The standard electrode potential (M$$^{3+}$$/M$$^{2+}$$) for V, Cr, Mn & Co are $$-0.26$$ V, $$-0.41$$ V, $$+1.57$$ V and $$+1.97$$ V, respectively. The metal ions which can liberate H$$_2$$ from a dilute acid are

We need to identify which M$$^{2+}$$ ions can liberate H$$_2$$ from dilute acid, given the standard electrode potentials E°(M$$^{3+}$$/M$$^{2+}$$).

Concept: For M$$^{2+}$$ to liberate H$$_2$$ from dilute acid, M$$^{2+}$$ must act as a reducing agent, i.e., it should be oxidized to M$$^{3+}$$ while reducing H$$^+$$ to H$$_2$$.

The reaction is: $$\text{M}^{2+} + \text{H}^+ \to \text{M}^{3+} + \frac{1}{2}\text{H}_2$$

For this reaction to be spontaneous, the overall cell potential must be positive:

$$E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} = E°(\text{H}^+/\text{H}_2) - E°(\text{M}^{3+}/\text{M}^{2+}) > 0$$

Since $$E°(\text{H}^+/\text{H}_2) = 0$$ V, we need:

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

Checking the given values:

V: E° = $$-0.26$$ V (negative) — V$$^{2+}$$ CAN liberate H$$_2$$

Cr: E° = $$-0.41$$ V (negative) — Cr$$^{2+}$$ CAN liberate H$$_2$$

Mn: E° = $$+1.57$$ V (positive) — Mn$$^{2+}$$ CANNOT liberate H$$_2$$

Co: E° = $$+1.97$$ V (positive) — Co$$^{2+}$$ CANNOT liberate H$$_2$$

Therefore, V$$^{2+}$$ and Cr$$^{2+}$$ can liberate H$$_2$$ from dilute acid.

The correct answer is Option 3: V$$^{2+}$$ and Cr$$^{2+}$$.