Question

The bond dissociation enthalpies for N2 and N2- are 945 kJ/mol and 765 kJ/mol respectively. (There is only a small difference between enthalpies and energies.) Using an argument based on MO theory, explain why N2- has a smaller bond dissociation energy than N2.

Answer 1

Answer on Question #55181 – Chemistry – General chemistry

Question:

The bond dissociation enthalpies for N2 and N2- are 945 kJ/mol and 765 kJ/mol respectively. (There is only a small difference between enthalpies and energies.) Using an argument based on MO theory, explain why N2- has a smaller bond dissociation energy than N2.

Answer:

In this case, we should draw MO diagram and calculate bond orders for $\mathsf{N}_2$ and $\mathsf{N}_2^-$:

$$\mathrm{Bond\ order} = (8 - 2)/2 = 3$$ $$\mathrm{Bond\ order} = (8 - 3)/2 = 2.5$$

As shown, in $\mathsf{N}_2^-$ one electron occupies antibonding orbital $\pi(2p_x)$ that decreases the bond order by 0.5 in comparison with $\mathsf{N}_2$. Moreover, the value of ratio for bond orders is close to that for energies found experimentally:

$$E(\mathrm{N}_2)/E(\mathrm{N}_2^-) = 945/765 = 1.24$$

and

$$\mathrm{Bond\ order}\ (\mathrm{N}_2)/\ \mathrm{Bond\ order}\ (\mathrm{N}_2^-) = 3/2.5 = 1.2$$

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