In which of the following ionization processe | vedclass

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

Question

(C) The bond order $(BO)$ is calculated as $\frac{1}{2}(N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibond

Vedclass · Accepted Answer

$NO \rightarrow NO^+$

Vedclass · Answer

(C) The bond order $(BO)$ is calculated as $\frac{1}{2}(N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.For $NO$ ($15$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. $BO = \frac{1}{2}(10 - 5) = 2.5$. It is paramagnetic.For $NO^+$ ($14$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. $BO = \frac{1}{2}(10 - 4) = 3.0$. It is diamagnetic.In the process $NO \rightarrow NO^+$,the electron is removed from the antibonding $\pi^*$ orbital,which increases the bond order from $2.5$ to $3.0$ and changes the magnetic behavior from paramagnetic to diamagnetic.

In which of the following ionization processes,the bond order has increased and the magnetic behaviour has changed?

Similar Questions

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

During the change of $O_{2}$ to $O_{2}^{-}$,the incoming electron goes to the orbital:

When two atomic orbitals combine,they form:

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

Arrange the following in the correct order of their bond orders.
$I. N_2, II. O_2, III. O_2^+, IV. O_2^-$

Vedclass Test Series

Exam Paper Generator

Online Exam Module