QJEEAdv16Ch18f9: JEE Advanced 2016 Chemistry - Class 12 JEEAdv16Ch18f9 Solution: Among [Ni(CO) 4 ], [NiCl 4 ] 2 , [Co(NH 3 ) 4 )Cl 2 ]Cl, Na 3…

For the given complexes:

In $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ : $\mathrm{Ni}$ (at. no. 28): $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^8 4 s^2$

$\mathrm{CO}$ being a strong field ligand causes the electrons to pair; hence, the complex diamagnetic.


$\text {In }\left[\mathrm{NiCl}_4\right]^{2-}: \mathrm{Ni}^{2+} \text { (at. No. 26): } 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^8$

There are two unpaired electrons, so the complex is paramagnetic.


$\begin{aligned}& \mathrm{In}\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}: \mathrm{Co}^{3+}\text {(at. No. 24): } 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^6\end{aligned}$


The ligand $\mathrm{NH}_3$ results in formation of low spin complex, which is diamagnetic.

$\text {In } \mathrm{Na}_3\left[\mathrm{CoF}_6\right]: \mathrm{Co}^{3+}\text {(at. No. 24): } 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^6$


$\mathrm{F}$ is a weak field ligand and results in the formation of high spin complex with four unpaired electrons, which is paramagnetic.

In $\mathrm{Na}_2 \mathrm{O}_2$ : The molecular orbital configuration of $\mathrm{O}_2^{2-}$ is

$\sigma 1 s^2, \sigma^* 1 \mathrm{~s}^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2,\left\{\begin{array}{l}\pi 2 p_x^2 \\\pi 2 p_y^2\end{array},\left\{\begin{array}{l}\pi * 2 p_x^2 \\\pi * 2 p_y^2\end{array}\right.\right.$

There is no unpaired electron, so it is diamagnetic.

In $\mathrm{CsO}_2$ : The molecular orbital configuration of $\mathrm{O}_2^{-}$ is

$\sigma 1 s^2, \sigma * 1 \mathrm{~s}^2, \sigma 2 s^2, \sigma * 2 s^2, \sigma 2 p_z^2,\left\{\begin{array}{l}\pi 2 p_x^2 \\\pi 2 p_y^2\end{array},\left\{\begin{array}{l}\pi * 2 p_x^2 \\\pi * 2 p_y^1\end{array}\right.\right.$

There is one unpaired electron, so it is paramagnetic. Hence, there are three paramagnetic compounds.