Identify the group-IV cation from the reactions

Group IV cations in classical qualitative analysis include ions like $\mathrm{Ni^{2+}}, \mathrm{Co^{2+}}, \mathrm{Zn^{2+}}, \mathrm{Mn^{2+}}$.

The first reaction is: $\mathrm{M^{2+} + H_2S \rightarrow A\;(black\ precipitate)}$

A black sulfide suggests either $\mathrm{NiS}$ or $\mathrm{CoS}$.

Use the aqua regia reaction

$\mathrm{A + aqua\ regia \rightarrow B + NOCl + S + H_2O}$

Sulfide dissolves in aqua regia, and the sulfide sulfur is oxidized to elemental sulfur. A very characteristic confirmatory sequence is for cobalt:

• $\mathrm{CoS}$ is black.
• On treatment with aqua regia, cobalt goes into solution as $\mathrm{Co^{2+}}$ (chloride medium).
• Then with $\mathrm{KNO_2}$ and acetic acid, cobalt forms potassium cobaltinitrite.

Identify complex $\mathrm{C}$

The reaction of cobalt(II) solution with $\mathrm{KNO_2}$ in acetic acid gives yellow precipitate of potassium hexanitritocobaltate(III): $\mathrm{K_3[Co(NO_2)_6]}$

So, $\mathrm{C = [Co(NO_2)_6]^{3-}}$ (present as its potassium salt).

Find oxidation state and d-electron count of Co in $C$

Let oxidation state of Co be $x$: $x + 6(-1) = -3$ $x = +3$

Therefore metal ion is $\mathrm{Co^{3+}}$.

Atomic number of Co = 27 $\mathrm{Co: [Ar]3d^74s^2}$ $\mathrm{Co^{3+}: 3d^6}$

Determine spin state

In $[\mathrm{Co(NO_2)_6}]^{3-}$, $\mathrm{NO_2^-}$ is a strong-field ligand. Thus for octahedral $\mathrm{Co^{3+}}$ ($d^6$), we get low-spin configuration: $t_{2g}^6 e_g^0$

Number of unpaired electrons: $n = 0$

Calculate spin-only magnetic moment

$\mu = \sqrt{n(n+2)}\;\mathrm{BM}$ $\mu = \sqrt{0(0+2)} = 0\;\mathrm{BM}$

Nearest integer

$\boxed{0}$

Compare with stored answer

Stored correct answer = $0$.

This matches the derived answer.