An organic compound, 2,3-dichlorobutane, exhibits stereoisomerism...

The given compound is 2,3-dichlorobutane, whose structure is $CH_3-CH(Cl)-CH(Cl)-CH_3$.

It contains two similar chiral carbon atoms ($n = 2$). Since the molecule can be divided into two equal halves, it is symmetrical.

For a symmetrical molecule with an even number of chiral centers ($n$), the number of optically active isomers is given by $a = 2^{n-1}$.

Substituting $n = 2$, we get $a = 2^{2-1} = 2$.

The number of meso forms is given by $m = 2^{\dfrac{n}{2}-1}$.

Substituting $n = 2$, we get $m = 2^{\dfrac{2}{2}-1} = 2^0 = 1$.

Thus, there are 2 optically active isomers (a pair of enantiomers) and 1 meso form.

Answer: 2 and 1