A) \[NaOH+HCl\xrightarrow{{}}NaCl+{{H}_{2}}O\]
B) \[BaC{{l}_{2}}+{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}BaS{{O}_{4}}+2HCl\]
C) \[CuS{{O}_{4}}+2{{H}_{2}}O\xrightarrow{{}}Cu{{(OH)}_{2}}+{{H}_{2}}S{{O}_{4}}\]
D) \[Zn+2HCl\xrightarrow{{}}ZnC{{l}_{2}}+{{H}_{2}}\]
Correct Answer: D
Solution :
[d] [a] and [b] are neutralisation reactions. The oxidation state of Cu is +2 in both reactant and product, and \[SO{{_{4}^{2-}}^{~}}\] ion does not change. In option [d] is a redox reaction. \[Zn\to Z{{n}^{2+}}+2{{e}^{-}}~\](Oxidation) \[2{{H}^{+}}+2{{e}^{-}}\to {{H}_{2}}\](Reduction)
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