Question

Which of the following species are stabilized by addition of electron?

• A. $C_2$
• B. $CN$
• C. $N_2$
• D. $F_2$

Answer 1

Answer: (A), (B)

The correct options are
A $C_2$
B $CN$

(A) When an electron is added to $C_2$ molecule, the bond order increases from $2$ to $2.5$ as the electron is added into $\sigma \left(2p_x\right)$ orbital which is bonding in nature. This leads to stabilization. For $C_2$

$1s^2\ast 1s^{2}2s^2\ast 2s^{2}2p{x}^{2}2p{y}^2$ Bond Order = $\frac{8-4}{2}$ = $2$

(B) When an electron is added to $CN$ molecule, the bond order increases from $2.5$ to $3$ as the electron is added to $\pi \left(2p_z\right)$ orbital which is bonding in nature. This leads to stabilization.

(C) When an electron is added to $N_2$ molecule, the bond order decreases from $3$ to $2.5$ as the electron is added to ${\pi }^{\ast }\left(2p_y\right)$ orbital which is antibonding in nature. This leads to destabilization. For $N_2$

$1s^2\ast 1s^{2}2s^2\ast 2s^{2}2p{x}^{2}2p{y}^{2}2p{z}^2$

Bond Order = $\frac{6-0}{2}$ = $3$

(D) When an electron is added to $F_2$ molecule, the bond order decreases from $1$ to $0.5$ as the electron is added to ${\sigma }^{\ast }\left(2p_x\right)$ orbital which is antibinding in nature. This leads to destabilization. For $F_2$,

$1s^2\ast 1s^{2}2s^2\ast 2s^{2}2p{z}^{2}2p{x}^{2}2p{y}^2\ast 2p{x}^2\ast 2p{y}^2$ Bond Order = $\frac{6-4}{2}$ = $1$.