Two pi and half sigma bonds are present in | vedclass

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

Question

(D) The bond order $(B.O.)$ is calculated using the formula: $B.O. = \frac{1}{2} (N_b - N_a)$.For $N_2^+$,the total electrons are $13$. The molecular

Vedclass · Accepted Answer

$N_2^+$

Vedclass · Answer

(D) The bond order $(B.O.)$ is calculated using the formula: $B.O. = \frac{1}{2} (N_b - N_a)$.For $N_2^+$,the total electrons are $13$. The molecular orbital configuration is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^1$.Here,$N_b = 9$ and $N_a = 4$. So,$B.O. = \frac{1}{2} (9 - 4) = 2.5$.In $N_2^+$,the triple bond structure of $N_2$ is modified by removing one electron from the bonding $\sigma 2p_z$ orbital,resulting in $2 \ \pi$ bonds and $0.5 \ \sigma$ bond.

Two $\pi$ and half $\sigma$ bonds are present in

Similar Questions

The number of antibonding electron pairs in $O_{2}^{2-}$ molecular ion on the basis of molecular orbital theory is (Atomic number of $O$ is $8$)

According to molecular orbital theory,the species among the following that does not exist is

The calculated bond order of superoxide ion $(O_{2}^{-})$ is

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

Thermal decomposition of $AgNO_3$ produces two paramagnetic gases. The total number of electrons present in the antibonding molecular orbitals of the gas that has the higher number of unpaired electrons is. . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module