Question

Which of the following coordination compounds have geometric isomers?

1. Square Planar: `Pt(NH_3)_2Cl_2`
2. Square Planar: `Cu(NH_3)_3Cl`
3. Octahedral: `Ni(NH_3)_3Cl_3^-`
4. Octahedral: `Co(NH_3)_5Cl_2^+`

Answer 1

Your second and fourth compounds... they don't exist.

(1)

• Platinum does form square planar complexes, so this is real.
• Since it has two equivalents for each of two different species, this can have cis/trans isomers, i.e. geometric isomers.

Therefore, diamminedichloroplatinum(II) has geometric isomers.

(3)

This looks real.

• `"Ni"^(2+)` has a common coordination number of `6`, and along with the three `"Cl"^(-)` ligands, the total charge is `-1`.

• This is an `Ma_3b_3` octahedral complex, so it can have two isomers only: fac and mer. Refer to this answer for more information.

• fac and mer isomers are actually diastereomers, which are a subset of geometric isomers (at least, according to this website!).

So, they count as geometric isomers, and triamminetrichloronickel(II) has a fac and mer geometric isomer.

---

(2)

I don't think your second compound is correct. `"Cu"^(+)` has 10 `3d` electrons, which would mean it fills the `d_(x^2-y^2)` antibonding orbitals, destabilizing the complex's `"Cu"-"L"` bonds.

However, `"Cu"^(2+)` has one less `3d` electron, giving less antibonding character to the `"Cu"-"L"` bond, making `["Cu"("NH"_3)_3"Cl"]^(+)` more (thermodynamically) stable than `"Cu"("NH"_3)_3"Cl"`.

Indeed, the common coordination number for `"Cu"^(2+)` is `4`, whereas that of `"Cu"^(+)` is `2`. Either way, this compound doesn't have two equivalents of each of two species, so it has no geometric isomers.

Oh yeah, and your compound probably doesn't exist either.

(4)

This compound is not real either. `"Co"^(3+)` in this compound would have a common coordination number of `6`, not `7`.

However, `"Co"("NH"_3)_4"Cl"_2` would exist. It does have geometric isomers, though it's `"Co"^(2+)` instead.

The trans isomer would have axial `"Cl"` ligands, and the cis isomer would exchange an axial `"Cl"` with an equatorial `"NH"_3`.