Question

The correct order of the calculated spin-only magnetic moments of complexes $\left(A\right)$ to $\left(D\right)$ is:
$\left(A\right)Ni(CO{)}_4$
$\left(B\right)\left[Ni\right(H_{2}O{)}_6]C{l}_2$
$\left(C\right)N{a}_2\left[Ni\right(CN{)}_4]$
$\left(D\right)PdC{l}_2(PP{h}_3{)}_2$

• A. $\left(C\right)\approx \left(D\right)<\left(B\right)<\left(A\right)$
• B. $\left(A\right)\approx \left(C\right)\approx \left(D\right)<\left(B\right)$
• C. $\left(C\right)<\left(D\right)<\left(B\right)<\left(A\right)$
• D. $\left(C\right)\approx \left(D\right)<\left(B\right)<\left(A\right)$

Answer 1

The correct option is B $\left(A\right)\approx \left(C\right)\approx \left(D\right)<\left(B\right)$

$\left(A\right)Ni(CO{)}_4$
$Ni$ is in the zero oxidation state i.e., it has a configuration of $3d^{8}4s^2.$ Since $CO$ is a strong field ligand, it causes the pairing of unpaired $3d$ electrons. So, number of unpaired electron is zero thus it is diamagnetic.
$\mu =0\text{B.M};\text{diamagnetic.}$

$\left(B\right)\left[Ni\right(H_{2}O{)}_6]C{l}_2$
$\left[Ni\right(H_{2}O{)}_6{]}^{2+}$
$N{i}^{2+}=3d^8$ with weak field ligand.
$\therefore \text{Number of unpaired}{e}^{-}=2$
$\mu =\sqrt{n(n+2)}$
$=\sqrt{2(2+2)}=2\sqrt{2}\text{B.M.}$

$\left(C\right)N{a}_2\left[Ni\right(CN{)}_4]$
$\left[Ni\right(CN{)}_4{]}^{2-}$
$N{i}^{2+}=3d^8$ with strong field ligand $C{N}^{-}$. so it will pair up $d$ electrons.
$\text{Number of unpaired}{e}^{-}=0$
$\therefore \mu =0\text{B.M.}$
$PdC{l}_2(PP{h}_3{)}_2$
$P{d}^{2+}=4d^8\left(\text{diamagnetic}\right)$
$\mu =0\text{B.M.}$

$\left(D\right)PdC{l}_2(PP{h}_3{)}_2$
$Pd$ is in the $+2$ oxidation state i.e., it has a configuration of $4d^{8}5s^0.$ Since $\left(PP{h}_3\right)$ is a strong field ligand, it causes the pairing of unpaired $3d$ electrons. so number of unpaired electron is zero thus it is diamagnetic.
$\mu =0\text{B.M.};\text{diamagnetic.}$

$\text{Hence, the correct order is}A\approx C\approx D<B$
$\text{Therefore, the correct option is D.}$