Question

In which one of the following reactions, $+2$ state in a reactant is reduced to $+1$ state?

• A. $Mg + S \rightarrow MgS$
• B. $2HgCl_2 + SnCl_2 \rightarrow Hg_2Cl_2 + SnCl_4$
• C. $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$
• D. $2K_4[Fe(CN)_6] + H_2O_2 \rightarrow 2K_3[Fe(CN)_6] + 2KOH$

Hint:

Identify the metal ions first. Mercury is unique as it forms both $Hg^{2+}$ (mercuric) and $Hg_2^{2+}$ (mercurous) ions, where the latter has an oxidation state of $+1$ per atom.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Reduction involves a decrease in the oxidation state of an element. We need to calculate the oxidation states of the central atoms in the reactants and products for each reaction.

Step 2: Detailed Explanation:
(A) $Mg$ ($0$) to $Mg^{2+}$ ($+2$): Oxidation.
(B) In $HgCl_2$, $Hg$ is in the $+2$ state. In $Hg_2Cl_2$, the $Hg$ atoms are in the $+1$ state ($2x + 2(-1) = 0 \implies x = +1$). This is a reduction from $+2$ to $+1$.
(C) $C$ in $CH_4$ is $-4$, in $CO_2$ it is $+4$: Oxidation.
(D) $Fe$ in $[Fe(CN)_6]^{4-}$ is $+2$, in $[Fe(CN)_6]^{3-}$ it is $+3$: Oxidation.
(E) $S$ in $H_2S$ is $-2$, in $S$ it is $0$: Oxidation.

Step 3: Final Answer:
In reaction (B), mercury is reduced from the $+2$ state to the $+1$ state.