this guy and this guy are the same. It's 1.5
volts. |
Cell voltage and Gibbs free energy .txt | Now, the reason that a D battery has more energy than AAA battery is not because a difference of cell voltage. In fact, the cell voltage for this guy and this guy are the same. It's 1.5
volts. You could check the label of the batteries. The reason this is more expensive is because it can do more work. But how can it do more work? |
Exothermic Reactions, Endothermic Reactions and BDE.txt | So let's begin by looking at the following diagram. So, we're basically taking two identical H atoms. We're combining them to form a Diatomic H two Model molecule. So, here are our two identical H atoms. So, this is a one s Atomic orbital of the first H atom and the second Atomic orbital of the second H atom. So we have an electron in each atomic orbital. |
Exothermic Reactions, Endothermic Reactions and BDE.txt | So, here are our two identical H atoms. So, this is a one s Atomic orbital of the first H atom and the second Atomic orbital of the second H atom. So we have an electron in each atomic orbital. So, if we combine two Atomic orbitals according to quantum mechanics, we're going to form two molecular orbitals. One will be the phi B, which is the Bonding molecular orbital. And the second will be the phi A, which is the Anti Bonding molecular orbital. |
Exothermic Reactions, Endothermic Reactions and BDE.txt | So, if we combine two Atomic orbitals according to quantum mechanics, we're going to form two molecular orbitals. One will be the phi B, which is the Bonding molecular orbital. And the second will be the phi A, which is the Anti Bonding molecular orbital. Now, the bonding molecular orbital is lower in energy. It's more stabilizing. And, in fact, this is the bond responsible for forming our covalent bond. |
Exothermic Reactions, Endothermic Reactions and BDE.txt | Now, the bonding molecular orbital is lower in energy. It's more stabilizing. And, in fact, this is the bond responsible for forming our covalent bond. So our electrons will be found in the Lower Energy bonding molecular orbital. So phi b. Now, phi A, or Antiboming Molecular Orbital, is responsible for breaking the bond. |
Exothermic Reactions, Endothermic Reactions and BDE.txt | So our electrons will be found in the Lower Energy bonding molecular orbital. So phi b. Now, phi A, or Antiboming Molecular Orbital, is responsible for breaking the bond. If the electrons are found within this bond, that means those electrons will play a role in Destabilizing our molecule in breaking that Covalent bond. Now, so electrons will be found in this bonding molecular orbital. And notice what happens. |
Exothermic Reactions, Endothermic Reactions and BDE.txt | If the electrons are found within this bond, that means those electrons will play a role in Destabilizing our molecule in breaking that Covalent bond. Now, so electrons will be found in this bonding molecular orbital. And notice what happens. Because these two Atomic orbitals are higher in energy than this molecular orbital, energy will Be lost. So, there is some change in energy that occurs when these two atomic orbitals form this molecular orbital or this molecular Diatomic molecule. Now, this energy is Released into the environment. |
Exothermic Reactions, Endothermic Reactions and BDE.txt | Because these two Atomic orbitals are higher in energy than this molecular orbital, energy will Be lost. So, there is some change in energy that occurs when these two atomic orbitals form this molecular orbital or this molecular Diatomic molecule. Now, this energy is Released into the environment. In other words, when my two H atoms form to create a bond, energy is released. And, in fact, any time we form bonds, energy will always Be Released into the environment. So let's look at this in a more simplified fashion. |
Exothermic Reactions, Endothermic Reactions and BDE.txt | In other words, when my two H atoms form to create a bond, energy is released. And, in fact, any time we form bonds, energy will always Be Released into the environment. So let's look at this in a more simplified fashion. So, here we have two H molecules. At some level, they react. They surmount this Activation barrier, and they form our Diatomic H two molecule, which is lower in energy than the initial. |
Exothermic Reactions, Endothermic Reactions and BDE.txt | So, here we have two H molecules. At some level, they react. They surmount this Activation barrier, and they form our Diatomic H two molecule, which is lower in energy than the initial. So our products are lower than Our reactants. And this change in energy is the same as the change in energy that we saw here. Another name for that is change in enthalpy. |
Exothermic Reactions, Endothermic Reactions and BDE.txt | So our products are lower than Our reactants. And this change in energy is the same as the change in energy that we saw here. Another name for that is change in enthalpy. So change in H. And from 1 mol, where whenever 1 mol of H reacts with Another Mole of H to form 1 Mol of H 2104, energy will Be Released Into The environment. And this is known as an exothermic reaction. In other words, exothermic reaction is A reaction in which the energy of products is lower than the energy of reactants, and the energy is released into the environment. |
Exothermic Reactions, Endothermic Reactions and BDE.txt | So change in H. And from 1 mol, where whenever 1 mol of H reacts with Another Mole of H to form 1 Mol of H 2104, energy will Be Released Into The environment. And this is known as an exothermic reaction. In other words, exothermic reaction is A reaction in which the energy of products is lower than the energy of reactants, and the energy is released into the environment. Now, let's look at the same Exact diagram. But now we're going to work backwards. So we basically want to begin with a diatomic H two molecule, and we want to somehow get to this H two. |
Exothermic Reactions, Endothermic Reactions and BDE.txt | Now, let's look at the same Exact diagram. But now we're going to work backwards. So we basically want to begin with a diatomic H two molecule, and we want to somehow get to this H two. So we essentially want to break this covalent bond and form two separate H molecules or h atoms. So that means since we go from a lower energy to a higher energy, we have to input energy to go from this guy to this guy. So we input energy to break the bond to form two individual H molecules. |
Exothermic Reactions, Endothermic Reactions and BDE.txt | So we essentially want to break this covalent bond and form two separate H molecules or h atoms. So that means since we go from a lower energy to a higher energy, we have to input energy to go from this guy to this guy. So we input energy to break the bond to form two individual H molecules. And this is known as an endothermic reaction. So once again, endothermic reaction is a reaction in which the energy of product is higher than the end of reactants and energy is used up or inputted into the system for that reaction to take place. So once again, anytime we have an endothermic reaction going this way, going the forward direction, we have an endothermic reaction going in the backward or reverse direction. |
Exothermic Reactions, Endothermic Reactions and BDE.txt | And this is known as an endothermic reaction. So once again, endothermic reaction is a reaction in which the energy of product is higher than the end of reactants and energy is used up or inputted into the system for that reaction to take place. So once again, anytime we have an endothermic reaction going this way, going the forward direction, we have an endothermic reaction going in the backward or reverse direction. So once again, to sum this information up, anytime we form a bond, energy is released. Anytime we want to break a bond, energy needs to be inputted into our system, into the molecule. Now, what is bond association energy? |
Exothermic Reactions, Endothermic Reactions and BDE.txt | So once again, to sum this information up, anytime we form a bond, energy is released. Anytime we want to break a bond, energy needs to be inputted into our system, into the molecule. Now, what is bond association energy? Well, bond association energy is the amount of energy required to break a bond. In other words, if we want to break this bond, we need to input some amount of energy. And this is known as the bond association energy. |
Exothermic Reactions, Endothermic Reactions and BDE.txt | Well, bond association energy is the amount of energy required to break a bond. In other words, if we want to break this bond, we need to input some amount of energy. And this is known as the bond association energy. So for example, let's compare two molecules or two bonds. That Ch bond and the CF bond. Now, Ch bond has a bond association of 103 kg/mol. |
Homo-Lumo Examples .txt | So in this reaction, we're going to identify the homo and luma orbitals of a few reactions. Now, before we look at that, let's recall what homo and luma orbitals are. Now, recall that a lewis acidbased reaction is simply a reaction between the highest occupied molecular orbital known as the homo of one kind compound, and the lowest unoccupied molecular orbital of a second compound known as the lumo. So we basically have a pair of electrons, grams, another molecule, or compound. So let's look at the following reaction. So here we have an amine. |
Homo-Lumo Examples .txt | So we basically have a pair of electrons, grams, another molecule, or compound. So let's look at the following reaction. So here we have an amine. We have ammonia that reacts with our H plus ion. So our lewis acid and lewis base. So here we have our molecular orbital that is occupied. |
Homo-Lumo Examples .txt | We have ammonia that reacts with our H plus ion. So our lewis acid and lewis base. So here we have our molecular orbital that is occupied. It has a pair of electrons which grab this unoccupied orbital, one s orbital. So this SP three hybridized molecular orbital is our homo. It's the highest occupied molecular orbital. |
Homo-Lumo Examples .txt | It has a pair of electrons which grab this unoccupied orbital, one s orbital. So this SP three hybridized molecular orbital is our homo. It's the highest occupied molecular orbital. While on this molecule, or actually atom, our lowest unoccupied molecular orbital is the one S orbital. Because we're missing an electron, we have a cat ion. So we have an MP, one S orbital. |
Homo-Lumo Examples .txt | While on this molecule, or actually atom, our lowest unoccupied molecular orbital is the one S orbital. Because we're missing an electron, we have a cat ion. So we have an MP, one S orbital. So this is our lumo, or lowest unoccupied molecular orbital. So when these two orbitals interact, so we have the SP three interacting with the one S, and we form the following two orbitals. We form the bonding molecular orbital and the anti bonding molecular orbital. |
Homo-Lumo Examples .txt | So this is our lumo, or lowest unoccupied molecular orbital. So when these two orbitals interact, so we have the SP three interacting with the one S, and we form the following two orbitals. We form the bonding molecular orbital and the anti bonding molecular orbital. Now, since this orbital is lower in energy, that means our electron pair coming from this highest occupied molecular orbital goes directly into this bonding molecular orbital. So let's look at a few more examples. Now, let's suppose we have example a. |
Homo-Lumo Examples .txt | Now, since this orbital is lower in energy, that means our electron pair coming from this highest occupied molecular orbital goes directly into this bonding molecular orbital. So let's look at a few more examples. Now, let's suppose we have example a. In this reaction, we have a hydroxide with a negative sign interacting with a sodium ion with a positive sign, and we form sodium hydroxide. So which one has the homo and which one has the lumo? So clearly, this is our base. |
Homo-Lumo Examples .txt | In this reaction, we have a hydroxide with a negative sign interacting with a sodium ion with a positive sign, and we form sodium hydroxide. So which one has the homo and which one has the lumo? So clearly, this is our base. Why? Well, because it has a pair of electrons that it can donate that it can use to grab another molecule, in this case, atom. So that means this pair of electrons, the orbital that is found in, must be the homo. |
Homo-Lumo Examples .txt | Why? Well, because it has a pair of electrons that it can donate that it can use to grab another molecule, in this case, atom. So that means this pair of electrons, the orbital that is found in, must be the homo. So the highest occupied molecular orbital is the field non bonding SP three hybridized orbital molecular orbital. So what's the lumo in our case? Well, if this is the homo, this must be the lumo. |
Homo-Lumo Examples .txt | So the highest occupied molecular orbital is the field non bonding SP three hybridized orbital molecular orbital. So what's the lumo in our case? Well, if this is the homo, this must be the lumo. So the lowest unoccupied molecular orbital is the empty three s orbital of our sodium atom. Let's move on to example two. Reaction two. |
Homo-Lumo Examples .txt | So the lowest unoccupied molecular orbital is the empty three s orbital of our sodium atom. Let's move on to example two. Reaction two. Here we have a hydronium acid reacting with our hydroxide base, forming two H, two O molecules. So which one is the homo and which one is the lumo? Now, this one is a bit tricky, and we'll see why in a second. |
Homo-Lumo Examples .txt | Here we have a hydronium acid reacting with our hydroxide base, forming two H, two O molecules. So which one is the homo and which one is the lumo? Now, this one is a bit tricky, and we'll see why in a second. So let's begin with homo. So the homo has the highest or is the highest occupied molecular orbital. And that means our hydroxide has a lone pair of electrons, so it must have the homo. |
Homo-Lumo Examples .txt | So let's begin with homo. So the homo has the highest or is the highest occupied molecular orbital. And that means our hydroxide has a lone pair of electrons, so it must have the homo. So our filled non bonding orbital of the oh is our homo and it's SP three hybridized. So it's the same exact homo that we saw in example A. Now, what's the lumo? |
Homo-Lumo Examples .txt | So our filled non bonding orbital of the oh is our homo and it's SP three hybridized. So it's the same exact homo that we saw in example A. Now, what's the lumo? Well, if we examine our hydronium ion, we see that every single orbital is taken. Every single bonding orbital is taken. We have the three bonds. |
Homo-Lumo Examples .txt | Well, if we examine our hydronium ion, we see that every single orbital is taken. Every single bonding orbital is taken. We have the three bonds. And we have a pair of electrons which aren't drawn. Here, let me fill them in. We have a pair of electrons on this oxygen. |
Homo-Lumo Examples .txt | And we have a pair of electrons which aren't drawn. Here, let me fill them in. We have a pair of electrons on this oxygen. So all four types of orbitals are filled. So that means if we don't have our bonding orbitals our Sigma bonding, we must use our Sigma antibodies. So the lowest unoccupied molecular orbital is the anti unoccupied Sigma antibonding orbital of one of the ho bonds. |
Homo-Lumo Examples .txt | So all four types of orbitals are filled. So that means if we don't have our bonding orbitals our Sigma bonding, we must use our Sigma antibodies. So the lowest unoccupied molecular orbital is the anti unoccupied Sigma antibonding orbital of one of the ho bonds. Now, let's go to example C. An example C.
We have an alkane reacting with our hydrobromic acid to form the following carbocation with a positive charge. And our bromine an ion. So which one is the homo? |
Homo-Lumo Examples .txt | Now, let's go to example C. An example C.
We have an alkane reacting with our hydrobromic acid to form the following carbocation with a positive charge. And our bromine an ion. So which one is the homo? Which one is the lumo? So which one of these is doing the donating? Which one is using its electron pair? |
Homo-Lumo Examples .txt | Which one is the lumo? So which one of these is doing the donating? Which one is using its electron pair? So clearly it's the alkene. This pair of electrons in the Pi bond is used to attract or take this H atom from our bromine. So that means our highest occupied molecular orbital is the pi bond. |
Homo-Lumo Examples .txt | So clearly it's the alkene. This pair of electrons in the Pi bond is used to attract or take this H atom from our bromine. So that means our highest occupied molecular orbital is the pi bond. The pi bonding of the carbon carbon double bond. So our homo is the pi bond. What about our lumo? |
Homo-Lumo Examples .txt | The pi bonding of the carbon carbon double bond. So our homo is the pi bond. What about our lumo? So once again, our lumo must be on this molecule. But notice that the Sigma bonding orbital is taking. So that means we must go to the next unoccupied orbital. |
Orthogonal Molecular Orbitals .txt | So here we have our two H atoms, the one S orbital and the one S orbital. We combine the two atomic orbitals to form two molecular orbitals, one molecular orbital. The bonding molecular orbital is lower in energy and is stabilizing, while the higher in energy and the destabilizing molecular orbital is called the antibonding molecular orbital. Now, what we haven't done so far is use our electrons. Remember, an electron is found in this atomic orbital, and one electron is found in this atomic orbital. Because we have two identical H atoms, and each are neutral. |
Orthogonal Molecular Orbitals .txt | Now, what we haven't done so far is use our electrons. Remember, an electron is found in this atomic orbital, and one electron is found in this atomic orbital. Because we have two identical H atoms, and each are neutral. So each have one neutron, one proton, and one electron. So here we have a positive one half spin. Here we have a negative one half spin. |
Orthogonal Molecular Orbitals .txt | So each have one neutron, one proton, and one electron. So here we have a positive one half spin. Here we have a negative one half spin. When these two guys combine, where would these two electrons want to go? In the higher energy or the lower energy? Remember, nature likes stabilizing states. |
Orthogonal Molecular Orbitals .txt | When these two guys combine, where would these two electrons want to go? In the higher energy or the lower energy? Remember, nature likes stabilizing states. They like low energy. So that means these two electrons will combine and will go into this bonding molecular orbital. Now, according to the poly exclusion principle, two things must happen. |
Orthogonal Molecular Orbitals .txt | They like low energy. So that means these two electrons will combine and will go into this bonding molecular orbital. Now, according to the poly exclusion principle, two things must happen. A maximum of two electrons should be found in this orbital, and these guys should have positive one half and negative one half. So opposite spins. And that's exactly what we have here. |
Orthogonal Molecular Orbitals .txt | A maximum of two electrons should be found in this orbital, and these guys should have positive one half and negative one half. So opposite spins. And that's exactly what we have here. So these electrons will not want to go into this orbital because this is higher in energy, and it causes the bond to destabilize or break apart. So, once again, this was combining two atomic one s orbitals to form two molecular orbitals. Now, let's combine one S orbital and a two p orbital. |
Orthogonal Molecular Orbitals .txt | So these electrons will not want to go into this orbital because this is higher in energy, and it causes the bond to destabilize or break apart. So, once again, this was combining two atomic one s orbitals to form two molecular orbitals. Now, let's combine one S orbital and a two p orbital. Remember, a two p orbital has this eight shape, right? Okay, so let's look at A. And let's look at B. |
Orthogonal Molecular Orbitals .txt | Remember, a two p orbital has this eight shape, right? Okay, so let's look at A. And let's look at B. Here I have two ways two potential ways in which our two atomic orbitals can interact in space. In part A, we have an orthogonal interaction. In part B, we have a non orthogonal interaction. |
Orthogonal Molecular Orbitals .txt | Here I have two ways two potential ways in which our two atomic orbitals can interact in space. In part A, we have an orthogonal interaction. In part B, we have a non orthogonal interaction. Orthogonal simply means perpendicular. So let's see which one of these is the correct interaction. In other words, which one of these will form molecular bonds, and which one of these will not form molecular bonds. |
Orthogonal Molecular Orbitals .txt | Orthogonal simply means perpendicular. So let's see which one of these is the correct interaction. In other words, which one of these will form molecular bonds, and which one of these will not form molecular bonds. So let's begin with A. Now, remember, this plus and minus does not mean charge. It means, for example, the plus means that we're combining the one s positive orbital and the two p positive orbital. |
Orthogonal Molecular Orbitals .txt | So let's begin with A. Now, remember, this plus and minus does not mean charge. It means, for example, the plus means that we're combining the one s positive orbital and the two p positive orbital. Now, when we combine the two positive orbitals, we get the following figure. When we combine the positive one S orbital and the negative one two p orbital, we get the following interaction. To get the negative two p, we simply switch it or flip it. |
Orthogonal Molecular Orbitals .txt | Now, when we combine the two positive orbitals, we get the following figure. When we combine the positive one S orbital and the negative one two p orbital, we get the following interaction. To get the negative two p, we simply switch it or flip it. So here we have our two interactions. So let's look at this guy. So, our positive will want to interact in a bonding way with the positive two p. So positive one s wants to interact with the positive sign of the two P orbital. |
Orthogonal Molecular Orbitals .txt | So here we have our two interactions. So let's look at this guy. So, our positive will want to interact in a bonding way with the positive two p. So positive one s wants to interact with the positive sign of the two P orbital. And likewise at the same time when these guys are interacting in a bonding way, these guys are interacting in an anti bonding way. And that's because we have a positive and a negative. Remember, positive and positive orbitals create bonding interactions. |
Orthogonal Molecular Orbitals .txt | And likewise at the same time when these guys are interacting in a bonding way, these guys are interacting in an anti bonding way. And that's because we have a positive and a negative. Remember, positive and positive orbitals create bonding interactions. Positive and negative create antibodies. So here we have a bonding and an antiboding. Let's go to this one. |
Orthogonal Molecular Orbitals .txt | Positive and negative create antibodies. So here we have a bonding and an antiboding. Let's go to this one. Here we have the same exact thing. Even though we flipped our two P orbital, we still have a bonding orbital between the positive one S and the positive two P and we have a negative interaction or an antibiotic interaction because we have a positive one S and a negative two P. So what happens when we have bonding and antiboding? Well, the bonding will exactly cancel out the antibonding. |
Orthogonal Molecular Orbitals .txt | Here we have the same exact thing. Even though we flipped our two P orbital, we still have a bonding orbital between the positive one S and the positive two P and we have a negative interaction or an antibiotic interaction because we have a positive one S and a negative two P. So what happens when we have bonding and antiboding? Well, the bonding will exactly cancel out the antibonding. And that means we will have a net interaction of zero. So orthogonal approach of orbitals. Or this guy does not allow for bonding because the bonding interactions are canceled out by the antibonding interaction. |
Orthogonal Molecular Orbitals .txt | And that means we will have a net interaction of zero. So orthogonal approach of orbitals. Or this guy does not allow for bonding because the bonding interactions are canceled out by the antibonding interaction. So there will be no interaction when our two atomic orbitals, the one S and the two P approach in this orthogonal fashion. So now let's look at part B. Now we have the following non orthogonal interaction. |
Orthogonal Molecular Orbitals .txt | So there will be no interaction when our two atomic orbitals, the one S and the two P approach in this orthogonal fashion. So now let's look at part B. Now we have the following non orthogonal interaction. So once again, let's show our pictures out. So we have the positive one S interact with the positive two P. So we keep the two orientations and we have the following picture. So this is one molecular orbital. |
Orthogonal Molecular Orbitals .txt | So once again, let's show our pictures out. So we have the positive one S interact with the positive two P. So we keep the two orientations and we have the following picture. So this is one molecular orbital. And now let's try to do the negative. So we have the one S positive and the two P negative. So we flip the two P and we have the following depiction. |
Orthogonal Molecular Orbitals .txt | And now let's try to do the negative. So we have the one S positive and the two P negative. So we flip the two P and we have the following depiction. Now we have a note or nodal plane symbolized by this black dash here. So we have the positive oneness interacts in an anti bonding fashion with the negative two P and we create this nodal plane which is once again simply our region where the electron density is zero. In other words, electrons will never will have a zero probability to be found in this region here. |
Orthogonal Molecular Orbitals .txt | Now we have a note or nodal plane symbolized by this black dash here. So we have the positive oneness interacts in an anti bonding fashion with the negative two P and we create this nodal plane which is once again simply our region where the electron density is zero. In other words, electrons will never will have a zero probability to be found in this region here. And notice that we don't have the same situation as we have here. In other words, we simply have bonding and then we have antibonding. So that means that this will be the correct interaction. |
Orthogonal Molecular Orbitals .txt | And notice that we don't have the same situation as we have here. In other words, we simply have bonding and then we have antibonding. So that means that this will be the correct interaction. This is how our two or how our one S and our two P orbitals will interact to form our two molecular orbitals. So we put in two atomic orbitals and we get out two molecular orbitals. And this is our picture here. |
Orthogonal Molecular Orbitals .txt | This is how our two or how our one S and our two P orbitals will interact to form our two molecular orbitals. So we put in two atomic orbitals and we get out two molecular orbitals. And this is our picture here. So this is our energy diagram. So this is our one S orbital. And remember, the two P orbital is slightly higher on the energy level. |
Cations, Anions and Isoelectronic Atoms .txt | So suppose we begin with a neutral atom in which the number of electrons equals number of protons. Now, what that basically means is that if we add up our charges due to our electrons, with the charges due to our protons, we're going to get a net charge or an overall charge of zero. Now, that's exactly what a neutral atom is. It's an atom. Atom in which our charge is zero. Well, now, suppose we take our atom and our neutral atom gains or loses electrons. |
Cations, Anions and Isoelectronic Atoms .txt | It's an atom. Atom in which our charge is zero. Well, now, suppose we take our atom and our neutral atom gains or loses electrons. Well, now we have a case in which the number of electrons is no longer equal to the number of protons. And so we're going to get a charge species called an ion. Now, whenever we're dealing with metals, be it transition metals, alkaline metals or alkaline earth metals, these guys tend to lose electrons because they can hold onto an electrons very tightly. |
Cations, Anions and Isoelectronic Atoms .txt | Well, now we have a case in which the number of electrons is no longer equal to the number of protons. And so we're going to get a charge species called an ion. Now, whenever we're dealing with metals, be it transition metals, alkaline metals or alkaline earth metals, these guys tend to lose electrons because they can hold onto an electrons very tightly. They're not very electronegative. And what that basically means is that they will form ions with positive charges. And these guys are called Cations. |
Cations, Anions and Isoelectronic Atoms .txt | They're not very electronegative. And what that basically means is that they will form ions with positive charges. And these guys are called Cations. Now. On the contrary. Nonmetal such as the halogens or oxygen or sulfur. |
Cations, Anions and Isoelectronic Atoms .txt | Now. On the contrary. Nonmetal such as the halogens or oxygen or sulfur. These guys have a very high affinity for electrons. And that means they will tend to take away electrons from other less electronegative atoms. And that means they will form ions in which there is a negative charge. |
Cations, Anions and Isoelectronic Atoms .txt | These guys have a very high affinity for electrons. And that means they will tend to take away electrons from other less electronegative atoms. And that means they will form ions in which there is a negative charge. And these guys are called nions. Now, let's look at the following illustration. Suppose we have a neutral atom x and this guy takes away the electron from some other atom forming an anion. |
Cations, Anions and Isoelectronic Atoms .txt | And these guys are called nions. Now, let's look at the following illustration. Suppose we have a neutral atom x and this guy takes away the electron from some other atom forming an anion. Well, this guy will simply have a negative charge represented in the following way now, most nonmetals follow this pathway. Now, suppose we have the reverse. Suppose we have an atom x that loses an electrons or loses an electron. |
Cations, Anions and Isoelectronic Atoms .txt | Well, this guy will simply have a negative charge represented in the following way now, most nonmetals follow this pathway. Now, suppose we have the reverse. Suppose we have an atom x that loses an electrons or loses an electron. Now, this guy will form a cation or an ion with a positive charge. Now, most metals follow this pathway. Now, one note about transition metals. |
Cations, Anions and Isoelectronic Atoms .txt | Now, this guy will form a cation or an ion with a positive charge. Now, most metals follow this pathway. Now, one note about transition metals. Transition metals. Whenever they lose electrons, they first lose electrons from the s orbital and therefore the d orbital. And that's because the S orbital is at a higher state than the D orbital. |
Cations, Anions and Isoelectronic Atoms .txt | Transition metals. Whenever they lose electrons, they first lose electrons from the s orbital and therefore the d orbital. And that's because the S orbital is at a higher state than the D orbital. And electrons lose or electrons leave the higher levels first before the lower levels. Now we're going to talk more about orbitals. The s orbitals and d orbitals in a future lecture. |
Cations, Anions and Isoelectronic Atoms .txt | And electrons lose or electrons leave the higher levels first before the lower levels. Now we're going to talk more about orbitals. The s orbitals and d orbitals in a future lecture. So stick around. Now, what happens to the size of our atom when it loses or gains an electron? Well, let's suppose we have a neutral atom that has two protons and two electrons. |
Cations, Anions and Isoelectronic Atoms .txt | So stick around. Now, what happens to the size of our atom when it loses or gains an electron? Well, let's suppose we have a neutral atom that has two protons and two electrons. And let's look at our illustration. So one electron on the atomo shell, one electron on the innermost shell, and two plus or two protons down there a nucleus. So its charge is zero. |
Cations, Anions and Isoelectronic Atoms .txt | And let's look at our illustration. So one electron on the atomo shell, one electron on the innermost shell, and two plus or two protons down there a nucleus. So its charge is zero. Now, suppose it loses an electron. Let's say it loses our atomos electron. Well, that means it's going to shrink in size. |
Cations, Anions and Isoelectronic Atoms .txt | Now, suppose it loses an electron. Let's say it loses our atomos electron. Well, that means it's going to shrink in size. And this is why. Remember, most atoms or atoms are generally composed of empty space. So when this guy disappears, all this empty space goes with it. |
Cations, Anions and Isoelectronic Atoms .txt | And this is why. Remember, most atoms or atoms are generally composed of empty space. So when this guy disappears, all this empty space goes with it. And so it becomes much smaller. So now the plus two charge will be greater than our negative one charge. And that means this electron will be pulled even closer to the nucleus, decreasing in size even further. |
Cations, Anions and Isoelectronic Atoms .txt | And so it becomes much smaller. So now the plus two charge will be greater than our negative one charge. And that means this electron will be pulled even closer to the nucleus, decreasing in size even further. So when we talk about cations, or when neutral atoms become can ions, there is a loss of electron, and this shrinks the element to a smaller size. Now, on the contrary, let's look at anions. So what happens to anion? |
Cations, Anions and Isoelectronic Atoms .txt | So when we talk about cations, or when neutral atoms become can ions, there is a loss of electron, and this shrinks the element to a smaller size. Now, on the contrary, let's look at anions. So what happens to anion? Suppose now we have a neutral atom with a plus one charge and a minus one charge, forming a neutral charge. And now suppose it gains an electron from some other atom, probably metal. Now what will happen is it will gain a new outer shell, forming this outer shell and forming this empty space in between. |
Cations, Anions and Isoelectronic Atoms .txt | Suppose now we have a neutral atom with a plus one charge and a minus one charge, forming a neutral charge. And now suppose it gains an electron from some other atom, probably metal. Now what will happen is it will gain a new outer shell, forming this outer shell and forming this empty space in between. And that means anions. When they gain electrons, this causes our atom or element, to grow in size. So cations, or the formation of cations, decrease the size of our atom, while the formation of anions increases the size of our atom. |
Cations, Anions and Isoelectronic Atoms .txt | And that means anions. When they gain electrons, this causes our atom or element, to grow in size. So cations, or the formation of cations, decrease the size of our atom, while the formation of anions increases the size of our atom. Now let's look at one last thing. Now isoelectronic atoms are those atoms that have the same number of electrons, but different number of protons. Suppose we have an atom with ten electrons and ten protons, and an atom with ten electrons and eleven protons. |
Rate Determining Step and Rate Law Part 1.txt | Now as of now we've only really spoken about oneway elementary actions in which reactants become products in a single forward step. Now we're going to look at complex reactions and we're going to find the rate laws for complex reactions. Now remember, in complex reactions reactants convert to products in more than one step. So the mechanism of our conversion will require more than one step. Now let's look at the following complex reaction. This is the overall reaction in which five molecules, two of these guys, one of this guy and two of these guys react to produce four moles of this guy and 1 mol of this guy. |
Rate Determining Step and Rate Law Part 1.txt | So the mechanism of our conversion will require more than one step. Now let's look at the following complex reaction. This is the overall reaction in which five molecules, two of these guys, one of this guy and two of these guys react to produce four moles of this guy and 1 mol of this guy. So let's break this overall net reaction into its individual steps. So step one is the following 1 mol of hydrogen peroxide reacts with 1 mol of this guy in a slow step producing 1 mol of hydroxide and 1 mol of hoi. Now this slow step will be important in determining the rate law and we'll see in a second why. |
Rate Determining Step and Rate Law Part 1.txt | So let's break this overall net reaction into its individual steps. So step one is the following 1 mol of hydrogen peroxide reacts with 1 mol of this guy in a slow step producing 1 mol of hydroxide and 1 mol of hoi. Now this slow step will be important in determining the rate law and we'll see in a second why. Now the second step is the following. This intermediate or this intermediate reacts with 1 mol reactants to produce again one molar hydroxide and one of our products. Now notice that this is one of our products. |
Rate Determining Step and Rate Law Part 1.txt | Now the second step is the following. This intermediate or this intermediate reacts with 1 mol reactants to produce again one molar hydroxide and one of our products. Now notice that this is one of our products. So step two is responsible for producing this product, namely this guy. And now these two guys, two moles of hydroxide finally react with two moles of hydronium to produce our four moles of water, the second product. So the second step is responsible for producing one of the products while the third step is responsible for producing the second type of product. |
Rate Determining Step and Rate Law Part 1.txt | So step two is responsible for producing this product, namely this guy. And now these two guys, two moles of hydroxide finally react with two moles of hydronium to produce our four moles of water, the second product. So the second step is responsible for producing one of the products while the third step is responsible for producing the second type of product. Now let's go back to this step. Now, relative to the rates of these two steps, this is a very, very slow step and in fact these steps can be assumed to be instantaneous very quick. So what's the significance of this slow step? |
Rate Determining Step and Rate Law Part 1.txt | Now let's go back to this step. Now, relative to the rates of these two steps, this is a very, very slow step and in fact these steps can be assumed to be instantaneous very quick. So what's the significance of this slow step? This low step is called a rate determining step. And this step limits the rate at which products are produced. And this equation or this reaction can be used to find the rate law. |
Rate Determining Step and Rate Law Part 1.txt | This low step is called a rate determining step. And this step limits the rate at which products are produced. And this equation or this reaction can be used to find the rate law. Now remember, the rate law can only really be found experimentally via results. But this is a second way with which you can find the rate law. But you still need to actually find the rate law using the results. |
Rate Determining Step and Rate Law Part 1.txt | Now remember, the rate law can only really be found experimentally via results. But this is a second way with which you can find the rate law. But you still need to actually find the rate law using the results. And then you can check the two and see if they coincide. And usually for the most part they will coincide. So this is a second way that you could find rate laws in complex reactions by using the rate law for the slow step, the rate determining step. |
Rate Determining Step and Rate Law Part 1.txt | And then you can check the two and see if they coincide. And usually for the most part they will coincide. So this is a second way that you could find rate laws in complex reactions by using the rate law for the slow step, the rate determining step. So since this is a bimolecular elementary reaction we can use the coefficients as the exponents. In other words, our rate of reaction is equal to KR rate constant times the concentration of hydrogen peroxide times the concentration of iodine. And each of the exponents is one because we have 1 mol of this guy and 1 mol of this guy react to produce these two intermediates. |
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