Question

why [CuCl4]2- has a flattened tetrahedral structure, not regular tetrahedron and [CoCl4]2- is tetrahedron?

Answer 1

It's probably a Jahn-Teller distortion.

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From the electron configuration of the atoms:

`"Cu": [Ar] 3d^10 4s^1`
`"Co": [Ar] 3d^7 4s^2`

Take away two electrons and you have the `+2` oxidation states for both:

`"Cu"^(2+): [Ar] 3d^9 4s^0`
`"Co"^(2+): [Ar] 3d^7 4s^0`

Therefore, we have a `d^9` complex in `["CuCl"_4]^(2-)`, and a `d^7` complex in `["CoCl"_4]^(2-)`.

Their ORIGINAL tetrahedral d-orbital splitting diagrams would look like:

But we can see that there are multiple ways to fill the orbitals in `"CuCl"_4^(2-)` and still get the same number of electrons in each orbital.

Three ways, actually, because the `t_2` orbitals are triply degenerate.

`ul(uarr darr)" "ul(uarr darr)" "ul(uarr color(white)(darr))` `" "bb((1))`

`ul(uarr darr)" "ul(uarr color(white)(darr))" "ul(uarr darr)` `" "bb((2))`

`ul(uarr color(white)(darr))" "ul(uarr darr)" "ul(uarr darr)` `" "bb((3))`

Therefore, to relieve this degeneracy (at least somewhat), a Jahn-Teller distortion occurs to descend in symmetry from tetrahedral `T_d` to trigonal pyramidal `C_(3v)`.

[This is analogous to the Jahn-Teller distortion that forces the octahedral `O_h` symmetry to descend to square bipyramidal `D_(4h)`.]

Here is how each irreducible representation correlates:

We have:

`E -> E`

`T_2 -> A_1 + E`

The orbitals will therefore split in energy as follows:

As it turns out, this is not going to completely lift the degeneracy. There is still a double degeneracy in the `e` orbitals highest in energy.

The transition metal complex distorts as follows:

and if the bonds stretch in the other direction, the orbitals will be in order of energy as `e, e, a_1` instead of `e,a_1,e`.