According to Molecular Orbital (MO) theory, the strength of a chemical bond is directly related to its bond order. A higher bond order indicates a stronger bond. The bond order is calculated as:

$\text{Bond Order}=\frac{1}{2}(\text{Number of electrons in bonding MOs}-\text{Number of electrons in antibonding MOs})$

When an electron is removed from a molecule, its bond order changes. If the removed electron comes from an antibonding molecular orbital, the bond order increases, leading to a stronger bond. If the removed electron comes from a bonding molecular orbital, the bond order decreases, leading to a weaker bond.

Let's analyze each molecule:

(A) NO (Nitric Oxide): Total electrons = 7 (N) + 8 (O) = 15 electrons.

MO configuration: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2p_z}^2({\pi }_{2p_x}^2={\pi }_{2p_y}^2){\pi }_{2p_x}^{\ast 1}$

Bond order = $\frac{1}{2}(10-5)=2.5$

The highest occupied molecular orbital (HOMO) is ${\pi }_{2p_x}^{\ast }$ (an antibonding orbital). Removing an electron from ${\pi }_{2p_x}^{\ast }$ will increase the bond order to $\frac{1}{2}(10-4)=3.0$. Thus, the bond becomes stronger.

(B) N${}_2$ (Nitrogen): Total electrons = 7 + 7 = 14 electrons.

MO configuration: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}({\pi }_{2p_x}^2={\pi }_{2p_y}^2){\sigma }_{2p_z}^2$

Bond order = $\frac{1}{2}(10-4)=3.0$

The HOMO is ${\sigma }_{2p_z}$ (a bonding orbital). Removing an electron from ${\sigma }_{2p_z}$ will decrease the bond order to $\frac{1}{2}(9-4)=2.5$. Thus, the bond becomes weaker.

(C) O${}_2$ (Oxygen): Total electrons = 8 + 8 = 16 electrons.

MO configuration: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2p_z}^2({\pi }_{2p_x}^2={\pi }_{2p_y}^2)({\pi }_{2p_x}^{\ast 1}={\pi }_{2p_y}^{\ast 1})$

Bond order = $\frac{1}{2}(10-6)=2.0$

The HOMO is ${\pi }_{2p_x}^{\ast }$ and ${\pi }_{2p_y}^{\ast }$ (antibonding orbitals). Removing an electron from an antibonding orbital will increase the bond order to $\frac{1}{2}(10-5)=2.5$. Thus, the bond becomes stronger.

(D) C${}_2$ (Carbon): Total electrons = 6 + 6 = 12 electrons.

MO configuration: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}({\pi }_{2p_x}^2={\pi }_{2p_y}^2)$

Bond order = $\frac{1}{2}(8-4)=2.0$

The HOMO is ${\pi }_{2p_x}$ and ${\pi }_{2p_y}$ (bonding orbitals). Removing an electron from a bonding orbital will decrease the bond order to $\frac{1}{2}(7-4)=1.5$. Thus, the bond becomes weaker.

(E) B${}_2$ (Boron): Total electrons = 5 + 5 = 10 electrons.

MO configuration: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}({\pi }_{2p_x}^1={\pi }_{2p_y}^1)$

Bond order = $\frac{1}{2}(6-4)=1.0$

The HOMO is ${\pi }_{2p_x}$ and ${\pi }_{2p_y}$ (bonding orbitals). Removing an electron from a bonding orbital will decrease the bond order to $\frac{1}{2}(5-4)=0.5$. Thus, the bond becomes weaker.

Therefore, for NO and O${}_2$, the bond becomes stronger upon removal of an electron.

Option Analysis:

• (A) (A), (B), (C) only: Incorrect, as N${}_2$ bond becomes weaker.
• (B) (B), (C), (E) only: Incorrect, as N${}_2$ and B${}_2$ bonds become weaker.
• (C) (A), (C) only: Correct, as NO and O${}_2$ bonds become stronger.
• (D) (D) only: Incorrect, as C${}_2$ bond becomes weaker.

Correct Answer: (C)