1. Idea of the reaction

An ion $M^{2+}$ will liberate hydrogen gas with dilute acid if it can reduce $H^+$ to $H_2$:

$2H^+ + 2e^- \rightarrow H_2$

So we check whether $M^{2+}$ is a sufficiently strong reducing agent, i.e. whether it gets oxidized to $M^{3+}$ easily:

$M^{2+} \rightarrow M^{3+} + e^-$

This depends on the standard reduction potential for:

$M^{3+} + e^- \rightarrow M^{2+}$

If $E^\circ(M^{3+}/M^{2+})$ is negative, then the reverse oxidation of $M^{2+}$ is favorable enough to reduce $H^+$ to $H_2$.

Using

$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$

with cathode reaction

$2H^+ + 2e^- \rightarrow H_2, \qquad E^\circ = 0.00\,V$

and anode involving oxidation of $M^{2+}$, we get effectively:

$E^\circ_{cell} = 0 - E^\circ(M^{3+}/M^{2+})$

Thus, reaction with acid occurs when:

$E^\circ(M^{3+}/M^{2+}) < 0$

2. Check each ion

(A) $Ti^{2+}$

For titanium:

$E^\circ(Ti^{3+}/Ti^{2+}) \approx -0.37\,V$

Hence,

$E^\circ_{cell} = 0 - (-0.37) = +0.37\,V$

So $Ti^{2+}$ reduces $H^+$ and liberates $H_2$.

(B) $V^{2+}$

For vanadium:

$E^\circ(V^{3+}/V^{2+}) \approx -0.26\,V$

Hence,

$E^\circ_{cell} = 0 - (-0.26) = +0.26\,V$

So $V^{2+}$ also liberates $H_2$.

(C) $Cr^{2+}$

For chromium:

$E^\circ(Cr^{3+}/Cr^{2+}) \approx -0.41\,V$

Hence,

$E^\circ_{cell} = 0 - (-0.41) = +0.41\,V$

So $Cr^{2+}$ also liberates $H_2$.

(D) $Mn^{2+}$

For manganese, the relevant oxidation would be:

$Mn^{2+} \rightarrow Mn^{3+} + e^-$

But

$E^\circ(Mn^{3+}/Mn^{2+}) \approx +1.51\,V$

Hence,

$E^\circ_{cell} = 0 - 1.51 = -1.51\,V$

This is not spontaneous. Therefore $Mn^{2+}$ does not reduce $H^+$ to $H_2$.

3. Conclusion

The ion which does not liberate hydrogen gas on reaction with dilute acids is:

$\boxed{Mn^{2+}}$

So the correct option is D.