1. Idea of the question

We must identify which ion among $Co^{3+}$, $Ti^{2+}$, $V^{2+}$ and $Cr^{2+}$ cannot liberate $H_2$ from dilute mineral acid.

For a metal ion $M^{n+}$ to liberate hydrogen from acid, it should be able to get oxidized further while reducing $H^+$ to $H_2$.

So we check whether the ion can be oxidized to a higher oxidation state easily.

2. Electronic configurations of the ions

• $Co^{3+}$ : $Co = [Ar]3d^74s^2$
  So $Co^{3+} = [Ar]3d^6$

• $Ti^{2+}$ : $Ti = [Ar]3d^24s^2$
  So $Ti^{2+} = [Ar]3d^2$

• $V^{2+}$ : $V = [Ar]3d^34s^2$
  So $V^{2+} = [Ar]3d^3$

• $Cr^{2+}$ : $Cr = [Ar]3d^54s^1$
  So $Cr^{2+} = [Ar]3d^4$

3. Which ion cannot liberate $H_2$?

If the ion is very stable in its present oxidation state, it will not be oxidized further and hence cannot reduce $H^+$ to $H_2$.

Among the given ions:

• $Ti^{2+} (d^2)$ can be oxidized to $Ti^{3+}$
• $V^{2+} (d^3)$ can be oxidized to $V^{3+}$
• $Cr^{2+} (d^4)$ can be oxidized to $Cr^{3+}$, which is especially stable
• $Co^{3+} (d^6)$ is already in a relatively high oxidation state and is not expected to act as a reducing reagent to liberate hydrogen from acid

Hence, the ion that cannot liberate $H_2$ is $Co^{3+}$.

4. Magnetic moment of $Co^{3+}$ in gaseous state

In gaseous state, pairing due to ligands is absent, so we use the free-ion configuration.

$Co^{3+} : 3d^6$

For free $d^6$, number of unpaired electrons $n = 4$.

Spin-only magnetic moment is

So,

Nearest integer:

5. Comparison with stored answer

Stored correct answer = $5$

Our derived answer = $5$

So they agree.