Which one of the following constitutes a group of the isoelectronic species?

Isoelectronic species are those chemical species (atoms, ions or molecules) that possess the same total number of electrons.

Step 1 Write the atomic numbers: C = 6, N = 7, O = 8.

Step 2 Compute the total electrons for every species in each option.
Extra electrons are added for a negative charge; electrons are removed for a positive charge.

$$C_2^{2-}: 2\times6+2 = 14$$
$$O_2^{-}: 2\times8+1 = 17$$
$$CO: 6+8 = 14$$
$$NO: 7+8 = 15$$
Different values → not isoelectronic.

$$NO^{+}: 7+8-1 = 14$$
$$C_2^{2-}: 2\times6+2 = 14$$
$$CN^{-}: 6+7+1 = 14$$
$$N_2: 2\times7 = 14$$
All have 14 electrons → isoelectronic group.

$$CN^{-}: 14$$  (from Case B)
$$N_2: 14$$
$$O_2^{2-}: 2\times8+2 = 18$$
$$C_2^{2-}: 14$$
Mixed values → not isoelectronic.

$$N_2: 14$$
$$O_2^{-}: 17$$
$$NO^{+}: 14$$
$$CO: 14$$
Mixed values → not isoelectronic.

Only the set in Option B has the same electron count (14) for every species.

Final Answer: Option B which is: $$NO^{+},\; C_2^{2-},\; CN^{-},\; N_2$$