The isomer will be cis if the distance between two chlorine atoms is less than twice the $C - Cl$ bond length. This is the case in the given complex as $2.8 A<2 \times$ $2 A$
Thus the given complex is cis square planar complex.
Hence, the correct option is $C$
$(en = NH_2CH_2CH_2NH_2)$
$(i)\, [Cr(NO_3)_3 (NH_3)_3]$ $(ii)\, K_3[Co(C_2O_4)_3]$
$(iii)\, K_3[CoCl_2(C_2O_4)_2]$ $(iv)\, [CoBrCl(en)_2]$