A) \[S{{c}^{3+}}\]
B) \[N{{i}^{2+}}\]
C) \[T{{i}^{4+}}\]
D) \[Z{{n}^{2+}}\]
Correct Answer: B
Solution :
The ions having unpaired electrons are coloured and ion having no unpaired electrons are colourless. The electronic configuration of ions are as: \[S{{c}^{3+}}=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{0}},4{{s}^{o}}\] \[N{{i}^{2+}}=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{8}},4{{s}^{0}}\] \[T{{i}^{4+}}=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{0}},4{{s}^{0}}\] \[Z{{n}^{2+}}=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{10}},4{{s}^{0}}\] From the above configurations, it is deal that \[\text{N}{{\text{i}}^{\text{2+}}}\]has two unpaired electrons Therefore, it will be coloured and others colourless.
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