In which of the following processes,the bond | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The electronic configuration of $NO$ ($15$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi

Vedclass · Accepted Answer

$NO \rightarrow NO^{+}$

Vedclass · Answer

(B) The electronic configuration of $NO$ ($15$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. It has one unpaired electron,so it is paramagnetic. Bond order $= (10-5)/2 = 2.5$. After losing one electron,$NO^{+}$ ($14$ electrons) has the configuration $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. It has no unpaired electrons,so it is diamagnetic. Bond order $= (10-4)/2 = 3$. Thus,in the process $NO \rightarrow NO^{+}$,the bond order increases from $2.5$ to $3$ and the character changes from paramagnetic to diamagnetic.

Column $I$	Column $II$
$(A)$ $B_2$	$(p)$ Paramagnetic
$(B)$ $N_2$	$(q)$ Undergoes oxidation
$(C)$ $O_2^{-}$	$(r)$ Undergoes reduction
$(D)$ $O_2$	$(s)$ Bond order $\geq 2$
	$(t)$ Mixing of '$s$' and '$p$' orbitals

In which of the following processes,the bond order increases and paramagnetic character changes to diamagnetic one?

Similar Questions

The correct order of bond length is :-

Among $KO_2$,$NO_2^-$,$BaO_2$ and $NO_2^+$,an unpaired electron is present in:

The correct statement regarding $SO_2$ molecule is

From elementary molecular orbital theory,the electronic configuration of the singly positive nitrogen molecular ion $N_2^+$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module