In which set of diatomic species the bond order is 2.5?

Bond order is one-half of the difference between the no.of electrons in the bonding molecular orbital and the no. of electrons in the antibonding molecular orbitals.

The total no.of electrons in ${\mathrm{N}}_2^{+},\mathrm{NO},\mathrm{CN}$ are 13, 15, and 13 respectively. 

$$\begin{gathered} \mathrm{NO}\left(15\right)=\mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{\ast }1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2{\mathrm{\sigma }}^{\ast }2{\mathrm{s}}^2 \mathrm{\pi }2{{\mathrm{p}}_{\mathrm{x}}}^2 \mathrm{\pi }2{{\mathrm{p}}_{\mathrm{y}}}^2 \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{x}}{ }^2 {\mathrm{\pi }}^{\ast }2{{\mathrm{p}}_{\mathrm{x}}}^1 \\ \begin{array}{rl}\mathrm{BO}& =\frac{10-5}{2}=2.5\\ \mathrm{CN}& = \mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{z}}^2 \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{x}}^1\\ \mathrm{BO}=\frac{9-4}{2}=2.5& \end{array} \\ {\mathrm{N}}_2^{+}\left(13{\mathrm{e}}^{-}\right)= \mathrm{\sigma }1\mathrm{s}² \mathrm{\sigma }*1\mathrm{s}² \mathrm{\sigma }2\mathrm{s}² \mathrm{\sigma }*2\mathrm{s}² \mathrm{\pi } 2{\mathrm{p}}_{\mathrm{y}}² \mathrm{\pi }2{\mathrm{p}}_{\mathrm{z}}² \mathrm{\sigma }2{{\mathrm{p}}_{\mathrm{x}}}^1 \\ \mathrm{BO}=\frac{9-4}{2}=2.5 \\ \mathrm{CN}\left(13{\mathrm{e}}^{-}\right): \mathrm{same} \mathrm{as} {\mathrm{N}}_2^{+} \end{gathered}$$

So, the bond order of ${\mathrm{N}}_2^{+},\mathrm{NO},\mathrm{CN}$ is 2.5.

So, option C) is correct.