In the previous section, we saw the bond enthalpies of diatomic molecules. We saw three examples related to homonuclear diatomic molecules. In this section, we will see the bond enthalpies of heteronuclear diatomic molecules
• H2, O2, and N2 are diatomic molecules
♦ To be precise, they are homonuclear diatomic molecules
■ Diatomic molecules are molecules composed of only two atoms
• Consider a diatomic molecule
♦ If both the two atoms are of the same element, it is called a homonuclear diatomic molecule
♦ If the two atoms are of different elements, it is called a heteronuclear diatomic molecule
• Let us see the bond enthalpy in the case of heteronuclear diatomic molecules. We will explain it using an example:
Example 4:
1. Take one mole of HCl molecules
• We want all those molecules to be separated into 'individual H and Cl atoms'
• Let us see how it can be done:
2. For our present discussion, we will represent the HCl molecule as H-Cl
• Where '-' represents the single covalent bond between the H and Cl atoms
3. We have taken 1 mole of HCl molecules
• So there will be one mole (6.023 × 1023 nos.) of 'H-Cl'
• So there will be 6.023 × 1023 nos. of 'H-Cl single bonds'
4. We must break all those H-Cl bonds
• Then only we will be able to convert all of the '1 mole HCl' into individual H and Cl atoms
■ The energy required to break one mole (6.023 × 1023 nos.) of 'H-Cl single bonds' is called 'bond enthalpy of H-Cl bond in hydrogen chloride molecule'
• We can write in any of the two ways:
♦ Bond enthalpy of H-Cl bond in hydrogen chloride molecule
OR
♦ H-Cl bond enthalpy in hydrogen chloride molecule
♦ The resulting H atoms must also be in gaseous state
♦ The resulting Cl atoms must also be in gaseous state
• Scientists have determined this energy using experimental methods. It is equal to 431.0 kJ mol-1
5. The process is represented as:
$\mathbf\small{\rm{HCl(g)\longrightarrow H(g)\,+Cl(g);\Delta_aH^\theta=431.0\,kJ\,mol^{-1}}}$
6. We can write the report as follows:
• Bond enthalpy of H-Cl bond in hydrogen chloride molecule is 431.0 kJ mol-1
OR
• H-Cl bond enthalpy in hydrogen chloride molecule is 431.0 kJ mol-1
• Next, let us see the bond enthalpy in the case of polyatomic molecules
• Polyatomic molecules are those molecules which have three or more atoms
Example 5:
1. Take one mole of H2O molecules
• We want all those molecules to be separated into 'individual H and O atoms'
• Let us see how it can be done:
2. For our present discussion, we will represent the H2O molecule as H-O-H
♦ See fig.4.15 in section 4.2
• Where '-' represents the single covalent bond between the H and O atoms
3. We have taken 1 mole of H2O molecules
• So there will be one mole (6.023 × 1023 nos.) of 'H-O-H'
4. In one molecule of H-O-H, there are two single bonds:
♦ The O-H bond on the left side
♦ The O-H bond on the right side
5. So in one mole of H2O, there will be:
♦ 6.023 × 1023 nos. of 'left side O-H bonds'
♦ 6.023 × 1023 nos. of 'right side O-H bonds'
6. We must break all those O-H bonds
• Then only we will be able to convert all of the '1 mole H2O' into individual H and O atoms
7. Now, there is a problem:
• The energy required to break the left side O-H bond
is not equal to
• The energy required to break the right side O-H bond
8. So we must split up the process. It can be explained in 3 steps:
(i) Consider 6.023 × 1023 nos. of H-O-H molecules
(ii) First we break the left side bond in each of them
We get two items:
♦ 6.023 × 1023 nos. of H
♦ 6.023 × 1023 nos. of O-H
• 502 kJ energy is required for this process
• So we can write:
$\mathbf\small{\rm{H_2O(g)\longrightarrow H(g)\,+OH(g);\Delta_aH^\theta=502.0\,kJ\,mol^{-1}}}$
(iii) Next we break the bond in 6.023 × 1023 nos. of O-H obtained in (ii)
• 427 kJ energy is required for this process
• So we can write:
$\mathbf\small{\rm{OH(g)\longrightarrow H(g)\,+O(g);\Delta_aH^\theta=427.0\,kJ\,mol^{-1}}}$
9. The difference in the two energies can be explained in 3 steps:
(i) First we break the left side O-H bond
• This gives H and O-H
• Thus one H is removed from the original H-O-H
(ii) Consider the two environments:
• The first 'O-H bond' is in a chemical environment in which, 'that bond is a part of H-O-H'
• The second 'O-H bond' is in a chemical environment in which, 'that bond is a part of O-H'
(iii) The two environments are clearly different
■ So the energies required will also be different
10. We want all 6.023 × 1023 nos. of H-O-H molecules to be separated into 'individual H and O atoms'
• Clearly, we will need '502 plus 427 kJ energy' for that
• These energies are called bond dissociation enthalpies
• These enthalpies are different from bond enthalpies that we were discussing about
11. So what is bond enthalpy in the case of H-O-H?
The answer can be written in 3 steps:
(i) Calculate the average (mean) of the two values: 502 and 427
• It is equal to 464.5
(ii) Since it is an average, we cannot simply write: 'bond enthalpy'
• We must write it as: 'average bond enthalpy'
(iii) So the final report is:
• The average bond enthalpy of the O-H bonds in water molecule is 464.5 kJ mol-1
OR
• Average O-H bond enthalpy in water molecule is 464.5 kJ mol-1
12. Now we can write an interesting point:
■ For a homonuclear diatomic molecule, the 'bond dissociation enthalpy' will be same as 'bond enthalpy'
• The reader may write the reason in his/her own notebooks
13. While breaking the bonds in H-O-H,
♦ The initial sample of H2O that we take, must be in gaseous state
♦ The resulting H atoms must also be in gaseous state
♦ The resulting O atoms must also be in gaseous state
14. So we have calculated the 'average bond enthalpy of the O-H bonds in water molecule'
• It is equal to 464.5 kJ mol-1
■ Can we use this value for O-H bonds in other molecules?
• The answer can be written by taking an example. It can be written in 3 steps:
(i) Consider the C2H5OH (ethanol) molecule
• It contains one O-H bond
(ii) Suppose that we have 6.023 × 1023 nos. of C2H5OH molecules
• There will be 6.023 × 1023 nos. of O-H bonds
(iii) The energy required to break all those bonds will not be equal to 464.5 kJ
• This is because, the two environments mentioned below are different:
♦ The chemical environment in which 'O-H bond is a part of ethanol'
♦ The chemical environment in which 'O-H bond is a part of water'
1. Strength of the bond
■ We must keep in mind that, bond enthalpy gives an idea about the 'strength of a bond'
• If the bond enthalpy of a bond is high, it is obvious that, the strength of that bond is high (a strong bond)
• If the bond enthalpy of a bond is low, it is obvious that, the strength of that bond is low (a weak bond)
2. Method of reporting:
• While reporting a bond enthalpy value, we must always include the following 3 items:
(i) The name of the two atoms between which the 'bond under consideration' is present
(ii) Whether the 'bond under consideration' is a single, double or triple bond
(iii) The molecule in which the 'bond under consideration' is present
• H2, O2, and N2 are diatomic molecules
♦ To be precise, they are homonuclear diatomic molecules
■ Diatomic molecules are molecules composed of only two atoms
• Consider a diatomic molecule
♦ If both the two atoms are of the same element, it is called a homonuclear diatomic molecule
♦ If the two atoms are of different elements, it is called a heteronuclear diatomic molecule
• Let us see the bond enthalpy in the case of heteronuclear diatomic molecules. We will explain it using an example:
Example 4:
1. Take one mole of HCl molecules
• We want all those molecules to be separated into 'individual H and Cl atoms'
• Let us see how it can be done:
2. For our present discussion, we will represent the HCl molecule as H-Cl
• Where '-' represents the single covalent bond between the H and Cl atoms
3. We have taken 1 mole of HCl molecules
• So there will be one mole (6.023 × 1023 nos.) of 'H-Cl'
• So there will be 6.023 × 1023 nos. of 'H-Cl single bonds'
4. We must break all those H-Cl bonds
• Then only we will be able to convert all of the '1 mole HCl' into individual H and Cl atoms
■ The energy required to break one mole (6.023 × 1023 nos.) of 'H-Cl single bonds' is called 'bond enthalpy of H-Cl bond in hydrogen chloride molecule'
• We can write in any of the two ways:
♦ Bond enthalpy of H-Cl bond in hydrogen chloride molecule
OR
♦ H-Cl bond enthalpy in hydrogen chloride molecule
• While breaking those bonds,
♦ The initial sample of HCl that we take, must be in gaseous state♦ The resulting H atoms must also be in gaseous state
♦ The resulting Cl atoms must also be in gaseous state
• Scientists have determined this energy using experimental methods. It is equal to 431.0 kJ mol-1
5. The process is represented as:
$\mathbf\small{\rm{HCl(g)\longrightarrow H(g)\,+Cl(g);\Delta_aH^\theta=431.0\,kJ\,mol^{-1}}}$
6. We can write the report as follows:
• Bond enthalpy of H-Cl bond in hydrogen chloride molecule is 431.0 kJ mol-1
OR
• H-Cl bond enthalpy in hydrogen chloride molecule is 431.0 kJ mol-1
• Next, let us see the bond enthalpy in the case of polyatomic molecules
• Polyatomic molecules are those molecules which have three or more atoms
Example 5:
1. Take one mole of H2O molecules
• We want all those molecules to be separated into 'individual H and O atoms'
• Let us see how it can be done:
2. For our present discussion, we will represent the H2O molecule as H-O-H
♦ See fig.4.15 in section 4.2
• Where '-' represents the single covalent bond between the H and O atoms
3. We have taken 1 mole of H2O molecules
• So there will be one mole (6.023 × 1023 nos.) of 'H-O-H'
4. In one molecule of H-O-H, there are two single bonds:
♦ The O-H bond on the left side
♦ The O-H bond on the right side
5. So in one mole of H2O, there will be:
♦ 6.023 × 1023 nos. of 'left side O-H bonds'
♦ 6.023 × 1023 nos. of 'right side O-H bonds'
6. We must break all those O-H bonds
• Then only we will be able to convert all of the '1 mole H2O' into individual H and O atoms
7. Now, there is a problem:
• The energy required to break the left side O-H bond
is not equal to
• The energy required to break the right side O-H bond
8. So we must split up the process. It can be explained in 3 steps:
(i) Consider 6.023 × 1023 nos. of H-O-H molecules
(ii) First we break the left side bond in each of them
We get two items:
♦ 6.023 × 1023 nos. of H
♦ 6.023 × 1023 nos. of O-H
• 502 kJ energy is required for this process
• So we can write:
$\mathbf\small{\rm{H_2O(g)\longrightarrow H(g)\,+OH(g);\Delta_aH^\theta=502.0\,kJ\,mol^{-1}}}$
(iii) Next we break the bond in 6.023 × 1023 nos. of O-H obtained in (ii)
• 427 kJ energy is required for this process
• So we can write:
$\mathbf\small{\rm{OH(g)\longrightarrow H(g)\,+O(g);\Delta_aH^\theta=427.0\,kJ\,mol^{-1}}}$
9. The difference in the two energies can be explained in 3 steps:
(i) First we break the left side O-H bond
• This gives H and O-H
• Thus one H is removed from the original H-O-H
(ii) Consider the two environments:
• The first 'O-H bond' is in a chemical environment in which, 'that bond is a part of H-O-H'
• The second 'O-H bond' is in a chemical environment in which, 'that bond is a part of O-H'
(iii) The two environments are clearly different
■ So the energies required will also be different
10. We want all 6.023 × 1023 nos. of H-O-H molecules to be separated into 'individual H and O atoms'
• Clearly, we will need '502 plus 427 kJ energy' for that
• These energies are called bond dissociation enthalpies
• These enthalpies are different from bond enthalpies that we were discussing about
11. So what is bond enthalpy in the case of H-O-H?
The answer can be written in 3 steps:
(i) Calculate the average (mean) of the two values: 502 and 427
• It is equal to 464.5
(ii) Since it is an average, we cannot simply write: 'bond enthalpy'
• We must write it as: 'average bond enthalpy'
(iii) So the final report is:
• The average bond enthalpy of the O-H bonds in water molecule is 464.5 kJ mol-1
OR
• Average O-H bond enthalpy in water molecule is 464.5 kJ mol-1
12. Now we can write an interesting point:
■ For a homonuclear diatomic molecule, the 'bond dissociation enthalpy' will be same as 'bond enthalpy'
• The reader may write the reason in his/her own notebooks
13. While breaking the bonds in H-O-H,
♦ The initial sample of H2O that we take, must be in gaseous state
♦ The resulting H atoms must also be in gaseous state
♦ The resulting O atoms must also be in gaseous state
14. So we have calculated the 'average bond enthalpy of the O-H bonds in water molecule'
• It is equal to 464.5 kJ mol-1
■ Can we use this value for O-H bonds in other molecules?
• The answer can be written by taking an example. It can be written in 3 steps:
(i) Consider the C2H5OH (ethanol) molecule
• It contains one O-H bond
(ii) Suppose that we have 6.023 × 1023 nos. of C2H5OH molecules
• There will be 6.023 × 1023 nos. of O-H bonds
(iii) The energy required to break all those bonds will not be equal to 464.5 kJ
• This is because, the two environments mentioned below are different:
♦ The chemical environment in which 'O-H bond is a part of ethanol'
♦ The chemical environment in which 'O-H bond is a part of water'
• We have completed a discussion on the basics of bond enthalpy. It is important to remember two points:
■ We must keep in mind that, bond enthalpy gives an idea about the 'strength of a bond'
• If the bond enthalpy of a bond is high, it is obvious that, the strength of that bond is high (a strong bond)
• If the bond enthalpy of a bond is low, it is obvious that, the strength of that bond is low (a weak bond)
2. Method of reporting:
• While reporting a bond enthalpy value, we must always include the following 3 items:
(i) The name of the two atoms between which the 'bond under consideration' is present
(ii) Whether the 'bond under consideration' is a single, double or triple bond
(iii) The molecule in which the 'bond under consideration' is present
Next we will see bond order. It is the 'number of bonds'. We will explain it using some examples:
Example 1:
1. Consider one H2 molecule
• It can be represented as H-H
♦ See fig.4.13 in section 4.2
• Where '-' represents the single covalent bond between the H atoms
2. Note down 2 points:
♦ It is a H2 molecule
♦ There is one covalent bond: H-H
3. We report the bond order using the 2 points written in (2):
■ In H2 molecule, the H-H bond order is 1
Example 2:
1. Consider one O2 molecule
• It can be represented as O=O
♦ See fig.4.16 in section 4.2
• Where '=' represents the double bond between the O atoms
2. Note down 2 points:
♦ It is a O2 molecule
♦ There are two covalent bonds: O=O
3. We report the bond order using the 2 points written in (2):
■ In O2 molecule, the O=O bond order is 2
Example 3:
1. Consider one N2 molecule
• It can be represented as N≡N
♦ See fig.4.17 in section 4.2
• Where '≡' represents the triple bond between the N atoms
2. Note down 2 points:
♦ It is a N2 molecule
♦ There are three covalent bonds: N≡N
3. We report the bond order using the 2 points written in (2):
■ In N2 molecule, the N≡N bond order is 3
Example 4:
1. Consider one CO molecule
• It can be represented as C≡O
♦ See fig.4.18 in section 4.2
• Where '≡' represents the triple bond between the C and O atoms
2. Note down 2 points:
♦ It is a CO molecule
♦ There are three covalent bonds: C≡O
3. We report the bond order using the 2 points written in (2):
■ In CO molecule, the C≡O bond order is 3
Example 5:
1. Consider one F2 molecule
• It can be represented as F-F
• Where '-' represents the single covalent bond between the F atoms
2. Note down 2 points:
♦ It is a F2 molecule
♦ There is one covalent bond: F-F
3. We report the bond order using the 2 points written in (2):
■ In F2 molecule, the F-F bond order is 1
Example 6:
1. Consider one O22- ion
• It's Lewis dot structure is shown in fig.4.43(a) below:
(Note that, in the previous examples, there was not much need to draw Lewis dot structures. This is because, they were simple molecules and we are familiar with their structures. However, whenever any doubt arises, the reader may draw the Lewis dot structure, or refer already drawn ones)
• From the Lewis dot structure, it is clear that, there is a single bond ('-') between the two O atoms
2. Note down 2 points:
♦ It is a O22- ion
♦ There is one covalent bond: O-O
3. We report the bond order using the 2 points written in (2):
■ In O22- ion, the O-O bond order is 1
• In example 5, F2 molecule has (7+7) = 14 valence electrons
♦ The bond order is 1
• In example 6, O22- ion has (6+6+2) = 14 valence electrons
♦ The bond order is 1
■ In general, isoelectronic species have the same bond orders
1. Consider one NO+ ion
• It's Lewis dot structure is shown in fig.4.43(b) above
• From the Lewis dot structure, it is clear that, there is a triple bond ('≡') between the N and O atoms
2. Note down 2 points:
♦ It is a NO+ ion
♦ There are three covalent bonds: N≡O
3. We report the bond order using the 2 points written in (2):
■ In NO+ ion, the N≡O bond order is 3
• In example 7, NO+ ion has (5+6-1) = 10 valence electrons
♦ The bond order is 3
• N2 molecule has (5+5) = 10 valence electrons
♦ The bond order is 3
• CO molecule has (4+6) = 10 valence electrons
♦ The bond order is 3
■ In general, isoelectronic species have the same bond orders
1. Consider one C2H2 molecule
• It can be represented as H-C≡C-H
• Where:
♦ '-' represents the single covalent bond between the C and H atoms
♦ '≡' represents the triple covalent bond between the C atoms
2. Note down 2 points:
♦ It is a C2H2 molecule
♦ There is one covalent bond: C-H
♦ There is one covalent bond: C-H
♦ There are three covalent bonds: C≡C
3. We report the bond order using the 4 points written in (2):
■ In C2H2 molecule:
♦ The C-H bond order is 1
♦ The C≡C bond order is 3
1. Strength of the bond
■ We must keep in mind that, bond order gives an idea about the 'strength of a bond'
• If the bond order is high, bond enthalpy will also be high
♦ Thus it will be a strong bond
• If the bond order is low, bond enthalpy will also be low
♦ Thus it will be a weak bond
2. Method of specification:
• While reporting a bond order, we must always include the following 3 items:
(i) The name of the two atoms between which the 'bond under consideration' is present
(ii) Whether the 'bond under consideration' is a single, double or triple bond
(iii) The molecule in which the 'bond under consideration' is present
How do you express the bond strength in terms of bond order ?
Solution:
1. An increase in bond order indicates that, greater number of bonds are present
2. Greater number of bonds indicates that the bond is stronger
• For example:
♦ A double bond is stronger than a single bond
♦ A triple bond is stronger than a double bond
3. So we can write:
♦ When bond order increases, bond strength increases
♦ When bond order decreases, bond strength decreases
Example 1:
1. Consider one H2 molecule
• It can be represented as H-H
♦ See fig.4.13 in section 4.2
• Where '-' represents the single covalent bond between the H atoms
2. Note down 2 points:
♦ It is a H2 molecule
♦ There is one covalent bond: H-H
3. We report the bond order using the 2 points written in (2):
■ In H2 molecule, the H-H bond order is 1
Example 2:
1. Consider one O2 molecule
• It can be represented as O=O
♦ See fig.4.16 in section 4.2
• Where '=' represents the double bond between the O atoms
2. Note down 2 points:
♦ It is a O2 molecule
♦ There are two covalent bonds: O=O
3. We report the bond order using the 2 points written in (2):
■ In O2 molecule, the O=O bond order is 2
Example 3:
1. Consider one N2 molecule
• It can be represented as N≡N
♦ See fig.4.17 in section 4.2
• Where '≡' represents the triple bond between the N atoms
2. Note down 2 points:
♦ It is a N2 molecule
♦ There are three covalent bonds: N≡N
3. We report the bond order using the 2 points written in (2):
■ In N2 molecule, the N≡N bond order is 3
Example 4:
1. Consider one CO molecule
• It can be represented as C≡O
♦ See fig.4.18 in section 4.2
• Where '≡' represents the triple bond between the C and O atoms
2. Note down 2 points:
♦ It is a CO molecule
♦ There are three covalent bonds: C≡O
3. We report the bond order using the 2 points written in (2):
■ In CO molecule, the C≡O bond order is 3
Example 5:
1. Consider one F2 molecule
• It can be represented as F-F
• Where '-' represents the single covalent bond between the F atoms
2. Note down 2 points:
♦ It is a F2 molecule
♦ There is one covalent bond: F-F
3. We report the bond order using the 2 points written in (2):
■ In F2 molecule, the F-F bond order is 1
Example 6:
1. Consider one O22- ion
• It's Lewis dot structure is shown in fig.4.43(a) below:
 |
| Fig.4.43 |
• From the Lewis dot structure, it is clear that, there is a single bond ('-') between the two O atoms
2. Note down 2 points:
♦ It is a O22- ion
♦ There is one covalent bond: O-O
3. We report the bond order using the 2 points written in (2):
■ In O22- ion, the O-O bond order is 1
Let us compare examples 5 and 6:
♦ The bond order is 1
• In example 6, O22- ion has (6+6+2) = 14 valence electrons
♦ The bond order is 1
■ In general, isoelectronic species have the same bond orders
Example 7:
• It's Lewis dot structure is shown in fig.4.43(b) above
• From the Lewis dot structure, it is clear that, there is a triple bond ('≡') between the N and O atoms
2. Note down 2 points:
♦ It is a NO+ ion
♦ There are three covalent bonds: N≡O
3. We report the bond order using the 2 points written in (2):
■ In NO+ ion, the N≡O bond order is 3
Let us compare example 7 with N2 and CO:
♦ The bond order is 3
• N2 molecule has (5+5) = 10 valence electrons
♦ The bond order is 3
• CO molecule has (4+6) = 10 valence electrons
♦ The bond order is 3
■ In general, isoelectronic species have the same bond orders
Example 8:
• It can be represented as H-C≡C-H
• Where:
♦ '-' represents the single covalent bond between the C and H atoms
♦ '≡' represents the triple covalent bond between the C atoms
2. Note down 2 points:
♦ It is a C2H2 molecule
♦ There is one covalent bond: C-H
♦ There is one covalent bond: C-H
♦ There are three covalent bonds: C≡C
3. We report the bond order using the 4 points written in (2):
■ In C2H2 molecule:
♦ The C-H bond order is 1
♦ The C≡C bond order is 3
• We have completed a discussion on the basics of bond order. It is important to remember two points:
■ We must keep in mind that, bond order gives an idea about the 'strength of a bond'
• If the bond order is high, bond enthalpy will also be high
♦ Thus it will be a strong bond
• If the bond order is low, bond enthalpy will also be low
♦ Thus it will be a weak bond
2. Method of specification:
• While reporting a bond order, we must always include the following 3 items:
(i) The name of the two atoms between which the 'bond under consideration' is present
(ii) Whether the 'bond under consideration' is a single, double or triple bond
(iii) The molecule in which the 'bond under consideration' is present
Solved example 4.4
Solution:
1. An increase in bond order indicates that, greater number of bonds are present
2. Greater number of bonds indicates that the bond is stronger
• For example:
♦ A double bond is stronger than a single bond
♦ A triple bond is stronger than a double bond
3. So we can write:
♦ When bond order increases, bond strength increases
♦ When bond order decreases, bond strength decreases
In the next section, we will see resonance structures
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