Question 32 of 32intermediate🔧 ApplyShort Answer3 marks

What is meant by the term bond order? Calculate the bond order of:

(i)

N₂

Answer

For N2\text{N}_2, the bond order is 3.

(ii)

O₂

Answer

For O2\text{O}_2, there are 10 bonding electrons (Nb=10N_b = 10) and 6 antibonding electrons (Na=6N_a = 6). Bond order = 12(10−6)=2\frac{1}{2} (10 - 6) = \mathbf{2}.

(iii)

O₂⁺

Answer

For O2+\text{O}_2^+, one electron is removed from the antibonding orbital of O2\text{O}_2, so Nb=10N_b = 10 and Na=5N_a = 5. Bond order = 12(10−5)=2.5\frac{1}{2} (10 - 5) = \mathbf{2.5}.

(iv)

O₂⁻

Answer

For O2−\text{O}_2^-, one electron is added to the antibonding orbital of O2\text{O}_2, so Nb=10N_b = 10 and Na=7N_a = 7. Bond order = 12(10−7)=1.5\frac{1}{2} (10 - 7) = \mathbf{1.5}.

Explanation

The term bond order is defined in the context as one half the difference between the number of electrons in bonding (NbN_b) and antibonding (NaN_a) molecular orbitals. For N2\text{N}_2, the context directly states the bond order is 3. For O2\text{O}_2, the context provides the electron counts: 10 bonding and 6 antibonding electrons, yielding a bond order of 2. For O2+\text{O}_2^+ and O2−\text{O}_2^-, we apply the same formula by adjusting the antibonding electron count (NaN_a) by -1 and +1 respectively, since adding or removing an electron from O2\text{O}_2 affects its antibonding π∗\pi^* orbitals.

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