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.

Please help me through this problem!

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TopNewestThe 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

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

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

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

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