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c5471ada-7910-4820-8d27-849301f4d715 | # Chemistry
requires heat for evaporation. At constant temperature of its boiling point Tb and at constant pressure:
We can represent the process of
H₂O(l) → H₂O(g); ∆vapH = + 40.79 kJ mol–1 evaporation as
H₂O(l) vaporisation
___________________________
1 mol
H₂O(g)
Amount of heat required to vaporize one mole of a liquid at constant temperature and under standard pressure (1 bar) is called its standard enthalpy of vaporization or molar enthalpy of vaporization, ∆vapH.
Sublimation is direct conversion of a solid into its vapour. Solid CO₂ or ‘dry ice’ sublimes at 195 K with ∆subH = 25.2 kJ mol–1; naphthalene sublimes slowly and for this ∆subH = 73.0 kJ mol–1.
∆subH is the change in enthalpy when one mole of a solid substance sublimes at a constant temperature and under standard pressure (1 bar).
The magnitude of the enthalpy change depends on the strength of the intermolecular interactions in the substance undergoing the phase transformations. For example, the strong hydrogen bonds between water molecules hold them tightly in liquid phase. For an organic liquid, such as acetone, the intermolecular dipole-dipole interactions are significantly weaker. Thus, it requires less heat to vaporise 1 mol of acetone than it does to vaporize 1 mol of water.
Table 5.1 gives values of standard enthalpy changes of fusion and vaporisation for some substances.
# Problem 5.8
Assuming the water vapour to be a perfect gas, calculate the internal energy change when 1 mol of water at 100°C and 1 bar pressure is converted to ice at 0°C. Given the enthalpy of fusion of ice is 6.00 kJ mol–1 and heat capacity of water is 4.2 J/g°C.
The change takes place as follows:
|Step - 1|1 mol H₂O (l, 100°C)|1 mol H₂O (l, 0°C)|Enthalpy change ∆H₁|
|---|---|---|---|
|Step - 2|1 mol H₂O (l, 0°C)|1 mol H₂O (s, 0°C)|Enthalpy change ∆H₂|
Total enthalpy change will be -
∆H = ∆H₁ + ∆H₂
∆H₁ = - (18 x 4.2 x 100) J mol–1 = - 7560 J mol–1 = - 7.56 kJ mol–1
∆H₂ = - 6.00 kJ mol–1 | 12 | 11 | Chemistry | 105 |
a307d1dd-8099-4ac1-9d18-38e612ca6f4e | # THERMODYNAMICS
# Table 5.2 standard molar enthalpies of Formation (∆f Hʸ) at 298K of a Few selected substances
Therefore, ∆H = - 7.56 kJ mol⁻¹ + (-6.00 kJ mol⁻¹) = -13.56 kJ mol⁻¹
There is negligible change in the volume during the change from liquid to solid state. Therefore, p∆v = ∆ng RT = 0
∆H = ∆U = - 13.56 kJ mol⁻¹
# (c) Standard Enthalpy of Formation
The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states is called standard molar enthalpy of Formation. Its symbol is ∆fH, where the subscript ‘f’ indicates that one mole of the compound in question has been formed in its standard state from its elements in their most stable states of aggregation. The reference state of an element is its most stable state of aggregation at 25°C and 1 bar pressure. For example, the reference state of dihydrogen is H₂ gas and those of dioxygen, carbon and sulphur are O₂ gas, Cgraphite and Srhombic respectively. Some reactions with standard molar enthalpies of formation are as follows. | 13 | 11 | Chemistry | 105 |
b457c574-0096-49cf-85c8-9d6533b92fcd | # Chemistry
H₂(g) + ½O₂ (g) → H₂O(1);
∆H = –285.8 kJ mol–1
Here, we can make use of standard enthalpy of formation and calculate the enthalpy change for the reaction. The following general equation can be used for the enthalpy change calculation.
∆ᵣH = ∑ ai∆fH (products) – ∑ bi∆fH (reactants)
where a and b represent the coefficients of the products and reactants in the balanced equation. Let us apply the above equation for decomposition of calcium carbonate. Here, coefficients ‘a’ and ‘b’ are 1 each. Therefore,
∆ᵣH = ∆fH = [CaO(s)] + ∆fH [CO₂(g)] – ∆fH = [CaCO₃(s)]
= 1 (–635.1 kJ mol–1) + 1(–393.5 kJ mol–1) – 1(–1206.9 kJ mol–1)
= 178.3 kJ mol–1
Thus, the decomposition of CaCO₃ (s) is an endothermic process and you have to heat it for getting the desired products.
# (d) Thermochemical Equations
A balanced chemical equation together with the value of its ∆ᵣH is called a thermochemical equation. We specify the physical state (along with allotropic state) of the substance in an equation. For example:
C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l);
∆ᵣH = – 1367 kJ mol–1
The above equation describes the combustion of liquid ethanol at constant temperature and pressure. The negative sign of enthalpy change indicates that this is an exothermic reaction.
It would be necessary to remember the following conventions regarding thermochemical equations.
1. The coefficients in a balanced thermochemical equation refer to the number of moles (never molecules) of reactants and products involved in the reaction.
2. The numerical value of ∆ᵣH refers to the number of moles of substances specified by an equation. Standard enthalpy change ∆rH will have units as kJ mol–1.
Suppose, you are a chemical engineer and want to know how much heat is required to decompose calcium carbonate to lime and carbon dioxide, with all the substances in their standard state.
CaCO₃(s) → CaO(s) + CO₂(g); ∆ᵣH = ? | 14 | 11 | Chemistry | 105 |
bbe85887-f368-4049-8539-0a7894b27edb | # THERMODYNAMICS
To illustrate the concept, let us consider the calculation of heat of reaction for the following reaction:
Fe2O3(s) + 3H2(g) → 2Fe(s) + 3H2O(l),
From the Table (5.2) of standard enthalpy of formation (∆fHf), we find:
- ∆fH(H2O,l) = –285.83 kJ mol–1;
- ∆fH(Fe2O3,s) = –824.2 kJ mol–1;
- Also ∆Hf(Fe, s) = 0 and ∆fH(H2, g) = 0 as per convention.
Then,
∆fH = 3(–285.83 kJ mol–1) – 1(–824.2 kJ mol–1)
= (–857.5 + 824.2) kJ mol–1
= –33.3 kJ mol–1
Note that the coefficients used in these calculations are pure numbers, which are equal to the respective stoichiometric coefficients. The unit for ∆rH is kJ mol–1, which means per mole of reaction. Once we balance the chemical equation in a particular way, as above, this defines the mole of reaction. If we had balanced the equation differently, for example:
1 Fe2O3(s) + 3H2(g) → 2Fe(s) + 3H2O(l)
then this amount of reaction would be one mole of reaction and ∆rH would be
∆fH = 3(–285.83 kJ mol–1)2 – 1(–824.2 kJ mol–1)2
= (–428.7 + 412.1) kJ mol–1
= –16.6 kJ mol–1 = ½ ∆rH1
It shows that enthalpy is an extensive quantity.
# 3.
When a chemical equation is reversed, the value of ∆rH is reversed in sign. For example:
N2(g) + 3H2(g) → 2NH3(g);
2NH3(g) → N2(g) + 3H2(g); ∆rH = –91.8 kJ mol–1
∆rH = +91.8 kJ mol–1
# (e) Hess’s Law of Constant Heat Summation
We know that enthalpy is a state function, therefore the change in enthalpy is independent of the path between initial state (reactants) and final state (products). In other words, enthalpy change for a reaction is the same whether it occurs in one step or in a series of steps. This may be stated as follows in the form of Hess’s Law:
If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.
Let us understand the importance of this law with the help of an example.
Consider the enthalpy change for the reaction:
C (graphite,s) + O2(g) → CO(g); ∆rH = ?
Although CO(g) is the major product, some CO2 gas is always produced in this reaction. Therefore, we cannot measure enthalpy change for the above reaction directly. However, if we can find some other reactions involving related species, it is possible to calculate the enthalpy change for the above reaction.
Let us consider the following reactions:
C (graphite,s) + O2(g) → CO2(g); ∆rH = –393.5 kJ mol–1 (i)
CO(g) + 2 O2(g) → CO2(g); ∆rH = –283.0 kJ mol–1 (ii)
We can combine the above two reactions in such a way so as to obtain the desired reaction. To get one mole of CO(g) on the right, we reverse equation (ii). In this, heat is absorbed instead of being released, so we change the sign of ∆rH value:
CO2(g) → CO(g) + O2(g); ∆rH = +283.0 kJ mol–1 (iii) | 15 | 11 | Chemistry | 105 |
238e0a77-5e2f-4824-bf2d-cc22c6aae186 | # Chemistry
Adding equation (i) and (iii), we get the desired equation,
Cgraphite (s) + 1/2 O2 (g) → CO2 (g);
C6H12O6 (g) + 6O2 (g) → 6CO2 (g) + 6H2O (l);
for which ∆rH = (– 393.5 + 283.0) = – 110.5 kJ mol–1
∆C6H12O6 = – 2802.0 kJ mol–1
Our body also generates energy from food by the same overall process as combustion, although the final products are produced after a series of complex bio-chemical reactions involving enzymes.
# Problem 5.9
The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, CO2 (g) and H2O (l) are produced and 3267.0 kJ of heat is liberated. Calculate the standard enthalpy of formation, ∆fH of benzene. Standard enthalpies of formation of CO2 (g) and H2O (l) are –393.5 kJ mol–1 and – 285.83 kJ mol–1 respectively.
# Solution
The formation reaction of benzene is given by:
6Cgraphite + 3H2 (g) → C6H6 (l);
∆fH = ? ... (i)
The enthalpy of combustion of 1 mol of benzene is:
C6H6 (l) + 15/2 O2 → 6CO2 (g) + 3H2O (l);
∆CH = – 3267 kJ mol–1... (ii)
The enthalpy of formation of 1 mol of CO2 (g) is:
Cgraphite + O2 (g) → CO2 (g);
∆fH = – 393.5 kJ mol–1... (iii)
The enthalpy of formation of 1 mol of H2O (l) is:
H2 (g) + 1/2 O2 (g) → H2O (l);
∆fH = – 285.83 kJ mol–1... (iv)
Multiplying eqn. (iii) by 6 and eqn. (iv) by 3 we get: | 16 | 11 | Chemistry | 105 |
05979ac1-610a-449c-a738-d8f32e82897a | # THERMODYNAMICS
In this case, the enthalpy of atomization is the same as the enthalpy of sublimation.
6C graphite + 6O₂(g) → 6CO₂(g); ∆f H = – 2361 kJ mol–1
3H2(g) + 3O2(g) → 3H2O(l); ∆f H = – 857.49 kJ mol–1
Summing up the above two equations:
6C graphite + 3H2(g) + 2O2(g) → 6CO₂(g) + 3H2O(l); ∆f H = – 3218.49 kJ mol–1...
Reversing equation (ii):
6CO2(g) + 3H2O(l) → 6C(graphite) + 6O2(g); ∆f H = – 3267.0 kJ mol–1...
Adding equations (v) and (vi), we get:
6C(graphite) + 3H2(g) → C6H6(l); ∆f H = – 48.51 kJ mol–1...
(b) Enthalpy of Atomization (symbol: ∆aH)
Consider the following example of atomization of dihydrogen:
H2(g) → 2H(g); ∆aH = 435.0 kJ mol–1
You can see that H atoms are formed by breaking H–H bonds in dihydrogen. The enthalpy change in this process is known as enthalpy of atomization, ∆aH. It is the enthalpy change on breaking one mole of bonds completely to obtain atoms in the gas phase.
In case of diatomic molecules, like dihydrogen (given above), the enthalpy of atomization is also the bond dissociation enthalpy. The other examples of enthalpy of atomization can be:
Cl2(g) → 2Cl(g); ∆Cl–ClH = 242 kJ mol–1
O2(g) → 2O(g); ∆O=OH = 428 kJ mol–1
In the case of polyatomic molecules, bond dissociation enthalpy is different for different bonds within the same molecule.
Polyatomic Molecules: Let us now consider a polyatomic molecule like methane, CH4. The overall thermochemical equation for its atomization reaction is given below:
CH4(g) → C(g) + 4H(g); ∆aH = 1665 kJ mol–1
Note that the products are only atoms of C and H in gaseous phase. Now see the following reaction:
Na(s) → Na(g); ∆aH = 108.4 kJ mol–1 | 17 | 11 | Chemistry | 105 |
2795a3fa-0d13-4c33-9355-3c763cd66a1d | # Chemistry
CH₄(g) → CH₃(g) + H(g); ∆bond H = +427 kJ mol–1 given in Table 5.3. The reaction enthalpies are very important quantities as these arise from the changes that accompany the breaking of old bonds and formation of the new bonds.
We can predict enthalpy of a reaction in gas phase, if we know different bond enthalpies.
Therefore,
CH₄(g) → C(g) + 4H(g); ∆Ha = 1665 kJ mol–1
In such cases we use mean bond enthalpy of C–H bond.
For example in CH₄, ∆C–HH is calculated as:
∆C–HH = ¼ (∆aH) = ¼ (1665 kJ mol–1) = 416 kJ mol–1
We find that mean C–H bond enthalpy in methane is 416 kJ/mol. It has been found that mean C–H bond enthalpies differ slightly from compound to compound, as in CH₃CH₂Cl, CH₃NO₂, etc., but it does not differ in a great deal*. Using Hess’s law, bond enthalpies can be calculated. Bond enthalpy values of some single and multiple bonds are
# Table 5.3(a) some mean single Bond enthalpies in kJ mol–1 at 298 K
|H|C|N|O|F|Si|P|S|Cl|Br|I|
|---|---|---|---|---|---|---|---|---|---|---|
|435.8|414|389|464|569|293|318|339|431|368|297|
|347|293|351|439|289|264|259|330|276|238| |
| |159|201|272|-|209|-|201|243|-| |
| | |138|184|368|351|-|205|-|201| |
| | |155|540|490|327|255|197|-|F| |
| | |176|213|226|360|289|213| |Si| |
| | | |213|230|331|272|213| |P| |
| | | | |213|251|213|-|S| | |
| | | | |243|218|209|Cl| | | |
| | | | |192| |180|Br| | | |
| | | | |151| | |I| | | |
# Table 5.3(b) some mean multiple Bond enthalpies in kJ mol–1 at 298 K
|N = N|418|
|---|---|
|C = C|611|
|O = O|498|
|N ≡ N|946|
|C ≡ C|837|
|C = N|615|
|C = O|741|
|C ≡ N|891|
|C ≡ O|1070|
* Note that symbol used for bond dissociation enthalpy and mean bond enthalpy is the same.
**If we use enthalpy of bond formation, (∆fHbond), which is the enthalpy change when one mole of a particular type of bond is formed from gaseous atom, then ∆fH = ∑ ∆fHbonds of products – ∑ ∆fHbonds of reactants | 18 | 11 | Chemistry | 105 |
3333970d-cbac-4647-9374-365ea8c23220 | # THERMODYNAMICS
(reactants and products) in the reaction are Nag + Na-1 in gaseous state.
2. ( ) → Na(g) + e(g), the ionization of sodium atoms, ionization enthalpy
(d) Lattice Enthalpy
∆iH = 496 kJ mol–1
3. 1 Cl2 (g) → Cl2 (g), the dissociation of chlorine, the reaction enthalpy is half the bond dissociation enthalpy.
NaCl(s) → Na(g) + Cl(g); ∆latticeH = +788 kJ mol–1
Since it is impossible to determine lattice enthalpies directly by experiment, we use an indirect method where we construct an enthalpy diagram called a Born-Haber cycle (Fig. 5.9).
Let us now calculate the lattice enthalpy of Na+Cl–(s) by following steps given below:
1. Na(s) → Na(g), sublimation of sodium metal, ∆subH = 108.4 kJ mol–1
The electron gain enthalpy, ∆egH = –348.6 kJ mol–1.
You have learnt about ionization enthalpy and electron gain enthalpy in Unit 3.
In fact, these terms have been taken from thermodynamics. Earlier terms, ionization energy and electron affinity were in practice in place of the above terms (see the box for justification).
# Ionization Energy and Electron Affinity
Ionization energy and electron affinity are defined at absolute zero. At any other temperature, heat capacities for the reactants and the products have to be taken into account.
Enthalpies of reactions for M(g) → M+(g) + e– (for ionization) and M(g) + e– → M–(g) (for electron gain) at temperature, T is
∆ᵣH(T) = ∆rH(0) + ∫0T ∆rCP dT
The value of CP for each species in the above reaction is 5/2 R (CV = 3/2R)
So, ∆ᵣCP = +5/2 R (for ionization)
∆ᵣCP = –5/2 R (for electron gain)
Therefore,
∆ᵣH (ionization enthalpy) = E0 (ionization energy) + 5/2 RT
∆H (electron gain enthalpy) = –A (electron affinity) – 5/2 RT
Fig. 5.9 Enthalpy diagram for lattice enthalpy of NaCl
5. Na(g) + Cl(g) → NaCl(s)
The sequence of steps is shown in Fig. 5.9, and is known as a Born-Haber cycle. | 19 | 11 | Chemistry | 105 |
615a235d-a5f0-4bf5-a540-4311b69f958c | # Chemistry
The importance of the cycle is that the sum of the enthalpy changes round a cycle is zero. Applying Hess’s law, we get,
∆latticeH = 411.2 + 108.4 + 121 + 496 – 348.6
∆latticeH = + 788 kJ
for NaCl(s) → Na+(g) + Cl–(g)
Internal energy is smaller by 2RT (because ∆n = 2) and is equal to + 783 kJ mol–1.
Now we use the value of lattice enthalpy to calculate enthalpy of solution from the expression:
∆solH = ∆latticeH + ∆hydH
For one mole of NaCl(s), lattice enthalpy = + 788 kJ mol–1 and ∆hydH = – 784 kJ mol–1 (from the literature)
∆solH = + 788 kJ mol–1 – 784 kJ mol–1 = + 4 kJ mol–1
The dissolution of NaCl(s) is accompanied by very little heat change.
# (e) Enthalpy of Solution (symbol: ∆solH)
Enthalpy of solution of a substance is the enthalpy change when one mole of it dissolves in a specified amount of solvent. The enthalpy of solution at infinite dilution is the enthalpy change observed on dissolving the substance in an infinite amount of solvent when the interactions between the ions (or solute molecules) are negligible.
When an ionic compound dissolves in a solvent, the ions leave their ordered positions on the crystal lattice. These are now more free in solution. But solvation of these ions (hydration in case solvent is water) also occurs at the same time. This is shown diagrammatically, for an ionic compound, AB (s)
HCl(g) + 10 aq. → HCl.10 aq.
∆H = –69.01 kJ / mol
# Let us consider the following set of enthalpy changes:
(S-1) HCl(g) + 25 aq. → HCl.25 aq.
∆H = –72.03 kJ / mol
(S-2) HCl(g) + 40 aq. → HCl.40 aq.
∆H = –72.79 kJ / mol
(S-3) HCl(g) + ∞ aq. → HCl. ∞ aq.
∆H = –74.85 kJ / mol
The values of ∆H show general dependence of the enthalpy of solution on amount of solvent. As more and more solvent is used, the enthalpy of solution approaches a limiting value, i.e., the value in infinitely dilute solution. For hydrochloric acid this value of ∆H is given above in equation (S-3).
2024-25 | 20 | 11 | Chemistry | 105 |
25316380-0138-4dff-a00c-c5cadf3b7984 | # THERMODYNAMICS
If we subtract the first equation (equation S-1) from the second equation (equation S-2) in the above set of equations, we obtain–
|HCl.25 aq. + 15 aq.|→|HCl.40 aq.|
|---|---|---|
|∆H = [ –72.79 – (–72.03)] kJ / mol|=|– 0.76 kJ / mol|
This value (–0.76kJ/mol) of ∆H is enthalpy of dilution. It is the heat withdrawn from the surroundings when additional solvent is added to the solution. The enthalpy of dilution of a solution is dependent on the original concentration of the solution and the amount of solvent added.
# 5.6 Spontaneity
The first law of thermodynamics tells us about the relationship between the heat absorbed and the work performed on or by a system. It puts no restrictions on the direction of heat flow. However, the flow of heat is unidirectional from higher temperature to lower temperature. In fact, all naturally occurring processes whether chemical or physical will tend to proceed spontaneously in one direction only. For example, a gas expanding to fill the available volume, burning carbon in dioxygen giving carbon dioxide.
But heat will not flow from colder body to warmer body on its own, the gas in a container will not spontaneously contract into one corner or carbon dioxide will not form carbon and dioxygen spontaneously. These and many other spontaneously occurring changes show unidirectional change. We may ask ‘what is the driving force of spontaneously occurring changes? What determines the direction of a spontaneous change? In this section, we shall establish some criterion for these processes whether these will take place or not.
Let us first understand what do we mean by spontaneous reaction or change? You may think by your common observation that spontaneous reaction is one which occurs immediately when contact is made between the reactants. Take the case of combination of hydrogen and oxygen. These gases may be mixed at room temperature and left for many years without observing any perceptible change. Although the reaction is taking place between them, it is at an extremely slow rate. It is still called spontaneous reaction. So spontaneity means ‘having the potential to proceed without the assistance of external agency’. However, it does not tell about the rate of the reaction or process. Another aspect of spontaneous reaction or process, as we see is that these cannot reverse their direction on their own. We may summarise it as follows: A spontaneous process is an irreversible process and may only be reversed by some external agency.
# (a) Is Decrease in Enthalpy a Criterion for Spontaneity?
If we examine the phenomenon like flow of water down hill or fall of a stone on to the ground, we find that there is a net decrease in potential energy in the direction of change. By analogy, we may be tempted to state that a chemical reaction is spontaneous in a given direction, because decrease in energy has taken place, as in the case of exothermic reactions. For example:
|1 N (g) + H (g)|=|NH (g);|
|---|---|---|
|2 2 2 3|∆ᵣ H|= – 46.1 kJ mol–1|
|H (g) + 1 Cl (g)|=|HCl (g);|
|2 2 2 2|∆ᵣ H|= – 92.32 kJ mol–1|
|H (g) + 1 O (g)|→|H O(l);|
|2 2 2 2|∆ H|= –285.8 kJ mol–1|
The decrease in enthalpy in passing from reactants to products may be shown for any exothermic reaction on an enthalpy diagram as shown in Fig. 5.10(a). Thus, the postulate that driving force for a chemical reaction may be due to decrease in energy sounds ‘reasonable’ as the basis of evidence so far!
Now let us examine the following reactions:
|1 N (g) + O (g)|→|NO (g);|
|---|---|---|
|2 2 2|∆²H|= +33.2 kJ mol –1|
|C(graphite, s) + 2 S(l)|→|CS₂(l);|
| | |∆ᵣ H = +128.5 kJ mol–1| | 21 | 11 | Chemistry | 105 |
87423d2e-0933-439e-938e-098e32e4e0b8 | # Chemistry
# 5.10 Enthalpy Diagrams
# (a) Enthalpy diagram for exothermic reactions
These reactions though endothermic, are spontaneous. The increase in enthalpy may be represented on an enthalpy diagram as shown in Fig. 5.10(b).
# (b) Enthalpy diagram for endothermic reactions
Therefore, it becomes obvious that while decrease in enthalpy may be a contributory factor for spontaneity, but it is not true for all cases.
# 5.11 Diffusion of Two Gases
Let us examine the process. Before partition, if we were to pick up the gas molecules from left container, we would be sure that these will be molecules of gas A and similarly if we were to pick up the gas molecules from right container, we would be sure that these will be molecules of gas B. But, if we were to pick up molecules from container when partition is removed, we are not sure whether the molecules picked are of gas A or gas B. We say that the system has become less predictable or more chaotic.
We may now formulate another postulate: in an isolated system, there is always a tendency for the systems’ energy to become more disordered or chaotic and this could be a criterion for spontaneous change!
At this point, we introduce another thermodynamic function, entropy denoted as S. The above mentioned disorder is the manifestation of entropy. | 22 | 11 | Chemistry | 105 |
2d311551-fade-492a-bfab-87d575c2ecc2 | # THERMODYNAMICS
picture, one can think of entropy as a measure of the degree of randomness or disorder in the system. The greater the disorder in an isolated system, the higher is the entropy. As far as a chemical reaction is concerned, this entropy change can be attributed to rearrangement of atoms or ions from one pattern in the reactants to another (in the products). If the structure of the products is very much disordered than that of the reactants, there will be a resultant increase in entropy. The change in entropy accompanying a chemical reaction may be estimated qualitatively by a consideration of the structures of the species taking part in the reaction. Decrease of regularity in structure would mean increase in entropy. For a given substance, the crystalline solid state is the state of lowest entropy (most ordered), The gaseous state is state of highest entropy.
Now let us try to quantify entropy. One way to calculate the degree of disorder or chaotic distribution of energy among molecules would be through statistical method which is beyond the scope of this treatment. Other way would be to relate this process to the heat involved in a process which would make entropy a thermodynamic concept. Entropy, like any other thermodynamic property such as internal energy U and enthalpy H is a state function and ∆S is independent of path.
Whenever heat is added to the system, it increases molecular motions causing increased randomness in the system. Thus heat (q) has randomising influence on the system. Can we then equate ∆S with q? Wait! Experience suggests us that the distribution of heat also depends on the temperature at which heat is added to the system. A system at higher temperature has greater randomness in it than one at lower temperature. Thus, temperature is the measure of average chaotic motion of particles in the system. Heat added to a system at lower temperature causes greater randomness than when the same quantity of heat is added to it at higher temperature. This suggests that the entropy change is inversely proportional to the temperature. ∆S is related with q and T for a reversible reaction as:
∆S = qrev / T
# Problem 5.10
Predict in which of the following, entropy increases/decreases:
1. A liquid crystallizes into a solid.
2. Temperature of a crystalline solid is raised from 0 K to 115 K.
3. 2NaHCO3 (s) → Na2CO3 (s) + CO2 (g) + H2O (g)
4. 2H2 (g) → 2H (g)
solution
1. After freezing, the molecules attain an ordered state and therefore, entropy decreases.
2. At 0 K, the constituent particles are static and entropy is minimum. If temperature is raised to 115 K, these begin to move and oscillate. | 23 | 11 | Chemistry | 105 |
ec1a3f4b-2f32-4673-b18e-8f8b1cc0419f | # Chemistry
About their equilibrium positions = 4980.6 JK–1 mol–1 in the lattice and system becomes more disordered, therefore entropy increases.
# (iii) Reactant, NaHCO3 is a solid and it has low entropy. Among products there are one solid and two gases. Therefore, the products represent a condition of higher entropy.
# (iv) Here one molecule gives two atoms i.e., number of particles increases leading to more disordered state. Two moles of H atoms have higher entropy than one mole of dihydrogen molecule.
# Problem 5.11
For oxidation of iron,
4Fe(s) + 3O2(g) → 2Fe2O3(s)
Entropy change is – 549.4 JK–1 mol–1 at 298 K. Despite negative entropy change of this reaction, why is the reaction spontaneous?
(∆rH for this reaction is –1648 × 103 J mol–1)
# Solution
One decides the spontaneity of a reaction by considering ∆Stotal (∆Ssys + ∆Ssurr). For calculating ∆Ssurr, we have to consider the heat absorbed by the surroundings which is equal to – ∆rH. At temperature T, entropy change of the surroundings is
–(–1648 × 103 J mol–1) = 5530 JK–1 mol–1
Thus, total entropy change for this reaction
∆rStotal = 5530 JK–1 mol–1 + (–549.4 JK–1 mol–1)
Gibbs function, G is an extensive property and a state function.
The change in Gibbs energy for the system, ∆Gsys can be written as
∆Gsys = ∆Hsys – T ∆Ssys – Ssys∆T
At constant temperature,
∴ ∆Gsys = ∆Hsys – T ∆Ssys
Usually the subscript ‘system’ is dropped and we simply write this equation as
∆G = ∆H – T ∆S
Thus, Gibbs energy change = enthalpy change – temperature × entropy change, and is referred to as the Gibbs equation, one of the most important equations in chemistry.
Here, we have considered both terms together for spontaneity: energy (in terms of ∆H) and entropy (∆S, a measure of disorder) as indicated earlier. Dimensionally if we analyse, we find that ∆G has units of energy because, both ∆H and the T∆S are energy terms, since T∆S = (K) (J/K) = J.
Now let us consider how ∆G is related to reaction spontaneity. | 24 | 11 | Chemistry | 105 |
cf7c87a6-e549-41cf-8754-a0e78735b17a | # THERMODYNAMICS
We know,
∆Stotal = ∆Ssys + ∆Ssurr
If the system is in thermal equilibrium with the surrounding, then the temperature of the surrounding is same as that of the system. Also, increase in enthalpy of the surrounding is equal to decrease in the enthalpy of the system.
Therefore, entropy change of surroundings,
ΔS = ΔHsurr / T = - ΔHsys / T
ΔStotal = ΔSsys + (−ΔHsys / T)
Rearranging the above equation:
TΔStotal = TΔSsys − ΔHsys
For spontaneous process, ΔStotal > 0, so
TΔSsys − ΔHsys > 0
−(ΔHsys − TΔSsys) > 0
Using equation 5.21, the above equation can be written as
−ΔG > 0
ΔG = ΔH − TΔSsys < 0 (5.22)
ΔH is the enthalpy change of a reaction, TΔSsys is the energy which is not available to do useful work. So ΔG is the net energy available to do useful work and is thus a measure of the ‘free energy’. For this reason, it is also known as the free energy of the reaction.
ΔG gives a criteria for spontaneity at constant pressure and temperature.
1. If ΔG is negative (< 0), the process is spontaneous.
2. If ΔG is positive (> 0), the process is non spontaneous.
Note: If a reaction has a positive enthalpy change and positive entropy change, it can be spontaneous when TΔS is large enough to outweigh ΔH. This can happen in two ways;
- (a) The positive entropy change of the system can be ‘small’ in which case T must be large.
- (b) The positive entropy change of the system can be ‘large’, in which case T may be small.
The former is one of the reasons why reactions are often carried out at high temperature. Table 5.4 summarises the effect of temperature on spontaneity of reactions.
# (d) Entropy and Second Law of Thermodynamics
We know that for an isolated system the change in energy remains constant. Therefore, increase in entropy in such systems is the natural direction of a spontaneous change. This, in fact, is the second law of thermodynamics. Like first law of thermodynamics, second law can also be stated in several ways. The second law of thermodynamics explains why spontaneous exothermic reactions are so common. In exothermic reactions heat released by the reaction increases the disorder of the surroundings and overall entropy change is positive which makes the reaction spontaneous.
# (e) Absolute Entropy and Third Law of Thermodynamics
Molecules of a substance may move in a straight line in any direction, they may spin like a top and the bonds in the molecules may stretch and compress. These motions of the molecule are called translational, rotational and vibrational motion respectively. When temperature of the system rises, these motions become more vigorous and entropy increases. On the other hand, when temperature is lowered, the entropy decreases. The entropy of any pure crystalline substance approaches zero as the temperature approaches absolute zero. This is called third law of thermodynamics. This is so because there is perfect order in a crystal at absolute zero. The statement is confined to pure crystalline solids because theoretical arguments and practical evidences have shown that entropy of solutions and super cooled liquids is not zero at 0 K. The importance of the third law lies in the fact that it permits the calculation of absolute values of entropy of pure substance from thermal data alone. For a pure substance, this can... | 25 | 11 | Chemistry | 105 |
6f508d5c-13b3-494d-b177-1d9303f69996 | # Chemistry
is minimum. If it is not, the system would be done by summing rev increments from 0 TK to 298 K. Standard entropies can be used to calculate standard entropy changes by a Hess’s law type of calculation.
# 5.7 Gibbs Energy Change and Equilibrium
We have seen how a knowledge of the sign and magnitude of the free energy change of a chemical reaction allows:
1. Prediction of the spontaneity of the chemical reaction.
2. Prediction of the useful work that could be extracted from it.
So far we have considered free energy changes in irreversible reactions. Let us now examine the free energy changes in reversible reactions.
‘Reversible’ under strict thermodynamic sense is a special way of carrying out a process such that system is at all times in perfect equilibrium with its surroundings. When applied to a chemical reaction, the term ‘reversible’ indicates that a given reaction can proceed in either direction simultaneously, so that a dynamic equilibrium is set up. This means that the reactions in both the directions should proceed with a decrease in free energy, which seems impossible. It is possible only if at equilibrium the free energy of the system is:
∆rG = 0
Gibbs energy for a reaction in which all reactants and products are in standard state, ∆rG is related to the equilibrium constant of the reaction as follows:
0 = ∆rG + RT ln K
or ∆rG = – RT ln K
or ∆rG = – 2.303 RT log K (5.23)
We also know that:
For strongly endothermic reactions, the value of ∆rH may be large and positive. In such a case, value of K will be much smaller than 1 and the reaction is unlikely to form much product. In case of exothermic reactions, ∆rH is large and negative, and ∆rG is likely to be large and negative too. In such cases, K will be much larger than 1. We may expect strongly exothermic reactions to have a large K, and hence can go to near completion. ∆rG also depends upon ∆rS, if the changes in the entropy of reaction is also taken into account, the value of K or extent of chemical reaction will also be affected, depending upon whether ∆rS is positive or negative.
# Table 5.4 Effect of Temperature on Spontaneity of Reactions
|∆rH|∆rS|∆rG|Description*|
|---|---|---|---|
|–|+|–|Reaction spontaneous at all temperatures|
|–|–|– (at low T)|Reaction spontaneous at low temperature|
|–|–|+ (at high T)|Reaction nonspontaneous at high temperature|
|+|+|+ (at low T)|Reaction nonspontaneous at low temperature|
|+|+|– (at high T)|Reaction spontaneous at high temperature|
|+|–|+ (at all T)|Reaction nonspontaneous at all temperatures|
* The term low temperature and high temperature are relative. For a particular reaction, high temperature could even mean room temperature. | 26 | 11 | Chemistry | 105 |
9d71e528-b02c-4b0e-afec-78399524b183 | # THERMODYNAMICS
(i) It is possible to obtain an estimate of ∆G from the measurement of ∆H and ∆S, = (–13 6 3 –1) = (2.303 × 8.314) (T) and then calculate K at any temperature = 2.38 JK–1 mol–1 298 K.
(ii) If K is measured directly in the laboratory, hence K = antilog 2.38 = 2.4 × 10². The value of ∆G at any other temperature can be calculated.
# Problem 5.14
At 60°C, dinitrogen tetroxide is 50 per cent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.
|N₂O₄(g)|2NO₂(g)|
|---|---|
|If N₂O₄ is 50% dissociated, the mole fraction of both the substances is given by|If N₂O₄ is 50% dissociated, the mole fraction of both the substances is given by|
|1 - 0.5|0.5|
|x|xNO₂ = 0.5|
|1 + 0.5|1 + 0.5|
– 2.303 (8.314 J K–1 mol–1) × (298 K) (log 2.47 × 10–29) = 163000 J mol–1 = 163 kJ mol–1.
# Problem 5.13
Find out the value of equilibrium constant for the following reaction at 298 K.
The equilibrium constant Kp is given by Kp = pNO2 = (1.5)(0.5) = 1.33 atm.
Standard Gibbs energy change, ∆ᵣG at the given temperature is –13.6 kJ mol–1.
We know, log K = ∆ᵣG = (–R T ln Kp) = (–8.314 J K–1 mol–1) × (333 K) × (2.303) × (0.1239) = –763.8 kJ mol–1. | 27 | 11 | Chemistry | 105 |
5eeafc98-8128-4bd0-bc47-193e3a8a2017 | # Chemistry
# Summary
Thermodynamics deals with energy changes in chemical or physical processes and enables us to study these changes quantitatively and to make useful predictions. For these purposes, we divide the universe into the system and the surroundings. Chemical or physical processes lead to evolution or absorption of heat (q), part of which may be converted into work (w). These quantities are related through the first law of thermodynamics via ∆U = q + w. ∆U, change in internal energy, depends on initial and final states only and is a state function, whereas q and w depend on the path and are not the state functions. We follow sign conventions of q and w by giving the positive sign to these quantities when these are added to the system. We can measure the transfer of heat from one system to another which causes the change in temperature. The magnitude of rise in temperature depends on the heat capacity (C) of a substance. Therefore, heat absorbed or evolved is q = C∆T. Work can be measured by w = –pₑₓ∆V, in case of expansion of gases. Under reversible process, we can put pₑₓ = p for infinitesimal changes in the volume making wrev = – p dV. In this condition, we can use gas equation, pV = nRT.
At constant volume, w = 0, then ∆U = qV, heat transfer at constant volume. But in study of chemical reactions, we usually have constant pressure. We define another state function enthalpy. Enthalpy change, ∆H = ∆U + ∆ngRT, can be found directly from the heat changes at constant pressure, ∆H = qp.
There are varieties of enthalpy changes. Changes of phase such as melting, vaporization and sublimation usually occur at constant temperature and can be characterized by enthalpy changes which are always positive. Enthalpy of formation, combustion and other enthalpy changes can be calculated using Hess’s law. Enthalpy change for chemical reactions can be determined by
ΔrH = Σ(aiΔfHproducts) - Σ(biΔfHreactions)
and in gaseous state by
ΔrH = Σ bond enthalpies of the reactants – Σ bond enthalpies of the products
First law of thermodynamics does not guide us about the direction of chemical reactions i.e., what is the driving force of a chemical reaction. For isolated systems, ∆U = 0. We define another state function, S, entropy for this purpose. Entropy is a measure of disorder or randomness. For a spontaneous change, total entropy change is positive. Therefore, for an isolated system, ∆U = 0, ∆S > 0, so entropy change distinguishes a spontaneous change, while energy change does not. Entropy changes can be measured by the equation ∆S = qrev for a reversible process. qrev is independent of path.
Chemical reactions are generally carried at constant pressure, so we define another state function Gibbs energy, G, which is related to entropy and enthalpy changes of the system by the equation:
ΔrG = ΔrH – TΔrS
For a spontaneous change, ΔGsys < 0 and at equilibrium, ΔGsys = 0. Standard Gibbs energy change is related to equilibrium constant by
ΔrG = – RT ln K.
K can be calculated from this equation, if we know ΔrG which can be found from. Temperature is an important factor in the equation. Many reactions which are non-spontaneous at low temperature, are made spontaneous at high temperature for systems having positive entropy of reaction.
2024-25 | 28 | 11 | Chemistry | 105 |
72cdc0a1-a776-4ea9-a276-0365411005f2 | # THERMODYNAMICS
# exercises
# 5.1
Choose the correct answer. A thermodynamic state function is a quantity
- (i) used to determine heat changes
- (ii) whose value is independent of path
- (iii) used to determine pressure volume work
- (iv) whose value depends on temperature only.
# 5.2
For the process to occur under adiabatic conditions, the correct condition is:
- (i) ∆T = 0
- (ii) ∆p = 0
- (iii) q = 0
- (iv) w = 0
# 5.3
The enthalpies of all elements in their standard states are:
- (i) unity
- (ii) zero
- (iii) < 0
- (iv) different for each element
# 5.4
∆U of combustion of methane is – X kJ mol–1. The value of ∆H is
- (i) = ∆U
- (ii) > ∆U
- (iii) < ∆U
- (iv) = 0
# 5.5
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol–1 –393.5 kJ mol–1, and –285.8 kJ mol–1 respectively. Enthalpy of formation of CH₄(g) will be
- (i) –74.8 kJ mol–1
- (ii) –52.27 kJ mol–1
- (iii) +74.8 kJ mol–1
- (iv) +52.26 kJ mol–1.
# 5.6
A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be
- (i) possible at high temperature
- (ii) possible only at low temperature
- (iii) not possible at any temperature
- (iv) possible at any temperature
# 5.7
In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?
# 5.8
The reaction of cyanamide, NH₂CN (s), with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be –742.7 kJ mol–1 at 298 K. Calculate enthalpy change for the reaction at 298 K.
NH₂CN(g) + 3 O₂(g) → N₂(g) + CO₂(g) + H₂O(l) | 29 | 11 | Chemistry | 105 |
f984e4de-7f57-4fcd-915c-ea3b2b6e5b98 | # 5.9
Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol–1 K–1.
# 5.10
Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at –10.0°C. ∆fusH = 6.03 kJ mol–1 at 0°C.
Cₚ [H₂O(l)] = 75.3 J mol–1 K–1
Cₚ [H₂O(s)] = 36.8 J mol–1 K–1
# 5.11
Enthalpy of combustion of carbon to CO₂ is –393.5 kJ mol–1. Calculate the heat released upon formation of 35.2 g of CO₂ from carbon and dioxygen gas.
# 5.12
Enthalpies of formation of CO(g), CO₂(g), N₂O(g) and N₂O₄(g) are –110, –393, 81 and 9.7 kJ mol–1 respectively. Find the value of ∆ᵣH for the reaction:
N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)
# 5.13
Given N₂(g) + 3H₂(g) → 2NH₃(g); ∆ᵣH = –92.4 kJ mol–1. What is the standard enthalpy of formation of NH₃ gas?
# 5.14
Calculate the standard enthalpy of formation of CH₃OH(l) from the following data:
CH₃OH(l) + 3 O₂(g) → CO(g) + 2H₂O(l); ∆rH = –726 kJ mol–1
C(graphite) + O₂(g) → CO₂(g); ∆cH = –393 kJ mol–1
H₂(g) + 1/2 O₂(g) → H₂O(l); ∆fH = –286 kJ mol–1.
# 5.15
Calculate the enthalpy change for the process CCl₄(g) → C(g) + 4 Cl(g) and calculate bond enthalpy of C – Cl in CCl₄(g).
∆vapH (CCl₄) = 30.5 kJ mol–1.
∆fH (CCl₄) = –135.5 kJ mol–1.
∆ₐH (C) = 715.0 kJ mol–1, where ∆aH is enthalpy of atomisation
∆ₐH (Cl₂) = 242 kJ mol–1
# 5.16
For an isolated system, ∆U = 0, what will be ∆S?
# 5.17
For the reaction at 298 K, 2A + B → C, ∆H = 400 kJ mol–1 and ∆S = 0.2 kJ K–1 mol–1. At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range?
# 5.18
For the reaction, 2 Cl(g) → Cl₂(g), what are the signs of ∆H and ∆S?
# 5.19
For the reaction 2 A(g) + B(g) → 2D(g), ∆U = –10.5 kJ and ∆S = –44.1 JK–1. Calculate ∆G for the reaction, and predict whether the reaction may occur spontaneously. | 30 | 11 | Chemistry | 105 |
9fda0b26-86bf-4b5c-bc51-6516724f922e | # THERMODYNAMICS
# 5.20
The equilibrium constant for a reaction is 10. What will be the value of ∆G?
R = 8.314 JK–1 mol–1, T = 300 K.
# 5.21
Comment on the thermodynamic stability of NO(g), given
1 N(g) + 1 O(g) → NO(g); ∆H = 90 kJ mol–1
2 NO(g) + 1 O2(g) → NO2(g): ∆rH = –74 kJ mol–1
# 5.22
Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ∆fH = –286 kJ mol–1. | 31 | 11 | Chemistry | 105 |
0be9a0cb-9f97-41fc-a964-843de5f07a0c | # Unit 4
# CHEMiCAL BOnDinG AnD MOLECULAR StRUCtURE
Scientists are constantly discovering new compounds, orderly arranging the facts about them, trying to explain with the existing knowledge, organising to modify the earlier views or evolve theories for explaining the newly observed facts.
# After studying this Unit, you will be able to
- understand Kössel-Lewis approach to chemical bonding;
- explain the octet rule and its limitations, draw Lewis structures of simple molecules;
- explain the formation of different types of bonds;
- describe the VSEPR theory and predict the geometry of simple molecules;
- explain the valence bond approach for the formation of covalent bonds;
- predict the directional properties of covalent bonds;
- explain the different types of hybridisation involving s, p and d orbitals and draw shapes of simple covalent molecules;
- describe the molecular orbital theory of homonuclear diatomic molecules;
- explain the concept of hydrogen bond.
Matter is made up of one or different type of elements. Under normal conditions no other element exists as an independent atom in nature, except noble gases. However, a group of atoms is found to exist together as one species having characteristic properties. Such a group of atoms is called a molecule. Obviously there must be some force which holds these constituent atoms together in the molecules. The attractive force which holds various constituents (atoms, ions, etc.) together in different chemical species is called a chemical bond. Since the formation of chemical compounds takes place as a result of combination of atoms of various elements in different ways, it raises many questions. Why do atoms combine? Why are only certain combinations possible? Why do some atoms combine while certain others do not? Why do molecules possess definite shapes? To answer such questions different theories and concepts have been put forward from time to time. These are Kössel-Lewis approach, Valence Shell Electron Pair Repulsion (VSEPR) Theory, Valence Bond (VB) Theory and Molecular Orbital (MO) Theory. The evolution of various theories of valence and the interpretation of the nature of chemical bonds have closely been related to the developments in the understanding of the structure of atom, the electronic configuration of elements and the periodic table. Every system tends to be more stable and bonding is nature’s way of lowering the energy of the system to attain stability.
2024-25 | 0 | 11 | Chemistry | 104 |
030616d1-0593-490a-a9d8-6ef41dceb40e | # CHEMICAL BONDING AND MOLECULAR STRUCTURE
# 4.1 KÖSSEL-LEWIS APPROACH
In order to explain the formation of chemical bond in terms of electrons, a number of attempts were made, but it was only in 1916 when Kössel and Lewis succeeded independently in giving a satisfactory explanation. They were the first to provide some logical explanation of valence which was based on the inertness of noble gases.
Lewis pictured the atom in terms of a positively charged ‘Kernel’ (the nucleus plus the inner electrons) and the outer shell that could accommodate a maximum of eight electrons. He further assumed that these eight electrons occupy the corners of a cube which surround the ‘Kernel’. Thus the single outer shell electron of sodium would occupy one corner of the cube, while in the case of a noble gas all the eight corners would be occupied. This octet of electrons represents a particularly stable electronic arrangement. Lewis postulated that atoms achieve the stable octet when they are linked by chemical bonds. In the case of sodium and chlorine, this can happen by the transfer of an electron from sodium to chlorine thereby giving the Na⁺ and Cl⁻ ions. In the case of other molecules like Cl₂, H₂, F₂, etc., the bond is formed by the sharing of a pair of electrons between the atoms. In the process each atom attains a stable outer octet of electrons.
# Lewis Symbols
In the formation of a molecule, only the outer shell electrons take part in chemical combination and they are known as valence electrons. The inner shell electrons are well protected and are generally not involved in the combination process. G.N. Lewis, an American chemist introduced simple notations to represent valence electrons in an atom. These notations are called Lewis symbols. For example, the Lewis symbols for the elements of second period are as under:
|Na|[Ne] 3s¹|Cl + e⁻ → Cl⁻|[Ne] 3s² 3p⁵|[Ne] 3s² 3p⁶ or [Ar]|
|---|---|---|---|---|
|Na⁺ + Cl⁻ → NaCl or Na⁺Cl⁻| | | | |
Similarly the formation of CaF₂ may be shown as:
|Ca|[Ar] 4s²|Ca²⁺ + 2e⁻| | |
|---|---|---|---|---|
|F + e⁻ → F⁻|[He] 2s² 2p⁵|[He] 2s² 2p⁶ or [Ne]|Ca²⁺ + 2F⁻ → CaF₂ or Ca²⁺(F⁻)₂| |
The bond formed, as a result of the electrostatic attraction between the positive and negative ions was termed as | 1 | 11 | Chemistry | 104 |
d83a7c1e-b268-4a62-b5f8-89b60121fd1b | # Chemistry
The electrovalent bond. The electrovalence is thus equal to the number of unit charge(s) on the ion. Thus, calcium is assigned a positive electrovalence of two, while chlorine a negative electrovalence of one. The Kössel’s postulations provide the basis for modern concepts regarding ion-formation by electron transfer and the formation of ionic crystalline compounds. His views have proved to be of great value in the understanding and systematisation of the ionic compounds. At the same time he did recognise the fact that a large number of compounds did not fit into these concepts.
# 4.1.1 Octet Rule
Kössel and Lewis in 1916 developed an important theory of chemical combination between atoms known as electronic theory of chemical bonding. According to this, atoms can combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing of valence electrons in order to have an octet in their valence shells. This is known as octet rule.
# 4.1.2 Covalent Bond
Langmuir (1919) refined the Lewis postulations by abandoning the idea of the stationary cubical arrangement of the octet, and by introducing the term covalent bond. The Lewis-Langmuir theory can be understood by considering the formation of the chlorine molecule, Cl₂. The Cl atom with electronic configuration, [Ne]3s² 3p⁵, is one electron short of the argon configuration. The formation of the Cl₂ molecule can be understood in terms of the sharing of a pair of electrons between the two chlorine atoms, each chlorine atom contributing one electron to the shared pair. In the process both atoms attain the outer shell octet of the nearest noble gas (i.e., argon).
Thus, when two atoms share one electron pair they are said to be joined by a single covalent bond. In many compounds we have multiple bonds between atoms. The formation of multiple bonds envisages sharing of more than one electron pair between two atoms. If two atoms share two pairs of electrons, the covalent bond between them is called a double bond. For example, in the carbon dioxide molecule, we have two double bonds between the carbon and oxygen atoms. Similarly, in ethene molecule the two carbon atoms are joined by a double bond.
Covalent bond between two Cl atoms: Cl – Cl
Double bonds in CO2 molecule | 2 | 11 | Chemistry | 104 |
e25783ba-f3e0-4125-839f-155e28721ea4 | # Chemical Bonding and Molecular Structure
number of valence electrons. For example, for the CO2– ion, the two negative charges indicate that there are two additional electrons than those provided by the neutral atoms. For NH+ ion, one positive charge indicates the loss of one electron from the group of neutral atoms.
C2H4 molecule when combining atoms share three electron pairs as in the case of two nitrogen atoms in the N2 molecule and the two carbon atoms in the ethyne molecule, a triple bond is formed.
In general the least electronegative atom occupies the central position in the molecule/ion. For example in the NF3 and CO2–, nitrogen and carbon are the central atoms whereas fluorine and oxygen occupy the terminal positions.
After accounting for the shared pairs of electrons for single bonds, the remaining electron pairs are either utilized for multiple bonding or remain as the lone pairs. The basic requirement being that each bonded atom gets an octet of electrons.
# 4.1.3 Lewis Representation of Simple Molecules (the Lewis Structures)
The Lewis dot structures provide a picture of bonding in molecules and ions in terms of the shared pairs of electrons and the octet rule. While such a picture may not explain the bonding and behaviour of a molecule completely, it does help in understanding the formation and properties of a molecule to a large extent. Writing of Lewis dot structures of molecules is, therefore, very useful. The Lewis dot structures can be written by adopting the following steps:
- The total number of electrons required for writing the structures are obtained by adding the valence electrons of the combining atoms. For example, in the CH4 molecule there are eight valence electrons available for bonding (4 from carbon and 4 from the four hydrogen atoms).
- For anions, each negative charge would mean addition of one electron. For cations, each positive charge would result in subtraction of one electron from the total.
* Each H atom attains the configuration of helium (a duplet of electrons)
2024-25 | 3 | 11 | Chemistry | 104 |
ffdd55aa-0eaa-4378-8eb2-5545adcacd82 | # Chemistry
# Problem 4.1
Write the Lewis dot structure of CO molecule.
# Solution
Step 1. Count the total number of valence electrons of carbon and oxygen atoms. The outer (valence) shell configurations of carbon and oxygen atoms are: 2s² 2p² and 2s² 2p⁴, respectively. The valence electrons available are 4 + 6 = 10.
Step 2. The skeletal structure of CO is written as: C O
Step 3. Draw a single bond (one shared electron pair) between C and O and complete the octet on O, the remaining two electrons are the lone pair on C. This does not complete the octet on carbon and hence we have to resort to multiple bonding (in this case a triple bond) between C and O atoms. This satisfies the octet rule condition for both atoms.
# 4.1.4 Formal Charge
Lewis dot structures, in general, do not represent the actual shapes of the molecules. In case of polyatomic ions, the net charge is possessed by the ion as a whole and not by a particular atom. It is, however, feasible to assign a formal charge on each atom. The formal charge of an atom in a polyatomic molecule or ion may be defined as the difference between the number of valence electrons of that atom in an isolated or free state and the number of electrons assigned to that atom in the Lewis structure. It is expressed as:
Formal charge (F.C.) on an atom in a Lewis structure = total number of valence electrons in the free atom - total number of non-bonding (lone pair) electrons - (1/2) total number of bonding (shared) electrons
# Problem 4.2
Write the Lewis structure of the nitrite ion, NO–.
# Solution
Step 1. Count the total number of valence electrons of the nitrogen atom, the oxygen atoms and the additional one negative charge (equal to one electron). N(2s² 2p³), O (2s² 2p⁴) 5 + (2 × 6) + 1 = 18 electrons.
Step 2. The skeletal structure of NO– is written as: O N O2
Step 3. Draw a single bond (one shared electron pair) between the nitrogen and oxygen atoms. | 4 | 11 | Chemistry | 104 |
db182065-2ffe-4342-85e1-2930ae0c3ce5 | # CHEMICAL BONDING AND MOLECULAR STRUCTURE
# 4.1.5 Limitations of the Octet Rule
The octet rule, though useful, is not universal. It is quite useful for understanding the structures of most of the organic compounds and it applies mainly to the second period elements of the periodic table. There are three types of exceptions to the octet rule.
# The incomplete octet of the central atom
In some compounds, the number of electrons surrounding the central atom is less than eight. This is especially the case with elements having less than four valence electrons. Examples are LiCl, BeH₂ and BCl₃. Li, Be and B have 1, 2 and 3 valence electrons only. Some other such compounds are AlCl₃ and BF₃.
# Odd-electron molecules
In molecules with an odd number of electrons like nitric oxide, NO and nitrogen dioxide, NO₂, the octet rule is not satisfied for all the atoms.
# The expanded octet
Elements in and beyond the third period of the periodic table have, apart from 3s and 3p orbitals, 3d orbitals also available for bonding. In a number of compounds of these elements there are more than eight valence electrons around the central atom. This is termed as the expanded octet. Obviously the octet rule does not apply in such cases. Some of the examples of such compounds are: PF₅, SF₆, H₂SO₄ and a number of coordination compounds.
We must understand that formal charges do not indicate real charge separation within the molecule. Indicating the charges on the atoms in the Lewis structure only helps in keeping track of the valence electrons in the molecule. Formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures for a given species. Generally the lowest energy structure is the one with the smallest formal charges on the atoms. The formal charge is a factor based on a pure covalent view of bonding in which electron pairs are shared equally by neighbouring atoms. | 5 | 11 | Chemistry | 104 |
2bced750-09c9-4d4e-94b3-fa2435729c22 | # Chemistry
Interestingly, sulphur also forms many compounds in which the octet rule is obeyed. In sulphur dichloride, the S atom has an octet of electrons around it.
Other drawbacks of the octet theory:
- It is clear that octet rule is based upon the chemical inertness of noble gases. However, some noble gases (for example xenon and krypton) also combine with oxygen and fluorine to form a number of compounds like XeF2, KrF2, XeOF2 etc.
- This theory does not account for the shape of molecules.
- It does not explain the relative stability of the molecules being totally silent about the energy of a molecule.
# 4.2 Ionic or Electrovalent Bond
From the Kössel and Lewis treatment of the formation of an ionic bond, it follows that the formation of ionic compounds would primarily depend upon:
- The ease of formation of the positive and negative ions from the respective neutral atoms;
- The arrangement of the positive and negative ions in the solid, that is, the lattice of the crystalline compound.
The formation of a positive ion involves ionization, i.e., removal of electron(s) from the neutral atom and that of the negative ion involves the addition of electron(s) to the neutral atom.
M(g) → M+(g) + e–;
Ionization enthalpy
X(g) + e– → X–(g);
Electron gain enthalpy
M+(g) + X–(g) → MX(s)
The electron gain enthalpy, ∆H, is the enthalpy change (Unit 3), when a gas phase atom in its ground state gains an electron. The electron gain process may be exothermic or endothermic. The ionization, on the other hand, is always endothermic. Electron gain enthalpy is:
The ionization enthalpy for Na+(g) formation from Na(g) is 495.8 kJ mol–1; while the electron gain enthalpy for the change Cl(g) + e– → Cl–(g) is, –348.7 kJ mol–1 only. The sum of the two, 147.1 kJ mol–1 is more than compensated for by the enthalpy of lattice formation of NaCl(s) (–788 kJ mol–1). Therefore, the energy released in the processes is more than the. | 6 | 11 | Chemistry | 104 |
39e08425-7686-4026-b1a6-d4de75a9d6c8 | # Chemical Bonding and Molecular Structure
energy absorbed. thus a qualitative measure of the stability of an ionic compound is provided by its enthalpy of lattice formation and not simply by achieving octet of electrons around the ionic species in gaseous state.
Since lattice enthalpy plays a key role in the formation of ionic compounds, it is important that we learn more about it.
# 4.2.1 Lattice Enthalpy
the Lattice Enthalpy of an ionic solid is defined as the energy required to completely separate one mole of a solid ionic compound into gaseous constituent ions. For example, the lattice enthalpy of NaCl is 788 kJ mol–1. This means that 788 kJ of energy is required to separate one mole of solid NaCl into one mole of Na+ (g) and one mole of Cl– (g) to an infinite distance.
This process involves both the attractive forces between ions of opposite charges and the repulsive forces between ions of like charge. The solid crystal being three-dimensional; it is not possible to calculate lattice enthalpy directly from the interaction of forces of attraction and repulsion only. Factors associated with the crystal geometry have to be included.
# 4.3 BOND PARAMETERS
# 4.3.1 Bond Length
Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule. Bond lengths are measured by spectroscopic, X-ray diffraction and electron-diffraction techniques about which you will learn in higher classes. Each atom of the bonded pair contributes to the bond length (Fig. 4.1). In the case of a covalent bond, the contribution from each atom is called the covalent radius of that atom.
the covalent radius is measured approximately as the radius of an atom’s core which is in contact with the core of an adjacent atom in a bonded situation. The covalent radius is half of the distance between two similar atoms joined by a covalent bond.
# Fig. 4.1
The bond length in a covalent molecule AB.
R = rA + rB (R is the bond length and rA and rB are the covalent radii of atoms A and B respectively)
# Fig. 4.2
Covalent and van der Waals radii in a chlorine molecule. The inner circles correspond to the size of the chlorine atom (rvdw and rc are van der Waals and covalent radii respectively).
r = 99 pm 198 pm
vdw = 180 pm
360 pm | 7 | 11 | Chemistry | 104 |
73a0e712-dde2-41e2-8a04-e0f28021b6b5 | # Chemistry
Some typical average bond lengths for single, double and triple bonds are shown in Table 4.2. Bond lengths for some common molecules are given in Table 4.3.
The covalent radii of some common elements are listed in Table 4.4.
# 4.3.2 Bond Angle
It is defined as the angle between the orbitals containing bonding electron pairs around the central atom in a molecule/complex ion. Bond angle is expressed in degree which can be experimentally determined by spectroscopic methods. It gives some idea regarding the distribution of orbitals around the central atom in a molecule/complex ion and hence it helps us in determining its shape. For example H–O–H bond angle in water can be represented as under:
# Table 4.2 Average Bond Lengths for Some Single, Double and Triple Bonds
|Covalent Bond|Bond type|Length (pm)|
|---|---|---|
|O–H| |96|
|C–H| |107|
|N–O| |136|
|C–O| |143|
|C–N| |143|
|C–C| |154|
|C=O| |121|
|N=O| |122|
|C=C| |133|
|C=N| |138|
|C≡N| |116|
|C≡C| |120|
# 4.3.3 Bond Enthalpy
It is defined as the amount of energy required to break one mole of bonds of a particular type between two atoms in a gaseous state. The unit of bond enthalpy is kJ mol–1. For example, the H – H bond enthalpy in hydrogen molecule is 435.8 kJ mol–1.
H2(g) → H(g) + H(g); ∆aH = 435.8 kJ mol–1
Similarly the bond enthalpy for molecules containing multiple bonds, for example O2 and N2 will be as under:
O2 (O = O) (g) → O(g) + O(g); ∆aH = 498 kJ mol–1
N2 (N ≡ N) (g) → N(g) + N(g); ∆aH = 946.0 kJ mol–1
It is important that larger the bond dissociation enthalpy, stronger will be the bond in the molecule. For a heteronuclear diatomic molecules like HCl, we have:
HCl (g) → H(g) + Cl (g); ∆aH = 431.0 kJ mol–1
In case of polyatomic molecules, the measurement of bond strength is more complicated. For example in case of H2O molecule, the enthalpy needed to break the two O – H bonds is not the same.
* The values cited are for single bonds, except where otherwise indicated in parenthesis. (See also Unit 3 for periodic trends). | 8 | 11 | Chemistry | 104 |
18cda8c6-c476-450b-8dbd-56a837d6140c | # CHEMICAL BONDING AND MOLECULAR STRUCTURE
H₂O(g) → H(g) + OH(g); ∆ₐH = 502 kJ mol–1
OH(g) → H(g) + O(g); ∆ₐH = 427 kJ mol–1
The difference in the ∆ₐH value shows that the second O – H bond undergoes some change because of changed chemical environment. This is the reason for some difference in energy of the same O – H bond in different molecules like C₂H₅OH (ethanol) and water. Therefore in polyatomic molecules the term mean or average bond enthalpy is used. It is obtained by dividing total bond dissociation enthalpy by the number of bonds broken as explained below in case of water molecule,
Average bond enthalpy = 502 + 427
= 464.5 kJ mol–1
Fig. 4.3 Resonance in the O₃ molecule (structures I and II represent the two canonical forms while the structure III is the resonance hybrid)
# 4.3.4 Bond Order
In the Lewis description of covalent bond, the Bond Order is given by the number of bonds between the two atoms in a molecule. The bond order, for example in H₂ (with a single shared electron pair), in O₂ (with two shared electron pairs) and in N₂ (with three shared electron pairs) is 1, 2, 3 respectively. Similarly in CO (three shared electron pairs between C and O) the bond order is 3. For N₂, bond order is 3 and its bond enthalpy is 946 kJ mol–1; being one of the highest for a diatomic molecule.
Isoelectronic molecules and ions have identical bond orders; for example, F₂ and O₂⁻ have bond order 1. N₂ and CO have bond order 3.
A general correlation useful for understanding the stabilities of molecules is that: with increase in bond order, bond enthalpy increases and bond length decreases.
# 4.3.5 Resonance Structures
It is often observed that a single Lewis structure is inadequate for the representation of a molecule in conformity with its experimentally determined parameters. For example, the ozone, O₃ molecule can be equally represented by the structures I and II shown below:
The two structures shown above constitute the canonical structures or resonance structures and their hybrid i.e., the III structure represents the structure of O₃ more accurately. This is also called resonance hybrid. Resonance is represented by a double headed arrow. | 9 | 11 | Chemistry | 104 |
dcf17b7f-89a5-4776-87e0-b854244471a1 | # Chemistry
Some of the other examples of resonance structures are provided by the carbonate ion and the carbon dioxide molecule.
# Problem 4.3
Explain the structure of CO2– ion in terms of resonance.
# Solution
The single Lewis structure based on the presence of two single bonds and one double bond between carbon and oxygen atoms is inadequate to represent the molecule accurately as it represents unequal bonds. According to the experimental findings, all carbon to oxygen bonds in CO2– are equivalent. Therefore, the carbonate ion is best described as a resonance hybrid of the canonical forms I, II, and III shown below.
Fig. 4.5 Resonance in CO2–, I, II and III represent the three canonical forms.
In general, it may be stated that:
- Resonance stabilizes the molecule as the energy of the resonance hybrid is less than the energy of any single canonical structure; and,
- Resonance averages the bond characteristics as a whole.
Thus the energy of the O3 resonance hybrid is lower than either of the two canonical forms I and II (Fig. 4.3).
Fig. 4.3 Resonance in CO2–, I, II and III represent the three canonical forms.
Many misconceptions are associated with resonance and the same need to be dispelled. You should remember that:
- The canonical forms have no real existence.
- The molecule does not exist for a certain fraction of time in one canonical form and for other fractions of time in other canonical forms.
- There is no such equilibrium between the canonical forms as we have between tautomeric forms (keto and enol) in tautomerism.
The molecule as such has a single structure which is the resonance hybrid of the canonical forms and which cannot as such be depicted by a single Lewis structure.
# Problem 4.4
Explain the structure of CO2 molecule.
# Solution
The experimentally determined carbon to oxygen bond length in CO2 is 115 pm. The lengths of a normal carbon to oxygen double bond (C=O) and carbon to oxygen triple bond (C≡O) are 121 pm and 110 pm respectively. The carbon-oxygen bond lengths in CO2 (115 pm) lie between the values for C=O and C≡O. Obviously, a single Lewis structure cannot depict this position and it becomes necessary to write more than one Lewis structures and to consider that the structure of CO2 is best described as a hybrid of the canonical or resonance forms I, II and III.
# 4.3.6 Polarity of Bonds
The existence of a hundred percent ionic or covalent bond represents an ideal situation. In reality no bond or a compound is either completely covalent or ionic. Even in case of covalent bond between two hydrogen atoms, there is some ionic character.
When covalent bond is formed between two similar atoms, for example in H2, O2, Cl2, N2 or F2, the shared pair of electrons is equally attracted by the two atoms. As a result... | 10 | 11 | Chemistry | 104 |
ab55d2ab-209b-462a-bb5e-2ca144ed436b | # CHEMICAL BONDING AND MOLECULAR STRUCTURE
Electron pair is situated exactly between the two identical nuclei. The bond so formed is called nonpolar covalent bond. Contrary to this in case of a heteronuclear molecule like HF, the shared electron pair between the two atoms gets displaced more towards fluorine since the electronegativity of fluorine (Unit 3) is far greater than that of hydrogen. The resultant covalent bond is a polar covalent bond.
As a result of polarisation, the molecule possesses the dipole moment (depicted below) which can be defined as the product of the magnitude of the charge and the distance between the centres of positive and negative charge. It is usually designated by a Greek letter ‘μ’. Mathematically, it is expressed as follows:
Dipole moment (μ) = charge (Q) × distance of separation (r)
Dipole moment is usually expressed in Debye units (D). The conversion factor is 1 D = 3.33564 × 10–30 C m where C is coulomb and m is meter.
Further dipole moment is a vector quantity and by convention it is depicted by a small arrow with tail on the negative centre and head pointing towards the positive centre. But in chemistry presence of dipole moment is represented by the crossed arrow (⟶) put on Lewis structure of the molecule. The cross is on positive end and arrow head is on the negative end. For example the dipole moment of HF may be represented as:
H F
This arrow symbolises the direction of the shift of electron density in the molecule. Note that the direction of crossed arrow is opposite to the conventional direction of dipole moment vector.
Peter Debye, the Dutch chemist received Nobel prize in 1936 for his work on X-ray diffraction and dipole moments. The magnitude of the dipole moment is given in Debye units in order to honour him.
Let us study an interesting case of NH3 and NF3 molecule. Both the molecules have pyramidal shape with a lone pair of electrons on nitrogen atom. Although fluorine is more electronegative than nitrogen, the resultant dipole moment is zero. | 11 | 11 | Chemistry | 104 |
88fcf488-71ce-46a5-a51e-5a48d86eced6 | # Chemistry
The dipole moment of NH₃ (4.90 × 10–30 C m) is greater than that of NF₃ (0.8 × 10–30 C m). This is because, in case of NH₃ the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the N – H bonds, whereas in NF₃ the orbital dipole is in the direction opposite to the resultant dipole moment of the three N–F bonds. The orbital dipole because of lone pair decreases the effect of the resultant N – F bond moments, which results in the low dipole moment of NF₃ as represented below:
The smaller the size of the cation and the larger the size of the anion, the greater the covalent character of an ionic bond.
The greater the charge on the cation, the greater the covalent character of the ionic bond.
For cations of the same size and charge, the one with electronic configuration (n-1)dⁿns⁰, typical of transition metals, is more polarising than the one with a noble gas configuration, ns² np⁶, typical of alkali and alkaline earth metal cations.
The cation polarises the anion, pulling the electronic charge toward itself and thereby increasing the electronic charge between the two. This is precisely what happens in a covalent bond, i.e., buildup of electron charge density between the nuclei. The polarising power of the cation, the polarisability of the anion and the extent of distortion (polarisation) of anion are the factors, which determine the percent covalent character of the ionic bond.
Dipole moments of some molecules are shown in Table 4.5.
# 4.4 The Valence Shell Electron Pair Repulsion (VSEPR) Theory
As already explained, Lewis concept is unable to explain the shapes of molecules. This theory provides a simple procedure to predict the shapes of covalent molecules. Sidgwick.
# Table 4.5 Dipole Moments of Selected Molecules
|Type of Molecule|Example|Dipole Moment, μ(D)|Geometry|
|---|---|---|---|
|Molecule (AB)|HF|1.78|linear|
| |HCl|1.07|linear|
| |HBr|0.79|linear|
| |Hl|0.38|linear|
| |H₂|0|linear|
|Molecule (AB₂)|H₂O|1.85|bent|
| |H₂S|0.95|bent|
| |CO₂|0|linear|
|Molecule (AB₃)|NH₃|1.47|trigonal-pyramidal|
| |NF₃|0.23|trigonal-pyramidal|
| |BF₃|0|trigonal-planar|
|Molecule (AB₄)|CH₄|0|tetrahedral|
| |CHCl₃|1.04|tetrahedral|
| |CCl₄|0|tetrahedral| | 12 | 11 | Chemistry | 104 |
99e4a5e8-9f01-430e-a4df-013713dc916d | # Chemical Bonding and Molecular Structure
and Powell in 1940, proposed a simple theory based on the repulsive interactions of the electron pairs in the valence shell of the atoms. It was further developed and redefined by Nyholm and Gillespie (1957).
For the prediction of geometrical shapes of molecules with the help of VSEPR theory, it is convenient to divide molecules into two categories as (i) molecules in which the central atom has no lone pair and (ii) molecules in which the central atom has one or more lone pairs.
Table 4.6 (page 114) shows the arrangement of electron pairs about a central atom A (without any lone pairs) and geometries of some molecules/ions of the type AB. Table 4.7 (page 115) shows shapes of some simple molecules and ions in which the central atom has one or more lone pairs. Table 4.8 (page 116) explains the reasons for the distortions in the geometry of the molecule.
As depicted in Table 4.6, in the compounds of AB₂, AB₃, AB₄, AB₅ and AB₆, the arrangement of electron pairs and the B atoms around the central atom A are: linear, trigonal planar, tetrahedral, trigonal-bipyramidal and octahedral, respectively. Such arrangement can be seen in the molecules like BF₃ (AB₃), CH₄ (AB₄) and PCl₅ (AB₅) as depicted below by their ball and stick models.
The repulsive interaction of electron pairs decrease in the order:
- Lone pair (lp) – Lone pair (lp) > Lone pair (lp) – Bond pair (bp) > Bond pair (bp) – Bond pair (bp)
Nyholm and Gillespie (1957) refined the VSEPR model by explaining the important difference between the lone pairs and bonding pairs of electrons. While the lone pairs are localised on the central atom, each bonded pair is shared between two atoms. As a result, the lone pair electrons in a molecule occupy more space as compared to the bonding pairs of electrons. This results in greater repulsion between lone pairs of electrons as compared to the lone pair - bond pair and bond pair - bond pair repulsions. These repulsion effects continue to be a subject of doubt and discussion.
The VSEPR Theory is able to predict geometry of a large number of molecules, especially the compounds of p-block elements accurately. It is also quite successful in determining the geometry quite-accurately even when the energy difference between possible structures is very small. The theoretical basis of the VSEPR theory regarding the effects of electron pair repulsions on molecular shapes is not clear. | 13 | 11 | Chemistry | 104 |
ae72ed65-32aa-4a66-b0b7-a70cc1ccfad0 | # Table 4.6 Geometry of Molecules in which the Central Atom has no Lone pair of Electrons
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2aadd0b8-96b2-48c8-8607-f27c38ba461b | # CHEMICAL BONDING AND MOLECULAR STRUCTURE
# 115
# Table 4.7
Shape (geometry) of Some Simple Molecules/ions with Central ions having One or More Lone pairs of Electrons (E).
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8d7b06c2-cc46-45e6-a1d5-60770600a5fd | # CHEMISTRY
# Table 4.8 Shapes of Molecules containing Bond pair and Lone pair
|Molecule type|no. of bonding pairs|no. of lone pairs|Arrangement of electrons|Shape|Reason for the shape acquired|
|---|---|---|---|---|---|
|AB₂E|4|1| |Bent|Theoretically the shape should have been triangular planar but actually it is found to be bent or v-shaped. The reason being the lone pair-bond pair repulsion is much more as compared to the bond pair-bond pair repulsion. So the angle is reduced to 119.5° from 120°.|
|AB₃E|3|1| |Trigonal pyramidal|Had there been a bp in place of lp the shape would have been tetrahedral but one lone pair is present and due to the repulsion between lp-bp (which is more than bp-bp repulsion) the angle between bond pairs is reduced to 107° from 109.5°.|
|AB₂E₂|2|2| |Bent|The shape should have been tetrahedral if there were all bp but two lp are present so the shape is distorted tetrahedral or angular. The reason is lp-lp repulsion is more than lp-bp repulsion which is more than bp-bp repulsion. Thus, the angle is reduced to 104.5° from 109.5°.|
|AB₄E|4|1| |See-saw|In (a) the lp is present at axial position so there are three lp—bp repulsions at 90°. In (b) the lp is in an equatorial position, and there are two lp—bp repulsions. Hence, arrangement (b) is more stable. The shape shown in (b) is described as a distorted tetrahedron, a folded square or a see-saw.| | 16 | 11 | Chemistry | 104 |
faa8ebf0-9266-4224-8ca9-a80d446970a2 | # Chemical Bonding and Molecular Structure
|Molecule type|no. of bonding pairs|no. of lone pairs|Arrangement of electrons|Shape|Reason for the shape acquired|
|---|---|---|---|---|---|
|AB₃E₂|3|2| |T-shape|In (a) the lp are at equatorial position so there are less lp-bp repulsions as compared to others in which the lp are at axial positions. So structure (a) is most stable. (T-shaped).|
# 4.5 VALENCE BOND THEORY
As we know that Lewis approach helps in writing the structure of molecules but it fails to explain the formation of chemical bond. It also does not give any reason for the difference in bond dissociation enthalpies and bond lengths in molecules like H (435.8 kJ mol⁻¹, 74 pm) and F (155 kJ mol⁻¹, 144 pm), although in both cases a single covalent bond is formed by the sharing of an electron pair between the respective atoms. It also gives no idea about the shapes of polyatomic molecules.
Similarly the VSEPR theory gives the geometry of simple molecules but theoretically, it does not explain them and also it has limited applications. To overcome these limitations the two important theories based on quantum mechanical principles are introduced. These are valence bond (VB) theory and molecular orbital (MO) theory.
Valence bond theory was introduced by Heitler and London (1927) and developed further by Pauling and others. A discussion of the VB theory in terms of these aspects is beyond the scope of this book. Therefore, for the sake of convenience, valence bond theory has been discussed in terms of qualitative and non-mathematical treatment only. To start with, let us consider the formation of hydrogen molecule which is the simplest of all molecules.
Consider two hydrogen atoms A and B approaching each other having nuclei NA and NB and electrons present in them are represented by eA and eB. When the two atoms are at large distance from each other, there is no interaction between them. As these two atoms approach each other, new attractive and repulsive forces begin to operate.
Attractive forces arise between:
1. nucleus of one atom and its own electron that is NA – eA and NB – eB. | 17 | 11 | Chemistry | 104 |
ccdcceef-b7f3-4f8c-955a-327ad468f547 | # Chemistry
(ii) nucleus of one atom and electron of other atom i.e., NA–eB, NB–eA.
Similarly repulsive forces arise between
1. electrons of two atoms like eA – eB,
2. nuclei of two atoms NA – NB.
Attractive forces tend to bring the two atoms close to each other whereas repulsive forces tend to push them apart (Fig. 4.7).
Since the energy gets released when the bond is formed between two hydrogen atoms, the hydrogen molecule is more stable than that of isolated hydrogen atoms. The energy so released is called as bond enthalpy, which is corresponding to minimum in the curve depicted in Fig. 4.8. Conversely, 435.8 kJ of energy is required to dissociate one mole of H₂ molecule.
H₂(g) + 435.8 kJ mol–1 → H(g) + H(g)
Fig. 4.8 The potential energy curve for the formation of H₂ molecule as a function of internuclear distance of the H atoms. The minimum in the curve corresponds to the most stable state of H₂.
# 4.5.1 Orbital Overlap Concept
In the formation of hydrogen molecule, there is a minimum energy state when two hydrogen atoms are so near that their atomic orbitals undergo partial interpenetration. This partial merging of atomic orbitals is called overlapping of atomic orbitals which results in the pairing of electrons. The extent of overlap decides the strength of a covalent bond. In general, greater the overlap the stronger is the bond formed between two atoms. Therefore, according to orbital overlap concept, the formation of a covalent bond between two atoms results by pairing of electrons present in the valence shell having opposite spins.
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b030c624-0d08-46a4-a9de-6bebdd666a7a | # CHEMICAL BONDING AND MOLECULAR STRUCTURE
# 4.5.2 Directional properties of Bonds
As we have already seen, the covalent bond is formed by overlapping of atomic orbitals. The molecule of hydrogen is formed due to the overlap of 1s-orbitals of two H atoms. In case of polyatomic molecules like CH₄, NH₃ and H₂O, the geometry of the molecules is also important in addition to the bond formation. For example, why is it so that CH₄ molecule has tetrahedral shape and HCH bond angles are 109.5°? Why is the shape of NH₃ molecule pyramidal?
The valence bond theory explains the shape, the formation and directional properties of bonds in polyatomic molecules like CH₄, NH₃ and H₂O, etc. in terms of overlap and hybridisation of atomic orbitals.
# 4.5.3 Overlapping of Atomic Orbitals
When orbitals of two atoms come close to form bond, their overlap may be positive, negative or zero depending upon the sign (phase) and direction of orientation of amplitude of orbital wave function in space. Positive and negative sign on boundary surface diagrams show the sign (phase) of orbital wave function and are not related to charge.
Orbitals forming bond should have same sign (phase) and orientation in space. This is called positive overlap. Various overlaps of s and p orbitals are depicted in the figures.
The criterion of overlap, as the main factor for the formation of covalent bonds applies uniformly to the homonuclear/heteronuclear diatomic molecules and polyatomic molecules. We know that the shapes of CH₄, NH₃, and H₂O molecules are tetrahedral, pyramidal and bent respectively. It would be therefore interesting to use VB theory to find out if these geometrical shapes can be explained in terms of the orbital overlaps.
Let us first consider the CH₄ (methane) molecule. The electronic configuration of carbon in its ground state is [He]2s² 2p² which in the excited state becomes [He] 2s¹ 2pₓ¹ 2py¹ 2pz¹. The energy required for this excitation is compensated by the release of energy due to overlap between the orbitals of carbon and the 1s orbitals of the four H atoms which are also singly occupied. This will result in the formation of four C-H bonds. It will, however, be observed that while the three p orbitals of carbon are at 90° to one another, the HCH angle for these will also be 90°. That is three C-H bonds will be oriented at 90° to one another. The 2s orbital of carbon and the 1s orbital of H are spherically symmetrical and they can overlap in any direction. Therefore the direction of the fourth C-H bond cannot be ascertained. This description does not fit in with the tetrahedral HCH angles of 109.5°.
Clearly, it follows that simple atomic orbital overlap does not account for the directional characteristics of bonds in CH₄. Using similar procedure and arguments, it can be seen that in the case of NH₃ and H₂O molecules, the HNH angles... | 19 | 11 | Chemistry | 104 |
3713cb59-3215-436b-977d-94e19dc98ada | # Chemistry
and HOH angles should be 90°. This is in disagreement with the actual bond angles of 107° and 104.5° in the NH₃ and H₂O molecules respectively.
# 4.5.4 Types of Overlapping and Nature of Covalent Bonds
The covalent bond may be classified into two types depending upon the types of overlapping:
1. Sigma (σ) bond
2. Pi (π) bond
# (i) Sigma (σ) bond
This type of covalent bond is formed by the end to end (head-on) overlap of bonding orbitals along the internuclear axis. This is called as head-on overlap or axial overlap. This can be formed by any one of the following types of combinations of atomic orbitals:
- s-s overlapping: In this case, there is overlap of two half filled s-orbitals along the internuclear axis as shown below:
- s-p overlapping: This type of overlap occurs between half filled s-orbitals of one atom and half filled p-orbitals of another atom.
- p-p overlapping: This type of overlap takes place between half filled p-orbitals of the two approaching atoms.
# 4.5.5 Strength of Sigma and Pi Bonds
Basically the strength of a bond depends upon the extent of overlapping. In case of sigma bond, the overlapping of orbitals takes place to a larger extent. Hence, it is stronger as compared to the pi bond where the extent of overlapping occurs to a smaller extent. Further, it is important to note that in the formation of multiple bonds between two atoms of a molecule, pi bond(s) is formed in addition to a sigma bond.
# 4.6 Hybridisation
In order to explain the characteristic geometrical shapes of polyatomic molecules like CH₄, NH₃ and H₂O etc., Pauling introduced the concept of hybridisation. According to him, the atomic orbitals combine to form a new set of equivalent orbitals known as hybrid orbitals. Unlike pure orbitals, the hybrid orbitals are used in bond formation. The phenomenon is known as hybridisation which can be defined as the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of a new set of orbitals of equivalent energies and shape. For example, when one 2s and three 2p-orbitals of carbon hybridise, there is the formation of four new sp³ hybrid orbitals.
# (ii) Pi (π) bond
In the formation of π bond, the atomic orbitals overlap in such a way that their axes remain parallel to each other and perpendicular to the internuclear axis. The orbitals formed due to sidewise overlapping consist of two saucer type charged clouds.
# Salient features of hybridisation:
1. The number of hybrid orbitals is equal to the number of the atomic orbitals that get hybridised.
2. The hybridised orbitals are always equivalent in energy and shape. | 20 | 11 | Chemistry | 104 |
e90a2ba6-8ec2-4ba9-84b3-f5f10aca4904 | # CHEMICAL BONDING AND MOLECULAR STRUCTURE
3. The hybrid orbitals are more effective in forming stable bonds than the pure atomic orbitals. One 2s and one 2p-orbital gets hybridised to form two sp hybridised orbitals. These two sp hybrid orbitals are oriented in opposite direction forming an angle of 180°. Each of the sp hybridised orbital overlaps with the 2p-orbital of chlorine axially and form two Be-Cl sigma bonds. This is shown in Fig. 4.10.
Therefore, the type of hybridisation indicates the geometry of the molecules.
# Important conditions for hybridisation
1. The orbitals present in the valence shell of the atom are hybridised.
2. The orbitals undergoing hybridisation should have almost equal energy.
3. Promotion of electron is not an essential condition prior to hybridisation.
4. It is not necessary that only half filled orbitals participate in hybridisation. In some cases, even filled orbitals of valence shell take part in hybridisation.
# 4.6.1 Types of Hybridisation
There are various types of hybridisation involving s, p and d orbitals. The different types of hybridisation are as under:
# (i) sp hybridisation
This type of hybridisation involves the mixing of one s and one p orbital resulting in the formation of two equivalent sp hybrid orbitals. The suitable orbitals for sp hybridisation are s and pz, if the hybrid orbitals are to lie along the z-axis. Each sp hybrid orbital has 50% s-character and 50% p-character. Such a molecule in which the central atom is sp-hybridised and linked directly to two other central atoms possesses linear geometry. This type of hybridisation is also known as diagonal hybridisation.
The two sp hybrids point in the opposite direction along the z-axis with projecting positive lobes and very small negative lobes, which provides more effective overlapping resulting in the formation of stronger bonds.
# Example of molecule having sp hybridisation
BeCl2: The ground state electronic configuration of Be is 1s²2s². In the exited state one of the 2s-electrons is promoted to vacant 2p orbital as shown in Fig. 4.11.
# Fig. 4.10
(a) Formation of sp hybrids from s and p orbitals; (b) Formation of the linear BeCl2 molecule
# Fig. 4.11
Formation of sp² hybrids and the BCl3 molecule | 21 | 11 | Chemistry | 104 |
4e1771f6-e884-4d07-9d1b-b0cfa9ede8e7 | # Chemistry
A result boron has three unpaired electrons. Ground state is 2S²2p¹2p¹2p¹ having three unpaired electrons in the sp³ hybrid orbitals. The three hybrid orbitals so formed are oriented in a trigonal planar arrangement and overlap with 2p orbitals of chlorine to form three B-Cl bonds. Therefore, in BCl (Fig. 4.11), the geometry is trigonal with ClBCl bond angle of 120°.
# (iii) sp³ hybridisation:
This type of hybridisation can be explained by taking the example of CH₄ molecule in which there is mixing of one 4s orbital and three p-orbitals to form four sp³ hybrid orbitals of equivalent energies and shape. There is 25% s-character and 75% p-character in each sp³ hybrid orbital. The four sp³ hybrid orbitals so formed are directed towards the four corners of the tetrahedron. The angle between sp³ hybrid orbitals is 109.5° as shown in Fig. 4.12.
In case of H₂O molecule, the four oxygen orbitals (one 2s and three 2p) undergo sp³ hybridisation forming four sp³ hybrid orbitals out of which two contain one electron each and the other two contain a pair of electrons. These four sp³ hybrid orbitals acquire a tetrahedral geometry, with two corners occupied by hydrogen atoms while the other two by the lone pairs. The bond angle in this case is reduced to 104.5° from 109.5° (Fig. 4.14) and the molecule thus acquires a V-shape or angular geometry.
Fig. 4.12 Formation of sp³ hybrids by the combination of s, pₓ, pᵧ and p𝓏 atomic orbitals of carbon and the formation of CH₄ molecule.
The structure of NH₃ and H₂O molecules can also be explained with the help of sp³ hybridisation. In NH₃, the valence shell (outer) electronic configuration of nitrogen in the.
Fig. 4.14 Formation of H₂O molecule. | 22 | 11 | Chemistry | 104 |
af656d3f-02eb-4d93-8d00-988dfef5c62e | # Chemical Bonding and Molecular Structure
# 4.6.2 Other Examples of sp³, sp² and sp Hybridisation
sp³ Hybridisation in C₂H₆ molecule: In ethane molecule both the carbon atoms assume sp³ hybrid state. One of the four sp³ hybrid orbitals of carbon atom overlaps axially with similar orbitals of other atom to form sp³-sp³ sigma bond while the other three hybrid orbitals of each carbon atom are used in forming sp³–s sigma bonds with hydrogen atoms as discussed in section 4.6.1(iii). Therefore in ethane C–C bond length is 154 pm and each C–H bond length is 109 pm.
sp² Hybridisation in C₂H₄: In the formation of ethene molecule, one of the sp² hybrid orbitals of carbon atom overlaps axially with sp² hybridised orbital of another carbon atom to form C–C sigma bond. While the other two sp² hybrid orbitals of each carbon atom are used for making sp²–s sigma bond with two hydrogen atoms. The unhybridised orbital (2pₓ or 2py) of one carbon atom overlaps sidewise with the similar orbital of the other carbon atom to form weak π bond, which consists of two equal electron clouds distributed above and below the plane of carbon and hydrogen atoms. Thus, in ethene molecule, the carbon-carbon bond consists of one sp²–sp² sigma bond and one pi (π) bond between p orbitals which are not used in the hybridisation and are perpendicular to the plane of molecule; the bond length 134 pm. The C–H bond is sp²–s sigma with bond length 108 pm. The H–C–H bond angle is 117.6° while the H–C–C angle is 121°. The formation of sigma and pi bonds in ethene is shown in Fig. 4.15.
Fig. 4.15 Formation of sigma and pi bonds in ethene
2024-25 | 23 | 11 | Chemistry | 104 |
36f566ac-8e4b-454f-a28c-29b3e9cb2c09 | # 4.6.3 Hybridisation of Elements involving d Orbitals
The elements present in the third period contain d orbitals in addition to s and p orbitals. The energy of the 3d orbitals are comparable to the energy of the 3s and 3p orbitals. The energy of 3d orbitals are also comparable to those of 4s and 4p orbitals. As a consequence, the hybridisation involving either 3s, 3p and 3d or 3d, 4s and 4p is possible. However, since the difference in energies of 3p and 4s orbitals is significant, no hybridisation involving 3p, 3d and 4s orbitals is possible.
The important hybridisation schemes involving s, p and d orbitals are summarised below:
|Shape of molecules/ions|Hybridisation type|Atomic orbitals|Examples|
|---|---|---|---|
|Square planar|dsp²|d+s+p(2)|[Ni(CN)₄]²⁻, [Pt(Cl)₄]²⁻|
|Trigonal bipyramidal|sp³d|s+p(3)+d|PF₅, PCl₅|
|Square pyramidal|sp³d²|s+p(3)+d(2)|BrF₅|
|Octahedral|sp³d²|s+p(3)+d(2)|SF₆, [CrF₆]³⁻|
| |d²sp³|d(2)+s+p(3)|[Co(NH₃)₆]³⁺|
(i) Formation of PCl₅ (sp³d hybridisation): The ground state and the excited state outer electronic configurations of phosphorus (Z=15) are represented below.
Fig.4.16 Formation of sigma and pi bonds in ethyne sp³d hybrid orbitals filled by electron pairs donated by five Cl atoms. | 24 | 11 | Chemistry | 104 |
79cf096d-912e-46d9-ad39-9a8f55c33bb4 | # CHEMICAL BONDING AND MOLECULAR STRUCTURE
Now the five orbitals (i.e., one s, three p and one d orbitals) are available for hybridisation to yield a set of five sp³d² hybrid orbitals which are directed towards the five corners of a trigonal bipyramidal as depicted in the Fig. 4.17.
Fig. 4.17 Trigonal bipyramidal geometry of PCl₅ molecule
It should be noted that all the bond angles in trigonal bipyramidal geometry are not equivalent. In PCl₅ the five sp³d orbitals of phosphorus overlap with the singly occupied p orbitals of chlorine atoms to form five P–Cl sigma bonds. Three P–Cl bonds lie in one plane and make an angle of 120° with each other; these bonds are termed as equatorial bonds. The remaining two P–Cl bonds–one lying above and the other lying below the equatorial plane, make an angle of 90° with the plane. These bonds are called axial bonds. As the axial bond pairs suffer more repulsive interaction from the equatorial bond pairs, therefore axial bonds have been found to be slightly longer and hence slightly weaker than the equatorial bonds; which makes PCl₅ molecule more reactive.
Fig. 4.18 Octahedral geometry of SF₆ molecule
# 4.7 MOLECULAR ORBITAL THEORY
Molecular orbital (MO) theory was developed by F. Hund and R.S. Mulliken in 1932. The salient features of this theory are:
1. The electrons in a molecule are present in the various molecular orbitals as the electrons of atoms are present in the various atomic orbitals.
2. The atomic orbitals of comparable energies and proper symmetry combine to form molecular orbitals.
3. While an electron in an atomic orbital is influenced by one nucleus, in a molecular orbital it is influenced by two or more nuclei depending upon the number of atoms in the molecule. Thus,
In SF₆ the central sulphur atom has the ground state outer electronic configuration 3s²3p⁴. In the exited state the available six orbitals i.e., one s, three p and two d are singly occupied by electrons. These orbitals hybridise to form six new sp³d² hybrid orbitals, which are projected towards the six corners of a regular octahedron in SF₆. | 25 | 11 | Chemistry | 104 |
4af97a27-60cc-4272-859c-935056e546c8 | # Chemistry
An atomic orbital is monocentric while a molecular orbital is polycentric.
1. The number of molecular orbitals formed is equal to the number of combining atomic orbitals. When two atomic orbitals combine, two molecular orbitals are formed. One is known as bonding molecular orbital while the other is called antibonding molecular orbital.
2. The bonding molecular orbital has lower energy and hence greater stability than the corresponding antibonding molecular orbital.
3. Just as the electron probability distribution around a nucleus in an atom is given by an atomic orbital, the electron probability distribution around a group of nuclei in a molecule is given by a molecular orbital.
4. The molecular orbitals like atomic orbitals are filled in accordance with the Aufbau principle obeying the Pauli’s exclusion principle and the Hund’s rule.
# 4.7.1 Formation of Molecular Orbitals
# Linear Combination of Atomic Orbitals (LCAO)
According to wave mechanics, the atomic orbitals can be expressed by wave functions (ψ’s) which represent the amplitude of the electron waves. These are obtained from the solution of Schrödinger wave equation. However, since it cannot be solved for any system containing more than one electron, molecular orbitals which are one electron wave functions for molecules are difficult to obtain directly from the solution of Schrödinger wave equation. To overcome this problem, an approximate method known as linear combination of atomic orbitals (LCAO) has been adopted.
Let us apply this method to the homonuclear diatomic hydrogen molecule. Consider the hydrogen molecule consisting of two atoms A and B. Each hydrogen atom in the ground state has one electron in 1s orbital. The atomic orbitals of these atoms may be represented by the wave functions.
|ψA|ψB|
|---|---|
|σ = ψA + ψB|σ* = ψA - ψB|
Qualitatively, the formation of molecular orbitals can be understood in terms of the constructive or destructive interference of the electron waves of the combining atoms. In the formation of bonding molecular orbital, the two electron waves of the bonding atoms reinforce each other due to constructive interference while in the formation of antibonding molecular orbital, the waves cancel each other out.
Fig. 4.19 Formation of bonding (σ) and antibonding (σ*) molecular orbitals by the linear combination of atomic orbitals ψA and ψB centered on two atoms A and B respectively. | 26 | 11 | Chemistry | 104 |
e93700f3-9592-47de-8858-550b651a6ea3 | # Chemical Bonding and Molecular Structure
Antibonding molecular orbital, the electron waves cancel each other due to destructive interference. As a result, the electron density in a bonding molecular orbital is located between the nuclei of the bonded atoms because of which the repulsion between the nuclei is very less while in case of an antibonding molecular orbital, most of the electron density is located away from the space between the nuclei. Infact, there is a nodal plane (on which the electron density is zero) between the nuclei and hence the repulsion between the nuclei is high. Electrons placed in a bonding molecular orbital tend to hold the nuclei together and stabilise a molecule. Therefore, a bonding molecular orbital always possesses lower energy than either of the atomic orbitals that have combined to form it. In contrast, the electrons placed in the antibonding molecular orbital destabilise the molecule. This is because the mutual repulsion of the electrons in this orbital is more than the attraction between the electrons and the nuclei, which causes a net increase in energy.
It may be noted that the energy of the antibonding orbital is raised above the energy of the parent atomic orbitals that have combined and the energy of the bonding orbital has been lowered than the parent orbitals. The total energy of two molecular orbitals, however, remains the same as that of two original atomic orbitals.
# 4.7.2 Conditions for the Combination of Atomic Orbitals
The linear combination of atomic orbitals to form molecular orbitals takes place only if the following conditions are satisfied:
1. The combining atomic orbitals must have the same or nearly the same energy. This means that 1s orbital can combine with another 1s orbital but not with 2s orbital because the energy of 2s orbital is appreciably higher than that of 1s orbital. This is not true if the atoms are very different.
2. The combining atomic orbitals must have the same symmetry about the molecular axis. By convention z-axis is taken as the molecular axis. It is important to note that atomic orbitals having same or nearly the same energy will not combine if they do not have the same symmetry. For example, 2pz orbital of one atom can combine with 2pz orbital of the other atom but not with the 2px or 2py orbitals because of their different symmetries.
3. The combining atomic orbitals must overlap to the maximum extent. Greater the extent of overlap, the greater will be the electron-density between the nuclei of a molecular orbital.
# 4.7.3 Types of Molecular Orbitals
Molecular orbitals of diatomic molecules are designated as σ (sigma), π (pi), δ (delta), etc. In this nomenclature, the sigma (σ) molecular orbitals are symmetrical around the bond-axis while pi (π) molecular orbitals are not symmetrical. For example, the linear combination of 1s orbitals centered on two nuclei produces two molecular orbitals which are symmetrical around the bond-axis. Such molecular orbitals are of the σ type and are designated as σ1s and σ*1s. If internuclear axis is taken to be in the z-direction, it can be seen that a linear combination of 2pz orbitals of two atoms also produces two sigma molecular orbitals designated as 2pz and *2pz. 2p Molecular orbitals obtained from 2px and 2py orbitals are not symmetrical around the bond axis because of the presence of positive lobes above and negative lobes below the molecular plane. Such molecular orbitals are labelled as π and π*.
A π bonding MO has larger electron density above and below the inter-nuclear axis. The π* antibonding MO has a node between the nuclei.
# 4.7.4 Energy Level Diagram for Molecular Orbitals
We have seen that 1s atomic orbitals on two atoms form two molecular orbitals designated as σ1s and σ*1s. In the same manner, the 2s and 2p atomic orbitals (eight atomic orbitals) | 27 | 11 | Chemistry | 104 |
f9288fd6-10c7-4c9d-a68a-3a6f1a732bda | # Chemistry
Fig. 4.20 Contours and energies of bonding and antibonding molecular orbitals formed through combinations of (a) 1s atomic orbitals; (b) 2pz atomic orbitals and (c) 2pₓ atomic orbitals.
on two atoms) give rise to the following eight molecular orbitals:
|Antibonding MOs|σ∗2s|σ∗2pz|π∗2pₓ|π∗2py|
|---|---|---|---|---|
|Bonding MOs|σ2s|σ2p|π2p|π2p|
| | |z|x|y|
The energy levels of these molecular orbitals have been determined experimentally from spectroscopic data for homonuclear diatomic molecules of second row elements of the periodic table. The increasing order of 2024-25 | 28 | 11 | Chemistry | 104 |
df6b12a6-9c34-4ac5-b567-8c5562de01f9 | # CHEMICAL BONDING AND MOLECULAR STRUCTURE
# energies of various molecular orbitals for O₂ and F₂
The rules discussed above regarding the stability of the molecule can be restated in terms of bond order as follows: A positive bond order (i.e., Nb > Na) means a stable molecule while a negative (i.e., Nb < Na) or zero (i.e., Nb = Na) bond order means an unstable molecule.
# Nature of the bond
Integral bond order values of 1, 2 or 3 correspond to single, double or triple bonds respectively as studied in the classical concept.
# Bond-length
The bond order between two atoms in a molecule may be taken as an approximate measure of the bond length. The bond length decreases as bond order increases.
# Magnetic nature
If all the molecular orbitals in a molecule are doubly occupied, the substance is diamagnetic (repelled by magnetic field). However, if one or more molecular orbitals are singly occupied it is paramagnetic (attracted by magnetic field), e.g., O₂ molecule.
# 4.8 BONDING in SOME HOMONUCLEAR DIATOMIC MOLECULES
In this section we shall discuss bonding in some homonuclear diatomic molecules.
# 1. Hydrogen molecule (H₂)
It is formed by the combination of two hydrogen atoms. Each hydrogen atom has one electron in 1s orbital. Therefore, in all there are two electrons in hydrogen molecule which are present in σ1s molecular orbital. So electronic configuration of hydrogen molecule is:
H₂ : (σ1s)²
# Bond order
The bond order of H₂ molecule can be calculated as given below:
Bond order = Nb - Na = 2 - 0 = 1
This means that the two hydrogen atoms are bonded together by a single covalent bond. The bond dissociation energy of hydrogen molecule has been found to be 438 kJ mol–1 and bond length equal to 74 pm. | 29 | 11 | Chemistry | 104 |
10bf992d-6396-4b83-a4b3-4d565f109182 | # Chemistry
An unpaired electron is present in hydrogen vapour phase. It is important to note that the double bond in C₂ consists of both pi bonds because of the presence of four electrons in two pi molecular orbitals. In most of the other molecules, a double bond is made up of a sigma bond and a pi bond. In a similar fashion, the bonding in N₂ molecule can be discussed.
# 2. Helium molecule (He₂)
The electronic configuration of helium atom is 1s². Each helium atom contains 2 electrons, therefore, in He₂ molecule there would be 4 electrons. These electrons will be accommodated in σ1s and σ*1s molecular orbitals leading to electronic configuration:
He: (σ1s)² (σ*1s)²
Bond order of He₂ is ½(2 – 2) = 0. He₂ molecule is therefore unstable and does not exist.
# 3. Lithium molecule (Li₂)
The electronic configuration of lithium is 1s², 2s¹. There are six electrons in Li₂. The electronic configuration of Li₂ molecule, therefore, is:
Li: (σ1s)² (σ*1s)² (σ2s)²
The above configuration is also written as KK(σ2s)² where KK represents the closed K shell structure (σ1s)² (σ*1s)².
From the electronic configuration of Li₂ molecule it is clear that there are four electrons present in bonding molecular orbitals and two electrons present in antibonding molecular orbitals. Its bond order, therefore, is ½ (4 – 2) = 1. It means that Li₂ molecule is stable and since it has no unpaired electrons it should be diamagnetic. Indeed diamagnetic Li₂ molecules are known to exist in the vapour phase.
# 4. Carbon molecule (C₂)
The electronic configuration of carbon is 1s² 2s² 2p². There are twelve electrons in C₂. The electronic configuration of C₂ molecule, therefore, is:
C₂: (σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (π2pₓ² ≡ π2pᵧ²)
The bond order of C₂ is ½ (8 – 4) = 2 and C₂ should be diamagnetic. Diamagnetic C₂ molecules have indeed been detected in the vapour phase.
# 5. Oxygen molecule (O₂)
The electronic configuration of oxygen atom is 1s² 2s² 2p⁴. Each oxygen atom has 8 electrons, hence, in O₂ molecule there are 16 electrons. The electronic configuration of O₂ molecule, therefore, is:
O₂: (σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (2pₓ)² (π2pₓ² ≡ π2pᵧ²) (π*2pₓ¹ ≡ π*2pᵧ¹)
From the electronic configuration of O₂ molecule it is clear that ten electrons are present in bonding molecular orbitals and six electrons are present in antibonding molecular orbitals. Its bond order, therefore, is:
Bond order = [Nb – Na] = [10 – 6] = 2
So in oxygen molecule, atoms are held by a double bond. Moreover, it may be noted that it contains two unpaired electrons in π*2p molecular orbitals, therefore, O₂ molecule should be paramagnetic, a prediction that corresponds to experimental observation. In this way, the theory successfully explains the paramagnetic nature of oxygen.
Similarly, the electronic configurations of other homonuclear diatomic molecules of the second row of the periodic table can be written. In Fig. 4.21 are given the molecular orbital occupancy and molecular properties for B₂ through Ne₂. The sequence of MOs and their electron population are shown. The bond energy, bond length, bond order, magnetic properties and valence electron configuration appear below the orbital diagrams. | 30 | 11 | Chemistry | 104 |
28dcab57-36cf-4643-9da2-5343862d905b | # Chemical Bonding and Molecular Structure
Fig. 4.21 MO occupancy and molecular properties for B₂ through Ne₂.
# 4.9 Hydrogen Bonding
Hydrogen bond is represented by a dotted line (– – –) while a solid line represents the covalent bond. Thus, hydrogen bond can be defined as the attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O or N) of another molecule.
# 4.9.1 Cause of Formation of Hydrogen Bond
When hydrogen is bonded to strongly electronegative element ‘X’, the electron pair shared between the two atoms moves far away from hydrogen atom. As a result, the hydrogen atom becomes highly electropositive with respect to the other atom ‘X’. Since there is displacement of electrons towards X, the hydrogen acquires fractional positive charge (δ+) while ‘X’ attains fractional negative charge (δ−). | 31 | 11 | Chemistry | 104 |
8e9e4f6f-d329-4696-b3a1-07f4f6a7506e | # 4.9.2 types of H-Bonds
There are two types of H-bonds
1. Intermolecular hydrogen bond
2. Intramolecular hydrogen bond
# (1) Intermolecular hydrogen bond :
It is formed between two different molecules of the same or different compounds. For example, water molecules, etc.
# (2) Intramolecular hydrogen bond :
It is formed when hydrogen atom is in between the two highly electronegative (F, O, N) atoms present within the same molecule. For example, in o-nitrophenol the hydrogen is in between the two oxygen atoms.
Fig. 4.22 Intramolecular hydrogen bonding in o-nitrophenol molecule
# SUMMARy
Kössel’s first insight into the mechanism of formation of electropositive and electronegative ions related the process to the attainment of noble gas configurations by the respective ions. Electrostatic attraction between ions is the cause for their stability. This gives the concept of electrovalency.
The first description of covalent bonding was provided by Lewis in terms of the sharing of electron pairs between atoms and he related the process to the attainment of noble gas configurations by reacting atoms as a result of sharing of electrons. The Lewis dot symbols show the number of valence electrons of the atoms of a given element and Lewis dot structures show pictorial representations of bonding in molecules.
An ionic compound is pictured as a three-dimensional aggregation of positive and negative ions in an ordered arrangement called the crystal lattice. In a crystalline solid there is a charge balance between the positive and negative ions. The crystal lattice is stabilized by the enthalpy of lattice formation.
While a single covalent bond is formed by sharing of an electron pair between two atoms, multiple bonds result from the sharing of two or three electron pairs. Some bonded atoms have additional pairs of electrons not involved in bonding. These are called lone-pairs of electrons. A Lewis dot structure shows the arrangement of bonded pairs and lone pairs around each atom in a molecule. Important parameters, associated with chemical bonds, like: bond length, bond angle, bond enthalpy, bond order and bond polarity have significant effect on the properties of compounds.
A number of molecules and polyatomic ions cannot be described accurately by a single Lewis structure and a number of descriptions (representations) based on the same skeletal structure are written and these taken together represent the molecule or ion. This is a very important and extremely useful concept called resonance. The contributing structures or canonical forms taken together constitute the resonance hybrid which represents the molecule or ion.
2024-25 | 32 | 11 | Chemistry | 104 |
bb58ea80-8702-478a-af5a-e6727b73f51b | # Chemical Bonding and Molecular Structure
The VSEPR model used for predicting the geometrical shapes of molecules is based on the assumption that electron pairs repel each other and, therefore, tend to remain as far apart as possible. According to this model, molecular geometry is determined by repulsions between lone pairs and lone pairs; lone pairs and bonding pairs and bonding pairs and bonding pairs. The order of these repulsions being: lp-lp > lp-bp > bp-bp.
The valence bond (vB) approach to covalent bonding is basically concerned with the energetics of covalent bond formation about which the Lewis and VSEPR models are silent. Basically, the VB theory discusses bond formation in terms of overlap of orbitals. For example, the formation of the H₂ molecule from two hydrogen atoms involves the overlap of the 1s orbitals of the two H atoms which are singly occupied. It is seen that the potential energy of the system gets lowered as the two H atoms come near to each other. At the equilibrium inter-nuclear distance (bond distance) the energy touches a minimum. Any attempt to bring the nuclei still closer results in a sudden increase in energy and consequent destabilization of the molecule. Because of orbital overlap, the electron density between the nuclei increases which helps in bringing them closer. It is however seen that the actual bond enthalpy and bond length values are not obtained by overlap alone and other variables have to be taken into account.
For explaining the characteristic shapes of polyatomic molecules, Pauling introduced the concept of hybridisation of atomic orbitals. sp, sp², sp³ hybridizations of atomic orbitals of Be, B, C, N and O are used to explain the formation and geometrical shapes of molecules like BeCl₂, BCl₃, CH₄, NH₃ and H₂O. They also explain the formation of multiple bonds in molecules like C₂H₂ and C₂H₄.
The molecular orbital (MO) theory describes bonding in terms of the combination and arrangement of atomic orbitals to form molecular orbitals that are associated with the molecule as a whole. The number of molecular orbitals are always equal to the number of atomic orbitals from which they are formed. Bonding molecular orbitals increase electron density between the nuclei and are lower in energy than the individual atomic orbitals. Antibonding molecular orbitals have a region of zero electron density between the nuclei and have more energy than the individual atomic orbitals.
The electronic configuration of the molecules is written by filling electrons in the molecular orbitals in the order of increasing energy levels. As in the case of atoms, the Pauli exclusion principle and Hund’s rule are applicable for the filling of molecular orbitals. Molecules are said to be stable if the number of electrons in bonding molecular orbitals is greater than that in antibonding molecular orbitals.
Hydrogen bond is formed when a hydrogen atom finds itself between two highly electronegative atoms such as F, O and N. It may be intermolecular (existing between two or more molecules of the same or different substances) or intramolecular (present within the same molecule). Hydrogen bonds have a powerful effect on the structure and properties of many compounds.
# EXERCISES
1. Explain the formation of a chemical bond.
2. Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.
3. Write Lewis symbols for the following atoms and ions: S and S²⁻; Al and Al³⁺; H and H⁻.
4. Draw the Lewis structures for the following molecules and ions: H₂S, SiCl₄, BeF₂, CO₂²⁻, HCOOH.
5. Define octet rule. Write its significance and limitations.
2024-25 | 33 | 11 | Chemistry | 104 |
cd0b96cf-9553-4648-a0c0-22a416fa7229 | # Chemistry
# 4.6
Write the favourable factors for the formation of ionic bond.
# 4.7
Discuss the shape of the following molecules using the VSEPR model: BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S, PH₃
# 4.8
Although geometries of NH₃ and H₂O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
# 4.9
How do you express the bond strength in terms of bond order?
# 4.10
Define the bond length.
# 4.11
Explain the important aspects of resonance with reference to the CO₂⁻ ion.
# 4.12
H₃PO₃ can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H₃PO₃? If not, give reasons for the same.
# 4.13
Write the resonance structures for SO₃, NO₂ and NO⁻.
# 4.14
Use Lewis symbols to show electron transfer between the following atoms to form cations and anions: (a) K and S (b) Ca and O (c) Al and N.
# 4.15
Although both CO₂ and H₂O are triatomic molecules, the shape of H₂O molecule is bent while that of CO₂ is linear. Explain this on the basis of dipole moment.
# 4.16
Write the significance/applications of dipole moment.
# 4.17
Define electronegativity. How does it differ from electron gain enthalpy?
# 4.18
Explain with the help of suitable example polar covalent bond.
# 4.19
Arrange the bonds in order of increasing ionic character in the molecules: LiF, K₂O, N₂, SO₂ and ClF₃.
# 4.20
The skeletal structure of CH₃COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.
# 4.21
Apart from tetrahedral geometry, another possible geometry for CH₄ is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH₄ is not square planar?
# 4.22
Explain why BeH₂ molecule has a zero dipole moment although the Be–H bonds are polar.
# 4.23
Which out of NH₃ and NF₃ has higher dipole moment and why?
# 4.24
What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp², sp³ hybrid orbitals.
# 4.25
Describe the change in hybridisation (if any) of the Al atom in the following reaction: AlCl₃ + Cl⁻ → AlCl₄⁻ | 34 | 11 | Chemistry | 104 |
dba07d09-84ac-4920-926e-f5aba3665bd9 | # Chemical Bonding and Molecular Structure
# 4.26
Is there any change in the hybridisation of B and N atoms as a result of the following reaction?
BF3 + NH3 → F3B-NH3
# 4.27
Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C2H4 and C2H2 molecules.
# 4.28
What is the total number of sigma and pi bonds in the following molecules?
- (a) C2H2
- (b) C2H4
# 4.29
Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why?
- (a) 1s and 1s
- (b) 1s and 2px
- (c) 2py and 2py
- (d) 1s and 2s
# 4.30
Which hybrid orbitals are used by carbon atoms in the following molecules?
- (a) CH3–CH3
- (b) CH3–CH=CH2
- (c) CH3–CH2–OH
- (d) CH3–CHO
- (e) CH3COOH
# 4.31
What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.
# 4.32
Distinguish between a sigma and a pi bond.
# 4.33
Explain the formation of H2 molecule on the basis of valence bond theory.
# 4.34
Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
# 4.35
Use molecular orbital theory to explain why the Be2 molecule does not exist.
# 4.36
Compare the relative stability of the following species and indicate their magnetic properties;
(superoxide) O2− (peroxide)
# 4.37
Write the significance of a plus and a minus sign shown in representing the orbitals.
# 4.38
Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to equatorial bonds?
# 4.39
Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
# 4.40
What is meant by the term bond order? Calculate the bond order of: N2, O2, O+ and O2−.
2024-25 | 35 | 11 | Chemistry | 104 |
eeeeb9a4-c722-40a6-b0a8-75b5506e2f62 | # HYDROCARBONS
# Hydrocarbons
After studying this unit, you will be able to:
- name hydrocarbons according to IUPAC system of nomenclature;
- recognise and write structures of isomers of alkanes, alkenes, alkynes and aromatic hydrocarbons;
- learn about various methods of preparation of hydrocarbons;
- distinguish between alkanes, alkenes, alkynes and aromatic hydrocarbons on the basis of physical and chemical properties;
- draw and differentiate between various conformations of ethane;
- appreciate the role of hydrocarbons as sources of energy and for other industrial applications;
- predict the formation of the addition products of unsymmetrical alkenes and alkynes on the basis of electronic mechanism;
- comprehend the structure of benzene, explain aromaticity and understand mechanism of electrophilic substitution reactions of benzene;
- predict the directive influence of substituents in monosubstituted benzene ring;
- learn about carcinogenicity and toxicity.
Hydrocarbons are the important sources of energy.
The term ‘hydrocarbon’ is self-explanatory which means compounds of carbon and hydrogen only. Hydrocarbons play a key role in our daily life. You must be familiar with the terms ‘LPG’ and ‘CNG’ used as fuels. LPG is the abbreviated form of liquified petroleum gas whereas CNG stands for compressed natural gas. Another term ‘LNG’ (liquified natural gas) is also in news these days. This is also a fuel and is obtained by liquifaction of natural gas. Petrol, diesel and kerosene oil are obtained by the fractional distillation of petroleum found under the earth’s crust. Coal gas is obtained by the destructive distillation of coal. Natural gas is found in upper strata during drilling of oil wells. The gas after compression is known as compressed natural gas. LPG is used as a domestic fuel with the least pollution. Kerosene oil is also used as a domestic fuel but it causes some pollution. Automobiles need fuels like petrol, diesel and CNG. Petrol and CNG operated automobiles cause less pollution. All these fuels contain mixture of hydrocarbons, which are sources of energy. Hydrocarbons are also used for the manufacture of polymers like polythene, polypropene, polystyrene etc. Higher hydrocarbons are used as solvents for paints. They are also used as the starting materials for manufacture of many dyes and drugs. Thus, you can well understand the importance of hydrocarbons in your daily life. In this unit, you will learn more about hydrocarbons.
# 9.1 Classification
Hydrocarbons are of different types. Depending upon the types of carbon-carbon bonds present, they can be classified into three main categories – (i) saturated | 0 | 11 | Chemistry | 203 |
eaf5d8ad-307e-4964-8dc7-8a9da06aad9d | # CHEMISTRY
(ii) unsaturated and (iii) aromatic of the general formula for alkane family hydrocarbons. Saturated hydrocarbons contain carbon-carbon and carbon-hydrogen single bonds. If different carbon atoms are joined together to form open chain of carbon atoms with single bonds, they are termed as alkanes as you have already studied in Unit 8. On the other hand, if carbon atoms form a closed chain or a ring, they are termed as cycloalkanes. Unsaturated hydrocarbons contain carbon-carbon multiple bonds – double bonds, triple bonds or both. Aromatic hydrocarbons are a special type of cyclic compounds. You can construct a large number of models of such molecules of both types (open chain and close chain) keeping in mind that carbon is tetravalent and hydrogen is monovalent. For making models of alkanes, you can use toothpicks for bonds and plasticine balls for atoms. For alkenes, alkynes and aromatic hydrocarbons, spring models can be constructed.
# 9.2 ALKANES
As already mentioned, alkanes are saturated open chain hydrocarbons containing carbon - carbon single bonds. Methane (CH4) is the first member of this family. Methane is a gas found in coal mines and marshy places. If you replace one hydrogen atom of methane by carbon and join the required number of hydrogens to satisfy the tetravalence of the other carbon atom, what do you get? You get C2H6. This hydrocarbon with molecular formula C2H6 is known as ethane. Thus you can consider C2H6 as derived from CH4 by replacing one hydrogen atom by -CH3 group. Go on constructing alkanes by doing this theoretical exercise i.e., replacing hydrogen atom by –CH3 group. The next molecules will be C3H8, C4H10 …
These hydrocarbons are inert under normal conditions as they do not react with acids, bases and other reagents. Hence, they were earlier known as paraffins (latin: parum, little; affinis, affinity). Can you think Butane (n- butane), (b.p. 273 K)
# 9.2.1 Nomenclature and Isomerism
You have already read about nomenclature of different classes of organic compounds in Unit 8. Nomenclature and isomerism in alkanes can further be understood with the help of a few more examples. Common names are given in parenthesis. First three alkanes – methane, ethane and propane have only one structure but higher alkanes can have more than one structure. Let us write structures for C4H10. Four carbon atoms of C4H10 can be joined either in a continuous chain or with a branched chain in the following two ways:
H H H<br/>
H—C—H replace any H by - CH3 H—C—C—H or C2H6 | 1 | 11 | Chemistry | 203 |
2bec45ce-5946-444c-bbef-a6b34ee9859c | # HYDROCARBONS
II isomers. It is also clear that structures I and III have continuous chain of carbon atoms but structures II, IV and V have a branched chain. Such structural isomers which differ in chain of carbon atoms are known as chain isomers. Thus, you have seen that C₄H₁₀ and C₅H₁₂ have two and three chain isomers respectively.
# Problem 9.1
In how many ways, you can join five carbon atoms and twelve hydrogen atoms of C₅H₁₂? They can be arranged in three ways as shown in structures III–V.
|III|IV|V|
|---|---|---|
|CH₃ – CH₂ – CH₂ – CH₂– CH₂– CH₃ Pentane (n-pentane) (b.p. 309 K)|2-Methylbutane (isopentane) (b.p. 301 K)|2,2-Dimethylpropane (neopentane) (b.p. 282.5 K)|
Based upon the number of carbon atoms attached to a carbon atom, the carbon atom is termed as primary (1°), secondary (2°), tertiary (3°) or quaternary (4°). Carbon atom attached to no other carbon atom as in methane or to only one carbon atom as in ethane is called primary carbon atom. Terminal carbon atoms are always primary. Carbon atom attached to two carbon atoms is known as secondary. Tertiary carbon is attached to three carbon atoms and neo or quaternary carbon is attached to four carbon atoms. Can you identify 1°, 2°, 3° and 4° carbon atoms in structures? | 2 | 11 | Chemistry | 203 |
40f2edf1-efc3-4627-af0d-af72f3623993 | # Chemistry
Structures I to V? If you go on constructing compounds. These groups or substituents are known as alkyl groups as they are derived from alkanes by removal of one hydrogen atom. General formula for alkyl groups is CₙH₂ₙ₊₁ (Unit 8).
In structures II, IV and V, you observed that –CH₃ group is attached to carbon atom numbered as 2. You will come across groups like –CH₃, –C₂H₅, –C₃H₇ etc. attached to carbon atoms in alkanes or other classes of.
# Problem 9.2
Write structures of different isomeric alkyl groups corresponding to the molecular formula C₅H₁₁. Write IUPAC names of alcohols obtained by attachment of –OH groups at different carbons of the chain.
# Solution
|Structures of – C₅H₁₁ group|Corresponding alcohols|Name of alcohol|
|---|---|---|
|(i) CH₃ – CH₂ – CH₂ – CH₂– CH₂|CH₃ – CH₂ – CH₂ – CH₂– OH|Pentan-1-ol|
|(ii) CH₃ – CH – CH₂ – CH₂ – CH₃|CH₃ – CH – CH₂ – CH₂– OH|Pentan-2-ol|
|(iii) CH₃ – CH₂ – CH – CH₂ – CH₃|CH₃ – CH₂ – CH – CH₂– OH|Pentan-3-ol|
| | |3-Methyl-butan-1-ol|
|(iv) CH₃ – CH – CH₂ – CH₂ – CH₃|CH₃ – CH – CH₂ – CH₂– OH|2-Methyl-butan-1-ol|
|(v) CH₃ – CH₂ – CH – CH₂ – CH₃|CH₃ – CH₂ – CH – CH₂– OH|2-Methyl-butan-2-ol|
|(vi) CH₃ – C – CH₂ – CH₃|CH₃ – C – CH₂ – CH₃|2,2-Dimethyl-propan-1-ol|
|(vii) CH₃ – C – CH₂ –|CH₃ – C – CH₂OH|3-Methyl-butan-2-ol|
|CH₃|CH₃|OH|
|(viii) CH₃ – CH – CH –CH₃| | | | 3 | 11 | Chemistry | 203 |
5163c859-18d0-47b8-84e0-14ffe03f14f0 | # HYDROCARBONS
# Table 9.1 Nomenclature of a Few Organic Compounds
|Structure and IUPAC Name|Remarks|
|---|---|
|(a) CH₃– CH – CH₂ – CH – CH₂ – CH₃ (4 – Ethyl – 2 – methylhexane)|Lowest sum and alphabetical arrangement|
|(b) 8CH –⁷CH –⁶CH –⁵CH – 4CH – 3C – 2CH – 1CH (3,3-Diethyl-5-isopropyl-4-methyloctane)|Lowest sum and alphabetical arrangement|
|(c) 1CH –2CH –3CH –4CH–5CH–6CH –7CH –8CH –9CH –10CH (5-sec– Butyl-4-isopropyldecane)|sec is not considered while arranging alphabetically; isopropyl is taken as one word|
|(d) 1CH3–2CH2–3CH2–4CH2–5CH–6CH2–7CH2–8CH2–9CH3 (5-(2,2– Dimethylpropyl)nonane)|Further numbering to the substituents of the side chain|
|(e) 1CH3 –2CH2 –3CH – 4CH2 – 5CH – 6CH2 – 7CH3 (3–Ethyl–5–methylheptane)|Alphabetical priority order|
# Problem 9.3
Important to write the correct structure from the given IUPAC name. To do this, first of all, the longest chain of carbon atoms corresponding to the parent alkane is written. Then after numbering it, the substituents are attached to the correct carbon atoms and finally valence of each carbon atom is satisfied by putting the correct number of hydrogen atoms. This can be clarified by writing the structure of 3-ethyl-2, 2–dimethylpentane in the following steps:
1. Draw the chain of five carbon atoms: C – C – C – C – C
2. Give number to carbon atoms: C¹– C²– C³– C⁴– C⁵
# Solution
1. (i) 2, 2, 4, 4-Tetramethylpentane
2. (ii) 3, 3-Dimethylpentane
3. (iii) 3,3-Di-tert-butyl -2, 2, 4, 4 - tetramethylpentane | 4 | 11 | Chemistry | 203 |
89637095-f198-4e41-8e66-31f472a11eb3 | # CHEMISTRY
iii) Attach ethyl group at carbon 3 and two methyl groups at carbon 2. Longest chain is of six carbon atoms and not that of five. Hence, correct name is 3-Methylhexane.
C¹ – 2C – 3C – 4C – 5C (ii) 7 6 5 4 3 2 1
CH₃– CH₂ – CH – CH₂ – CH – CH₂ – CH
3
iv) Satisfy the valence of each carbon atom by putting requisite number of hydrogen atoms: Numbering is to be started from the end which gives lower number to ethyl group. Hence, correct name is 3-ethyl-5-methylheptane.
CH₃ – C – CH – CH₂ – CH₃
# 9.2.2 Preparation
Petroleum and natural gas are the main sources of alkanes. However, alkanes can be prepared by following methods:
1. From unsaturated hydrocarbons
Dihydrogen gas adds to alkenes and alkynes in the presence of finely divided catalysts like platinum, palladium or nickel to form alkanes. This process is called hydrogenation. These metals adsorb dihydrogen gas on their surfaces and activate the hydrogen – hydrogen bond. Platinum and palladium catalyse the reaction at room temperature but relatively higher temperature and pressure are required with nickel catalysts.
(i) CH₃ – CH₂ – CH – C – CH– CH – CH₂ =CH₂ + H₂ Pt/Pd/Ni CH₃−CH₃
Ethene Propane (9.1)
CH₂–CH=CH₂ + H₂ Pt/Pd/Ni CH₃−CH₂CH₃
Propane Propane
(ii) CH₃ – CH – CH₂ – CH₂ – CH – CH₃ (9.2)
# Problem 9.5
Write structures for each of the following compounds. Why are the given names incorrect? Write correct IUPAC names.
1. (i) 2-Ethylpentane
2. (ii) 5-Ethyl – 3-methylheptane
Solution
(i) CH₃ – CH – CH₂– CH₂ – CH₃
CH–C1+H₂ Zn,H+ CH₄+HC1 (9.4)
Chloromethane Methane | 5 | 11 | Chemistry | 203 |
c5f4a557-1dd3-4c2c-932e-9a1583a83fd0 | # HYDROCARBONS
# 301
C₂H₅–C1+H₂ Zn,H+ C₂H₆+HC1 alkane containing even number of carbon atoms at the anode.
Chloroethane Ethane (9.5)
− + 2CH COO Na + 2H O
CH₃CH₂CH₂C1 + H₂ Zn,H CH₃CH₂CH₃+CH1 3 2
1-Chloropropane Propane Sodium acetate (9.6)
↓ Electrolysts
ii) Alkyl halides on treatment with sodium metal in dry ethereal (free from moisture) solution give higher alkanes. This reaction is known as Wurtz reaction and is used for the preparation of higher alkanes containing even number of carbon atoms.
CH atoms.
3Br+2Na+BrCH3 dry ether CH₃+2Na
Bromomenthane Ethane (9.7)
C₂H₅Br+2Na+BrC₂H₅ dry ether C₂H₅–C₂H Acetate ion Acetate Methyl free radical
iii) H C + CH H free radical radical
iv) 3 3 3C–CH3↑
What will happen if two different alkyl halides are taken?
3. From carboxylic acids
i) Sodium salts of carboxylic acids on heating with soda lime (mixture of sodium hydroxide and calcium oxide) give alkanes containing one carbon atom less than the carboxylic acid. This process of elimination of carbon dioxide from a carboxylic acid is known as decarboxylation.
CH COO– Na⁺+NaOH CaO CH +Na CO
3 ∆ 4 2 3
Sodium ethanoate
Problem 9.6
Sodium salt of which acid will be needed for the preparation of propane? Write chemical equation for the reaction.
Solution
Butanoic acid,
CH₃CH₂CH₂COO–Na⁺+ NaOH CaO
CH CH CH +Na CO
3 2 3 2 3
ii) Kolbe’s electrolytic method: An aqueous solution of sodium or potassium salt of a carboxylic acid on electrolysis gives
2024-25 | 6 | 11 | Chemistry | 203 |
638ac222-eba3-4463-8064-fa8f05873a4b | # Chemistry
Polar and, hence, hydrophobic in nature. It is generally observed that in relation to solubility of substances in solvents, polar substances are soluble in polar solvents, whereas the non-polar ones in non-polar solvents i.e., like dissolves like.
Boiling point (b.p.) of different alkanes are given in Table 9.2 from which it is clear that there is a steady increase in boiling point with increase in molecular mass. This is due to the fact that the intermolecular van der Waals forces increase with increase of the molecular size or the surface area of the molecule.
You can make an interesting observation by having a look on the boiling points of three isomeric pentanes viz., (pentane, 2-methylbutane and 2,2-dimethylpropane). It is observed (Table 9.2) that pentane having a continuous chain of five carbon atoms has the highest boiling point (309.1K) whereas 2,2 – dimethylpropane boils at 282.5K. With increase in number of branched chains, the molecule attains the shape of a sphere. This results in smaller area of contact and therefore weak intermolecular forces between spherical molecules, which are overcome at relatively lower temperatures.
# Chemical properties
As already mentioned, alkanes are generally inert towards acids, bases, oxidising and reducing agents. However, they undergo the following reactions under certain conditions.
# 1. Substitution reactions
One or more hydrogen atoms of alkanes can be replaced by halogens, nitro group and sulphonic acid group. Halogenation takes place either at higher temperature (573-773 K) or in the presence of diffused sunlight or ultraviolet light. Lower alkanes do not undergo nitration and sulphonation reactions. These reactions in which hydrogen atoms of alkanes are substituted are known as substitution reactions. As an example, chlorination of methane is given below:
# Halogenation
CH4 + Cl2 hv → CH3Cl + HCl
Chloromethane (9.10)
CH3Cl + hv → CH2Cl2 + HCl
Dichloromethane (9.11)
CH2Cl2 + hv → CHCl3 + HCl
Trichloromethane (9.12)
CHCl3 + Cl2 hv → CCl4 + HCl
Tetrachloromethane (9.13)
# Table 9.2 Variation of Melting Point and Boiling Point in Alkanes
|Molecular formula|Name|Molecular mass/u|b.p./(K)|m.p./(K)|
|---|---|---|---|---|
|CH4|Methane|16|111.0|90.5|
|C2H6|Ethane|30|184.4|101.0|
|C3H8|Propane|44|230.9|85.3|
|C4H10|Butane|58|272.4|134.6|
|C4H10|2-Methylpropane|58|261.0|114.7|
|C5H12|Pentane|72|309.1|143.3|
|C5H12|2-Methylbutane|72|300.9|113.1|
|C5H12|2,2-Dimethylpropane|72|282.5|256.4|
|C6H14|Hexane|86|341.9|178.5|
|C7H16|Heptane|100|371.4|182.4|
|C8H18|Octane|114|398.7|216.2|
|C9H20|Nonane|128|423.8|222.0|
|C10H22|Decane|142|447.1|243.3|
|C20H42|Eicosane|282|615.0|236.2| | 7 | 11 | Chemistry | 203 |
d8769485-cdb9-4138-9b8b-b3b1f3497c76 | # HYDROCARBONS
CH₃–CH₃ + C1₂ hv CH₃–CH₂C1 + HC1
Chloroethane (9.14)
It is found that the rate of reaction of alkanes with halogens is F₂ > Cl₂ > Br₂ > I₂. Rate of replacement of hydrogens of alkanes is: 3° > 2° > 1°. Fluorination is too violent to be controlled. Iodination is very slow and a reversible reaction. It can be carried out in the presence of oxidizing agents like HIO₃ or HNO₃.
CH₄ + I₂ CH₃I + HI (9.15)
HIO + 5HI → 3I + 3H₂O (9.16)
Halogenation is supposed to proceed via free radical chain mechanism involving three steps namely initiation, propagation and termination as given below:
# Mechanism
1. Initiation: The reaction is initiated by homolysis of chlorine molecule in the presence of light or heat. The Cl–Cl bond is weaker than the C–C and C–H bond and hence, is easiest to break.
C1–C1 hv . homolysis CH₃ + C1
Chlorine free radicals
2. Propagation: Chlorine free radical attacks the methane molecule and takes the reaction in the forward direction by breaking the C-H bond to generate methyl free radical with the formation of H-Cl.
1. (a) CH₄ + Cl₁ hv + Cl → CH₃ + H–Cl
The methyl radical thus obtained attacks the second molecule of chlorine to form CH₃–Cl with the liberation of another chlorine free radical by homolysis of chlorine molecule.
2. (b) CH₃ + Cl–Cl hv CH₃–Cl + Cl
The chlorine and methyl free radicals generated above repeat steps (a) and (b) respectively and thereby setup a chain of reactions. The propagation steps (a) and (b) are those which directly give principal products, but many other propagation steps are possible and may occur. Two such steps given below explain how more highly haloginated products are formed.
CH₃ + Cl → CH₂Cl + HCl
CH₂Cl + Cl–Cl → CHCl₂ + Cl
# 2. Combustion
Alkanes on heating in the presence of air or dioxygen are completely oxidized to carbon dioxide and water with the evolution of large amount of heat.
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l); ΔH = −890 kJ mol⁻¹ (9.17)
C₄H₁₀(g) + 13/2O₂(g) → 4CO₂(g) + 5H₂O(l); ΔH = −2875.84 kJ mol⁻¹ (9.18)
The general combustion equation for any alkane is:
CₙH₂ₙ₊₂ + ⌈3n + 1⌉ O₂ → nCO₂ + (n + 1)H₂O (9.19)
Due to the evolution of large amount of heat during combustion, alkanes are used as fuels.
During incomplete combustion of alkanes with insufficient amount of air or dioxygen, carbon black is formed which is used in the manufacture of ink, printer ink, black pigments and as filters. | 8 | 11 | Chemistry | 203 |
13d86196-1315-4595-bfca-c39dd871c393 | # CHEMISTRY
CH4 + O2 incomplete C(s) + 2H2O(l) pressure in the presence of oxides of vanadium, molybdenum or chromium supported over alumina get dehydrogenated and cyclised to benzene and its homologues. This reaction is known as aromatization or reforming.
# 3. Controlled oxidation
Alkanes on heating with a regulated supply of dioxygen or air at high pressure and in the presence of suitable catalysts give a variety of oxidation products.
1. 2CH4 + O2 → 2CH3OH <br /> Methanol
2. CH4 + O2 → HCHO + H2O <br /> Toluene (C7H8) is methyl derivative of benzene. Which alkane do you suggest for preparation of toluene?
3. 2CH3CH3 + 3O2 → 2CH3COOH <br /> Ethanoic acid + 2H2O
4. Ordinarily alkanes resist oxidation but alkanes having tertiary H atom can be oxidized to corresponding alcohols by potassium permanganate.
5. CH3(CH3)2CH3 + KMnO4 → 2-Methylpropane-2-ol
# 6. Reaction with steam
Methane reacts with steam at 1273 K in the presence of nickel catalyst to form carbon monoxide and dihydrogen. This method is used for industrial preparation of dihydrogen gas.
CH4 + H2O → CO + 3H2 <br /> Ni
# 7. Pyrolysis
Higher alkanes on heating to higher temperature decompose into lower alkanes, alkenes etc. Such a decomposition reaction into smaller fragments by the application of heat is called pyrolysis or cracking.
Pyrolysis of alkanes is believed to be a free radical reaction. Preparation of oil gas or petrol gas from kerosene oil or petrol involves the principle of pyrolysis. For example, dodecane, a constituent of kerosene oil on heating to 973K in the presence of platinum, palladium or nickel gives a mixture of heptane and pentene.
# 5. Aromatization
n-Alkanes having six or more carbon atoms on heating to 773K at 10-20 atmospheric pressure in the presence of suitable catalysts can undergo aromatization.
|C12H26|Pt/Pd/Ni|C7H16|C5H10|Other Products|
|---|---|---|---|---|
|Dodecane| |Heptane|Pentene| | | 9 | 11 | Chemistry | 203 |
8119eeeb-9d11-407c-a3a4-21ef7b87bfd2 | # 9.2.4 Conformations
Alkanes contain carbon-carbon sigma (σ) bonds. Electron distribution of the sigma molecular orbital is symmetrical around the internuclear axis of the C–C bond which is not disturbed due to rotation about its axis. This permits free rotation about C–C single bond. This rotation results into different spatial arrangements of atoms in space which can change into one another. Such spatial arrangements of atoms which can be converted into one another by rotation around a C-C single bond are called conformations or conformers or rotamers. Alkanes can thus have infinite number of conformations by rotation around C-C single bonds. However, it may be remembered that rotation around a C-C single bond is not completely free. It is hindered by a small energy barrier of 1-20 kJ mol–1 due to weak repulsive interaction between the adjacent bonds. Such a type of repulsive interaction is called torsional strain.
# Conformations of Ethane
Ethane molecule (C2H6) contains a carbon – carbon single bond with each carbon atom attached to three hydrogen atoms. Considering the ball and stick model of ethane, keep one carbon atom stationary and rotate the other carbon atom around the C-C axis. This rotation results into infinite number of spatial arrangements of hydrogen atoms attached to one carbon atom with respect to the hydrogen atoms attached to the other carbon atom. These are called conformational isomers (conformers). Thus there are infinite number of conformations of ethane. However, there are two extreme cases. One such conformation in which hydrogen atoms attached to two carbons are as closed together as possible is called eclipsed conformation and the other in which hydrogens are as far apart as possible is known as the staggered conformation. Any other intermediate conformation is called a skew conformation. It may be remembered that in all the conformations, the bond angles and the bond lengths remain the same. Eclipsed and the staggered conformations can be represented by sawhorse and Newman projections.
# 1. Sawhorse Projections
In this projection, the molecule is viewed along the molecular axis. It is then projected on paper by drawing the central C–C bond as a somewhat longer straight line. Upper end of the line is slightly tilted towards right or left hand side. The front carbon is shown at the lower end of the line, whereas the rear carbon is shown at the upper end. Each carbon has three lines attached to it corresponding to three hydrogen atoms. The lines are inclined at an angle of 120° to each other. Sawhorse projections of eclipsed and staggered conformations of ethane are depicted in Fig. 9.2.
# 2. Newman Projections
In this projection, the molecule is viewed at the C–C bond head on. The carbon atom nearer to the eye is represented by a point. Three hydrogen atoms attached to the front carbon atom are shown by three lines drawn at an angle of 120° to each other. The rear carbon atom (the carbon atom away from the eye) is represented by a circle and the three hydrogen atoms are shown attached to it by the shorter lines drawn at an angle of 120° to each other. The Newman’s projections are depicted in Fig. 9.3.
Fig. 9.2 Sawhorse projections of ethane
Fig. 9.3 Newman’s projections of ethane | 10 | 11 | Chemistry | 203 |
75ed8dab-b8b7-433b-8e35-d20a4077113f | # CHEMISTRY
Relative stability of conformations: As mentioned earlier, in staggered form of ethane, the electron clouds of carbon-hydrogen bonds are as far apart as possible. Thus, there are minimum repulsive forces, minimum energy and maximum stability of the molecule. On the other hand, when the staggered form changes into the eclipsed form, the electron clouds of the carbon – hydrogen bonds come closer to each other resulting in increase in electron cloud repulsions. To check the increased repulsive forces, molecule will have to possess more energy and thus has lesser stability. As already mentioned, the repulsive interaction between the electron clouds, which affects stability of a conformation, is called torsional strain. Magnitude of torsional strain depends upon the angle of rotation about C–C bond. This angle is also called dihedral angle or torsional angle. Of all the conformations of ethane, the staggered form has the least torsional strain and the eclipsed form, the maximum torsional strain. Therefore, staggered conformation is more stable than the eclipsed conformation. Hence, molecule largely remains in staggered conformation or we can say that it is preferred conformation. Thus it may be inferred that rotation around C–C bond in ethane is not completely free. The energy difference between the two extreme forms is of the order of 12.5 kJ mol–1, which is very small. Even at ordinary temperatures, the ethane molecule gains thermal or kinetic energy sufficient enough to overcome this energy barrier of 12.5 kJ mol–1 through intermolecular collisions. Thus, it can be said that rotation about carbon-carbon single bond in ethane is almost free for all practical purposes. It has not been possible to separate and isolate different conformational isomers of ethane.
# 9.3 Alkenes
Alkenes are unsaturated hydrocarbons containing at least one double bond. What should be the general formula of alkenes? If there is one double bond between two carbon atoms in alkenes, they must possess two hydrogen atoms less than alkanes. Hence, general formula for alkenes is CₙH₂ₙ. Alkenes are also known as olefins (oil forming) since the longest chain of carbon atoms containing the double bond is selected. Numbering of the chain is done from the end which is nearer to
# 9.3.1 Structure of Double Bond
Carbon-carbon double bond in alkenes consists of one strong sigma (σ) bond (bond enthalpy about 397 kJ mol–1) due to head-on overlapping of sp² hybridised orbitals and one weak pi (π) bond (bond enthalpy about 284 kJ mol–1) obtained by lateral or sideways overlapping of the two 2p orbitals of the two carbon atoms. The double bond is shorter in bond length (134 pm) than the C–C single bond (154 pm). You have already read that the pi (π) bond is a weaker bond due to poor sideways overlapping between the two 2p orbitals. Thus, the presence of the pi (π) bond makes alkenes behave as sources of loosely held mobile electrons. Therefore, alkenes are easily attacked by reagents or compounds which are in search of electrons. Such reagents are called electrophilic reagents. The presence of weaker π-bond makes alkenes unstable molecules in comparison to alkanes and thus, alkenes can be changed into single bond compounds by combining with the electrophilic reagents. Strength of the double bond (bond enthalpy, 681 kJ mol–1) is greater than that of a carbon-carbon single bond in ethane (bond enthalpy, 348 kJ mol–1). Orbital diagrams of ethene molecule are shown in Figs. 9.4 and 9.5.
Fig. 9.4 Orbital picture of ethene depicting σ bonds only
# 9.3.2 Nomenclature
For nomenclature of alkenes in IUPAC system, the longest chain of carbon atoms containing the double bond is selected. Numbering of the chain is done from the end which is nearer to | 11 | 11 | Chemistry | 203 |
a3a265bb-12bf-4538-b004-aee5c6fcc47c | # HYDROCARBONS
Fig. 9.5 Orbital picture of ethene showing formation of (a) π-bond, (b) π-cloud and (c) bond angles and bond lengths the double bond. The suffix ‘ene’ replaces ‘ane’ solution of alkanes. It may be remembered that first member of alkene series is: CH₂ (replacing n by 1 in CₙH₂ₙ) known as methene but has a very short life. As already mentioned, first stable member of alkene series is C₂H₄ known as ethylene (common) or ethene (IUPAC).
# IUPAC names of a few members of alkenes are given below:
|structure|iUPAC name|
|---|---|
|CH₃ – CH = CH₂|Propene|
|CH₃ – CH₂ – CH = CH₂|But – 1 - ene|
|CH₃ – CH = CH–CH₃|But-2-ene|
|CH₂ = CH – CH = CH₂|Buta – 1,3 - diene|
|CH₂ = C – CH₃|2-Methylprop-1-ene|
|CH₂ = CH – CH – CH₃|3-Methylbut-1-ene|
# Problem 9.8
Calculate number of sigma (σ) and pi (π) bonds in the above structures (i-iv).
solution
σ bonds : 33, π bonds : 2
σ bonds : 17, π bonds : 4
σ bonds : 23, π bond : 1
σ bonds : 41, π bond : 1
# 9.3.3 Isomerism
Alkenes show both structural isomerism and geometrical isomerism.
Structural isomerism: As in alkanes, ethene (C₂H₄) and propene (C₃H₆) can have only one structure but alkenes higher than propene have different structures. Alkenes possessing C₄H₈ as molecular formula can be written in the following three ways:
# I.
CH₂ = CH – CH₂ – CH₃
But-1-ene (C₄H₈)
# II.
CH₃ – CH = CH – CH₃
But-2-ene (C₄H₈) | 12 | 11 | Chemistry | 203 |
6629d06c-2461-40be-a159-f0c35db8307c | # CHEMISTRY
In (a), the two identical atoms i.e., both the X or both the Y lie on the same side of the double bond but in (b) the two X or two Y lie across the double bond or on the opposite sides of the double bond. This results in different geometry of (a) and (b) i.e. disposition of atoms or groups in space in the two arrangements is different. Therefore, they are stereoisomers. They would have the same geometry if atoms or groups around C=C bond can be rotated but rotation around C=C bond is not free. It is restricted. For understanding this concept, take two pieces of strong cardboards and join them with the help of two nails. Hold one cardboard in your one hand and try to rotate the other. Can you really rotate the other cardboard? The answer is no. The rotation is restricted. This illustrates that the restricted rotation of atoms or groups around the doubly bonded carbon atoms gives rise to different geometries of such compounds. The stereoisomers of this type are called geometrical isomers. The isomer of the type (a), in which two identical atoms or groups lie on the same side of the double bond is called cis isomer and the other isomer of the type (b), in which identical atoms or groups lie on the opposite sides of the double bond is called trans isomer. Thus cis and trans isomers have the same structure but have different configuration (arrangement of atoms or groups in space). Due to different arrangement of atoms or groups in space, these isomers differ in their properties like melting point, boiling point, dipole moment, solubility etc.
# Problem 9.9
Write structures and IUPAC names of different structural isomers of alkenes corresponding to C5H10.
# Solution
|(a)|CH2 = CH – CH2 – CH2 – CH3|Pent-1-ene|
|---|---|---|
|(b)|CH3 – CH = CH – CH2 – CH3|Pent-2-ene|
|(c)|CH3 – C = CH – CH3|2-Methylbut-2-ene|
|(d)|CH3 – CH – CH = CH2|3-Methylbut-1-ene|
|(e)|CH2 = C – CH2 – CH3|2-Methylbut-1-ene|
# Geometrical isomerism:
Doubly bonded carbon atoms have to satisfy the remaining two valences by joining with two atoms or groups. If the two atoms or groups attached to each carbon atom are different, they can be represented by YX C = C XY like structure. YX C = C XY can be represented in space in the following two ways:
Cis form of alkene is found to be more polar than the trans form. For example, dipole moment of cis-but-2-ene is 0.33 Debye, whereas, dipole moment of the trans form is almost zero or it can be said that. | 13 | 11 | Chemistry | 203 |
50a6a7df-1fd7-47de-9f44-3b8b85b8e1d5 | # HYDROCARBONS
trans-but-2-ene is non-polar. This can be understood by drawing geometries of the two forms as given below from which it is clear that in the trans-but-2-ene, the two methyl groups are in opposite directions, Therefore, dipole moments of C-CH₃ bonds cancel, thus making the trans form non-polar.
(ii) CH₂ = CBr₂
(iii) C6H5CH = CH – CH3
(iv) CH₃CH = CClCH₃
solution (iii) and (iv). In structures (i) and (ii), two identical groups are attached to one of the doubly bonded carbon atom.
# 9.3.4 Preparation
1. From alkynes: Alkynes on partial reduction with calculated amount of dihydrogen in the presence of palladised charcoal partially deactivated with poisons like sulphur compounds or quinoline give alkenes. Partially deactivated palladised charcoal is known as Lindlar’s catalyst. Alkenes thus obtained are having cis geometry. However, alkynes on reduction with sodium in liquid ammonia form trans alkenes.
# Problem 9.10
Draw cis and trans isomers of the following compounds. Also write their IUPAC names:
(i) CHCl = CHCl (9.30)
(ii) C2H5CCH3 = CCH3C2H5 (9.31)
(iii) CH≡CH + H₂ Pd/C CH₂=CH₂ (9.32)
Ethyne Ethene
(iv) CH₃–C≡CH + H₂ Pd/C CH₃–CH=CH₂ (9.33)
Propyne Propene
Will propene thus obtained show geometrical isomerism? Think for the reason in support of your answer.
# Problem 9.11
Which of the following compounds will show cis-trans isomerism?
(i) (CH₃)₂C = CH – C₂H₅
2. From alkyl halides: Alkyl halides (R-X) on heating with alcoholic potash (potassium hydroxide dissolved in alcohol, say, | 14 | 11 | Chemistry | 203 |
93989963-3d4e-4202-9d8c-62b9862f29a9 | # Chemistry
Ethanol) eliminate one molecule of halogen takes out one hydrogen atom from the acid to form alkenes. This reaction is known as dehydrohalogenation i.e., removal of halogen acid. This is example of β-elimination reaction, since hydrogen atom is eliminated from the β carbon atom (carbon atom next to the carbon to which halogen is attached). (9.37)
# 9.3.5 Properties
# Physical properties
Alkenes as a class resemble alkanes in physical properties, except in types of isomerism and difference in polar nature. The first three members are gases, the next fourteen are liquids and the higher ones are solids. Ethene is a colourless gas with a faint sweet smell. All other alkenes are colourless and odourless, insoluble in water but fairly soluble in non-polar solvents like benzene, petroleum ether. They show a regular increase in boiling point with increase in size i.e., every – CH₂ group added increases boiling point by 20–30 K. Like alkanes, straight chain alkenes have higher boiling point than isomeric branched chain compounds.
# From vicinal dihalides:
Dihalides in which two halogen atoms are attached to two adjacent carbon atoms are known as vicinal dihalides. Vicinal dihalides on treatment with zinc metal lose a molecule of ZnX₂ to form an alkene. This reaction is known as dehalogenation.
|CH2Br–CH2Br + Zn|CH2=CH2 + ZnBr2|
|---|---|
|CH3CHBr–CH2Br + Zn|CH3CH=CH2 + ZnBr2|
# Chemical properties
Alkenes are the rich source of loosely held pi (π) electrons, due to which they show addition reactions in which the electrophiles add on to the carbon-carbon double bond to form the addition products. Some reagents also add by free radical mechanism. There are cases when under special conditions, alkenes also undergo free radical substitution reactions. Oxidation and ozonolysis reactions are also quite prominent in alkenes. A brief description of different reactions of alkenes is given below:
1. Addition of dihydrogen: Alkenes add up one molecule of dihydrogen gas in the presence of finely divided nickel, palladium or platinum to form alkanes (Section 9.2.2).
2. Addition of halogens: Halogens like bromine or chlorine add up to alkene to form vicinal dihalides. However, iodine does not show addition reaction under. | 15 | 11 | Chemistry | 203 |
96959e37-5548-46e0-abd1-12fe0b414adb | # HYDROCARBONS
The reddish orange colour of bromine solution in carbon tetrachloride is discharged when bromine adds up to an unsaturation site. This reaction is used as a test for unsaturation. Addition of halogens to alkenes is an example of electrophilic addition reaction (9.42) involving cyclic halonium ion formation which you will study in higher classes.
Markovnikov, a Russian chemist made a generalisation in 1869 after studying such reactions in detail. These generalisations led Markovnikov to frame a rule called Markovnikov rule. The rule states that negative part of the addendum (adding molecule) gets attached to that carbon atom which possesses lesser number of hydrogen atoms.
|(i1) CH: CH-CH2 + Cl2 ~ CH: CH-CH2| | |
|---|---|---|
|Fropene|Dichloropropane|1,2-|
Thus according to this rule, product i.e., 2-bromopropane is expected. In actual practice, this is the principal product of the reaction. This generalisation of Markovnikov rule can be better understood in terms of mechanism of the reaction.
# Mechanism
Hydrogen bromide provides an electrophile, H+, which attacks the double bond to form carbocation as shown below:
|(a) less stable|(b) more stable|
|---|---|
|primary carbocation|secondary carbocation|
(i) The secondary carbocation (b) is more stable than the primary carbocation (a), therefore, the former predominates because it is formed at a faster rate.
(ii) The carbocation (b) is attacked by Br- ion to form the product as follows:
|CH = CH + H–Br|CH3–CH–Br|
|---|---|
|(9.40)| |
CH3–CH=CH–CH3 + HBr CH3–CH–CHCH3
(ii) The addition reaction of HBr to unsymmetrical alkenes (Markovnikov Rule) How will H–Br add to propene? The two possible products are I and II.
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955a1711-89cd-40c3-a2b4-fcb3e744117f | # CHEMISTRY
# Anti Markovnikov addition or peroxide effect or Kharash effect
In the presence of peroxide, addition of HBr to unsymmetrical alkenes like propene takes place contrary to the Markovnikov rule. This happens only with HBr but not with HCl and HI. This addition reaction was observed by M.S. Kharash and F.R. Mayo in 1933 at the University of Chicago. This reaction is known as peroxide or Kharash effect or addition reaction anti to Markovnikov rule.
CH3–CH=CH2 + HBr (C6H5CO)2O2 → CH3–CH2–Br
1–Bromopropane (9.43)
The secondary free radical obtained in the above mechanism (step iii) is more stable than the primary. This explains the formation of 1-bromopropane as the major product. It may be noted that the peroxide effect is not observed in addition of HCl and HI. This may be due to the fact that the H–Cl bond being stronger (430.5 kJ mol–1) than H–Br bond (363.7 kJ mol–1), is not cleaved by the free radical, whereas the H–I bond is weaker (296.8 kJ mol–1) and iodine free radicals combine to form iodine molecules instead of adding to the double bond.
# Mechanism:
Peroxide effect proceeds via free radical chain mechanism as given below:
(i) Homolysis
(ii) C6H5 + H–Br → C6H3 + Br
# Problem 9.12
Write IUPAC names of the products obtained by addition reactions of HBr to hex-1-ene
(i) in the absence of peroxide and
(ii) in the presence of peroxide.
Solution
4. Addition of sulphuric acid: Cold concentrated sulphuric acid adds to alkenes in accordance with Markovnikov rule to form alkyl hydrogen sulphate by the electrophilic addition reaction. | 17 | 11 | Chemistry | 203 |
bda78d18-4a68-4ed5-9c5a-dd931fe89546 | # HYDROCARBONS
ketones and/or acids depending upon the nature of the alkene and the experimental conditions
(9.49)
|CH₃ – CH=CH–CH₃|KMnO4/H+|2CH₃COOH|
|---|---|---|
|(9.44)|But-2-ene|Ethanoic acid|
(9.50)
# 7. Ozonolysis
Ozonolysis of alkenes involves the addition of ozone molecule to alkene to form ozonide, and then cleavage of the ozonide by Zn-H₂O to smaller molecules. This reaction is highly useful in detecting the position of the double bond in alkenes or other unsaturated compounds.
(9.45)
# 5. Addition of water
In the presence of a few drops of concentrated sulphuric acid, alkenes react with water to form alcohols, in accordance with the Markovnikov rule.
(9.46)
# 6. Oxidation
Alkenes on reaction with cold, dilute, aqueous solution of potassium permanganate (Baeyer’s reagent) produce vicinal glycols. Decolorisation of KMnO₄ solution is used as a test for unsaturation.
(9.52)
(9.47)
# 8. Polymerisation
You are familiar with polythene bags and polythene sheets. Polythene is obtained by the combination of large number of ethene molecules at high temperature, high pressure and in the presence of a catalyst. The large molecules thus obtained are called polymers. This reaction is known as polymerisation.
(9.48)
b) Acidic potassium permanganate or acidic potassium dichromate oxidises alkenes to
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9ad86a17-2b67-47d9-ac1a-2a30be03ef3f | # Chemistry
monomers. Other alkenes also undergo polymerisation.
n(CH2=CH2) High temp./pressure — (CH2–CH2) Catalyst Polythene (9.53)
The position of the triple bond is indicated by the first triply bonded carbon. Common and IUPAC names of a few members of alkyne series are given in Table 9.2.
n(CH3–CH=CH2) High temp./pressure — (CH2–CH3) Catalyst n CH3 Polypropene (9.54)
Polymers are used for the manufacture of plastic bags, squeeze bottles, refrigerator dishes, toys, pipes, radio and T.V. cabinets etc. Polypropene is used for the manufacture of milk crates, plastic buckets and other moulded articles. Though these materials have now become common, excessive use of polythene and polypropylene is a matter of great concern for all of us.
# 9.4 Alkynes
Like alkenes, alkynes are also unsaturated hydrocarbons. They contain at least one triple bond between two carbon atoms. The number of hydrogen atoms is still less in alkynes as compared to alkenes or alkanes. Their general formula is CnH2n–2.
The first stable member of alkyne series is ethyne which is popularly known as acetylene. Acetylene is used for arc welding purposes in the form of oxyacetylene flame obtained by mixing acetylene with oxygen gas. Alkynes are starting materials for a large number of organic compounds. Hence, it is interesting to study this class of organic compounds.
# 9.4.1 Nomenclature and Isomerism
In common system, alkynes are named as derivatives of acetylene. In IUPAC system, they are named as derivatives of the corresponding alkanes replacing ‘ane’ by the suffix ‘yne’.
|Value of n|Formula|Structure|Common name|IUPAC name|
|---|---|---|---|---|
|2|C2H2|H-C≡CH|Acetylene|Ethyne|
|3|C3H4|CH3-C≡CH|Methylacetylene|Propyne|
|4|C4H6|CH3CH2-C≡CH|Ethylacetylene|But-1-yne|
|4|C4H6|CH3-C≡C-CH3|Dimethylacetylene|But-2-yne|
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42f9a07d-e34d-4508-8909-0b56c144653f | # HYDROCARBONS
# 9.4.2 Structure of Triple Bond
Ethyne is the simplest molecule of alkyne series. Structure of ethyne is shown in Fig. 9.6.
Each carbon atom of ethyne has two sp hybridised orbitals. Carbon-carbon sigma (σ) bond is obtained by the head-on overlapping of the two sp hybridised orbitals of the two carbon atoms. The remaining sp hybridised orbital of each carbon atom undergoes overlapping along the internuclear axis with the 1s orbital of each of the two hydrogen atoms forming two C-H sigma bonds. H-C-C bond angle is of 180°. Each carbon has two unhybridised p orbitals which are perpendicular to each other as well as to the plane of the C-C sigma bond. The 2p orbitals of one carbon atom are parallel to the 2p orbitals of the other carbon atom, which undergo lateral or sideways overlapping to form two pi (π) bonds between two carbon atoms. Thus ethyne molecule consists of one C–C σ bond, two C–H σ bonds and two C–C π bonds. The strength of C≡C bond (bond enthalpy 823 kJ mol⁻¹) is more than those of C=C bond (bond enthalpy 681 kJ mol–1) and C–C bond (bond enthalpy 348 kJ mol–1). The C≡C bond length is shorter (120 pm) than those of C=C (133 pm) and C–C (154 pm). Electron cloud between two carbon atoms is cylindrically symmetrical about the internuclear axis. Thus, ethyne is a linear molecule.
# 9.4.3 Preparation
1. From calcium carbide: On industrial scale, ethyne is prepared by treating calcium carbide with water. Calcium carbide is prepared by heating quick lime with coke. Quick lime can be obtained by heating limestone as shown in the following reactions:
CaCO3
Δ
CaO
+
CO2 | 20 | 11 | Chemistry | 203 |
2ed5e504-47fd-4d21-a197-da631effdd6b | CaO + 3C CaC₂ + CO (9.56)
Calcium
carbide
CaC₂ + 2H₂O Ca(OH)₂ + C₂H₂ (9.57)
# 2. From vicinal dihalides :
Vicinal dihalides on treatment with alcoholic potassium hydroxide undergo dehydrohalogenation. One molecule of hydrogen halide is eliminated to form alkenyl halide which on treatment with sodamide gives alkyne.
# 9.4.4 Properties
# Physical properties
Physical properties of alkynes follow the same trend of alkenes and alkanes. First three members are gases, the next eight are liquids and the higher ones are solids. All alkynes are colourless. Ethylene has characteristic odour. Other members are odourless. Alkynes are weakly polar in nature. They are lighter than water and immiscible with water but soluble in organic solvents like ethers, carbon tetrachloride and benzene. Their melting point, boiling point and density increase with increase in molar mass.
# Chemical properties
Alkynes show acidic nature, addition reactions and polymerisation reactions as follows :
# A. Acidic character of alkyne:
Sodium metal and sodamide (NaNH₂) are strong bases. They react with ethyne to form sodium acetylide with the liberation of dihydrogen gas. These reactions have not been observed in case of ethene and ethane thus indicating that ethyne is acidic in nature in comparison to ethene and ethane. Why is it so? Has it something to do with their structures and the hybridisation? You have read that hydrogen atoms in ethyne are attached to the sp hybridised carbon atoms whereas they are attached to sp² hybridised carbon atoms in ethene and sp³ hybridised carbons in ethane. Due to the maximum percentage of s character (50%), the sp hybridised orbitals of carbon atoms in ethyne molecules have highest electronegativity; hence, these attract the shared electron pair of the C-H bond of ethyne to a greater extent than that of the sp² hybridised orbitals of carbon in ethene and the sp³ hybridised orbital of carbon in ethane. Thus in ethyne, hydrogen atoms can be liberated as protons more easily as compared to ethene and ethane. Hence, hydrogen atoms of ethyne attached to triply bonded carbon atom are acidic in nature. You may note that the hydrogen atoms attached to the triply bonded carbons are acidic but not all the hydrogen atoms of alkynes.
HC ≡ CH + Na → HC ≡ C–Na⁺ + 1/₂H₂
Monosodium ethynide (9.59)
HC ≡ C–Na + Na → Na⁺ C–Na⁺ ≡ C–Na⁺ + 1/₂H₂
Disodium ethynide (9.60)
CH₃ – C ≡ C – H + Na⁺ NH₂
↓
CH₃ – C ≡ C– Na⁺ + NH₃
Sodium propynide (9.61)
These reactions are not shown by alkenes and alkanes, hence used for distinction between alkynes, alkenes and alkanes. What about the above reactions with but-1-yne and but-2-yne? Alkanes, alkenes and alkynes follow the following trend in their acidic behaviour :
i) CH ≡ CH > H C –CH > CH –CH
2 2 3 3
ii) HC ≡ CH > CH₃ –C≡ CH >> CH₃ –C≡C–CH₃
# b. Addition reactions:
Alkynes contain a triple bond, so they add up, two molecules of dihydrogen, halogen, hydrogen halides etc. Formation of the addition product takes place according to the following steps. | 21 | 11 | Chemistry | 203 |
aaa5de6c-5819-463b-952b-ab7d8d6eea28 | # HYDROCARBONS
The addition product formed depends upon stability of vinylic cation. Addition in unsymmetrical alkynes takes place according to Markovnikov rule. Majority of the reactions of alkynes are the examples of electrophilic addition reactions. A few addition reactions are given below:
1. Addition of dihydrogen
|HC≡CH + H₂|Pt/Pd/Ni|[H₂C= CH₂]|
|---|---|---|
|H2 CH₃–CH₃|Like alkanes and alkenes, alkynes are also immiscible and do not react with water.|Like alkanes and alkenes, alkynes are also immiscible and do not react with water.|
(9.62)
|CH3–C≡CH + H2|Pt/Pd/Ni|[CH3–CH=CH2]|
|---|---|---|
|Propyne|Propene|CH3–CH2–CH3|
|Propane|Propane|Propane|
(9.63)
2. Addition of halogens
(9.64)
Reddish orange colour of the solution of bromine in carbon tetrachloride is decolourised. This is used as a test for unsaturation.
3. Addition of hydrogen halides
(9.68)
Two molecules of hydrogen halides (HCl, HBr, HI) add to alkynes to form gem dihalides (in which two halogens are attached to the same carbon atom)
|H–C≡C–H + H–Br|[CH2 = CH–Br]|→ CHBr2|
|---|---|---|
|Bromoethene|Bromoethene|CH3|
|1,1-Dibromoethane|1,1-Dibromoethane|(9.65)|
4. Polymerisation
(v)
(a) Linear polymerisation: Under suitable conditions, linear polymerisation of ethyne takes place to produce polyacetylene or polyethyne which is a high molecular weight polyene containing repeating units of (CH = CH – CH = CH) and can be represented as —
(CH = CH – CH = CH)n — Under special conditions, this polymer conducts electricity.
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407a729d-66ae-4d7a-bcdb-fe92779c47d5 | # CHEMISTRY
Thin film of polyacetylene can be used as electrodes in batteries. These films are good conductors, lighter and cheaper than the metal conductors.
# (b) Cyclic polymerisation:
Ethyne on passing through red hot iron tube at 873K undergoes cyclic polymerization. Three molecules polymerise to form benzene, which is the starting molecule for the preparation of derivatives of benzene, dyes, drugs and large number of other organic compounds. This is the best route for entering from aliphatic to aromatic compounds as discussed below:
|Benzene|Toluene|Naphthalene|
|---|---|---|
|(9.69)|(9.69)|(9.69)|
|Biphenyl| | |
# Problem 9.14
How will you convert ethanoic acid into benzene?
# Solution
The nomenclature and isomerism of aromatic hydrocarbons has already been discussed in Unit 8. All six hydrogen atoms in benzene are equivalent; so it forms one and only one type of monosubstituted product. When two hydrogen atoms in benzene are replaced by two similar or different monovalent atoms or groups, three different position isomers are possible. The 1, 2 or 1, 6 is known as the ortho (o–), the 1, 3 or 1, 5 as meta (m–) and the 1, 4 as para (p–) disubstituted compounds. A few examples of derivatives of benzene are given below:
# 9.5 Aromatic Hydrocarbon
These hydrocarbons are also known as ‘arenes’. Since most of them possess pleasant odour (Greek; aroma meaning pleasant smelling), the class of compounds was named as ‘aromatic compounds’. Most of such compounds were found to contain benzene ring. Benzene ring is highly unsaturated.
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dd4865af-e6d7-45a8-bc00-4a8dec5ed549 | # HYDROCARBONS
Friedrich August Kekulé, a German chemist was born in 1829 at Darmsdt in Germany. He became Professor in 1856 and Fellow of Royal Society in 1875. He made major contribution to structural organic chemistry by proposing in 1858 that carbon atoms can join to one another to form chains and later in 1865, he found an answer to the challenging problem of benzene structure by suggesting that these chains can close to form rings. He gave the dynamic structural formula to benzene which forms the basis for its modern electronic structure. He described the discovery of benzene structure later as:
“I was sitting writing at my textbook, but the work did not progress; my thoughts were elsewhere. I turned my chair to the fire, and dozed. Again the atoms were gambolling before my eyes. This time the smaller groups kept modestly in the background. My mental eye, rendered more acute by repeated visions of this kind, could now distinguish larger structures of manifold conformations; long rows, sometimes more closely fitted together; all twisting and turning in snake like motion. But look! What was that? One of the snakes had seized hold of its own tail, and the form whirled mockingly before my eyes. As if by a flash of lightning I woke;.... I spent the rest of the night working out the consequences of the hypothesis. Let us learn to dream, gentlemen, and then perhaps we shall learn the truth but let us beware of making our dreams public before they have been approved by the waking mind.” (1890).
One hundred years later, on the occasion of Kekulé’s centenary celebrations a group of compounds having polybenzenoid structures have been named as Kekulenes.
was further found to produce one and only one monosubstituted derivative which indicated that all the six carbon and six hydrogen atoms of benzene are identical. On the basis of this observation August Kekulé in 1865 proposed the following structure for benzene having cyclic arrangement of six carbon atoms with alternate single and double bonds and one hydrogen atom attached to each carbon atom.
|1,3 Dimethylbenzene|1,4-Dimethylbenzene|
|---|---|
|(m-Xylene)|(p-Xylene)|
# 9.5.2 Structure of Benzene
Benzene was isolated by Michael Faraday in 1825. The molecular formula of benzene, C₆H₆, indicates a high degree of unsaturation. This molecular formula did not account for its relationship to corresponding alkanes, alkenes and alkynes which you have studied in earlier sections of this unit. What do you think about its possible structure? Due to its unique properties and unusual stability, it took several years to assign its structure. Benzene was found to be a stable molecule and found to form a triozonide which indicates the presence of three double bonds. Benzene indicates the possibility of two isomeric 1,2-dibromobenzenes. In one of the isomers, the bromine atoms are attached to the doubly bonded carbon atoms whereas in the other, they are attached to the singly bonded carbons.
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aac5820e-c778-4850-8008-c6b17adbcfe6 | # CHEMISTRY
However, benzene was found to form only one ortho disubstituted product. This problem was overcome by Kekulé by suggesting the concept of oscillating nature of double bonds in benzene as given below.
The unhybridised p orbital of carbon atoms are close enough to form a π bond by lateral overlap. There are two equal possibilities of forming three π bonds by overlap of p orbitals of C₁ –C₂, C₃ – C₄, C₅ – C₆ or C₂ – C₃, C₄ – C₅, C₆ – C₁ respectively as shown in the following figures.
# Resonance and stability of benzene
According to Valence Bond Theory, the concept of oscillating double bonds in benzene is now explained by resonance. Benzene is a hybrid of various resonating structures. The two structures, A and B given by Kekulé are the main contributing structures. The hybrid structure is represented by inserting a circle or a dotted circle in the hexagon as shown in (C). The circle represents the six electrons which are delocalised between the six carbon atoms of the benzene ring.
(A)
(B)
(C)
The orbital overlapping gives us better picture about the structure of benzene. All the six carbon atoms in benzene are sp² hybridized. Two sp² hybrid orbitals of each carbon atom overlap with sp² hybrid orbitals of adjacent carbon atoms to form six C—C sigma bonds which are in the hexagonal plane. The remaining sp² hybrid orbital of each carbon atom overlaps with s orbital of a hydrogen atom to form six C—H sigma bonds. Each carbon atom is now left with one localized π bond.
Structures shown in Fig. 9.7(a) and (b) correspond to two Kekulé’s structure with localised π bonds. The internuclear distance. | 25 | 11 | Chemistry | 203 |
ce428e2c-352d-48ce-99c2-420f4786c734 | # HYDROCARBONS
between all the carbon atoms in the ring has been determined by the X-ray diffraction to be the same; there is equal probability for the p orbital of each carbon atom to overlap with the p orbitals of adjacent carbon atoms [Fig. 9.7 (c)]. This can be represented in the form of two doughtnuts (rings) of electron clouds [Fig. 9.7 (d)], one above and one below the plane of the hexagonal ring as shown below:
(electron cloud)
Fig. 9.7 (c) Fig. 9.7 (d)
The six π electrons are thus delocalised and can move freely about the six carbon nuclei, instead of any two as shown in Fig. 9.6 (a) or (b). The delocalised π electron cloud is attracted more strongly by the nuclei of the carbon atoms than the electron cloud localised between two carbon atoms. Therefore, presence of delocalised π electrons in benzene makes it more stable than the hypothetical cyclohexatriene.
X-Ray diffraction data reveals that benzene is a planar molecule. Had any one of the above structures of benzene (A or B) been correct, two types of C—C bond lengths were expected. However, X-ray data indicates that all the six C—C bond lengths are of the same order (139 pm) which is intermediate between C—C single bond (154 pm) and C—C double bond (133 pm). Thus the absence of pure double bond in benzene accounts for the reluctance of benzene to show addition reactions under normal conditions, thus explaining the unusual behaviour of benzene.
# 9.5.4 Preparation of Benzene
Benzene is commercially isolated from coal tar. However, it may be prepared in the laboratory by the following methods:
- (i) Cyclic polymerisation of ethyne: (Section 9.4.4)
- (ii) Decarboxylation of aromatic acids: Sodium salt of benzoic acid on heating with sodalime gives benzene.
# 9.5.3 Aromaticity
Benzene was considered as parent ‘aromatic’ compound. Now, the name is applied to all the ring systems whether or not having benzene ring, possessing following characteristics. | 26 | 11 | Chemistry | 203 |
a38b5c1f-225f-4928-b001-30eea1f5fef0 | # CHEMISTRY
# 9.5.5 Properties
# Physical properties
Aromatic hydrocarbons are non-polar molecules and are usually colourless liquids or solids with a characteristic aroma. You are also familiar with naphthalene balls which are used in toilets and for preservation of clothes because of unique smell of the compound and the moth repellent property. Aromatic hydrocarbons are immiscible with water but are readily miscible with organic solvents. They burn with sooty flame.
# Chemical properties
Arenes are characterised by electrophilic substitution reactions. However, under special conditions they can also undergo addition and oxidation reactions.
# Electrophilic substitution reactions
The common electrophilic substitution reactions of arenes are nitration, halogenation, sulphonation, Friedel Craft’s alkylation and acylation reactions in which attacking reagent is an electrophile (E⁺).
# (i) Nitration
A nitro group is introduced into benzene ring when benzene is heated with a mixture of concentrated nitric acid and concentrated sulphuric acid (nitrating mixture).
# (ii) Halogenation
Arenes react with halogens in the presence of a Lewis acid like anhydrous FeCl₃, FeBr₃ or AlCl₃ to yield haloarenes.
Chlorobenzene
# (iii) Sulphonation
The replacement of a hydrogen atom by a sulphonic acid group in a ring is called sulphonation. It is carried out by heating benzene with fuming sulphuric acid (oleum).
# (iv) Friedel-Crafts alkylation reaction
When benzene is treated with an alkyl halide in the presence of anhydrous aluminium chloride, alkylbenene is formed.
# (v) Friedel-Crafts acylation reaction
The reaction of benzene with an acyl halide or acid anhydride in the presence of Lewis acids (AlCl₃) yields acyl benzene.
# Why do we get isopropyl benzene on treating benzene with 1-chloropropane instead of n-propyl benzene?
Nitrobenzene | 27 | 11 | Chemistry | 203 |
8e679f61-4d07-40c6-a4d5-ce1da2f54522 | # HYDROCARBONS
(9.77) In the case of nitration, the electrophile, nitronium ion, is produced by transfer of a proton (from sulphuric acid) to nitric acid in the following manner:
# (9.78) Step I
If excess of electrophilic reagent is used, further substitution reaction may take place in which other hydrogen atoms of benzene ring may also be successively replaced by the electrophile. For example, benzene on treatment with excess of chlorine in the presence of anhydrous AlCl₃ can be chlorinated to hexachlorobenzene (C₆Cl₆).
Protonated nitric acid
Nitronium ion
It is interesting to note that in the process of generation of nitronium ion, sulphuric acid serves as an acid and nitric acid as a base. Thus, it is a simple acid-base equilibrium.
# (b) Formation of Carbocation (arenium ion):
Attack of electrophile results in the formation of σ-complex or arenium ion in which one of the carbon is sp³ hybridised.
# Mechanism of electrophilic substitution reactions:
According to experimental evidences, SE (S = substitution; E = electrophilic) reactions are supposed to proceed via the following three steps:
1. Generation of the electrophile
2. Formation of carbocation intermediate
3. Removal of proton from the carbocation intermediate
(a) Generation of electrophile E⊕: During chlorination, alkylation and acylation of benzene, anhydrous AlCl₃, being a Lewis acid helps in generation of the electrophile Cl⊕, R⊕, RC⊕O (acylium ion) respectively by combining with the attacking reagent.
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83ee499f-1134-4d2a-a530-69a46b747a42 | # CHEMISTRY
Sigma complex or arenium ion loses its aromatic character because delocalisation of electrons stops at sp³ hybridised carbon.
(c) Removal of proton: To restore the aromatic character, σ-complex releases proton from sp³ hybridised carbon on attack by [AlCl₄]– (in case of halogenation, alkylation and acylation) and [HSO₄]– (in case of nitration).
# 9.5.6 directive influence of a functional group in monosubstituted benzene
When monosubstituted benzene is subjected to further substitution, three possible disubstituted products are not formed in equal amounts. Two types of behaviour are observed. Either ortho and para products or meta product is predominantly formed. It has also been observed that this behaviour depends on the nature of the substituent already present in the benzene ring and not on the nature of the entering group. This is known as directive influence of substituents. Reasons for ortho/para or meta directive nature of groups are discussed below:
# Addition reactions
Under vigorous conditions, i.e., at high temperature and/or pressure in the presence of nickel catalyst, hydrogenation of benzene gives cyclohexane.
Cyclohexane (9.80)
Under ultra-violet light, three chlorine molecules add to benzene to produce benzene hexachloride, C₆H₆Cl₆ which is also called gammaxane.
Benzene hexachloride (BHC) (9.81)
It is clear from the above resonating structures that the electron density is more on o – and p – positions. Hence, the substitution takes place mainly at these positions. However, it may be noted that –I effect of – OH group also operates due to which the electron density on ortho and para positions of the benzene ring is slightly reduced. But the overall electron density increases at these positions of the ring due to resonance. Therefore, –OH group activates the benzene ring for the attack by.
Combustion: When heated in air, benzene burns with sooty flame producing CO₂ and H₂O.
C₆H₆ + 15 O₂ → 6CO₂ + 3H₂O (9.82)
General combustion reaction for any hydrocarbon may be given by the following. | 29 | 11 | Chemistry | 203 |
41f8e802-0aa5-4b37-b6c1-6455d00774e3 | # HYDROCARBONS
an electrophile. Other examples of activating groups are –NH₂, –NHR, –NHCOCH₃, –OCH₃, –CH₃, –C₂H₅, etc.
In the case of aryl halides, halogens are moderately deactivating. Because of their strong – I effect, overall electron density on benzene ring decreases. It makes further substitution difficult. However, due to resonance the electron density on o– and p– positions is comparatively less than that at meta position. Hence, the electrophile attacks on comparatively electron rich meta position resulting in meta substitution.
# 9.6 Carcinogenicity and Toxicity
Benzene and polynuclear hydrocarbons containing more than two benzene rings fused together are toxic and said to possess cancer producing (carcinogenic) property. Such polynuclear hydrocarbons are formed on incomplete combustion of organic materials like tobacco, coal and petroleum. They enter into human body and undergo various biochemical reactions and finally damage DNA and cause cancer. Some of the carcinogenic hydrocarbons are given below (see box).
Meta directing group: The groups which direct the incoming group to meta position are called meta directing groups. Some examples of meta directing groups are –NO₂, –CN, –CHO, –COR, –COOH, –COOR, –SO₃H, etc.
Let us take the example of nitro group. Nitro group reduces the electron density in the benzene ring due to its strong–I effect. Nitrobenzene is a resonance hybrid of the following structures. | 30 | 11 | Chemistry | 203 |
8d210854-67a2-496f-be98-f34bc8ea5a6f | # CHEMISTRY
# sUMMary
Hydrocarbons are the compounds of carbon and hydrogen only. Hydrocarbons are mainly obtained from coal and petroleum, which are the major sources of energy. Petrochemicals are the prominent starting materials used for the manufacture of a large number of commercially important products. LPG (liquefied petroleum gas) and CNG (compressed natural gas), the main sources of energy for domestic fuels and the automobile industry, are obtained from petroleum. Hydrocarbons are classified as open chain saturated (alkanes) and unsaturated (alkenes and alkynes), cyclic (alicyclic) and aromatic, according to their structure.
The important reactions of alkanes are free radical substitution, combustion, oxidation and aromatization. Alkenes and alkynes undergo addition reactions, which are mainly electrophilic additions. Aromatic hydrocarbons, despite having unsaturation, undergo mainly electrophilic substitution reactions. These undergo addition reactions only under special conditions.
Alkanes show conformational isomerism due to free rotation along the C–C sigma bonds. Out of staggered and the eclipsed conformations of ethane, staggered conformation is more stable as hydrogen atoms are farthest apart. Alkenes exhibit geometrical (cis-trans) isomerism due to restricted rotation around the carbon–carbon double bond. Benzene and benzenoid compounds show aromatic character. Aromaticity, the property of being aromatic is possessed by compounds having specific electronic structure characterised by Hückel (4n+2)π electron rule. The nature of groups or substituents attached to benzene ring is responsible for activation or deactivation of the benzene ring towards further electrophilic substitution and also for orientation of the incoming group. Some of the polynuclear hydrocarbons having fused benzene ring system have carcinogenic property.
# ExErcisEs
1. How do you account for the formation of ethane during chlorination of methane?
2. Write IUPAC names of the following compounds:
- (a) CH₃CH=C(CH₃)₂
- (b) CH₂=CH-C≡C-CH₃
- (c)
- (d) –CH₂–CH₂–CH=CH₂
- (e) CH₂ –CH (CH₃)₂
- (f) CH₃(CH₂)₄ CH (CH₂)₃CH₃
- (g) CH₃ – CH = CH – CH₂ – CH = CH – CH – CH₂ – CH = CH₂
|
C₂H₅
3. For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated:
- (a) C₄H₈ (one double bond)
- (b) C₅H₈ (one triple bond)
4. Write IUPAC names of the products obtained by the ozonolysis of the following compounds:
- (i) Pent-2-ene
- (ii) 3,4-Dimethylhept-3-ene
- (iii) 2-Ethylbut-1-ene
- (iv) 1-Phenylbut-1-ene
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61908b66-c6bd-4338-b6ef-9e320106fcba | # HYDROCARBONS
# 9.5
An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of ‘A’.
# 9.6
An alkene ‘A’ contains three C – C, eight C – H σ bonds and one C – C π bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘A’.
# 9.7
Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?
# 9.8
Write chemical equations for combustion reaction of the following hydrocarbons:
- (i) Butane
- (ii) Pentene
- (iii) Hexyne
- (iv) Toluene
# 9.9
Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?
# 9.10
Why is benzene extra ordinarily stable though it contains three double bonds?
# 9.11
What are the necessary conditions for any system to be aromatic?
# 9.12
Explain why the following systems are not aromatic?
- (i)
- (ii)
- (iii)
# 9.13
How will you convert benzene into:
- (i) p-nitrobromobenzene
- (ii) m-nitrochlorobenzene
- (iii) p-nitrotoluene
- (iv) acetophenone?
# 9.14
In the alkane H₃C – CH₂ – C(CH₃)₂– CH₂– CH(CH₃)₂, identify 1°, 2°, 3° carbon atoms and give the number of H atoms bonded to each one of these.
# 9.15
What effect does branching of an alkane chain has on its boiling point?
# 9.16
Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.
# 9.17
Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene?
# 9.18
Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour.
# 9.19
Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?
# 9.20
How would you convert the following compounds into benzene?
- (i) Ethyne
- (ii) Ethene
- (iii) Hexane
# 9.21
Write structures of all the alkenes which on hydrogenation give 2-methylbutane.
# 9.22
Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E⁺:
- (a) Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene
- (b) Toluene, p-H₃C – C₆H₄– NO₂, p-O₂N – C₆H₄ – NO₂.
# 9.23
Out of benzene, m–dinitrobenzene and toluene which will undergo nitration most easily and why?
# 9.24
Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.
# 9.25
Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example.
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40f6d06c-a401-4664-b792-f1e132a11b31 | # REDOX REACTIONS
Where there is oxidation, there is always reduction – Chemistry is essentially a study of redox systems.
After studying this unit you will be able to:
- identify redox reactions as a class of reactions in which oxidation and reduction reactions occur simultaneously;
- define the terms oxidation, reduction, oxidant (oxidising agent) and reductant (reducing agent);
- explain mechanism of redox reactions by electron transfer process;
- use the concept of oxidation number to identify oxidant and reductant in a reaction;
- classify redox reaction into combination (synthesis), decomposition, displacement and disproportionation reactions;
- suggest a comparative order among various reductants and oxidants;
- balance chemical equations using (i) oxidation number (ii) half reaction method;
- learn the concept of redox reactions in terms of electrode processes.
# 7.1 CLASSICAL IDEA OF REDOX REACTIONS – OXIDATION AND REDUCTION REACTIONS
Originally, the term oxidation was used to describe the addition of oxygen to an element or a compound. Because of the presence of dioxygen in the atmosphere (~20%), many elements combine with it and this is the principal reason why they commonly occur on the earth in the form of their oxides. The following reactions represent oxidation processes according to the limited definition of oxidation:
2 Mg (s) + O₂ (g) → 2 MgO (s) (7.1)
S (s) + O₂ (g) → SO₂ (g) (7.2) | 0 | 11 | Chemistry | 201 |
eddce72e-19b4-4704-847c-0bd9f0467dbb | # CHEMISTRY
In reactions (7.1) and (7.2), the elements magnesium and sulphur are oxidised on account of addition of oxygen to them. Similarly, methane is oxidised owing to the addition of oxygen to it.
CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (l) (7.3)
A careful examination of reaction (7.3) in which hydrogen has been replaced by oxygen prompted chemists to reinterpret oxidation in terms of removal of hydrogen from it and, therefore, the scope of term oxidation was broadened to include the removal of hydrogen from a substance. The following illustration is another reaction where removal of hydrogen can also be cited as an oxidation reaction.
2 H₂S(g) + O₂ (g) → 2 S (s) + 2 H₂O (l) (7.4)
As knowledge of chemists grew, it was natural to extend the term oxidation for reactions similar to (7.1 to 7.4), which do not involve oxygen but other electronegative elements. The oxidation of magnesium with fluorine, chlorine and sulphur etc. occurs according to the following reactions:
Mg (s) + F (g) → MgF₂ (s) (7.5)
Mg (s) + Cl₂ (g) → MgCl₂ (s) (7.6)
Mg (s) + S (s) → MgS (s) (7.7)
Incorporating the reactions (7.5 to 7.7) within the fold of oxidation reactions encouraged chemists to consider not only the removal of hydrogen as oxidation, but also the removal of electropositive elements as oxidation. Thus the reaction:
2K₄ Fe(CN)₆ + H₂O₂ (aq) → 2K₃Fe(CN)₆ + 2 KOH (aq)
is interpreted as oxidation due to the removal of electropositive element potassium from potassium ferrocyanide before it changes to potassium ferricyanide. To summarise, the term “oxidation” is defined as the addition of oxygen/electronegative element to a substance or removal of hydrogen/electropositive element from a substance.
In the beginning, reduction was considered as removal of oxygen from a compound. However, the term reduction has been broadened these days to include removal of oxygen/electronegative element from a substance or addition of hydrogen/electropositive element to a substance.
# Problem 7.1
In the reactions given below, identify the species undergoing oxidation and reduction:
1. H₂S (g) + Cl₂ (g) → 2 HCl (g) + S (s)
2. 3Fe₃O₄ (s) + 8 Al (s) → 9 Fe (s) + 4Al₂O₃ (s)
3. 2 Na (s) + H₂ (g) → 2 NaH (s)
# Solution
1. H₂S is oxidised because a more electronegative element, chlorine is added to hydrogen (or a more electropositive element, hydrogen has been removed from S). Chlorine is reduced due to addition of hydrogen to it.
2. Aluminium is oxidised because oxygen is added to it. Ferrous ferric oxide is reduced. | 1 | 11 | Chemistry | 201 |
fb49b0ec-6fa0-4cff-ac84-6bac1f7b7aad | # REDOX REACTIONS
(Fe3O4) is reduced because oxygen has been removed from it.
(iii) With the careful application of the concept of electronegativity only we may infer that sodium is oxidised and hydrogen is reduced.
Reaction (iii) chosen here prompts us to think in terms of another way to define redox reactions.
# 7.2 REDOX REACTIONS IN TERMS OF ELECTRON TRANSFER REACTIONS
We have already learnt that the reactions are redox reactions because in each of these reactions sodium is oxidised due to the addition of either oxygen or more electronegative element to sodium. Simultaneously, chlorine, oxygen and sulphur are reduced because to each of these, the electropositive element sodium has been added.
As an illustration, we may further elaborate one of these, say, the formation of sodium chloride.
2 Na(s) → 2 Na+(g) + 2e–
Cl2(g) + 2e– → 2 Cl–(g)
Each of the above steps is called a half reaction, which explicitly shows involvement of electrons. Sum of the half reactions gives the overall reaction:
2 Na(s) + Cl2(g) → 2 NaCl(s) (7.12)
4 Na(s) + O2(g) → 2 Na2O(s) (7.13)
2 Na(s) + S(s) → Na2S(s) (7.14)
Reactions 7.12 to 7.14 suggest that half reactions that involve loss of electrons are called oxidation reactions. Similarly, the half reactions that involve gain of electrons are called reduction reactions.
It may not be out of context to mention here that the new way of defining oxidation and reduction has been achieved only by establishing a correlation between the behaviour of species as per the classical idea and their interplay in electron-transfer change.
In reactions (7.12 to 7.14) sodium, which is oxidised, acts as a reducing agent because it donates electron to each of the elements interacting with it and thus helps in reducing them. Chlorine, oxygen and sulphur are reduced and act as oxidising agents because these accept electrons from sodium.
To summarise, we may mention that:
- Oxidation: Loss of electron(s) by any species.
- Reduction: Gain of electron(s) by any species.
- Oxidising agent: Acceptor of electron(s).
- Reducing agent: Donor of electron(s).
# Problem 7.2
Justify that the reaction:
2 Na(s) + H2(g) → 2 NaH(s) is a redox change.
# Solution
Since in the above reaction the compound formed is an ionic compound, which may also be represented as Na+H–(s), this suggests that one half reaction in this process is:
2 Na(s) → 2 Na+(g) + 2e– | 2 | 11 | Chemistry | 201 |
2124a8fd-8ef2-42e6-8332-590c19e56b0c | # Chemistry
and the other half reaction is: H₂ (g) + 2e– → 2 H–(g)
This splitting of the reaction under examination into two half reactions automatically reveals that here sodium is oxidised and hydrogen is reduced, therefore, the complete reaction is a redox change.
# 7.2.1 Competitive Electron Transfer Reactions
Place a strip of metallic zinc in an aqueous solution of copper nitrate as shown in Fig. 7.1, for about one hour. You may notice that the strip becomes coated with reddish metallic copper and the blue colour of the solution disappears. Formation of Zn²⁺ ions among the products can easily be judged when the blue colour of the solution due to Cu²⁺ has disappeared. If hydrogen sulphide gas is passed through the colourless solution containing Zn²⁺ ions, appearance of white zinc sulphide, ZnS can be seen on making the solution alkaline with ammonia.
The reaction between metallic zinc and the aqueous solution of copper nitrate is:
Zn(s) + Cu²⁺ (aq) → Zn²⁺ (aq) + Cu(s) (7.15)
In reaction (7.15), zinc has lost electrons to form Zn²⁺ and, therefore, zinc is oxidised. Evidently, now if zinc is oxidised, releasing electrons, something must be reduced, accepting the electrons lost by zinc. Copper ion is reduced by gaining electrons from the zinc.
Reaction (7.15) may be rewritten as:
Fig. 7.1 Redox reaction between zinc and aqueous solution of copper nitrate occurring in a beaker.
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2ba884bc-53a3-4ece-8d0c-67851ded4c80 | # REDOX REACTIONS
Fig. 7.2 Redox reaction between copper and aqueous solution of silver nitrate occurring in a beaker.
At equilibrium, chemical tests reveal that both Ni²⁺(aq) and Co²⁺(aq) are present at moderate concentrations. In this case, neither the reactants [Co(s) and Ni²⁺(aq)] nor the products [Co²⁺(aq) and Ni (s)] are greatly favoured.
This competition for release of electrons incidently reminds us of the competition for release of protons among acids. The similarity suggests that we might develop a table in which metals and their ions are listed on the basis of their tendency to release electrons just as we do in the case of acids to indicate the strength of the acids. As a matter of fact we have already made certain comparisons.
By comparison we have come to know that zinc releases electrons to copper and copper releases electrons to silver and, therefore, the electron releasing tendency of the metals is in the order: Zn > Cu > Ag. We would love to make our list more vast and design a metal activity series or electrochemical series.
The competition for electrons between various metals helps us to design a class of cells, named as Galvanic cells in which the chemical reactions become the source of electrical energy. We would study more about these cells in Class XII.
# 7.3 OXIDATION NUMBER
A less obvious example of electron transfer is realised when hydrogen combines with oxygen to form water by the reaction:
2H₂(g) + O₂(g) → 2H₂O (l) (7.18)
Though not simple in its approach, yet we can visualise the H atom as going from a neutral (zero) state in H₂ to a positive state in H₂O, the O atom goes from a zero state in O₂ to a dinegative state in H₂O. It is assumed that there is an electron transfer from H to O and consequently H₂ is oxidised and O₂ is reduced.
It may be emphasised that the assumption of electron transfer is made for book-keeping purpose only and it will become obvious at a later stage in this unit that it leads to the simple description of redox reactions.
Oxidation number denotes the oxidation state of an element in a compound ascertained according to a set of rules formulated on the basis that electron pair. | 4 | 11 | Chemistry | 201 |
6f21fdeb-9927-4f16-9b2f-b50a44df4367 | # Chemistry
In a covalent bond, the electron belongs entirely to the more electronegative element. It is not always possible to remember or make out easily in a compound/ion, which element is more electronegative than the other. Therefore, a set of rules has been formulated to determine the oxidation number of an element in a compound/ion. If two or more than two atoms of an element are present in the molecule/ion such as Na₂S₂O₃/Cr₂O₂⁻, the oxidation number of the atom of that element will then be the average of the oxidation number of all the atoms of that element. We may at this stage, state the rules for the calculation of oxidation number. These rules are:
1. In elements, in the free or the uncombined state, each atom bears an oxidation number of zero. Evidently, each atom in H₂, O₂, Cl₂, O₃, P₄, S₈, Na, Mg, Al has the oxidation number zero.
2. For ions composed of only one atom, the oxidation number is equal to the charge on the ion. Thus Na⁺ ion has an oxidation number of +1, Mg²⁺ ion, +2, Fe³⁺ ion, +3, Cl⁻ ion, –1, O²⁻ ion, –2; and so on. In their compounds, all alkali metals have oxidation number of +1, and all alkaline earth metals have an oxidation number of +2. Aluminium is regarded to have an oxidation number of +3 in all its compounds.
3. The oxidation number of oxygen in most compounds is –2. However, we come across two kinds of exceptions here. One arises in the case of peroxides and superoxides, the compounds of oxygen in which oxygen atoms are directly linked to each other. While in peroxides (e.g., H₂O₂, Na₂O₂), each oxygen atom is assigned an oxidation number of –1, in superoxides (e.g., KO₂, RbO₂) each oxygen atom is assigned an oxidation number of –(½). The second exception appears rarely, i.e., when oxygen is bonded to fluorine. In such compounds e.g., oxygen difluoride (OF₂) and dioxygen difluoride (O₂F₂), the oxygen is assigned an oxidation number of +2 and +1, respectively. The number assigned to oxygen will depend upon the bonding state of oxygen but this number would now be a positive figure only.
4. The algebraic sum of the oxidation number of all the atoms in a compound must be zero. In a polyatomic ion, the algebraic sum of all the oxidation numbers of atoms of the ion must equal the charge on the ion. Thus, the sum of oxidation number of three oxygen atoms and one carbon atom in the carbonate ion, (CO₃)²⁻ must equal –2. By the application of the above rules, we can find out the oxidation number of the desired element in a molecule or in an ion. It is clear that the metallic elements have positive oxidation numbers and nonmetallic elements have positive or negative oxidation numbers. The atoms of transition elements usually display several positive oxidation states. The highest oxidation number of a representative element is the group number for the first two groups and the group number minus 10 (following the long form of periodic table) for the other groups. Thus, it implies that the highest value of oxidation number exhibited by an atom of an element generally increases across the period in the periodic table. In the third period, the highest value of oxidation number changes from 1 to 7 as indicated below in the compounds of the elements.
A term that is often used interchangeably with the oxidation number is the oxidation state. Thus in CO₂, the oxidation state of carbon is +4, that is also its oxidation number and similarly the oxidation state as well as oxidation number of oxygen is –2. This implies that the oxidation number denotes the oxidation state of an element in a compound. | 5 | 11 | Chemistry | 201 |
75a8db73-d026-41b2-86c4-086e1c2cdd69 | # REDOX REACTIONS
|Group|1|2|13|
|---|---|---|---|
|Element|Na|Mg|Al|
|Compound|NaCl|MgSO₄|AlF₃|
|Highest oxidation number state of the group element|+1|+2|+3|
The oxidation number/state of a metal in a compound is sometimes presented according to the notation given by German chemist, Alfred Stock. It is popularly known as Stock notation. According to this, the oxidation number is expressed by putting a Roman numeral representing the oxidation number in parenthesis after the symbol of the metal in the molecular formula. Thus aurous chloride and auric chloride are written as Au(I)Cl and Au(III)Cl₃. Similarly, stannous chloride and stannic chloride are written as Sn(II)Cl₂ and Sn(IV)Cl₄. This change in oxidation number implies change in oxidation state, which in turn helps to identify whether the species is present in oxidised form or reduced form. Thus, Hg₂(I)Cl₂ is the reduced form of Hg(II)Cl₂.
# Problem 7.3
Using Stock notation, represent the following compounds: HAuCl₄, Tl₂O, FeO, Fe₂O₃, CuI, CuO, MnO and MnO₂.
# Solution
By applying various rules of calculating the oxidation number of the desired element in a compound, the oxidation number of each metallic element in its compound is as follows:
|Compound|Oxidation Number|
|---|---|
|HAuCl₄|Au has 3|
|Tl₂O|Tl has 1|
|FeO|Fe has 2|
|Fe₂O₃|Fe has 3|
|CuI|Cu has 1|
|CuO|Cu has 2|
|MnO|Mn has 2|
|MnO₂|Mn has 4|
Therefore, these compounds may be represented as: HAu(III)Cl₄, Tl₂(I)O, Fe(II)O, Fe₂(III)O₃, Cu(I)I, Cu(II)O, Mn(II)O, Mn(IV)O₂.
# 2024-25
|Group|14|15|16|17|
|---|---|---|---|---|
|Element|Si|P|S|Cl|
|Compound|SiCl₄|P₄O₁₀|SF₆|HClO|
|Highest oxidation number state of the group element|+4|+5|+6|+7|
The idea of oxidation number has been invariably applied to define oxidation, reduction, oxidising agent (oxidant), reducing agent (reductant) and the redox reaction. To summarise, we may say that:
- Oxidation: An increase in the oxidation number of the element in the given substance.
- Reduction: A decrease in the oxidation number of the element in the given substance.
- Oxidising agent: A reagent which can increase the oxidation number of an element in a given substance. These reagents are called as oxidants also.
- Reducing agent: A reagent which lowers the oxidation number of an element in a given substance. These reagents are also called as reductants.
- Redox reactions: Reactions which involve change in oxidation number of the interacting species.
# Problem 7.4
Justify that the reaction: 2Cu₂O(s) + Cu₂S(s) → 6Cu(s) + SO₂(g) is a redox reaction. Identify the species oxidised/reduced, which acts as an oxidant and which acts as a reductant.
# Solution
Let us assign oxidation number to each of the species in the reaction under examination. This results into:
|Species|Oxidation Number|
|---|---|
|2Cu₂O(s)|+1|
|Cu₂S(s)|–2|
|6Cu(s)|0|
|SO₂|+4|
We therefore, conclude that in this reaction copper is reduced from +1 state to zero oxidation state and sulphur is oxidised from –2 state to +4 state. The above reaction is thus a redox reaction. | 6 | 11 | Chemistry | 201 |
f6efc7fd-b646-4699-bb28-a74a02dd1593 | # CHEMISTRY
Further, Cu2O helps sulphur in Cu2S to increase its oxidation number, therefore, Cu(I) is an oxidant; and sulphur of Cu2S helps copper both in Cu2S itself and Cu2O to decrease its oxidation number; therefore, sulphur of Cu2S is reductant.
# 7.3.1 Types of Redox Reactions
# 1. Combination reactions
A combination reaction may be denoted in the manner:
A + B → C
Either A and B or both A and B must be in the elemental form for such a reaction to be a redox reaction. All combustion reactions, which make use of elemental dioxygen, as well as other reactions involving elements other than dioxygen, are redox reactions. Some important examples of this category are:
|0|0|+4|–2| | | |
|---|---|---|---|---|---|---|
|C(s)|+ O2 (g)|CO2(g)|(7.24)| | | |
|0|0|+2|–3| | | |
|3Mg(s)|+ N2(g)|Mg3N2(s)|(7.25)| | | |
|–4|+1|0|+4|–2|+1|–2|
|CH4(g)|+ 2O2(g)|CO2(g)|+ 2H2O (l)|(7.29)| | |
# 2. Decomposition reactions
Decomposition reactions are the opposite of combination reactions. Precisely, a decomposition reaction leads to the breakdown of a compound into two or more components at least one of which must be in the elemental state. Examples of this class of reactions are:
|+1|–2|0|0| | |
|---|---|---|---|---|---|
|2H2O (l)|2H2 (g)|+ O2(g)|(7.26)| | |
|+1|–1|0|0| | |
|2NaH (s)|2Na (s)|+ H2(g)|(7.27)| | |
|+1|+5|–2|+1|–1|0|
|2KClO3 (s)|2KCl (s)|+ 3O2(g)|(7.28)| | |
It may carefully be noted that there is no change in the oxidation number of hydrogen in methane under combination reactions and that of potassium in potassium chlorate in reaction (7.28). This may also be noted that the non-metal displacement redox reactions include hydrogen displacement and a rarely occurring reaction involving oxygen displacement. | 7 | 11 | Chemistry | 201 |
945de0e4-7219-4d6e-80b1-9bfa0b09d8d6 | # REDOX REACTIONS
All alkali metals and some alkaline earth metals (Ca, Sr, and Ba) which are very good reductants, will displace hydrogen from cold water.
0 +1 –2 +1 –2 +1 0
2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g) (7.33)
0 +1 –2 +2 –2 +1 0
Ca(s) + 2H₂O(l) → Ca(OH)₂(aq) + H₂(g) (7.34)
Less active metals such as magnesium and iron react with steam to produce dihydrogen gas:
0 +1 –2 +2 –2 +1 0
Mg(s) + 2H₂O(l) → Mg(OH)₂(s) + H₂(g) (7.35)
0 +1 –2 +3 –2 0
2Fe(s) + 3H₂O(l) → Fe₂O₃(s) + 3H₂(g) (7.36)
Many metals, including those which do not react with cold water, are capable of displacing hydrogen from acids. Dihydrogen from acids may even be produced by such metals which do not react with steam. Cadmium and tin are the examples of such metals. A few examples for the displacement of hydrogen from acids are:
0 +1 –1 +2 –1 0
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g) (7.37)
0 +1 –1 +2 –1 0
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g) (7.38)
0 +1 –1 +2 –1 0
Fe(s) + 2HCl(aq) → FeCl₂(aq) + H₂(g) (7.39)
Reactions (7.37 to 7.39) are used to prepare dihydrogen gas in the laboratory. Here, the reactivity of metals is reflected in the rate of hydrogen gas evolution, which is the slowest for the least active metal Fe, and the fastest for the most reactive metal, Mg.
Very less active metals, which may occur in the native state such as silver (Ag), and gold (Au) do not react even with hydrochloric acid. In section (7.2.1) we have already discussed that the metals – zinc (Zn), copper (Cu) and silver (Ag) through tendency to lose electrons show their reducing activity in the order Zn > Cu > Ag. Like metals, activity series also exists for the halogens. The power of these elements as oxidising agents decreases as we move down from fluorine to iodine in group 17 of the periodic table. This implies that fluorine is so reactive that it can replace chloride, bromide and iodide ions in solution. In fact, fluorine is so reactive that it attacks water and displaces the oxygen of water:
0 +1 –2 0 +1 –1 0
2H₂O(l) + 2F₂(g) → 4HF(aq) + O₂(g) (7.40)
It is for this reason that the displacement reactions of chlorine, bromine and iodine using fluorine are not generally carried out in aqueous solution. On the other hand, chlorine can displace bromide and iodide ions in an aqueous solution as shown below:
0 +1 –1 +1 –1 0
Cl₂(g) + 2KBr(aq) → 2KCl(aq) + Br₂(l) (7.41)
0 +1 –1 +1 –1 0
Cl₂(g) + 2KI(aq) → 2KCl(aq) + I₂(s) (7.42)
As Br₂ and I₂ are coloured and dissolve in CCl₄, can easily be identified from the colour of the solution. The above reactions can be written in ionic form as:
0 –1 –1 0
Cl₂(g) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(l) (7.41a)
0 –1 –1 0
Cl₂(g) + 2I⁻(aq) → 2Cl⁻(aq) + I₂(s) (7.42b)
Reactions (7.41) and (7.42) form the basis of identifying Br⁻ and I⁻ in the laboratory through the test popularly known as ‘Layer Test’. It may not be out of place to mention here that bromine likewise can displace iodide ion in solution:
0 –1 –1 0
Br₂(l) + 2I⁻(aq) → 2Br⁻(aq) + I₂(s) (7.43)
The halogen displacement reactions have a direct industrial application. The recovery of halogens from their halides requires an oxidation process, which is represented by:
2X⁻ → X₂ + 2e⁻ (7.44)
here X denotes a halogen element. Whereas chemical means are available to oxidise Cl⁻, Br⁻ and I⁻, as fluorine is the strongest oxidising. | 8 | 11 | Chemistry | 201 |
e3cfcc33-4bcb-4ebe-86f9-1ffe33d2b161 | # Chemistry
agent; there is no way to convert F– ions to F₂ by chemical means. The only way to achieve F₂ from F– is to oxidise electrolytically, the details of which you will study at a later stage.
Fluorine shows deviation from this behaviour when it reacts with alkali. The reaction that takes place in the case of fluorine is as follows:
2 F₂(g) + 2OH–(aq) → 2 F–(aq) + OF₂(g) + H₂O(l) (7.49)
# 4. Disproportionation reactions
Disproportionation reactions are a special type of redox reactions. In a disproportionation reaction an element in one oxidation state is simultaneously oxidised and reduced. One of the reacting substances in a disproportionation reaction always contains an element that can exist in at least three oxidation states. The element in the form of reacting substance is in the intermediate oxidation state; and both higher and lower oxidation states of that element are formed in the reaction. The decomposition of hydrogen peroxide is a familiar example of the reaction, where oxygen experiences disproportionation.
2H₂O₂(aq) → 2H₂O(l) + O₂(g) (7.45)
Here the oxygen of peroxide, which is present in –1 state, is converted to zero oxidation state in O₂ and decreases to –2 oxidation state in H₂O.
Phosphorous, sulphur and chlorine undergo disproportionation in the alkaline medium as shown below:
P₄(s) + 3OH–(aq) + 3H₂O(l) → PH₃(g) + 3H₂PO₂–(aq) (7.46)
S₈(s) + 12 OH–(aq) → 4S²–(aq) + 2S₂O₃²–(aq) + 6H₂O(l) (7.47)
Cl₂(g) + 2 OH–(aq) → ClO–(aq) + Cl–(aq) + H₂O(l) (7.48)
The reaction (7.48) describes the formation of household bleaching agents. The hypochlorite ion (ClO–) formed in the reaction oxidises the colour-bearing stains of the substances to colourless compounds. It is of interest to mention here that whereas bromine and iodine follow the same trend as exhibited by chlorine in reaction (7.48),
# Problem 7.5
Which of the following species, do not show disproportionation reaction and why?
ClO–, ClO₃–, ClO₂– and ClO₄–
Also write reaction for each of the species that disproportionates.
# Solution
Among the oxoanions of chlorine listed above, ClO₄– does not disproportionate because in this oxoanion chlorine is present in its highest oxidation state that is, +7. The disproportionation reactions for the other three oxoanions of chlorine are as follows:
3ClO– → 2Cl– + ClO₂–
6ClO₃– → 4Cl– + 2ClO₄–
4ClO₂– → Cl– + 3ClO₄–
# Problem 7.6
Suggest a scheme of classification of the following redox reactions:
1. N₂(g) + O₂(g) → 2 NO(g)
2. 2Pb(NO₃)₂(s) → 2PbO(s) + 4 NO(g) + O₂(g)
3. NaH(s) + H₂O(l) → NaOH(aq) + H₂(g)
4. 2NO₂(g) + 2OH–(aq) → NO–(aq) + NO₂–(aq) + H₂O(l) | 9 | 11 | Chemistry | 201 |
5e3665b5-8464-48ea-9d40-a33ee8a6c742 | # REDOX REACTIONS
Solution reaction (c), hydrogen of water has been displaced by hydride ion into dihydrogen gas. Therefore, this may be called as displacement redox reaction. The reaction (d) involves disproportionation of NO₂ (+4 state) into NO– (+3 state) and NO₂ (+5 state). Therefore reaction (d) is an example of disproportionation redox reaction.
# The Paradox of Fractional Oxidation Number
Sometimes, we come across with certain compounds in which the oxidation number of a particular element in the compound is in fraction. Examples are:
- C₃O₂ [where oxidation number of carbon is (4/3)],
- Br₃O₈ [where oxidation number of bromine is (16/3)]
- Na₂S₄O₆ (where oxidation number of sulphur is 2.5).
We know that the idea of fractional oxidation number is unconvincing to us, because electrons are never shared/transferred in fraction. Actually this fractional oxidation state is the average oxidation state of the element under examination and the structural parameters reveal that the element for whom fractional oxidation state is realised is present in different oxidation states.
# Structure of the species
C₃O₂, Br₃O₈ and S₄O₂– reveal the following bonding situations:
Structure of C₃O₂ (carbon suboxide)
O = C = C* = C = O
Structure of Br₃O₈ (tribromooctaoxide)
Structure of S₄O₂– (tetrathionate ion)
The element marked with asterisk in each species is exhibiting the different oxidation state (oxidation number) from rest of the atoms of the same element in each of the species. This reveals that in C₃O₂, two carbon atoms are present in +2 oxidation state each, whereas the third one is present in zero oxidation state and the average is 4/3. However, the realistic picture is +2 for two terminal carbons and zero for the middle carbon. Likewise in Br₃O₈, each of the two terminal bromine atoms are present in +6 oxidation state and the middle bromine is present in +4 oxidation state. Once again the average, that is different from reality, is 16/3. In the same fashion, in the species S₄O₂–, each of the two extreme sulphurs exhibits oxidation state of +5 and the two middle sulphurs as zero. The average of four oxidation numbers of sulphurs of the S₄O₆²– is 2.5, whereas the reality being +5, 0, 0 and +5 oxidation number respectively for each sulphur.
We may thus, in general, conclude that the idea of fractional oxidation state should be taken with care and the reality is revealed by the structures only. Further, whenever we come across with fractional oxidation state of any particular element in any species, we must understand that this is the average oxidation number only. In reality (revealed by structures only), the element in that particular species is present in more than one whole number oxidation states. Fe₃O₄, Mn₃O₄, Pb₃O₄ are some of the other examples of the compounds, which are mixed oxides, where we come across with fractional oxidation states of the metal atom. However, the oxidation states may be in fraction as in O⁺ and O²– where it is +½ and –½ respectively.
2024-25 | 10 | 11 | Chemistry | 201 |
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