WBJEE 2026 Physics and Chemistry Paper

Concept:The complex ion is $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right]^{+}$; find its coordination number, oxidation state, $d$-electron count, and unpaired electrons.Explanation:Bromide $\mathrm{Br}^{-}$ lies outside the coordination sphere, so the complex ion is $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right]^{+}$.$\mathrm{NH}_{3}$ is neutral and $\mathrm{SO}_{4}^{2-}$ contributes $-2$.Let the oxidation state of $\mathrm{Co}$ be $x$.$$x + 0 + (-2) = +1 \Rightarrow x = +3.$$$\mathrm{Co}$ has atomic number 27, so $\mathrm{Co}^{3+}$ has the configuration $3d^6$.In an octahedral field with strong ligands like $\mathrm{NH}_{3}$ and $\mathrm{SO}_{4}^{2-}$, $3d^6$ gives $t_{2g}^6 e_g^0$, hence no unpaired electrons.Coordination number: $5 + 1 = 6$.Answer:Coordination number = 6, oxidation number = 3, number of $d$ electrons = 6, number of unpaired $d$ electrons = 0.Correct option: A.