A) \[{{[Ti\,\,{{(en)}_{2}}{{(N{{H}_{3}})}_{2}}]}^{4+}}\]
B) \[{{[Cr\,\,{{(N{{H}_{3}})}_{6}}]}^{3+}}\]
C) \[{{[Zn\,\,{{(N{{H}_{3}})}_{6}}]}^{2+}}\]
D) \[{{[Sc\,\,{{({{H}_{2}}O)}_{3}}{{(N{{H}_{3}})}_{3}}]}^{3+}}\] (At. no. Zn = 30, Sc = 21, Ti = 22, Cr = 24)
Correct Answer: B
Solution :
[b] Since \[C{{r}^{3+}}\] in the complex has unpaired electrons in the d orbital, hence will be coloured \[Ti=[Ar]\,\,3{{d}^{2}}4{{s}^{2}};\] \[T{{i}^{4+}}=3{{d}^{0}}\] \[Cr=[Ar]\,\,3{{d}^{5}}4{{s}^{1}};\] \[C{{r}^{3+}}=3{{d}^{3}}\] \[Zn=[Ar]\,\,3{{d}^{10}}4{{s}^{2}};\] \[Z{{n}^{2+}}=3{{d}^{10}}\] \[Sc=[Ar]\,\,3{{d}^{1}}4{{s}^{2}};\] \[S{{c}^{3+}}=3{{d}^{0}}\]
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