In which change do both the bond order and th | vedclass

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

Question

(D) Let us analyze the changes in bond order $(BO)$ and number of unpaired electrons $(n)$: $1$. $N_2 (14e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2

Vedclass · Accepted Answer

$O_2^{-} 	o O_2$

Vedclass · Answer

(D) Let us analyze the changes in bond order $(BO)$ and number of unpaired electrons $(n)$: $1$. $N_2 (14e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. $BO = 3, n = 0$. $N_2^{-2} (16e^-): \dots \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. $BO = 2, n = 2$. (Both decrease/increase differently) $2$. $B_2 (10e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1 = \pi 2p_y^1$. $BO = 1, n = 2$. $B_2^{-} (11e^-): \dots \pi 2p_x^2 = \pi 2p_y^1$. $BO = 1.5, n = 1$. $3$. $O_2^{-} (17e^-): \dots \sigma 2p_z^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. $BO = 1.5, n = 1$. $O_2^{2-} (18e^-): \dots \pi^* 2p_x^2 = \pi^* 2p_y^2$. $BO = 1, n = 0$. $4$. $O_2^{-} (17e^-): \dots \sigma 2p_z^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. $BO = 1.5, n = 1$. $O_2 (16e^-): \dots \sigma 2p_z^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. $BO = 2, n = 2$. In the change $O_2^{-} 	o O_2$,the bond order increases from $1.5$ to $2$ and the number of unpaired electrons increases from $1$ to $2$.

In which change do both the bond order and the number of unpaired electrons increase?

Similar Questions

Which of the following pairs of species have the same bond order?

The correct order of bond orders of $C_{2}^{2-}$,$N_{2}^{2-}$,and $O_{2}^{2-}$ is,respectively.

Which one of the following molecules is paramagnetic?

Compare the relative stability of the following species and indicate their magnetic properties: $O_2$,$O_2^+$,$O_2^-$ (superoxide),$O_2^{2-}$ (peroxide).

In which of the following is the $O-O$ bond distance minimum?

Vedclass Test Series

Exam Paper Generator

Online Exam Module