[nothincomin]

I'm reading my textbook, but it doesn't go into too much depth of isomers like cis-trans isomers.
For ex how do you know if the following have isomers or not?
a) [Cr(CO)3(NH3)3]3+
b) [Pd(CO)2(H2O)Cl]+
c) [Ni(CN)3H2O]-1
d) [Zn(NO2)2Cl2]
e) [Co(H2O)(C2O4)2Cl]+
f) [Cr(NH3)Br3]

[Schrödinger]

Draw the compound, visualise the structure and try all different permutations.

Let's take b) for example. I think the geometry would be square planar since CO is a strong ligand. Hence, the 2 CO molecules can be either on the same edge of the square(cis) or diagonally opposite (trans)

[nothincomin]

Yes it would be square planar, but isn't the formula for square planar complexes MA2B2 in order for it to have isomerism?

[Schrödinger]

I think I know what you're problem is.

Always try to understand the concept first, then try to use tips to remember it. This example is a very typical one. Cis and trans occur when you have atleast one pair of similar groups.


When you have a molecule of MA₂BC type, when A's are on the same side of the square, it is cis (with reference to A) and when they are diagonally opposite, it is trans

When you have a molecule of MA₂B₂ type, when the A's and B's are on the same side, it is cis and when 2 A's are diagonally opposite (that automatically puts the B's diagonally opposite too), it is trans.

You need to remember that cis and trans refer to positions of similar groups. It doesn't matter whether you have 1 pair or 2 pairs of similar groups.

I think now you will see that remembering the MA₂B₂ thingy is just absurd

[nothincomin]

Ok, so for F) [Cr(NH3)Br3], there wouldn't be any isomers even though you have Br3 since it would end up being the same.

[ardbeg]

Cr(NH3)Br3 seems unusual as most Cr(III) complexes are six coordinate and octahedral.  Maybe its a typo or error.