# How many geometric isomers are there for the following compound? [Mn(CO)_2Cl_2Br_2]^+?

Mar 16, 2016

I would use the Bailar method for this one, which is a way to systematically determine the number of isomers, including enantiomers.

(Geometric isomers contain the same connectivities and the same atoms, but different spatial arrangements, so enantiomers can still be geometric isomers of non-enantiomers.)

Here's the overall process:

1. Start by labeling the ligands $a - c$, in this order: top axial, bottom axial, rear-right equatorial, front-left equatorial, front-right equatorial, rear-left equatorial. Assign each new ligand $a$, $b$, $c$, etc. to monodentate ligands, and $A$, $B$, $C$, etc. to polydentate ligands. Identical ligands are the same letter, and polydentate ligands must NEVER be trans to each other. It's sterically unfavorable because it's too far of a reach.

2. Use the notation $\text{M} \left\langlet u\right\rangle \left\langlev w\right\rangle \left\langlex y\right\rangle$, where $t$ and $u$ are trans to each other and $\text{M}$ is your central metal. Switch ligands around until you've discovered all the possible placements of each ligand. This is best illustrated with an example...

3. Check for enantiomers after drawing your isomers out. If you do NOT see a mirror plane, then it is an enantiomer. This can be difficult to do.

1) 2)

1) $\text{M} \left\langlea a\right\rangle \left\langleb b\right\rangle \left\langlec c\right\rangle$ (starting point)
2) $\text{M} \left\langlea a\right\rangle \left\langleb c\right\rangle \left\langleb c\right\rangle$ (start from 1, switch $b$ with $c$)
3) $\text{M} \left\langlea b\right\rangle \left\langlea b\right\rangle \left\langlec c\right\rangle$ (start from 1, switch $a$ with $b$)
4) $\text{M} \left\langlea c\right\rangle \left\langleb b\right\rangle \left\langlea c\right\rangle$ (start from 1, switch $a$ with $c$)

Note that on (4), to make it easier to compare isomers, I will instead use $\text{M} \left\langlea c\right\rangle \left\langlea c\right\rangle \left\langleb b\right\rangle$. That way, the carbonyl ligands will be in the rear, just like in (3) and (5) (we'll get to (5) soon).

Now, do we have all of them? Let's think about it. We should see, in those brackets, $a a$, $a b$, $a c$, $b b$, $b c$, and $c c$.

...Yeah, notice how we forgot this?

5) $\text{M} \left\langlea b\right\rangle \left\langlea c\right\rangle \left\langleb c\right\rangle$ (start from 2, switch $a$ with $b$)

That should be all of them. Now let's draw these out (remember that ligands in the same bracket are trans to each other). 3)

Note that all of these have mirror planes except for (5), which is a pair of enantiomers. Let's check a reference here to make sure we got all of them... Since according to Table 9.6, an $\text{M} {a}_{2} {b}_{2} {c}_{2}$ metal-ligand complex will have $6$ stereoisomers, within which contains $1$ pair of enantiomers, we got this right!

Four nonchiral, and one chiral pair, for a total of $6$ geometric isomers (though the enantiomers are not geometric isomers of each other, they are with respect to the nonchiral isomers).