Of the following sets which one does NOT contain isoelectronic species?

For a set of species to be isoelectronic, every member must possess the same total number of electrons.

Atomic numbers required:
$$B=5,\; C=6,\; N=7,\; O=8,\; P=15,\; S=16,\; Cl=17$$

Case A: $$PO_4^{3-},\; SO_4^{2-},\; ClO_4^-$$

Total electrons

$$PO_4^{3-}: 15 + 4(8) + 3 = 50$$
$$SO_4^{2-}: 16 + 4(8) + 2 = 50$$
$$ClO_4^- : 17 + 4(8) + 1 = 50$$

All three have $$50$$ electrons ⇒ isoelectronic.

Case B: $$CN^- ,\; N_2 ,\; C_2^{2-}$$

$$CN^- : 6 + 7 + 1 = 14$$
$$N_2 : 2(7) = 14$$
$$C_2^{2-}: 2(6) + 2 = 14$$

All three have $$14$$ electrons ⇒ isoelectronic.

Case C: $$SO_3^{2-},\; CO_3^{2-},\; NO_3^-$$

$$SO_3^{2-}: 16 + 3(8) + 2 = 16 + 24 + 2 = 42$$
$$CO_3^{2-}: 6 + 3(8) + 2 = 6 + 24 + 2 = 32$$
$$NO_3^- : 7 + 3(8) + 1 = 7 + 24 + 1 = 32$$

The first ion has $$42$$ electrons whereas the other two have $$32$$ electrons. Hence the trio is not isoelectronic.

Case D: $$BO_3^{3-},\; CO_3^{2-},\; NO_3^-$$

$$BO_3^{3-}: 5 + 3(8) + 3 = 5 + 24 + 3 = 32$$
$$CO_3^{2-}: 6 + 24 + 2 = 32$$
$$NO_3^- : 7 + 24 + 1 = 32$$

All three have $$32$$ electrons ⇒ isoelectronic.

Therefore, among the given sets, the only one that does not contain isoelectronic species is:

Option C which is: $$SO_3^{2-},\; CO_3^{2-},\; NO_3^-$$