The correct order of the calculated spin-only magnetic moments of complexes $\left(\mathrm{A}\right)$ to $\left(\mathrm{D}\right)$ is

$\left(\mathrm{a}\right)$ $\text{Ni}(\text{CO}{)}_4$ $\left(\mathrm{b}\right)$ $[\text{Ni}\left({\mathrm{H}}_2\mathrm{O}{)}_6\right]\text{C}{\text{l}}_2$ $\left(\mathrm{c}\right)$ $\text{N}{\text{a}}_2\left[\text{Ni}\right(\text{CN}{)}_4]$ $\left(\mathrm{d}\right)$ $\text{PdC}{\text{l}}_2{\left(\text{PP}{\text{h}}_3\right)}_2$

• A

  $\left(\mathrm{c}\right)<\left(\mathrm{d}\right)<\left(\mathrm{b}\right)<\left(\mathrm{a}\right)$
• B

  $\left(\mathrm{a}\right)\approx \left(\mathrm{c}\right)<\left(\mathrm{b}\right)\approx \left(\mathrm{d}\right)$
• Correct answer
  C

  $\left(\mathrm{a}\right)\approx \left(\mathrm{c}\right)\approx \left(\mathrm{d}\right)<\left(\mathrm{b}\right)$
• D

  $\left(\mathrm{c}\right)\approx \left(\mathrm{d}\right)<\left(\mathrm{b}\right)<\left(\mathrm{a}\right)$

Solution

$\left(\mathrm{a}\right)$ $\text{Ni}(\text{CO}{)}_4$

$\text{Ni}={\text{3d}}^{10}{\text{4s}}^0\left[\text{S F L}\right]\text{ ;}\text{μ}=0$

$\left(\mathrm{b}\right)$ $[\text{Ni}\left({\mathrm{H}}_2\mathrm{O}{)}_6\right]\text{C}{\text{l}}_2$

${\text{Ni}}^{2+}={\text{3d}}^8$

Weak field ligand



No. of unpaired electron $=2$

$\mathrm{\mu }=\sqrt{\mathrm{n}\left(\mathrm{n}+2\right)}=\sqrt{2\left(2+2\right)}=\sqrt{8}\mathrm{ }\mathrm{B}.\mathrm{M}$

$\left(\mathrm{c}\right)$ $\text{N}{\text{a}}_2[\text{Ni}\left(\text{CN}{)}_4\right]$

$\text{N}{\text{i}}^{2+}–\text{3}{\text{d}}^8\left(\text{S F L}\right)$

pairing take place (n = 0)

$\mathrm{\mu }=0$

$\left(\mathrm{d}\right)$ $\text{PdC}{\text{l}}_2{\left(\text{PP}{\text{h}}_3\right)}_2$

$\text{P}{\text{d}}^{2+}=\text{4}{\text{d}}^8$

square planar

$\mathrm{\mu }=0$

Hence, the correct answer is option: 'C'.