Principle involved

An ion will liberate hydrogen from dilute acid if it can reduce $\mathrm{H}^+$ to $\mathrm{H}_2$.

The relevant half-reaction is: $2\mathrm{H}^+ + 2e^- \rightarrow \mathrm{H}_2$ with standard reduction potential: $E^\circ = 0.00\,\text{V}$

So, if a metal ion in a lower oxidation state can be oxidized easily, i.e. its corresponding reduction potential $\mathrm{M}^{3+} + e^- \rightarrow \mathrm{M}^{2+}$ is negative enough in reverse direction, then $\mathrm{M}^{2+}$ can act as a reducing agent and reduce $\mathrm{H}^+$ to $\mathrm{H}_2$.

Check each ion

We examine whether $\mathrm{M}^{2+}$ can be oxidized to $\mathrm{M}^{3+}$ while reducing $\mathrm{H}^+$.

(i) $\mathrm{Ti}^{2+}$

Relevant reduction potential: $\mathrm{Ti}^{3+} + e^- \rightarrow \mathrm{Ti}^{2+}, \quad E^\circ = -0.37\,\text{V}$

Therefore, oxidation of $\mathrm{Ti}^{2+}$ to $\mathrm{Ti}^{3+}$ has: $E^\circ_{\text{ox}} = +0.37\,\text{V}$

Combining with $2\mathrm{H}^+ + 2e^- \rightarrow \mathrm{H}_2, \quad E^\circ = 0.00\,\text{V}$ gives positive overall tendency.

Hence, $\mathrm{Ti}^{2+}$ can liberate hydrogen from dilute acid.

(ii) $\mathrm{Cr}^{2+}$

Relevant reduction potential: $\mathrm{Cr}^{3+} + e^- \rightarrow \mathrm{Cr}^{2+}, \quad E^\circ = -0.41\,\text{V}$

So oxidation of $\mathrm{Cr}^{2+}$ to $\mathrm{Cr}^{3+}$ has: $E^\circ_{\text{ox}} = +0.41\,\text{V}$

This means $\mathrm{Cr}^{2+}$ is a sufficiently strong reducing agent to reduce $\mathrm{H}^+$ to $\mathrm{H}_2$.

Hence, $\mathrm{Cr}^{2+}$ can liberate hydrogen.

(iii) $\mathrm{V}^{2+}$

Relevant reduction potential: $\mathrm{V}^{3+} + e^- \rightarrow \mathrm{V}^{2+}, \quad E^\circ = -0.26\,\text{V}$

Therefore, oxidation of $\mathrm{V}^{2+}$ to $\mathrm{V}^{3+}$ has: $E^\circ_{\text{ox}} = +0.26\,\text{V}$

Again, this allows reduction of $\mathrm{H}^+$ to $\mathrm{H}_2$.

Hence, $\mathrm{V}^{2+}$ can liberate hydrogen.

Count the ions

All three ions: $\mathrm{Ti}^{2+},\; \mathrm{Cr}^{2+},\; \mathrm{V}^{2+}$ can liberate hydrogen from dilute acid.

Therefore, the number of such ions is: $3$

Option matching

$\boxed{\text{Option C: } 3}$