A) \[2N{{H}_{3}}+{{H}_{2}}S{{O}_{4}}\rightleftharpoons 2N{{H}_{4}}^{+}+S{{O}_{4}}^{2-}\]
B) \[N{{H}_{3}}+C{{H}_{3}}COOH\rightleftharpoons N{{H}_{4}}^{+}+C{{H}_{3}}CO{{O}^{-}}\]
C) \[{{H}_{2}}O+C{{H}_{3}}COOH\rightleftharpoons {{H}_{3}}{{O}^{+}}+C{{H}_{3}}CO{{O}^{-}}\]
D) \[{{[Cu{{({{H}_{2}}O)}_{4}}]}^{2+}}+4N{{H}_{3}}\rightleftharpoons {{[Cu{{(N{{H}_{3}})}_{4}}]}^{2+}}+4{{H}_{2}}O\]
Correct Answer: D
Solution :
[d] \[{{[Cu{{({{H}_{2}}O)}_{4}}]}^{2+}}+4N{{H}_{3}}\rightleftharpoons {{[Cu{{(N{{H}_{3}})}_{4}}]}^{2+}}+4{{H}_{2}}O\] involves lose and gain of electrons. \[{{H}_{2}}O\]is coordinated to Cu by donating electrons (LHS). It is then removed by withdrawing electrons.
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