Question

Match the ionization processes listed in column-I with the changes observed as listed in column-II. For this, use the codes given below:

	Column-I		Column-II
(a)	$N_2\to N_2^{+}$	(p)	Bond order increases and magnetic property is changed
(b)	$O_2^{+}\to O_2^{2+}$	(q)	Bond order decreases and magnetic property is not changed
(c)	$B_2\to B_2^{+}$	(r)	Bond order increases and magnetic property is not changed
(d)	$N{O}^{-}\to NO$	(s)	Bond order decreases and magnetic property is changed

Note: Here change in magnetic property refers to change from diamagnetic to paramagnetic or paramagnetic to diamagnetic.

• A. A - s, B - p, C - r, D - q
• B. A - s, B - p, C - q, D - r
• C. A - r, B - q, C - s, D - p
• D. A - p, B - s, C - q, D - r

Answer 1

Answer: (B)

A - s, B - p, C - q, D - r

(a) $N_2=\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2=\pi 2p_y^2\sigma 2p_z^2$
$B.O=\frac{10-4}{2}=3;n=0\left(D\right)$
$N_2^{+}=\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2=\pi 2p_y^2\sigma 2p_z^1$
$B.O=\frac{9-4}{2}=2.5;n=1\left(P\right)$

It matches to (s).

(b) $O_2^{+}=\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2=\pi 2p_y^2{x}^{\ast }2p_x^1={\pi }^{\ast }2p_y^0$
$B.O=\frac{10-5}{2}=2.5;n=1\left(P\right)$
$O_2^{2+}=\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2=\pi 2p_y^2{\pi }^{\ast }2p_x^0={\pi }^{\ast }2p_y^0$
$B.O=\frac{10-4}{2}=3;n=0\left(D\right)$

It matches to (p).

(c) $B_2=\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^1=\pi 2p_y^1$
$B.O=\frac{6-4}{2}=1;n=2\left(P\right)$
$B_2^{+}=\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2=\pi 2p_y^1\sigma 2p_z^0$
$=\frac{5-4}{2}=1/2;n=1\left(P\right)$
It matches to (q).

(d) $N{O}^{-}=\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2\sigma 2p_z^2\pi 2p_x^2=\pi 2p_y^2{\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^1$
$B.O=\frac{10-6}{2}=2.0;n=2\left(P\right)$
$NO=\sigma 1s^2{\sigma }^{\ast }2s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2=\pi 2p_y^2{\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^0$
$B.O=\frac{10-5}{2}=2.5;n=1\left(P\right)$

It matches to (r).

Here, P and D means paramagnetic and diamagnetic respectively.

Therefore, the correct match is given by option B.