1. Principle involved

   A metal liberates $\mathrm{H_2}$ from mineral acids if it can displace hydrogen from acid: $M + 2H^+ \to M^{2+} + H_2$

   This depends on the metal's position in the electrochemical/reactivity series.

   • Metals above hydrogen in the series generally liberate $\mathrm{H_2}$ from dilute mineral acids.
   • Metals below hydrogen do not liberate $\mathrm{H_2}$ easily.

2. Check the given metals

   Standard reduction potentials:

   • $\mathrm{Cu^{2+}/Cu} = +0.34\,V$
   • $\mathrm{Ni^{2+}/Ni} \approx -0.25\,V$
   • $\mathrm{Zn^{2+}/Zn} = -0.76\,V$
   • $\mathrm{Mn^{2+}/Mn} \approx -1.18\,V$

   The more negative the reduction potential, the greater the tendency of the metal to get oxidized and hence to liberate $\mathrm{H_2}$ from acids.

3. Compare tendency to liberate $\mathrm{H_2}$

   • $\mathrm{Mn}$: very high tendency
   • $\mathrm{Zn}$: high tendency
   • $\mathrm{Ni}$: moderate tendency
   • $\mathrm{Cu}$: least tendency

   Copper lies below hydrogen in the reactivity series, so it does not liberate $\mathrm{H_2}$ from non-oxidizing mineral acids like dilute $\mathrm{HCl}$ or dilute $\mathrm{H_2SO_4}$.

4. Evaluate options

   • A: Cu — least tendency ✅
   • B: Mn — readily liberates $\mathrm{H_2}$ ❌
   • C: Ni — can liberate $\mathrm{H_2}$ ❌
   • D: Zn — readily liberates $\mathrm{H_2}$ ❌

Therefore, the correct answer is: $\boxed{\text{A: Cu}}$