Bonding in which of the following diatomic molecule(s) become(s) stronger, on the basis of MO Theory, by removal of an electron?
(A) NO
(B) $$N_2$$
(C) $$O_2$$
(D) $$C_2$$
(E) $$B_2$$
Choose the most appropriate answer from the options given below :

A molecule's bond becomes stronger when an electron is removed if that electron was in an antibonding molecular orbital, since removing it increases the bond order.

Bond order = $$\frac{1}{2}$$(bonding electrons - antibonding electrons)

Let us analyze each molecule:

(A) NO: Total electrons = 15. Electronic configuration ends with one electron in $$\pi^*_{2p}$$ (antibonding). Removing this antibonding electron increases bond order from 2.5 to 3. Bond becomes stronger.

(B) $$N_2$$: Total electrons = 14. All bonding MOs are filled up to $$\sigma_{2p}$$, no antibonding electrons in the highest occupied MO. The last electron is in $$\sigma_{2p}$$ (bonding). Removing it decreases bond order from 3 to 2.5. Bond becomes weaker.

(C) $$O_2$$: Total electrons = 16. Has 2 electrons in $$\pi^*_{2p}$$ (antibonding). Removing one antibonding electron increases bond order from 2 to 2.5. Bond becomes stronger.

(D) $$C_2$$: Total electrons = 12. Last electrons are in $$\pi_{2p}$$ (bonding). Removing a bonding electron decreases bond order from 2 to 1.5. Bond becomes weaker.

(E) $$B_2$$: Total electrons = 10. Last electrons are in $$\pi_{2p}$$ (bonding). Removing a bonding electron decreases bond order from 1 to 0.5. Bond becomes weaker.

Only NO and $$O_2$$ have their highest occupied electrons in antibonding orbitals, so only A and C become stronger upon electron removal.

The correct answer is Option C: A, C only.