JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Coordination Compounds

$[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_6{]}^{3+}$ has Co${}^{3+}$ ($3d^6$); the strong-field NH${}_3$ pairs the $d$ electrons and frees two inner $3d$ orbitals, giving an inner-orbital $d^{2}s{p}^3$ complex. $[\mathrm{MnC}{\mathrm{l}}_6{]}^{3-}$ has Mn${}^{3+}$ ($3d^4$) with weak-field Cl${}^{-}$, so no pairing occurs and the outer $4d$ orbitals are used, giving an outer-orbital $s{p}^3{d}^2$ complex; $[\mathrm{Fe}{\mathrm{F}}_6{]}^{3-}$ (Fe${}^{3+}$, $3d^5$, weak-field F${}^{-}$) is likewise outer-orbital $s{p}^3{d}^2$. $[\mathrm{Co}{\mathrm{F}}_6{]}^{3-}$ with weak-field F${}^{-}$ is outer-orbital $s{p}^3{d}^2$, not $d^{2}s{p}^3$, so its description is wrong, and $[\mathrm{Ni}(\mathrm{CN}{)}_4{]}^{2-}$ (Ni${}^{2+}$, $3d^8$, strong-field CN${}^{-}$) is square planar $ds{p}^2$, not $s{p}^3$. The correctly described species are $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_6{]}^{3+}$, $[\mathrm{MnC}{\mathrm{l}}_6{]}^{3-}$ and $[\mathrm{Fe}{\mathrm{F}}_6{]}^{3-}$.