The number of possible optical isomers for the complexes MA2B2 with sp3 and dsp2 hybridised metal at

Q: Why doesn't a tetrahedral complex with two identical ligands show optical isomerism?

Because you can always pass a plane through the central metal and the two unique ligands (or one set of identical ligands), which will perfectly bisect the other two identical ligands, creating a plane of symmetry.

Q: Can a square planar complex ever be optically active?

Standard square planar complexes with simple unidentate ligands are always achiral because the plane containing the molecule itself is a plane of symmetry. Optical activity is only possible in highly specific cases with bulky, chiral ligands that force the molecule out of planarity.

Q: What about $MABCD$ tetrahedral complexes? Do they show optical isomerism?

Yes! Just like a carbon atom attached to four different groups, a tetrahedral metal complex with four different unidentate ligands ($MABCD$) lacks any plane of symmetry and is therefore chiral and optically active.

Q: Why do we check for a plane of symmetry instead of just drawing mirror images?

Checking for a plane of symmetry is a much faster and more reliable mental shortcut. If a molecule has a plane of symmetry, it is mathematically guaranteed to be superimposable on its mirror image (achiral).

Q: Does the existence of cis-trans isomerism guarantee optical isomerism?

No. Cis-trans isomerism is a type of geometrical isomerism (diastereomers), which is entirely different from optical isomerism (enantiomers). A molecule can have geometrical isomers while all of them remain optically inactive, as seen in square planar complexes.

Question

The number of possible optical isomers for the complexes $MA_2B_2$ with $sp^3$ and $dsp^2$ hybridised metal atom, respectively, is
Note : $A$ and $B$ are unidentate neutral and unidentate monoanionic ligands, respectively.

Options:

(A) 0 and 0

(B) 0 and 1

(C) 2 and 2

(D) 0 and 2

Sigma Percentile · Accepted Answer

### The Essence of Optical Isomerism

Before we dive into the specific geometries, we must establish the golden rule of optical isomerism: **Chirality**. For any molecule to exhibit optical isomerism, it must be chiral. This means its mirror image must be non-superimposable on the original molecule.

In simpler terms, if you can find a **Plane of Symmetry (POS)** or a **Center of Symmetry (COS)** within the molecule, it is achiral and will not show optical isomerism. Our mission is to hunt for these elements of symmetry in the given $MA_2B_2$ complexes.

### Analyzing the Tetrahedral Geometry ($sp^3$)

When a central metal atom undergoes $sp^3$ hybridization, it adopts a **tetrahedral geometry**. Imagine the central metal $M$ sitting at the core of a tetrahedron, with the four ligands ($A, A, B, B$) occupying the four corners.

At first glance, a tetrahedral molecule with different ligands might seem like a good candidate for chirality. However, look closely at the $MA_2B_2$ arrangement. If we pass an imaginary plane directly through the central metal $M$ and the two $A$ ligands, this plane will perfectly bisect the angle formed by the two $B$ ligands.

Because the two $B$ ligands are identical, one $B$ reflects perfectly onto the other $B$ across this plane. This is a clear **Plane of Symmetry**. Since the molecule possesses a POS, it is superimposable on its mirror image. Therefore, it is optically inactive, yielding **0 optical isomers**.

### Analyzing the Square Planar Geometry ($dsp^2$)

Now, let's shift our focus to the $dsp^2$ hybridization, which results in a **square planar geometry**. In this arrangement, the central metal $M$ and all four ligands lie flat in the exact same 2D plane.

For an $MA_2B_2$ square planar complex, we can have two geometrical isomers: **cis** (where identical ligands are adjacent) and **trans** (where identical ligands are opposite).

But what about optical isomerism? There is a massive catch here. Because all the atoms lie in a single plane, that **molecular plane itself acts as a Plane of Symmetry**! If you were to slice the molecule horizontally through this plane, the top half (which is essentially nothing) perfectly mirrors the bottom half.

Because of this inherent molecular plane of symmetry, all standard square planar complexes are achiral. Thus, both the *cis* and *trans* forms are optically inactive, yielding **0 optical isomers**.

### The Final Verdict

We have thoroughly analyzed both hybridizations. The tetrahedral $MA_2B_2$ complex has a bisecting plane of symmetry, and the square planar $MA_2B_2$ complex has a molecular plane of symmetry.

In both scenarios, the molecules are achiral. Therefore, the number of possible optical isomers is **0 and 0**. The correct option is undeniably **(a)**.

Animated Solution for Chemistry - Coordination Compounds: The number of possible optical isomers for the complexes $M A_{2} B_{2}$ with $s p^{3}$ and $d s p^{2}$ hybridised metal atom, respectively, is Note : $A$ and $B$ are unidentate neutral and unidentate monoanionic ligands, respectively.

Select Answer:

Visualized Solution

$M A_{2} B_{2}$ Complexes

Condition for Optical Isomerism

$s p^{3}$ Hybridization

Symmetry in Tetrahedral $M A_{2} B_{2}$

$d s p^{2}$ Hybridization

Symmetry in Square Planar

Final Conclusion

The Way Forward

The Sigma Insight: Nomenclature, Isomerism, Importance and Werner's Theory

The Essence of Optical Isomerism

Analyzing the Tetrahedral Geometry ( $s p^{3}$ )

Analyzing the Square Planar Geometry ( $d s p^{2}$ )

The Final Verdict

Similar Questions

The number of optical isomers possible for $[C r (C_{2} O_{4})_{3}]^{3 -}$ is ...

The number of geometric isomers possible for the complex $[CoL_{2} Cl_{2}]^{-}$ ( $L = H_{2} NCH_{2} CH_{2} O^{-}$ ) is

Which of the following has an optical isomer?

The total number of isomers for a square planar complex $[M (F) (C l) (S C N) (N O_{2})]$ is

Which one of the following complexes shows optical isomerism?

Which of the following complex species is not expected to exhibit optical isomerism?

The number of geometrical isomers found in the metal complexes $[P tC l_{2} (N H_{3})_{2}]$ , $[N i (C O)_{4}]$ , $[R u (H_{2} O)_{3} C l_{3}]$ and $[C o C l_{2} (N H_{3})_{4}]^{+}$ respectively, are

The total number of all possible isomers for the square planar complex with formula $K [M (N C S) (N O_{2}) (g l y)]$ is ______ ( $M =$ metal ion and $g l y = N H_{2} C H_{2} C O O^{-}$ )

Which of the following compounds shows optical isomerism ?

Which one of the following has an optical isomer? (en = ethylenediamine)