`CuF_(2)` contains `Cu^(2+)` ions surrounded by six `F^(-)` ions, four `F^(-)` being at a distance of `0.193` nm other two at a distance of `0.227` nm. Thus, it should have the configuration.

Options

A. `(d_(z^(2)))^(1) (d_(x^(2)-y^(2)))^(2)`
B. `(d_(z^(2)))^(2) (d_(x^(2)-y^(2)))^(1)`
C. `(d_(xy))^(1) (d_(yz))^(1) (d_(xz))^(1)`
D. `(d_(xy))^(2) (d_(yz))^(2) (d_(xz))^(0)`

Answer

B

Explanation

Planar complexes can be regarded as distorted octahedral complexes. Application of crytal field theory shows that an octahedral arrangement of ligands interacts more strongly with `e_(g)` orbitals than with `t_(2g)` orbitals. Therefore an asymmetric occupation of the two `e_(g)` orbitals results in a distorted ocahedral arrangement of ligands in transition metal complex. The two possible distributions of the electrons in the `e_(g)` orbitals are shown below for a transition metal ion having nne 3d electrons (thus three electrons in the `e_(g)` orbitals).

Consider the arrangement of two electrons in the `d_(z^(2))` orbital and one electron in the `d_(x^(2)-y^(2))` orbital. The two electrons in the former orbital will interact more strongly with the ligands directed along the z axis than the one elecron in the latter with ligands lying along the x and y axes. Consequently the ligands in teh xy plane are drawn closer to the central transition metal ion than the ligands directed along the z axis, i.e., for short and two long bonds would be formed.If the occupation of the two `e_(g)` orbitals is reversed it is result. Many octahedrally coordinated compounds of `Cu^(2+)` (nine 3d electrons) do indeed have four short bonds and two long ones. Thus, the configuration `(d_(z^(2)))^(1)(d_(x^(2)-y^(2)))^(1)` is preferred to the alternative `(d_(z^(2)))^(1)` `(d_(x^(2)-y^(2)))^(2)`.