Question

Pick out the incorrect statement?

• A. $N_2$ has greater dissociation energy than $N_2^{+}$
• B. $O_2$ has lower dissociation energy than $O_2^{+}$
• C. Bond length in $N_2^{+}$ is less than $N_2$
• D. Bond length in $N{O}^{+}$ is less than in $NO$

Answer 1

The correct option is C Bond length in $N_2^{+}$ is less than $N_2$

(a) The electronic configuration of
$N_2=\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\pi }_{2p_x}{)}^2\right({\pi }_{2p_y}{)}^2({\sigma }_{2p_z}{)}^2$
Thus, electronic configuration of
$N_2^{+}=\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\pi }_{2p_x}{)}^2\right({\pi }_{2p_y}{)}^2({\sigma }_{2p_z}{)}^2$

So bond order of $N_2=\frac{10-4}{2}=3$
Bond order of $N_2^{+}=\frac{9-5}{2}=2.5$

Since bond order of $N_2$ is more compared to $N_2^{+}$, thus, thus $N_2$ has greater dissociation energy than $N_2^{+}$
(b) The electronic configuration of
$O_2=\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\sigma }_{2p_z}{)}^2\right({\pi }_{2p_x}{)}^2\left({\pi }_{2p_y}{)}^2\right({\pi }_{2p_x}^{\ast }{)}^1({\pi }_{2p_y}^{\ast }{)}^1$
The electronic configuration of
$O_2^{+}=\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\sigma }_{2p_z}{)}^2\right({\pi }_{2p_x}{)}^2\left({\pi }_{2p_y}{)}^2\right({\pi }_{2p_x}^{\ast }{)}^1$
So the bond order of $O_2=\frac{10-6}{2}=2$
Bond order of $O_2^{+}=\frac{10-5}{2}=2.5$
Since bond order of $O_2$ is less than that of $O_2^{+}$, thus $O_2$ has lower dissociation energy than $O_2^{+}$.

(c) Bond order of $N_2$ is more compared to $N_2^{+}$ and again we know, bond length $\propto \frac{1}{\text{bond order}}$ .
SO bond length of $N_2^{+}$ is more than that of $N_2$.
(d) The electronic confiuration of
$NO=\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\pi }_{2p_x}{)}^2\right({\pi }_{2p_y}{)}^2\left({\sigma }_{2p_z}{)}^2\right({\pi }_{2p_x}^{\ast }{)}^1$
The electronic confiuration of
$N{O}^{+}=\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\pi }_{2p_x}{)}^2\right({\pi }_{2p_y}{)}^2({\sigma }_{2p_z}{)}^2$

The bond order of $NO=\frac{10-5}{2}=2.5$
​​​​​​​The bond order of $N{O}^{+}=\frac{10-4}{2}=3$

Since bond order of $N{O}^{+}$ is more, thus bond length in $N{O}^{+}$ is less than in $NO$