Question

In which of the following ionisation processes has the bond order increased and the magnetic behaviour changed?

• A. $C_2\to C_2^{+}$
• B. $NO\to N{O}^{+}$
• C. $O_2\to O_2^{+}$
• D. $N_2\to N_2^{+}$

Answer 1

The correct option is B $NO\to N{O}^{+}$

(A) The electronic configuration of
$C_2=\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\pi }_{2p_x}{)}^2\right({\pi }_{2p_y}{)}^2$
$C_2^{+}=\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\pi }_{2p_x}{)}^2\right({\pi }_{2p_y}{)}^1$
Since one electron is removed from the bonding molecular orbital, bond order decreases. $C_2$ is diamagnetic and $C_2^{+}$ is paramagnetic in nature.

(B) The electronic configuration of
$NO=\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\pi }_{2p_x}{)}^2\right({\pi }_{2p_y}{)}^2\left({\sigma }_{2p_z}{)}^2\right({\pi }_{2p_x}^{\ast }{)}^1$
$N{O}^{+}=\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\pi }_{2p_x}{)}^2\right({\pi }_{2p_y}{)}^2({\sigma }_{2p_z}{)}^2$
Since one electron is removed from the antibonding molecular orbital, the bond order of $N{O}^{+}$ is more than that of $NO$. $N{O}^{+}$ does not have any unpaired electrons and so it is diamagnetic in nature, and $NO$ is paramagnatic in nature.

(C) The electronic configuration of
$O_2=\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\sigma }_{2p_z}{)}^2\right({\pi }_{2p_x}{)}^2\left({\pi }_{2p_y}{)}^2\right({\pi }_{2p_x}^{\ast }{)}^1({\pi }_{2p_y}^{\ast }{)}^1$
$O_2^{+}=\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\sigma }_{2p_z}{)}^2\right({\pi }_{2p_x}{)}^2\left({\pi }_{2p_y}{)}^2\right({\pi }_{2p_x}^{\ast }{)}^1$
Since one electron is removed from the antibonding molecular orbital, the bond order of $O_2^{+}$ is more compared to $O_2$, but both are paramagnetic in nature.

(D) The electronic configuration of
$N_2=\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\sigma }_{2p_z}{)}^2\right({\pi }_{2p_x}{)}^2({\pi }_{2p_y}{)}^2$
$N_2^{+}=\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\sigma }_{2p_z}{)}^2\right({\pi }_{2p_x}{)}^2({\pi }_{2p_y}{)}^1$
Since one electron is removed from the bonding molecular orbital, the bond order decreases. $N_2$ is diamagnetic and $N_2^{+}$ is paramagnetic in nature.