JEE Main 2025 Apr 2 Shift 2, Chemistry Q25: Colour, Magnetism And Stability Of Co-ordination Compounds

Step 1: Find $n$ by counting diamagnetic complexes.

${\mathrm{K}}_2[\mathrm{NiC}{\mathrm{l}}_4]$: $\mathrm{N}{\mathrm{i}}^{2+}$ ($d^8$) with $\mathrm{C}{\mathrm{l}}^{-}$ (weak field) - tetrahedral, $s{p}^3$, two unpaired electrons $\Rightarrow$ paramagnetic.

$[\mathrm{Zn}({\mathrm{H}}_2\mathrm{O}{)}_6]\mathrm{C}{\mathrm{l}}_2$: $\mathrm{Z}{\mathrm{n}}^{2+}$ is $d^{10}$, all electrons paired $\Rightarrow$ diamagnetic.

${\mathrm{K}}_3[\mathrm{Mn}(\mathrm{CN}{)}_6]$: $\mathrm{M}{\mathrm{n}}^{3+}$ ($d^4$) with strong-field $\mathrm{C}{\mathrm{N}}^{-}$ - low-spin $t_{2g}^4$, two unpaired electrons $\Rightarrow$ paramagnetic.

$[\mathrm{Cu}(\mathrm{PP}{\mathrm{h}}_3{)}_3\mathrm{I}]$: $\mathrm{C}{\mathrm{u}}^{+}$ is $d^{10}$, all electrons paired $\Rightarrow$ diamagnetic.

Number of diamagnetic complexes $=2$, so $n=2$.

Step 2: Identify M with the least enthalpy of atomisation. Among Ni, Zn, Mn and Cu, Zn has the lowest enthalpy of atomisation (its fully filled $3d^{10}4s^2$ configuration contributes no unpaired electrons to metallic bonding).

Step 3: Spin-only moment of ${\mathrm{Zn}}^{2+}$. ${\mathrm{Zn}}^{2+}$: $[\mathrm{Ar}]3d^{10}$, so the number of unpaired electrons is $0$.

${\mu }_{\text{spin-only}}=\sqrt{0(0+2)}=0{\mu }_B$.

$\therefore$ Nearest integer $=0$.