# Coordination Compounds

I had an acquaintance with this question of coordination compounds which I find difficult to solve:

We have a coordination compound of ML$$_3$$ type structure where the ligand L is

.

Now find the total number of stereoisomers.

Note by Sai Krishna Attaluri
1 year ago

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The ligand $$L$$ stated above is a methyl substituted derivative of ethylenediamine ( $$\ce {NH2 - CH2 - CH2 - NH2}$$ or $$\ce {en}$$ ). This ligand is a bidentate chelating ligand where both the nitrogen atoms donate their electrons to the metal ion. Thus the ligand has a coordination number $$6$$ and not $$3$$ and it looks like a propeller, like this example

A complex of $$\ce {[Fe (Me-en)3]}^{3+}$$

As you can see this has three chiral centers and no restricted rotation sites. So by using the three formulas of finding the number of stereoisomers when number of chiral centers is odd as listed below

\begin{align} & \text{No. of enantiomers } = 2^{n-1} - 2^{{(n-1)}/{2}} \\ & \text{No. of meso compounds } = 2^{{(n-1)}/{2}} \\ & \text{No. of optical stereoisomers } = 2^{n-1} \end{align}

We respectively get

\begin{align} & \text{No. of enantiomers } = 2 \\ & \text{No. of meso compounds } = 2 \\ & \text{No. of optical stereoisomers } = 4 \end{align}

- 1 year ago

That's true but I faced problem in counting the cases of geometrical isomers in each case above.

Is my answer correct? Do you know what the answer is? I didn't make a count btw, just went with the formula.

- 1 year ago

The answer is 24 stereoisomers. Your answer must be correct regarding optical isomers as I had tried it (in another way) and got the same number of optical isomers but I am not certain.

Yes, I got the answer. The number of optical isomers is $$4$$ along with that if you also account the position of methyl group on each ligand, you can have a possible of $$6$$ configurations wrt position of methyl group, thus $$4 \times 6 = 24$$. But this is the total number of all possible isomers.

- 1 year ago