Which of the following statements is not true about meso compounds?

They have an internal mirror plane.

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They are optically active.

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They are optically inactive.

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They are achiral.

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Solution

The correct option is B They are optically active.
Meso compounds are optically inactive due to internal compensation, caused because of equal and opposite rotation of plane polarised light by two adjacent chiral carbons in the molecule.
Hence, the correct option is $B$

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Q. Match List - I with List II and select the correct answer using the code given below the lists.

List -I	List -II
A. Meso compounds	1. An equimolar mixture of enantiomer
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C. Diastereoisomers	3. Non superimposable mirror images
D. Racemic mixture	4. An optically inactive compound whose molecules are achiral even though they contain chiral centre

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Which of the following statements is correct regarding the chirality (optical isomerism) of the cis and trans isomers of the type $M (a a)_{2} b_{2}$ (M stands for a metal, a and b are achiral ligands and aa is a bidentate ligand) ?

S1: Geometrical isomerism along double bond in cycloalkenes is possible with at least eight atoms in the ring.

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can have two meso forms which are mirror images and enantiomers of each other.

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Q. Consider the following statements.
I. Meso form is optically inactive due to the absence of dissymmetry in the molecule.
II. Compound will be optically active if it is a dissymmetric molecule.
III. Geometrical isomers are stereoisomers.
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Which of the following statements is not true about meso compounds?

The correct option is B They are optically active.Meso compounds are optically inactive due to internal compensation, caused because of equal and opposite rotation of plane polarised light by two adjacent chiral carbons in the molecule.Hence, the correct option is B

The correct option is B They are optically active.
Meso compounds are optically inactive due to internal compensation, caused because of equal and opposite rotation of plane polarised light by two adjacent chiral carbons in the molecule.
Hence, the correct option is $B$