A) \[NaCl+KN{{O}_{3}}\xrightarrow[{}]{{}}NaN{{O}_{3}}+KCl\]
B) \[Ca{{C}_{2}}{{O}_{4}}+2HCl\xrightarrow[{}]{{}}CaC{{l}_{2}}+{{H}_{2}}{{C}_{2}}{{O}_{4}}\]
C) \[Ca{{(OH)}_{2}}+2N{{H}_{4}}Cl\xrightarrow[{}]{{}}CaC{{l}_{2}}+2N{{H}_{3}}\] \[+2{{H}_{2}}O\]
D) \[2K[Ag{{(CN)}_{2}}]+Zn\xrightarrow[{}]{{}}2Ag\] \[+{{K}_{2}}[Zn{{(CN)}_{4}}]\]
Correct Answer: D
Solution :
[a] \[\overset{+1\,\,\,\,\,\,\,\,\,-1}{\mathop{NaCl}}\,+\overset{+1\,\,\,\,\,\,\,\,\,\,\,-1}{\mathop{KN{{O}_{3}}}}\,\xrightarrow[{}]{{}}\overset{+1\,\,\,\,\,\,\,\,\,\,\,\,\,-1}{\mathop{NaN{{O}_{3}}}}\,+\overset{+1\,\,\,\,\,\,-1}{\mathop{KCl}}\,\] [b] \[\overset{+2\,\,\,\,\,\,\,\,\,\,\,\,\,\,-2}{\mathop{Ca{{C}_{2}}{{O}_{4}}}}\,+\overset{+1\,\,\,\,\,\,\,\,\,\,\,-1}{\mathop{2HCl}}\,\xrightarrow[{}]{{}}\overset{+2\,\,\,\,\,\,\,\,\,\,\,\,\,-1}{\mathop{CaC{{l}_{2}}}}\,+\overset{+1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+2}{\mathop{{{H}_{2}}{{C}_{2}}{{O}_{4}}}}\,\] [c] \[\overset{+2\,\,\,\,\,\,\,\,\,\,\,\,\,\,-1}{\mathop{Ca{{(OH)}_{2}}}}\,+\overset{\,\,\,-3+1\,\,\,\,\,\,\,-1}{\mathop{2N{{H}_{4}}Cl}}\,\xrightarrow[{}]{{}}\overset{+2\,\,\,\,\,\,\,\,\,\,\,\,\,-1}{\mathop{CaC{{l}_{2}}}}\,+\overset{-3\,\,\,\,\,\,+1}{\mathop{2N{{H}_{3}}}}\,\] \[+\overset{+1\,\,\,\,\,\,\,-2}{\mathop{2{{H}_{2}}O}}\,\] in all these cases during reaction, there is no change in oxidation state of ion or molecule or constituent atom, these are simply ionic reactions. [d] \[2K[\overset{+1}{\mathop{Ag}}\,{{(CN)}_{2}}]+\overset{0}{\mathop{Zn}}\,\xrightarrow[{}]{{}}\overset{0}{\mathop{2Ag}}\,+{{K}_{2}}[\overset{+2}{\mathop{Zn}}\,{{(CN)}_{4}}]\] \[A{{g}^{+}}\xrightarrow[{}]{{}}Ag\]gaining of\[{{e}^{-}},\]reduction \[Zn\xrightarrow[{}]{{}}Z{{n}^{2+}}\]loss of\[{{e}^{-}},\]oxidation Therefore, it is a redox reaction.
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