Identify the incorrect statements about `[Cu(NH_(3))_(4)]^(2+)`

Options

A. The complex is tetrahedral
B. The complex is square planar
C. `Cu^(2+)` in the complex is `dsp^(2)` hybridised
D. `Cu^(2+)` in the complex is `sp^(2)` hybridised

Answer

To identify the incorrect statements about the complex ion \([Cu(NH_3)_4]^{2+}\), we will analyze the oxidation state of copper, its electronic configuration, hybridization, and geometry.

### Step-by-Step Solution:

1. **Determine the oxidation state of Cu in \([Cu(NH_3)_4]^{2+}\)**:
- Let the oxidation state of Cu be \(x\).
- Each ammonia (\(NH_3\)) is a neutral ligand, contributing 0 to the charge.
- The overall charge of the complex is \(+2\).
- Therefore, we can write the equation:
\[ x + 0 = +2 \implies x = +2 \]
- Thus, the oxidation state of Cu is \(+2\).

**Hint**: Remember that neutral ligands do not affect the overall charge of the complex.

2. **Write the electronic configuration of Cu**:
- The atomic number of Cu is 29. Its ground state electronic configuration is:
\[ [Ar] \, 3d^{10} \, 4s^1 \]
- In the \(+2\) oxidation state, Cu loses two electrons, typically from the \(4s\) and \(3d\) orbitals:
\[ [Ar] \, 3d^9 \]

**Hint**: For transition metals, the \(4s\) electrons are usually lost before the \(3d\) electrons.

3. **Determine the hybridization and geometry**:
- Ammonia (\(NH_3\)) is a strong field ligand and causes pairing of electrons in the \(3d\) subshell.
- The electronic configuration after pairing would be:
\[ [Ar] \, 3d^8 \, 4s^0 \, 4p^1 \]
- The hybridization can be determined by the number of orbitals involved:
- \(3d\) (2 orbitals) + \(4s\) (0 orbitals) + \(4p\) (1 orbital) = 4 orbitals
- This corresponds to \(dsp^2\) hybridization, leading to a square planar geometry.

**Hint**: The type of hybridization can often be inferred from the number of ligands and their arrangement around the central metal ion.

4. **Analyze the statements**:
- **Statement A**: The complex is tetrahedral. **(Incorrect)**
- We established that the geometry is square planar due to \(dsp^2\) hybridization.
- **Statement B**: The complex is square planar. **(Correct)**
- **Statement C**: Cu\(^{2+}\) in the complex is \(dsp^2\) hybridized. **(Correct)**
- **Statement D**: Cu\(^{2+}\) in the complex is \(sp^2\) hybridized. **(Incorrect)**
- The correct hybridization is \(dsp^2\), not \(sp^2\).

### Conclusion:
The incorrect statements about \([Cu(NH_3)_4]^{2+}\) are:
- A: The complex is tetrahedral.
- D: Cu\(^{2+}\) in the complex is \(sp^2\) hybridized.