(c) : Number of electrons in each species are given below
N2 = 14 CN– = 14
O2– = 17 C22– = 14
NO+ = 14 O22– = 18
CO = 14 NO = 15
It is quite evident from the above that NO+,C22–, CN–, N2 and CO are isoelectronic in nature.
Hence option (c) is correct.