A) \[Ca{{(OH)}_{2}}+2HCl\xrightarrow[{}]{{}}CaC{{l}_{2}}+2{{H}_{2}}O\]
B) \[BaC{{l}_{2}}+MgS{{O}_{4}}\xrightarrow[{}]{{}}BaS{{O}_{4}}+MgC{{l}_{2}}\]
C) \[2{{S}_{2}}O_{7}^{2-}+{{H}_{2}}O\xrightarrow[{}]{{}}4SO_{4}^{2-}+4{{H}^{+}}\]
D) \[C{{u}_{2}}S+2FeO\xrightarrow[{}]{{}}2Cu+2Fe+S{{O}_{2}}\]
Correct Answer: D
Solution :
Only in reaction [d], the oxidation number of the elements undergo a change. \[\overset{+1}{\mathop{C{{u}_{2}}}}\,\overset{-2}{\mathop{S}}\,\text{ }+\overset{+2}{\mathop{2FeO}}\,\xrightarrow[{}]{{}}\overset{0}{\mathop{2Cu}}\,+\overset{0}{\mathop{2Fe}}\,+\overset{+4}{\mathop{S{{O}_{2}}}}\,\] ie, oxidation number of Cu decreases from\[+1\]to 0 and that of Fe decreases from +2 to zero while that of S increases from -2 to 4. Therefore, it is a redox reaction.
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