Concept: The strength of a chemical bond is directly related to its bond order. According to Molecular Orbital (MO) theory, bond order is calculated as half the difference between the number of electrons in bonding molecular orbitals ($N_b$) and antibonding molecular orbitals ($N_a$). Removal of an electron can either increase or decrease the bond order, depending on whether the electron is removed from an antibonding or a bonding molecular orbital, respectively.

Why (C) is correct: We need to analyze the effect of removing an electron on the bond order for each molecule.

• (i) NO (Nitric Oxide): Total electrons = 7 (N) + 8 (O) = 15. The MO configuration is $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\sigma }_{2p_z}{)}^2({\pi }_{2p_x}{)}^2({\pi }_{2p_y}{)}^2({\pi }_{2p_x}^{\ast }{)}^1$. Bond order = $\frac{1}{2}(N_b-N_a)=\frac{1}{2}(10-5)=2.5$. The highest occupied molecular orbital (HOMO) is an antibonding orbital (${\pi }_{2p_x}^{\ast }$). Removing an electron from this antibonding orbital will increase the bond order to $\frac{1}{2}(10-4)=3.0$. Thus, bonding becomes stronger.
• (ii) N${}_2$ (Nitrogen): Total electrons = 7 + 7 = 14. The MO configuration is $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\pi }_{2p_x}{)}^2({\pi }_{2p_y}{)}^2({\sigma }_{2p_z}{)}^2$. Bond order = $\frac{1}{2}(10-4)=3.0$. The HOMO is a bonding orbital (${\sigma }_{2p_z}$). Removing an electron from this bonding orbital will decrease the bond order to $\frac{1}{2}(9-4)=2.5$. Thus, bonding becomes weaker.
• (iii) O${}_2$ (Oxygen): Total electrons = 8 + 8 = 16. The MO configuration is $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\sigma }_{2p_z}{)}^2({\pi }_{2p_x}{)}^2({\pi }_{2p_y}{)}^2({\pi }_{2p_x}^{\ast }{)}^1({\pi }_{2p_y}^{\ast }{)}^1$. Bond order = $\frac{1}{2}(10-6)=2.0$. The HOMO is an antibonding orbital (${\pi }_{2p_x}^{\ast }$ or ${\pi }_{2p_y}^{\ast }$). Removing an electron from this antibonding orbital will increase the bond order to $\frac{1}{2}(10-5)=2.5$. Thus, bonding becomes stronger.
• (iv) C${}_2$ (Carbon): Total electrons = 6 + 6 = 12. The MO configuration is $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\pi }_{2p_x}{)}^2({\pi }_{2p_y}{)}^2$. Bond order = $\frac{1}{2}(8-4)=2.0$. The HOMO is a bonding orbital (${\pi }_{2p_x}$ or ${\pi }_{2p_y}$). Removing an electron from this bonding orbital will decrease the bond order to $\frac{1}{2}(7-4)=1.5$. Thus, bonding becomes weaker.
• (v) B${}_2$ (Boron): Total electrons = 5 + 5 = 10. The MO configuration is $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\pi }_{2p_x}{)}^1({\pi }_{2p_y}{)}^1$. Bond order = $\frac{1}{2}(6-4)=1.0$. The HOMO is a bonding orbital (${\pi }_{2p_x}$ or ${\pi }_{2p_y}$). Removing an electron from this bonding orbital will decrease the bond order to $\frac{1}{2}(5-4)=0.5$. Thus, bonding becomes weaker.

Therefore, bonding becomes stronger for NO and O${}_2$.

Option Analysis:

• A) (i), (ii), (iii) only: Incorrect, as N${}_2$ bonding becomes weaker.
• B) (ii), (iii), (v) only: Incorrect, as N${}_2$ and B${}_2$ bonding becomes weaker.
• C) (i), (iii) only: Correct, as NO and O${}_2$ bonding becomes stronger.
• D) (iv) only: Incorrect, as C${}_2$ bonding becomes weaker.

Correct Answer: (C)