A) \[{{[Sc{{({{H}_{2}}O)}_{6}}]}^{3+}}\]
B) \[{{[Ti{{(N{{H}_{3}})}_{6}}]}^{4+}}\]
C) \[{{[V{{(N{{H}_{3}})}_{6}}]}^{3+}}\]
D) \[{{[Zn{{(N{{H}_{3}})}_{6}}]}^{2+}}\]
Correct Answer: C
Solution :
[c] The absorption of visible light and hence coloured nature of the transition metal cation is due to the promotion of one or more unpaired - d - electron from a lower to higher level withing same d-subshell. Hence higher will be the number of unpaired electron higher will be the absorpion in visible light. The electronic configuration of the given elements is \[S{{c}^{3+}}(18)=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{0}}4{{s}^{0}}-\]no Unpaired \[{{e}^{-}}\]. \[T{{i}^{4+}}(18)=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{0}}4{{s}^{0}}\]- no Unpaired \[{{e}^{-}}\]. \[{{V}^{3+}}\left( 20 \right)=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{2}}4{{s}^{0}}\]- Two Unpaired \[{{e}^{-}}\]. \[Z{{n}^{2+}}(28)=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}4{{s}^{0}}\]- no Unpaired \[{{e}^{-}}\]. hence \[{{[V{{(N{{H}_{3}})}_{6}}]}^{3+}}\] will most likely absorb visible light.
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