Which of the following are square planar complexes ?
(i) `[AuCl_(4)]^(-)`
(ii) `[PtCl_(4)]^(2-)`
(iii) `[Mn(Br)_(4)]^(2-)`
(iv) `[Cu (NH_(3) )_(4) ]^(2+)`
Select the correct answer using the codes given below: -

Options

A. i and ii only
B. ii and iii only
C. ii and iv only
D. i, ii, and iv only

Answer

To determine which of the given complexes are square planar, we will analyze each complex based on the oxidation state of the metal and the nature of the ligands involved.

### Step-by-Step Solution:

1. **Analyze `[AuCl_(4)]^(-)`**:
- Gold (Au) has an electronic configuration of [Xe] 4f14 5d10 6s1.
- In the complex `[AuCl_(4)]^(-)`, Au is in the +3 oxidation state, leading to an electronic configuration of 5d8.
- Chloride (Cl) is a weak field ligand, but due to the presence of a 5d orbital, pairing occurs.
- The geometry is determined to be dsp², which is characteristic of square planar complexes.
- **Conclusion**: `[AuCl_(4)]^(-)` is a square planar complex.

2. **Analyze `[PtCl_(4)]^(2-)`**:
- Platinum (Pt) has an electronic configuration of [Xe] 4f14 5d9 6s1.
- In the complex `[PtCl_(4)]^(2-)`, Pt is in the +2 oxidation state, leading to an electronic configuration of 5d8.
- Similar to Au, Cl is a weak ligand, but pairing occurs in the 5d orbital.
- The geometry is also dsp², indicating a square planar arrangement.
- **Conclusion**: `[PtCl_(4)]^(2-)` is a square planar complex.

3. **Analyze `[Mn(Br)_(4)]^(2-)`**:
- Manganese (Mn) has an electronic configuration of [Ar] 3d5 4s2.
- In the complex `[Mn(Br)_(4)]^(2-)`, Mn is in the +2 oxidation state, leading to an electronic configuration of 3d5.
- Bromide (Br) is a weak field ligand, and no pairing occurs in the 3d orbitals.
- The geometry is sp³, which is characteristic of tetrahedral complexes.
- **Conclusion**: `[Mn(Br)_(4)]^(2-)` is NOT a square planar complex.

4. **Analyze `[Cu(NH3)_(4)]^(2+)`**:
- Copper (Cu) has an electronic configuration of [Ar] 3d10 4s1.
- In the complex `[Cu(NH3)_(4)]^(2+)`, Cu is in the +2 oxidation state, leading to an electronic configuration of 3d9.
- Ammonia (NH3) is a strong field ligand, leading to pairing in the 3d orbitals.
- The geometry is dsp², indicating a square planar arrangement.
- **Conclusion**: `[Cu(NH3)_(4)]^(2+)` is a square planar complex.

### Final Answer:
The complexes that are square planar are:
- (i) `[AuCl_(4)]^(-)`
- (ii) `[PtCl_(4)]^(2-)`
- (iv) `[Cu(NH3)_(4)]^(2+)`

Thus, the correct answer is **Option D: (i), (ii), and (iv)**.