The bond order and magnetic property of acetylide ion are same as that of:

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Given: We compare the bond order and magnetic property of acetylide ion, $\mathrm{C}_2^{2-}$ .

Find: Which given species has the same bond order and magnetic property.

For $\mathrm{C}_2^{2-}$ , total electrons are $14$ . Its molecular orbital configuration is:

\sigma(1s)^2\, \sigma^*(1s)^2\, \sigma(2s)^2\, \sigma^*(2s)^2\, \sigma(2p_z)^2\, \pi(2p_x)^2\, \pi(2p_y)^2

Now calculate bond order:

\text{Bond Order} = \frac{\text{bonding electrons} - \text{antibonding electrons}}{2}

Here, bonding electrons $= 10$ and antibonding electrons $= 4$ , so

\text{Bond Order} = \frac{10 - 4}{2} = 3

All electrons are paired, so $\mathrm{C}_2^{2-}$ is diamagnetic.

Among the given species, $\mathrm{NO}^+$ also has $14$ electrons, bond order $3$ , and is diamagnetic.

Therefore, the correct option is A, i.e. $\mathrm{NO}^+$ .

The solution shows option C, but the worked explanation and final answer clearly conclude $\mathrm{NO}^+$ , so the worked solution is taken as authoritative.

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