A) \[{{[Ni{{(N{{O}_{2}})}_{6}}]}^{4}}<{{[Ni{{(N{{H}_{3}})}_{6}}]}^{2+}}<{{[Ni{{({{H}_{2}}O)}_{6}}]}^{2+}}\]
B) \[{{[Ni{{(N{{O}_{2}})}_{6}}]}^{4}}<{{[Ni{{({{H}_{2}}O)}_{6}}]}^{2+}}<{{[Ni{{(N{{H}_{3}})}_{6}}]}^{2+}}\]
C) \[{{[Ni{{({{H}_{2}}O)}_{6}}]}^{2+}}<{{[Ni{{(N{{H}_{3}})}_{6}}]}^{2+}}<{{[Ni{{(N{{O}_{2}})}_{6}}]}^{4-}}\]
D) \[{{[Ni{{(N{{H}_{3}})}_{6}}]}^{2+}}<{{[Ni{{({{H}_{2}}O)}_{6}}]}^{2+}}<{{[Ni{{(N{{O}_{2}})}_{6}}]}^{4-}}\]
Correct Answer: A
Solution :
When central metal ion is same (here \[N{{i}^{2+}}\]), the absorption of colour depends on the, ligand. According to spectrochemical series, various ligands are as follows \[{{I}^{-}}<B{{r}^{-}}<C{{l}^{-}}<NO_{3}^{-}<{{F}^{-}}<{{H}_{2}}O<N{{H}_{3}}\] \[<NO_{2}^{-}<C{{N}^{-}}<CO\] Thus, \[{{H}_{2}}O\] is weakest ligand among these, therefore, the absorbed energy will be lowest in \[{{[Ni{{({{H}_{2}}O)}_{6}}]}^{2+}},\]so it will absorb highest wavelength (red light) \[\left( E\propto \frac{1}{\lambda } \right)\] \[\therefore \] the order of increasing wavelength is \[{{[Ni{{(N{{O}_{2}})}_{6}}]}^{4-}}<{{[Ni{{(N{{H}_{3}})}_{6}}]}^{2+}}<{{[Ni{{({{H}_{2}}O)}_{6}}]}^{2+}}\]
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