 Solution

The average Pt oxidation state is $+3$, since six bromides balance the two metal atoms. A natural mixed valence assignment is alternating Pt(II), $d^8$, and Pt(IV), $d^6$, rather than two equivalent Pt(III) sites. Use neutral square-planar $\mathrm{P}{\mathrm{t}}^{\mathrm{II}}(\mathrm{N}{\mathrm{H}}_3{)}_2\mathrm{B}{\mathrm{r}}_2$ units and octahedral $\mathrm{P}{\mathrm{t}}^{\mathrm{IV}}(\mathrm{N}{\mathrm{H}}_3{)}_2\mathrm{B}{\mathrm{r}}_4$ units. At every site place two ammonia and two bromide ligands in an equatorial plane; the Pt(IV) sites have two additional axial bromides pointing toward the adjacent Pt(II) sites.

Along the chain the connectivity isEvery Pt–Br–Pt bridge is asymmetric: short to Pt(IV), long to Pt(II). The latter contact is a secondary axial interaction, consistent with the preference of Pt(II) for a square-planar primary coordination sphere. The Pt(IV) site is smaller and more strongly attracts the bridging bromides. A drawing with all axial distances identical would obscure the distinct local environments.

There are four equatorial bromides and two axial bridging bromides per two-Pt repeat, giving precisely $\mathrm{P}{\mathrm{t}}_2(\mathrm{N}{\mathrm{H}}_3{)}_4\mathrm{B}{\mathrm{r}}_6$. A direct Pt–Pt bond is not required. This is a proposed structure consistent with the stated composition and two environments; the formula alone does not determine every equatorial cis/trans arrangement or numerical bond length.