Question

Fill in the blanks with appropriate choice.
Bond order of $N_2^{+}$ is P while that of $N_2$ is Q .
Bond order of $O_2^{+}$ is R while that of $O_2$ is S .
$N—N$ bond distance T, when $N_2$ changes to $N_2^{+}$ and when $O_2$ changes to $O_2^{+}$, the $O—O$ bond distance U .

	P	Q	R	S	T	U
(a)	2	2.5	2.5	1	increases	decreases
(b)	2.5	3	2	1.5	decreases	increases
(c)	3	2	1.5	1	increases	decreases
(d)	2.5	3	2.5	2	increases	decreases

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (D)

$N_2^{+}:\left(\sigma 1s^2\right)\left({\sigma }^{\ast }1s^2\right)\left(\sigma 2s^2\right)\left({\sigma }^{\ast }2s^2\right)\left(\pi 2p_x^2=\pi 2p_y^2\right)\left(\sigma 2p_z^1\right)$
$B.O.=\frac{1}{2}\times (9-4)=2.5$
$N_2:B.O.=\frac{1}{2}\times (10-4)=3$
$O_2^{+}:\left(\sigma 1s^2\right)\left({\sigma }^{\ast }1s^2\right)\left(\sigma 2s^2\right)\left({\sigma }^{\ast }2s^2\right)\left(\sigma 2p_z^2\right)\left(\pi 2p_x^2=\pi 2p_y^2\right)$
$\left({\pi }^{\ast }2p_x^1\right)$
$B.O.=\frac{1}{2}\times (10-5)=2.5$
$O_2:B.O.=\frac{1}{2}\times (10-6)=2$
Since $N_2^{+}$ has lower bond order than $N_2$, bond length of $N—N$ in $N_2^{+}$ increases. In $O_2^{+}$ , bond order increases from $2$ to $2.5$ hence, bond length decreases.