Bond order of O_2 is | vedclass

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

Question

(A) The electronic configuration of $O_2$ ($16$ electrons) is: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) The electronic configuration of $O_2$ ($16$ electrons) is: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$. The number of bonding electrons $(N_b)$ $= 10$. The number of antibonding electrons $(N_a)$ $= 6$. Bond order $= \frac{1}{2} [N_b - N_a] = \frac{1}{2} [10 - 6] = 2$.

Bond order of $O_2$ is

Similar Questions

Bond order of $C_2$ vapour is :

Give features of Molecular Orbital $(MO)$ theory:

During the change of $O_2$ to $O_2^-$ ion,the electron adds to which one of the following orbitals?

The compound $(SiH_3)_3N$ is expected to be

The lowest $O-O$ bond length in the following molecules is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module