Question

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

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

Answer 1

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

$\left[Pd\right(PP{h}_3{)}_{2}C{l}_2]$ : Here $Pd$ is in $+2$ oxidation state and the configuration of $P{d}^{2+}$ is $\left[Kr\right]4d^8$. As the Crystal Field Stabilisation Energy (CFSE) value for $Pd$ is very high, all the electrons will be paired and hence the magnetic moment for this complex will be zero.

$\left[Ni\right(CO{)}_4]$: Here $Ni$ is in $0$ oxidation state and the configuration of $Ni$ is $\left[Ar\right]3d^{8}4s_2$. Here the ligand is carbonyl, which is a strong field ligand. So, all the electrons will be paired and hence the magnetic moment for this complex will be zero.

$\left[Ni\right(CN{)}_4{]}^{2-}$: Here $Ni$ is in $+2$ oxidation state and the configuration of $N{i}^{2+}$ is $\left[Ar\right]3d^8$. Here the ligand is cyanide, which is a strong field ligand. So, all the electrons will be paired and hence the magnetic moment for this complex will be zero.

$\left[Ni\right(H_{2}O{)}_6{]}^{2+}$: Here $Ni$ is in $+2$ oxidation state and the configuration of $N{i}^{2+}$ is $\left[Ar\right]3d^8$. Here the ligand is water, which is a weak field ligand. So, all the electrons will not be paired and there are two unpaired electrons in this complex. Hence, the magnetic moment for this complex will be $\sqrt{8}\text{BM}$.