Chemistry - CHEMICAL BONDING Question with Solution | TestHub

To determine the species with a different bond order than CO, we first find the bond order of CO.

CO has 14 electrons (6 from C + 8 from O).

Its molecular orbital configuration is ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\pi }_{2p}^4{\sigma }_{2p}^2$.

Bond order = $\frac{1}{2}$ (Number of bonding electrons - Number of antibonding electrons)

Bond order of CO = $\frac{1}{2}(10-4)=\frac{1}{2}(6)=3$.

Now, let's calculate the bond order for each option:

1. ${\mathrm{NO}}^{-}$: 7 (N) + 8 (O) + 1 (charge) = 16 electrons.

Configuration: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2p}^2{\pi }_{2p}^4{\pi }_{2p}^{\ast 2}$.

Bond order = $\frac{1}{2}(10-6)=\frac{1}{2}(4)=2$.

2. ${\mathrm{NO}}^{+}$: 7 (N) + 8 (O) - 1 (charge) = 14 electrons.

This is isoelectronic with CO.

Bond order = $\frac{1}{2}(10-4)=3$.

3. ${\mathrm{CN}}^{-}$: 6 (C) + 7 (N) + 1 (charge) = 14 electrons.

This is isoelectronic with CO.

Bond order = $\frac{1}{2}(10-4)=3$.

4. ${\mathrm{N}}_2$: 7 (N) + 7 (N) = 14 electrons.

This is isoelectronic with CO.

Bond order = $\frac{1}{2}(10-4)=3$.

The species having a bond order different from that in CO is ${\mathrm{NO}}^{-}$, with a bond order of 2.

The final answer is $\boxed{N{O}^{-}}$.