Which one of the following pairs of species have the same bond order?

For homonuclear and heteronuclear diatomic species, Molecular Orbital Theory (MOT) gives

Bond order $$= \dfrac{N_b - N_a}{2}$$, where $$N_b$$ is the number of electrons in bonding M.O.’s and $$N_a$$ is the number of electrons in antibonding M.O.’s.

Because the ordering of $$\sigma_{2p}$$ and $$\pi_{2p}$$ orbitals becomes regular (same as $$O_2$$) once the total electron count reaches $$14$$ or more, we can simply compare total electrons to known templates:

• $$10$$ e⁻ (like $$B_2$$) → B.O. $$=1$$     • $$12$$ e⁻ (like $$O_2^{2+}$$ or $$CN^{+}$$) → B.O. $$=2$$
• $$13$$ e⁻ (like $$CN$$) → B.O. $$=2.5$$
• $$14$$ e⁻ (like $$N_2,\,CO,\,CN^{-},\,NO^{+}$$) → B.O. $$=3$$
• $$15$$ e⁻ (like $$O_2^{+}$$) → B.O. $$=2.5$$
• $$16$$ e⁻ (like $$O_2$$) → B.O. $$=2$$
• $$17$$ e⁻ (like $$O_2^{-}$$) → B.O. $$=1.5$$

Hence only $$CN^{-}$$ and $$NO^{+}$$ possess the same bond order (three).

Option A which is: $$CN^{-}$$ and $$NO^{+}$$