Question

Which of the complex is paramagnetic ?

• A. $K_2\left[Ni{\left(CN\right)}_4\right]$
• B. $K_4\left[Fe{\left(CN\right)}_6\right]$
• C. $\left[Co{\left(N{H}_3\right)}_6\right]C{l}_3$
• D. $\left[Cr{\left(H_{2}O\right)}_6\right]S{O}_4$

Answer 1

The correct option is D $\left[Cr{\left(H_{2}O\right)}_6\right]S{O}_4$

Option (D) is correct.

$\left[Cr{\left(H_{2}O\right)}_6\right]S{O}_4$ is paramagnetic.
The electronic configurations of $Ni,Fe,Co$ and $Cr$ in the given complexes are ${\left(3d\right)}^8,{\left(3d\right)}^6,{\left(3d\right)}^6$ and ${\left(3d\right)}^4$ respectively.
The $t_{2g}$ and $e_g$ configuration for the complex ions are-

A) $Ni\left(II\right)\Rightarrow e_g^4,t_{2g}^4$ $\Rightarrow n=0$

B) $Fe\left(II\right)\Rightarrow t_{2g}^6,e_g^0$ $\Rightarrow n=0$

C) $Co\left(III\right)\Rightarrow t_{2g}^6{e}_g^0$ $\Rightarrow n=0$

D) $Cr\left(II\right)\Rightarrow t_{2g}^3{e}_g^1$ $\Rightarrow n=4$

Where n is the no. of unpaired electrons.

These involve $ds{p}^2,d^{2}s{p}^3,d^{2}s{p}^3$ and $s{p}^3{d}^2-$ hybridization respectively.

Ni, Fe, and Co contains strong field ligand. The complex $\left[Cr{\left(H_{2}O\right)}_6\right]S{O}_4$ will be paramagnetic as $Cr$ has weak field ligand as water. Thus contains four unpaired electrons and hence paramagnetic in nature.

Option D is the correct answer.