Correct Answer – D
Gold will dissolve in `1 M HCl` if it gets oxidized and ` H+` ions fo `HCl` are reduced :
` Au(s) + 3H^(+) (aq.) rarr Au^(3+) (aq.) + 3/2 H_2(g) `
But this is not possible because ` E_(Au//Au^(3+))^(Ө) (-1.42 V)` is less than ` E_(H^+//H_(2))^Ө (0.00V)`.
In other words, the standard reduction potential of gold is greater than zero . Therefore, the reduced from of gold, i.e., Au (s) is mor stable than the reduced from of hydrogen (i.e., `H_2`) i.e., gold cannot be oxidized by ` H^(+)` ions to give ` H_2`. Hence, gold does not dissolve in ` 1M HCl` solution.
Since ` E_(Au^(3+) //Au)^(Ө) (1.42 V)` is greater than ` E_(H^+ //H_2)^Ө` (0.00 V), hydrogen gas passed through the salt solution of gold will reduce auric ions and metallic gold will be percipitated.

Consider the half reaction, 

2H⁺(aq) + 2e^– → H₂(g); E° = 0.0V

Au³⁺(aq) + 3e^– → Au(s); E° = 1.50V

Since E° (1.50V) for (Au³⁺/Au) is higher than that H⁺/-\(\frac{1}{2}\)H₂ (0.0V), therefore, Au³⁺ can be more easily reduced than H⁺ ions Au³⁺ can be reduced to Au metal by H₂ but H⁺ cannot oxidize metallic gold to Au³⁺ ions. In other words, metallic gold does not dissolve in 1M HCl instead H₂ gas can reduce gold salt to metallic gold.