What is the bond order of O_2^- (in .5)? | vedclass

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

Question

(A) The total number of electrons in $O_2^-$ is $8 + 8 + 1 = 17$. According to Molecular Orbital Theory,the electronic configuration is: $KK \sigma 2s

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) The total number of electrons in $O_2^-$ is $8 + 8 + 1 = 17$. According to Molecular Orbital Theory,the electronic configuration is: $KK \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 = \pi 2p_y^2 \pi^* 2p_x^2 = \pi^* 2p_y^1$. Number of bonding electrons $(N_b)$ = $8$. Number of antibonding electrons $(N_a)$ = $7$. Bond Order $(B.O.)$ = $\frac{1}{2}(N_b - N_a) = \frac{8 - 7}{2} = 0.5$.

What is the bond order of $O_2^-$ (in $.5$)?

Similar Questions

Which statements are correct for the peroxide ion?

Which one of the following molecules is expected to exhibit diamagnetic behaviour?

The number of electrons present in bonding and antibonding orbitals in $O_2^{2-}$ is respectively

How do you express the bond strength in terms of bond order?

As per molecular orbital theory,the pair of molecules which do not exist is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module