Question

When \[{H_2}S\] is passed through \[Hg_2^{2 + }\]​, we get:
(A) \[HgS\]
(B) \[HgS + H{g_2}S\]
(C) \[HgS + Hg\]
(D) \[H{g_2}S\]

Answer 1

Hint: Disproportionation is a chemical transformation of a molecule into two or more other products, usually a redox reaction. In essence, a molecule with an intermediate oxidation state is split into two new compounds with higher and lower oxidation states. Thus, a species undergoes simultaneous reduction and oxidation to produce two distinct products.

Complete Step by Step Solution:
As we know that the definition of \[{H_2}S\]: When drilling for and producing crude oil and natural gas, as well as in wastewater treatment and utility facilities and sewers, hydrogen sulphide \[({H_2}S)\] is a gas that is frequently encountered. The breakdown of organic molecules by microorganisms in the absence of oxygen results in the production of the gas.

Now, look at the symbol \[Hg_2^{2 + }\], As we know that Mercury \[(I)\] compounds often undergo disproportionation, producing black metallic mercury and mercury \[(II)\] compounds.

The middle oxidation state of an element will probably be less stable than the higher and lower oxidation levels, depending on whether the element's three oxidation states are listed in increasing or decreasing order. The entire process must be capable of thermodynamic stability.

\[Hg_2^{2 + } + \;{H_2}S\; \to \;HgS\; + \;Hg\; + \;2{H^ + }\]
Therefore, when \[{H_2}S\] is passed through \[Hg_2^{2 + }\]​then we obtain: \[HgS + Hg\]
Thus, the correct option is: (C) \[HgS + Hg\].

Note: It should be noted that the element in the middle oxidation state should be less stable and the upper and lower oxidation states that have to be significantly more stable when three oxidation states of an element are written in an increasing or decreasing order. Additionally, the total process must be thermodynamically possible.