The total number of possible isomers for $$[Pt(NH_3)_4 Cl_2]Br_2$$ is ___.

The compound is written as $$[Pt(NH_3)_4Cl_2]Br_2$$.
Square brackets show the ligands that are directly bonded to the metal; ions written outside act only as counter-ions. Because both $$Cl^-$$ and $$Br^-$$ can serve as ligands, an exchange between the inner and outer sphere is possible. Such exchanges give ionisation isomers.

Step 1 - Enumerate all ionisation isomers
Keep the overall composition $$Pt(NH_3)_4Cl_2Br_2$$ fixed and move 0, 1, or 2 $$Br^-$$ ions into the coordination sphere:

  • Case I : $$[Pt(NH_3)_4Cl_2]^{2+}\;+\;2Br^-$$   (both $$Cl^-$$ inside, none $$Br^-$$ inside)
  • Case II: $$[Pt(NH_3)_4ClBr]^{2+}\;+\;Br^-+Cl^-$$ (one $$Cl^-$$ and one $$Br^-$$ inside)
  • Case III: $$[Pt(NH_3)_4Br_2]^{2+}\;+\;2Cl^-$$ (both $$Br^-$$ inside, none $$Cl^-$$ inside)

Step 2 - Geometrical (cis / trans) possibilities for each case
Platinum(IV) is six-coordinate; the geometry is octahedral.

  • For $$[Pt(NH_3)_4Cl_2]^{2+}$$ (type $$Ma_4b_2$$) we can place the two identical $$Cl^-$$ ligands
    - trans (opposite)   or  - cis (adjacent).   ⇒ 2 isomers

  • For $$[Pt(NH_3)_4Br_2]^{2+}$$ (again $$Ma_4b_2$$) we likewise get
    - trans   or  - cis.   ⇒ 2 isomers

  • For $$[Pt(NH_3)_4ClBr]^{2+}$$ (type $$Ma_4bc$$) the two different halides can be
    - trans to each other   or  - cis to each other.   ⇒ 2 isomers

No optical isomers arise because each of the above geometries possesses at least one mirror plane.

Step 3 - Total count
Case I gives 2, Case II gives 2, Case III gives 2   ⇒  $$2+2+2 = 6$$ isomers in all.

Hence, the total number of possible isomers for $$[Pt(NH_3)_4Cl_2]Br_2$$ is 6.