1. Criterion for oxidation by $\mathrm{NO_3^-}$

If a metal $M$ is oxidized by $\mathrm{NO_3^-}$, then:

• $\mathrm{NO_3^-}$ acts as the oxidizing agent and gets reduced.
• The metal acts as the reducing agent and gets oxidized.

Given reduction potential: $E^\circ(\mathrm{NO_3^-\ reduction})=+0.96\,\text{V}$

For spontaneity, $E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}>0$

Here cathode is nitrate reduction, so: $E^\circ_{\text{cell}}=0.96-E^\circ(\text{metal ion/metal reduction})$

Thus the metal will be oxidized if: $0.96-E^\circ(\text{metal ion/metal})>0$ $\Rightarrow E^\circ(\text{metal ion/metal})<0.96\,\text{V}$

So we compare each metal's reduction potential with $0.96\,\text{V}$.

2. Check each metal

(i) Vanadium

$\mathrm{V^{2+}+2e^-\to V},\quad E^\circ=-1.19\,\text{V}$ $E^\circ_{\text{cell}}=0.96-(-1.19)=2.15\,\text{V}>0$ So $\mathrm{V}$ is oxidized by $\mathrm{NO_3^-}$.

(ii) Iron

$\mathrm{Fe^{3+}+3e^-\to Fe},\quad E^\circ=-0.04\,\text{V}$ $E^\circ_{\text{cell}}=0.96-(-0.04)=1.00\,\text{V}>0$ So $\mathrm{Fe}$ is oxidized by $\mathrm{NO_3^-}$.

(iii) Gold

$\mathrm{Au^{3+}+3e^-\to Au},\quad E^\circ=+1.40\,\text{V}$ $E^\circ_{\text{cell}}=0.96-1.40=-0.44\,\text{V}<0$ So $\mathrm{Au}$ is not oxidized by $\mathrm{NO_3^-}$.

(iv) Mercury

$\mathrm{Hg^{2+}+2e^-\to Hg},\quad E^\circ=+0.86\,\text{V}$ $E^\circ_{\text{cell}}=0.96-0.86=0.10\,\text{V}>0$ So $\mathrm{Hg}$ is oxidized by $\mathrm{NO_3^-}$.

3. Metals oxidized by $\mathrm{NO_3^-}$

The metals are: $\boxed{\mathrm{V,\,Fe,\,Hg}}$

Now match with options:

• A: $\mathrm{V}$ and $\mathrm{Hg}$ ✅
• B: $\mathrm{Hg}$ and $\mathrm{Fe}$ ✅
• C: $\mathrm{Fe}$ and $\mathrm{Au}$ ❌
• D: $\mathrm{Fe}$ and $\mathrm{V}$ ✅

Hence the correct options are: $\boxed{A,\ B,\ D}$