The general criterion for a spontaneous redox reaction is that the reduction potential of the reducing agent (which gets oxidized) must be lower than the reduction potential of the oxidizing agent (which gets reduced). Here, the reduction of NO₃^- has an E⁰ of +0.96 V. For a metal ion to be oxidized by NO₃^-, the E⁰ of the metal ion’s reduction must be less than +0.96 V.

Let's check each given ion to see if it can be oxidized:

• V²⁺ (E⁰ = -1.19 V): This potential is significantly lower than +0.96 V, thus vanadium can be oxidized by nitrate, as it is much easier to reduce NO₃^- than to reduce V²⁺ to V.
• Fe³⁺ (E⁰ = -0.04 V): This potential is also lower than +0.96 V, so iron can be oxidized by nitrate.
• Au³⁺ (E⁰ = +1.40 V): Since the potential for Au³⁺ is higher than the potential for nitrate reduction, gold cannot be oxidized by nitrate. It means nitrate cannot provide sufficient potential to reduce Au³⁺.
• Hg²⁺ (E⁰ = +0.86 V): This potential is close but still lower than +0.96 V, therefore mercury can theoretically be oxidized by nitrate, though it is only slightly easier to reduce NO₃^- than to reduce Hg²⁺.

Given this analysis:

• Option A (V and Hg): Correct, as both V and Hg have lower reduction potentials than the nitrate ion.
• Option B (Hg and Fe): Correct, as analyzed above, both can indeed be oxidized by nitrate.
• Option C (Fe and Au): Incorrect because Au³⁺ has a higher potential and thus cannot be oxidized by nitrate.
• Option D (Fe and V): Correct, given that both V and Fe have lower potentials than nitrate reduction.

The correct answer includes Option A, Option B, and Option D.