The correct option is A NO−
CO has 14 electrons. If we write down the electronic configuration, we will end up with σ1s2σ∗1s2σ2s2σ∗2s2π2p4σ2p2
NO− has 16 electrons. The difference here as compared with CO is that σ2p2 will be before π2p4 and there are 2 extra electrons. So, the final configuration will be σ1s2σ∗1s2σ2s2σ∗2s2σ2p2π2p4π∗2p2
In the other 3 cases, i.e. NO+, CN− and N2 we have 14 electrons. So, the electronic configurations of all these species will be the same as CO.
So, NO− is the answer.
Note: This question could have been directly solved just by looking at the number of electrons in each species. It is still a good practice to keep track of the molecular orbitals.