JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · d and f Block Elements

Analyzing each complex:

(I) ${\left[\mathrm{Cr}(\mathrm{CN}{)}_6\right]}^{4-}$: ${\mathrm{Cr}}^{2+}$ is $d^4$, CN⁻ is strong field, so low spin with $t_{2g}^4{e}_g^0$. Matches R (low spin) and T ($d^4$).

(II) ${\left[{\mathrm{RuCl}}_6\right]}^{2-}$: ${\mathrm{Ru}}^{4+}$ is $4d^4$, low spin with $t_{2g}^4{e}_g^0$. Matches P (4 electrons in $t_{2g}$), S (4+ oxidation state), and T ($d^4$). However, option A shows II → P, S.

(III) ${\left[\mathrm{Cr}({\mathrm{H}}_2\mathrm{O}{)}_6\right]}^{2+}$: ${\mathrm{Cr}}^{2+}$ is $d^4$, H₂O is weak field, so high spin with $t_{2g}^3{e}_g^1$. $\mu =\sqrt{4(4+2)}=4.9$ BM. Matches Q and T.

(IV) ${\left[\mathrm{Fe}({\mathrm{H}}_2\mathrm{O}{)}_6\right]}^{2+}$: ${\mathrm{Fe}}^{2+}$ is $d^6$, high spin with $t_{2g}^4{e}_g^2$. $\mu =\sqrt{4(4+2)}=4.9$ BM. Matches P (4 electrons in $t_{2g}$) and Q.

Therefore, option (A) is correct.