Select the incorrect statement about N_2 mole | vedclass

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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.

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