# The correct order of the calculated spin-only magnetic moments of complexes (A) to (D) is : (A) $\inline Ni(CO)_{4}$ (B) $\inline \left [ Ni(H_{2}O)_{6} \right ]Cl_{2}$ (C) $\inline Na_{2}\left [ Ni(CN)_{4} \right ]$ (D) $\inline PdCl_{2}(PPh_{3})_{2}$ Option: 1      Option: 2  Option: 3  Option: 4

As we have learnt,

$[Ni(CO)_4]$ contains $Ni(0)$ which has a $d^{10}$ configuration having $n=0$ and $\mu=0$

$[Ni(H_2O)_6]^{2+}$contains $Ni^{2+}$ which has a $d^8$ configuration. In a weak ligand field, it has $n=2$ and hence has $\mu=\sqrt8 \ BM$

$[Ni(CN)]_4$contains $Ni^{2+}$which has a $d^8$ configuration. In a strong ligand field, these electrons get paired up and hence $n=0$ and $\mu =0$

$[Pd(PPh_3)_2Cl_2]$ contains $Pd^{2+}$which has a $d^8$ configuration. These electrons get paired in the presence of strong ligand like $PPh_3$ and hence, $n=0$ and correspondingly $\mu=0$.

Therefore, the correct order of spin only magnetic moment of the complexes is

$(B)>(A)=(C)=(D)$

Therefore, Option(4) is correct.

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