Select the incorrect statement about N_2 mole | vedclass

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

Question

(D) The electronic configuration of $N_2$ ($14$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma

Vedclass · Accepted Answer

It is diamagnetic.

Vedclass · Answer

(D) The electronic configuration of $N_2$ ($14$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. Bond order = $\frac{1}{2}(N_b - N_a) = \frac{1}{2}(10 - 4) = 3$. Since all electrons are paired,$N_2$ is diamagnetic. $O_2$ ($16$ electrons) has a bond order of $2$ and is paramagnetic due to two unpaired electrons in $\pi^* 2p$ orbitals. $N_2$ has $10$ bonding electrons,while $O_2$ has $8$ bonding electrons. Therefore,the statement '$N_2$ is diamagnetic' is correct,and the statement '$N_2$ is paramagnetic' would be incorrect. Given the options,the statement '$N_2$ is diamagnetic' is correct,but if the option was written as 'It is dimagnetic' (a typo for diamagnetic),it is still the intended correct property. However,all statements $A$,$B$,$C$,and $D$ are actually correct descriptions of $N_2$. If we assume the question asks for the incorrect statement,there might be a typo in the provided options. Based on standard chemistry,$N_2$ is diamagnetic,so option $D$ is a correct statement.

Select the incorrect statement about $N_2$ molecule.

Similar Questions

Why is $KO_{2}$ paramagnetic?

Which of the following pairs of species have unpaired electrons in antibonding Molecular Orbital?

Which of the following is not paramagnetic?

How many molecular orbitals filled with one or more electrons are present in $N_2^+$?

Which of the following orbitals cannot form a $\pi$-bond as well as a $\delta$-bond?

Vedclass Test Series

Exam Paper Generator

Online Exam Module