The complex(es), which can exhibit the type of isomerism shown by [Pt(NH$$_3$$)$$_2$$Br$$_2$$], is(are)

[en = H$$_2$$NCH$$_2$$CH$$_2$$NH$$_2$$]

The complex $$[Pt(NH_3)_2Br_2]$$ is square-planar with the composition $$MA_2B_2$$ (here $$A = NH_3$$ and $$B = Br^-$$).

Square-planar $$MA_2B_2$$ species can exist in two geometrical arrangements:
  • cis : the two $$A$$ ligands are adjacent.
  • trans : the two $$A$$ ligands are opposite.
Hence the “type of isomerism” referred to in the question is cis-trans (geometrical) isomerism.

We must therefore look for complexes in the options that are capable of showing the same cis-trans geometrical isomerism.

Option A : $$[Pt(en)(SCN)_2]$$
  • Coordination number = 4, geometry = square-planar.
  • The bidentate ligand $$en$$ occupies two adjacent positions compulsorily (it must be cis).
  • The two $$SCN^-$$ ligands are forced into the remaining two adjacent positions, so only one spatial arrangement is possible.
  • Therefore no cis-trans isomerism is possible. (The complex can show linkage isomerism through $$S$$ vs $$N$$ binding, but that is a different type.)
  ⇒ Does not match the required isomerism.

Option B : $$[Zn(NH_3)_2Cl_2]$$
  • $$Zn^{2+}$$ with coordination number 4 usually forms a tetrahedral complex.
  • A tetrahedral $$MA_2B_2$$ arrangement is symmetric; cis and trans positions are indistinguishable.
  • Therefore no geometrical isomerism exists.
  ⇒ Does not match the required isomerism.

Option C : $$[Pt(NH_3)_2Cl_4]$$
  • Coordination number = 6, oxidation state $$+4$$, geometry = octahedral.
  • General formula $$MA_2B_4$$ (two $$NH_3$$ and four $$Cl^-$$).
  • In an octahedron, two distinct arrangements exist:
    - cis : the two $$NH_3$$ ligands are 90° apart.
    - trans : the two $$NH_3$$ ligands are 180° apart.
  • Thus the complex does show cis-trans geometrical isomerism.
  ⇒ Matches the required type.

Option D : $$[Cr(en)_2(H_2O)(SO_4)]^{+}$$
  • Coordination number = 6, geometry = octahedral.
  • Two bidentate $$en$$ ligands occupy four sites; $$H_2O$$ and $$SO_4^{2-}$$ occupy the remaining two sites.
  • For the monodentate pair ($$H_2O$$ vs $$SO_4^{2-}$$) we can have:
    - cis : $$H_2O$$ adjacent to $$SO_4^{2-}$$.
    - trans : $$H_2O$$ opposite $$SO_4^{2-}$$.
  • Hence the complex exhibits cis-trans geometrical isomerism; in addition it can be optically active, but that does not affect the present question.
  ⇒ Matches the required type.

Therefore, the complexes capable of showing the same cis-trans geometrical isomerism as $$[Pt(NH_3)_2Br_2]$$ are:

Option C which is: $$[Pt(NH_3)_2Cl_4]$$
Option D which is: $$[Cr(en)_2(H_2O)(SO_4)]^{+}$$

Answer: Option C and Option D.