The coordination compound has five stereoisomers (geometric plus optical isomers). Select all that apply...

A:NH4[Pt(NH3)Cl3]
B: tris(ethylenediamine)cobalt(III) sulfate
C: tetraaquadichloronickel(IV) chloride
D: [Co(NH3)2(en)Cl2]Cl
E: [Co(en)2Cl2]Br
F: sodium hexacyanoferrate(II)
None of these compounds

To determine the number of stereoisomers for each compound, we need to analyze the presence of chiral centers or geometric isomers.

A: NH4[Pt(NH3)Cl3]
This compound does not have any chiral centers or geometric isomers, so it does not have any stereoisomers.

B: tris(ethylenediamine)cobalt(III) sulfate
This compound has three chiral centers, so it has 2^3 = 8 possible stereoisomers (including both geometric and optical isomers).

C: tetraaquadichloronickel(IV) chloride
This compound does not have any chiral centers or geometric isomers, so it does not have any stereoisomers.

D: [Co(NH3)2(en)Cl2]Cl
This compound has two chiral centers, so it has 2^2 = 4 possible stereoisomers (including both geometric and optical isomers).

E: [Co(en)2Cl2]Br
This compound has one chiral center, so it has 2^1 = 2 possible stereoisomers (including both geometric and optical isomers).

F: sodium hexacyanoferrate(II)
This compound does not have any chiral centers or geometric isomers, so it does not have any stereoisomers.

Based on this analysis, the compounds with stereoisomers are:
- B: tris(ethylenediamine)cobalt(III) sulfate (8 stereoisomers)
- D: [Co(NH3)2(en)Cl2]Cl (4 stereoisomers)
- E: [Co(en)2Cl2]Br (2 stereoisomers)

Therefore, the answer is B, D, and E.

To determine the number of stereoisomers for each coordination compound, we need to assess the presence of geometric and optical isomers.

Geometric isomers arise when there is a difference in the spatial arrangement around a bond due to restricted rotation. Optical isomers result from the presence of chiral centers in the molecule, which create non-superimposable mirror images.

Let's evaluate each compound:

A: NH4[Pt(NH3)Cl3]
This compound does not have any chiral centers or double bonds, so it has no geometric or optical isomers.

B: tris(ethylenediamine)cobalt(III) sulfate
This compound contains three ethylenediamine (en) ligands, each of which can form geometric isomers. Additionally, there is a cobalt center (Co3+) that could potentially be a chiral center. Therefore, this compound has both geometric and optical isomers.

C: tetraaquadichloronickel(IV) chloride
This compound has four aqua (H2O) ligands, meaning they can exhibit geometric isomerism. However, there are no chiral centers, so no optical isomers are present.

D: [Co(NH3)2(en)Cl2]Cl
This complex contains two different ligands, NH3 (ammine) and en (ethylenediamine). The presence of both these ligands allows for the formation of geometric isomers. Additionally, the presence of the en ligand creates a chiral center, leading to the presence of optical isomers.

E: [Co(en)2Cl2]Br
Similar to compound D, this coordination compound contains the ligand en, allowing for both geometric and optical isomerism.

F: sodium hexacyanoferrate(II)
This compound does not possess any chiral centers or double bonds, so it has no geometric or optical isomers.

Based on this analysis, the compounds that have both geometric and optical isomers are B and D. Consequently, the correct answer is B and D.