A) \[P{{b}^{2+}},C{{u}^{2+}}\]
B) \[Z{{n}^{2+}},A{{l}^{3+}}\]
C) \[C{{u}^{2+}},Z{{n}^{2+}}\]
D) \[A{{l}^{3+}},C{{u}^{2+}}\]
Correct Answer: B
Solution :
: \[Z{{n}^{2+}}\] ion reacts with aq. \[NaOH\] and \[N{{H}_{3}}\]and forms a coordination compound. \[Z{{n}^{2+}}+4NaOH\to {{[Zn{{(OH)}_{4}}]}^{2-}}+4N{{a}^{+}}\] \[Z{{n}^{2+}}+4N{{H}_{3}}\to {{[Zn{{(N{{H}_{3}})}_{4}}]}^{2+}}\] But, \[A{{l}^{3+}}\] ion reacts only with \[NaOH\]. \[A{{l}^{3+}}+4NaOH\to [A]{{(OH)}_{4}}{{]}^{-}}+4N{{a}^{+}}\]
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