Explanation:

Let's analyze each option one by one to determine which one will have a meso-isomer.

a) 2, 3-Dichloropentane:
The compound has two chiral centers (carbon atoms with four different substituents), which means it can potentially have four stereoisomers.
However, since it has two identical chlorine atoms in the 2nd and 3rd positions, these two stereoisomers will be superimposable mirror images of each other and, therefore, not meso-isomers. Hence, option 'a' does not have a meso-isomer.

b) 2, 3-Dichlorobutane:
Similar to the previous compound, this one also has two chiral centers. However, in this case, the chlorine atoms are present in the 2nd and 3rd positions, which are not adjacent to each other.
This arrangement allows for the possibility of having a meso-isomer. To form a meso-isomer, the molecule needs to have an internal plane of symmetry. In this case, if we switch the substituents on the chiral centers, we can see that the molecule can be divided into two identical halves, which indicates the presence of a plane of symmetry.
Therefore, option 'b' is the correct answer as it has a meso-isomer.

c) 2-Chlorobutane:
This compound has only one chiral center, which means it can only have two stereoisomers. Since it does not have multiple chiral centers or a plane of symmetry, it cannot have a meso-isomer.

d) 2-Hydroxypropanoic acid:
This compound also has only one chiral center. Although it has a hydroxyl group (-OH) attached to it, it does not have a plane of symmetry. Therefore, it cannot have a meso-isomer.

In conclusion, out of the given options, only option 'b' (2, 3-Dichlorobutane) has a meso-isomer.