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Coordination Compounds question

2023 · 13 Apr · Shift 2 · Q4

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Coordination Compounds
2023 · 13 Apr · Shift 2 · Q4

Coordination Compounds question

2023 · 13 Apr · Shift 2 · Q4

JEE MainChemistryCoordination CompoundsMCQ+4 / −1

The total number of stereoisomers for the complex

\left[\mathrm{Cr}(o x)_{2} \mathrm{ClBr}\right]^{3-}

(where

o x=

oxalate) is :

A
1
B
3
C
2
D
4

View written solutionFree

Correct answer: B

Identify the geometry and ligand type

The complex is $[\mathrm{Cr}(ox)_2ClBr]^{3-}$ where each $ox$ (oxalate) is a bidentate ligand.

So chromium is 6-coordinate, hence the complex is octahedral.

There are:

two identical bidentate ligands: $(ox)_2$
two different monodentate ligands: $Cl$ and $Br$

Thus the complex is of type: $[M(AA)_2ab]$ where $AA$ is a symmetrical bidentate ligand and $a,b$ are different monodentates.

Possible geometrical isomers

For an octahedral complex of type $[M(AA)_2ab]$ , the two monodentate ligands $a$ and $b$ can be arranged in two ways:

cis: $Cl$ and $Br$ adjacent
trans: $Cl$ and $Br$ opposite

So there are 2 geometrical isomers.

Check optical activity of each geometrical isomer

(i) cis- $[\mathrm{Cr}(ox)_2ClBr]^{3-}$

With two bidentate ligands in octahedral geometry, the cis form becomes chiral. It exists as a pair of enantiomers:

$\Delta$
$\Lambda$

So cis-form gives 2 stereoisomers.

(ii) trans- $[\mathrm{Cr}(ox)_2ClBr]^{3-}$

The trans form has a symmetry element and is achiral. So it gives only 1 stereoisomer.

Total number of stereoisomers

Therefore, $\text{Total stereoisomers} = 2\text{ (from cis)} + 1\text{ (from trans)} = 3$

Match with options

The correct option is: $\boxed{3}$ which is Option B.

Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

Hence, they agree.

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