Which of the following is not formed when H$$_2$$S reacts with acidic K$$_2$$Cr$$_2$$O$$_7$$ solution?

The reaction involves hydrogen sulfide (H₂S) reacting with acidic potassium dichromate (K₂Cr₂O₇) solution. Potassium dichromate in acidic medium acts as a strong oxidizing agent, and hydrogen sulfide is a reducing agent. The typical reaction proceeds as follows.

First, the dichromate ion (Cr₂O₇²⁻) is reduced to chromium(III) ions (Cr³⁺) in acidic medium. The reduction half-reaction is:

$$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$$

Hydrogen sulfide (H₂S) is oxidized to elemental sulfur (S). The oxidation half-reaction is:

$$\text{H}_2\text{S} \rightarrow \text{S} + 2\text{H}^+ + 2\text{e}^-$$

To balance the electrons, multiply the oxidation half-reaction by 3:

$$3\text{H}_2\text{S} \rightarrow 3\text{S} + 6\text{H}^+ + 6\text{e}^-$$

Now, add this to the reduction half-reaction:

$$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- + 3\text{H}_2\text{S} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 3\text{S} + 6\text{H}^+ + 6\text{e}^-$$

Cancel the common terms: 6e⁻ on both sides, and 6H⁺ from the right with 6H⁺ from the left (leaving 8H⁺ on the left):

$$\text{Cr}_2\text{O}_7^{2-} + 8\text{H}^+ + 3\text{H}_2\text{S} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 3\text{S}$$

This is the net ionic equation. To write the complete molecular equation, include the potassium ions from K₂Cr₂O₇ and the sulfate ions from the acid (H₂SO₄). The acid provides H⁺ ions, and each H₂SO₄ provides 2H⁺. Since 8H⁺ are needed, 4 molecules of H₂SO₄ are required, which also provide 4 SO₄²⁻ ions.

The potassium ions (2K⁺ from K₂Cr₂O₇) and chromium ions (2Cr³⁺) will combine with sulfate ions. The 2Cr³⁺ ions require 3 SO₄²⁻ ions to form Cr₂(SO₄)₃, and the remaining SO₄²⁻ ion (from the 4 provided) combines with 2K⁺ to form K₂SO₄.

Thus, the balanced molecular equation is:

$$\text{K}_2\text{Cr}_2\text{O}_7 + 4\text{H}_2\text{SO}_4 + 3\text{H}_2\text{S} \rightarrow \text{K}_2\text{SO}_4 + \text{Cr}_2(\text{SO}_4)_3 + 7\text{H}_2\text{O} + 3\text{S}$$

The products are potassium sulfate (K₂SO₄), chromium(III) sulfate (Cr₂(SO₄)₃), water (H₂O), and elemental sulfur (S).

Now, examining the options:

A. K₂SO₄ is formed, as shown in the products.

B. Cr₂(SO₄)₃ is formed, as chromium is in the +3 oxidation state.

C. S (elemental sulfur) is formed from the oxidation of H₂S.

D. CrSO₄ would imply chromium(II) sulfate, where chromium is in the +2 oxidation state. However, in the reaction, chromium is reduced to +3, not +2, so CrSO₄ is not formed.

Hence, the correct answer is Option D.