Which of the following complex ions is not expected to absorve visible light ?

Options

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

Answer

B

Explanation

To explain why complexes are coloured, we need to look at the effect of the ligand charges on the energies of the d orbitals. Because the d electrons are negatively charged, they are repelled by negatively charged ligands. Thus, their orbital energies are higher in the complex than in the free metal ion. But not all the d orbitals are raised in energy by the same amount. The `d_(z^(2))` and `d_(x^(2)-y^(2))` orbitals, which point directly at the ligands are raised in energy to a greater extent than the `d_(xy)` , `d_(xz)` and `d_(yz)` orbitals, which point between the ligands. This energy spliting between the two sets of sets of d orbitals is called the rystal field splitting and is represented by the Greek letter `Delta`. In general, the crystal field splitting energy `Delta` corresponds to wavelengths of light in the visible region of the spectrum and the colours of complexes can therefore be attributed to electronic transitions between the lower the lower and higher energy sets of d orbitals. In octahedral complexes (1), (3) and (4), the d-d trandition by absorbing visible light is possible as they all have a vacancy in the higher energy `e_(g)` set. In square planar `[Ni(CN)_(4)]^(2-)` , all the lower energy d orbitals `(d_(xz), d_(yz),d_(z^(2)))` and `d_(xy))` are completely filled and vacant `d_(x^(2)-y^(2)` orbital, is too high in energy to absorb visible light.