The two structures shown (left and right) are stereoisomers of tartaric acid drawn with opposite wedge patterns at both carbons. Which statements are correct?

In the left structure the centres are $R$ and $S$; inverting both gives $S$ and $R$ in the right structure. Because the two ends of tartaric acid are identical, (2S,3R) renumbered from the other end is (2R,3S). Both drawings are therefore the meso acid, which has an internal mirror plane, is achiral and is optically inactive. A compound that is superimposable on its mirror image cannot have an enantiomer, so (C) is wrong; the right-hand drawing is simply the left one viewed from the other side.