The shape of `[Cu(NH_(3))_(4)]SO_(4)`​ is :

Options

A. Square planar
B. Pyramidal
C. Octahedral
D. Tetrahedral

Explanation

To determine the shape of the complex ion \([Cu(NH_3)_4]SO_4\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Complex Ion**:
The compound consists of a copper complex \([Cu(NH_3)_4]^{2+}\) and a sulfate ion \(SO_4^{2-}\). The sulfate ion carries a charge of -2, which means the copper complex must have a charge of +2.

2. **Determine the Oxidation State of Copper**:
Since \(NH_3\) is a neutral ligand (it has no charge), the oxidation state of copper in the complex can be determined directly from the charge of the complex. Therefore, the oxidation state of Cu in \([Cu(NH_3)_4]^{2+}\) is +2.

3. **Electron Configuration of Copper**:
The ground-state electron configuration of copper (Cu) is \([Ar] 3d^{10} 4s^1\). When copper is in the +2 oxidation state, it loses two electrons: one from the 4s orbital and one from the 3d orbital. This results in the configuration \([Ar] 3d^9\).

4. **Hybridization**:
The coordination number of the complex is 4 (due to the four ammonia ligands). In this case, ammonia is a strong field ligand, which causes pairing of electrons in the d-orbitals. The 3d orbitals will be involved in hybridization. The hybridization for a coordination number of 4 with strong field ligands is typically \(dsp^2\).

5. **Shape of the Complex**:
The \(dsp^2\) hybridization leads to a square planar geometry. Therefore, the shape of the complex \([Cu(NH_3)_4]^{2+}\) is square planar.

6. **Magnetic Properties**:
Since the electron configuration of Cu in the +2 state is \(3d^9\), there is one unpaired electron. This indicates that the complex is paramagnetic.

### Final Answer:
The shape of \([Cu(NH_3)_4]SO_4\) is **square planar**.

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