(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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