This molecule has a plane of symmetry. Imagine a vertical plane go through $\ce{C}$-3 atom of pentane chain. One side of that plane is the mirror image of other side. Hence, this is not optically active molecule by definition. It is called meso-isomer. If you look at (R,S)-configuration and number the chains as they are drown, first compound is (2R,4S)-2,4-dichloropentane and second compound is (2S,4R)-2,4-dichloropentane. But, since each have plane of symmetry, you may number second compound from opposite direction. Then it is (2R,4S)-2,4-dichloropentane. Therefore, they are identical.
The two enantiomers relevant to this meso-isomer are (2R,4R)-2,4-dichloropentane and (2S,4S)-2,4-dichloropentane.
To understand more on enantiomers and relevant meso-isomer, study halogenation of cis-/trans-2-butene or 1,2-diphenylethene (stilbene).