Which of the following is paramagnetic ?

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

Question

$\mathrm{CN}^{-}, \mathrm{CO}$ and $\mathrm{NO}^{+}$ are isoelectronic with 14 electrons each and there is no unpaired electrons in the MO configurati

Vedclass · Accepted Answer

$O_{2}^-$

Vedclass · Answer

$\mathrm{CN}^{-}, \mathrm{CO}$ and $\mathrm{NO}^{+}$ are isoelectronic with 14 electrons each and there is no unpaired electrons in the MO configuration of these species.

So these are diamagnetic. $\mathrm{O}_{2}^{*}$ is paramagnetic due to the presence of one unpaired electron.

$\left(\mathrm{CN}^{-}, \mathrm{CO}, \mathrm{NO}^{+}\right) 14=\sigma \mathrm{s}^{2}, \sigma^{*} \mathrm{L}^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2} \sigma 2 p_{x}^{2}$

$\pi 2 p_{y}^{2} \approx \pi 2 p_{z}^{2} \quad$ (No unpaired electron, diamagnetic)

$\mathrm{O}_{2}^{-}=\sigma \mathrm{s}^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{x}^{2}$

$\pi 2 p_{x}^{2} \approx \pi 2 p_{z}^{2}, \pi^{*} 2 p_{y}^{2}=\pi^{*} 2 p_{z}^{1}$

(one unpaired electron, paramagnetic)

Similar Questions

Consider the ions/molecule

$O _{2}^{+}, O _{2}, O _{2}^{-}, O _{2}^{2-}$

For increasing bond order the correct option is ..... .

According to the molecular orbital theory, the bond order in ${C_2}$ molecule is

State resemblance and differentence - both sigma and pai bond

In which of the following process, the value of bond order & magnetic nature does not change?