Idea of isoelectronic species

Isoelectronic species have the same total number of electrons.

Count electrons in each species

Atomic numbers:

• $C = 6$
• $N = 7$
• $O = 8$

Check each option

Option A: $C_2^{2-}, O_2^{-}, CO, NO$

• $C_2^{2-}$: $2(6) + 2 = 14$
• $O_2^{-}$: $2(8) + 1 = 17$
• $CO$: $6 + 8 = 14$
• $NO$: $7 + 8 = 15$

Electron counts are $14, 17, 14, 15$.

Not all same, so A is incorrect.

Option B: $NO^{+}, C_2^{2-}, CN^-, N_2$

• $NO^{+}$: $7 + 8 - 1 = 14$
• $C_2^{2-}$: $2(6) + 2 = 14$
• $CN^-$: $6 + 7 + 1 = 14$
• $N_2$: $2(7) = 14$

Electron counts are $14, 14, 14, 14$.

All are same, so B is correct.

Option C: $CN^-, N_2, O_2^{2-}, C_2^{2-}$

• $CN^-$: $6 + 7 + 1 = 14$
• $N_2$: $2(7) = 14$
• $O_2^{2-}$: $2(8) + 2 = 18$
• $C_2^{2-}$: $2(6) + 2 = 14$

Electron counts are $14, 14, 18, 14$.

Not all same, so C is incorrect.

Option D: $N_2, O_2^{-}, NO^{+}, CO$

• $N_2$: $14$
• $O_2^{-}$: $17$
• $NO^{+}$: $14$
• $CO$: $14$

Electron counts are $14, 17, 14, 14$.

Not all same, so D is incorrect.

Final answer

The group of isoelectronic species is: $\boxed{B}$

Comparison with stored answer

Stored correct answer is B, which matches my result.