Chemistry - Coordination Chemistry Question with Solution | TestHub

Identify $d^3$ configurations: Both $[Cr(N{H}_3{)}_6]C{l}_3$ ($C{r}^{3+}$ = $d^3$) and $[V(H_{2}O{)}_6]S{O}_4$ ($V^{2+}$ = $d^3$) have three unpaired electrons in their $t_{2g}$ orbitals.

Inner orbital formation: Since $d^3$ ions always have two empty $d$-orbitals available ($t_{2g}{}^3{e}_g{}^0$), they naturally undergo $d^{2}s{p}^3$ hybridization without requiring pairing.

Analyze other complexes: $[Ni(N{H}_3{)}_6{]}^{2+}$ ($d^8$) forms an outer orbital complex ($s{p}^3{d}^2$), while $K_2[PtC{l}_6]$ ($5d^6$) is completely diamagnetic due to strong pairing.

Conclusion: Only $[Cr(N{H}_3{)}_6]C{l}_3$ and $[V(H_{2}O{)}_6]S{O}_4$ satisfy both conditions simultaneously.