The correct order for the wavelength of absorption in the visible region is

Options

A. `[Ni(NO_(2))_(6)]^(4-) lt [Ni(NH_(3))_(6)]^(2+) lt [Ni(H_(2)O)_(6)]^(2+)`
B. `[Ni(NO_(2))_(6)]^(4-) lt [Ni(H_(2)O)_(6)]^(2+) lt [Ni(NH_(3))_(6)]^(2+)`
C. `[Ni(H_(2)O)_(6)]^(2+) lt [Ni(NH_(3))_(6)]^(2+) lt [Ni(NO_(2))_(6)]^(4-)`
D. `[Ni(NH_(3))_(6)]^(2+) lt [Ni(H_(2)O)_(6)]^(2+) lt [Ni(NO_(2))_(6)]^(4-)`

Answer

A

Explanation

When central metal ion is same (here `Ni^(2+)`), the absorption of colour depends on the ligand. According to spectrochemical series, various ligands are as follows
`I^(-) lt Br^(-) lt Cl^(-) lt NO_(3)^(-) lt F^(-) ltH_(2)O lt NH_(3) lt NO_(2)^(-) lt CN^(-) lt CO`.
Thus `H_(2)O` is the 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) `E porp (1)/(lamda)`
`therefore` order of increasing wavelength is
`[Ni(NO_(2))_(6)]^(4-) lt [Ni(NH_(3))_(6)]^(2+) lt [Ni(H_(2)O)_(6)]^(2+)`