| (I). \[[Ti{{(N{{O}_{3}})}_{4}}]\] |
| (II) \[{{[Cu(NC-C{{H}_{3}})]}^{\oplus }}Cl{{O}_{4}}^{\Theta }\] |
| (III) \[{{[Cr{{(N{{H}_{3}})}_{6}}]}^{3+}}3B{{F}_{4}}^{\Theta }\] |
| (IV) \[N{{a}_{3}}[V{{F}_{6}}]\] |
A) I, II
B) III, IV
C) II, III, IV
D) All
Correct Answer: B
Solution :
| [b] (I) \[[Ti{{(N{{O}_{3}})}_{4}}]\] |
| \[Ti=[Ar]\,\,3{{d}^{2}}\,\,4{{s}^{2}}\] |
| \[T{{i}^{4+}}=[Ar]\,\,3{{d}^{0}}\] |
| No unpaired electron. So it is colourless. |
| (II) \[{{[Cu(NCC{{H}_{3}})]}^{\oplus }}\] |
| \[Cu=[Ar]\,\,3{{d}^{10}}4{{s}^{1}}\] |
| \[+1\] |
| \[Cu=[Ar]3{{d}^{10}}\] |
| No unpaired electrons. So it is colourless. |
| (III) \[{{[Cr{{(N{{H}_{3}})}_{6}}]}^{3+}}\] |
| \[Cr=[Ar]\,\,3{{d}^{5}}4{{s}^{1}}\] |
| \[+3\] |
| \[Cr=[Ar]\,3{{d}^{3}}\] |
| It has 3 unpaired electrons. So it is coloured. |
| (IV) \[N{{a}_{3}}[V{{F}_{6}}]\] |
| \[V=[Ar]\,3{{d}^{3}}4{{s}^{2}}\] |
| \[{{V}^{3+}}=[Ar]\,3{{d}^{2}}\] |
| It has 2 unpaired electron. So it is coloured. |
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