Which of the following will have three stereoisomeric forms?
(i) `[Cr(NO_(3))_(3)(NH_(3))_(3)]`
(ii) `K_(3)[Co(C_(2)O_(4))_(3)]`
(iii) `K_(3)[CoCl_(2)(C_(2)O_(4))_(2)]`
(iv) `[CoBrCl(en)_(2)]`

Options

A. (iii) and (iv)
B. (i), (iii) and (iv)
C. (iv) only
D. all four

Explanation

To determine which of the given coordination compounds can exhibit three stereoisomeric forms, we need to analyze each compound based on its geometry and the nature of its ligands. Stereoisomers can include both geometrical (cis/trans) and optical isomers.

### Step-by-Step Solution:

1. **Identify the Coordination Number and Geometry:**
- The coordination number for all the given compounds is 6, indicating an octahedral geometry.

2. **Analyze Each Compound:**
- **(i) `[Cr(NO3)3(NH3)3]`:**
- This is an octahedral complex with three bidentate ligands (NO3) and three monodentate ligands (NH3).
- It can exhibit facial (fac) and meridional (mer) isomers.
- However, it does not have optical isomers because the fac and mer forms are not chiral.
- **Total Isomers: 2 (fac and mer)**

- **(ii) `K3[Co(C2O4)3]`:**
- This complex has three bidentate oxalate ligands.
- It can exhibit geometrical isomerism but does not have optical isomers since it is symmetric.
- **Total Isomers: 1 (only one structure)**

- **(iii) `K3[CoCl2(C2O4)2]`:**
- This complex has two bidentate oxalate ligands and two monodentate chloride ligands.
- It can exhibit cis and trans forms.
- The cis form can be optically active (two forms: D and L), while the trans form is not optically active.
- **Total Isomers: 3 (1 trans + 2 optical cis)**

- **(iv) `[CoBrCl(en)2]`:**
- This complex has two bidentate ethylenediamine (en) ligands and two monodentate ligands (Br and Cl).
- It can exhibit cis and trans forms.
- Both cis forms can be optically active (two forms: D and L), while the trans form is not optically active.
- **Total Isomers: 3 (1 trans + 2 optical cis)**

3. **Conclusion:**
- The compounds that exhibit three stereoisomeric forms are:
- (iii) `K3[CoCl2(C2O4)2]`
- (iv) `[CoBrCl(en)2]`

### Final Answer:
The compounds that will have three stereoisomeric forms are **(iii) and (iv)**.