The shape of `[Cu(NH_3)_4]^(2+) is square planar, `Cu^(2+)` in this complex is

Options

A. `sp^3` hybridised
B. `dsp^2` hybridised
C. `sp^3d` hybridised
D. `sp^3d^2` hybridised

Answer

To determine the hybridization of Cu²⁺ in the complex \([Cu(NH_3)_4]^{2+}\) and confirm its square planar shape, we can follow these steps:

### Step 1: Identify the Coordination Number
The coordination number of a metal complex is determined by the number of ligands attached to the central metal ion. In this case, there are four ammonia (\(NH_3\)) ligands coordinated to the copper ion.

**Hint:** Count the number of ligands attached to the metal ion to find the coordination number.

### Step 2: Determine the Geometry of the Complex
The problem states that the shape of the complex is square planar. Square planar geometry is typically associated with a coordination number of 4.

**Hint:** Recall that square planar geometry is common for d⁸ metal ions, particularly in complexes with a coordination number of 4.

### Step 3: Identify the Type of Complex
Since the geometry is square planar, we can classify this complex as an inner orbital complex. Inner orbital complexes typically involve the use of d-orbitals from the inner shell of the metal ion.

**Hint:** Remember that inner orbital complexes often involve the hybridization of d-orbitals from the same shell as the valence electrons.

### Step 4: Determine the Hybridization
For square planar complexes, the hybridization can be either \(dsp^2\) or \(d^2sp^3\). However, since we are dealing with a d⁸ metal ion (Cu²⁺), the appropriate hybridization is \(dsp^2\).

- Copper in its +2 oxidation state has the electronic configuration of \([Ar] 3d^9\).
- When forming the complex, one of the 3d electrons is used for bonding, resulting in \(d^8\) configuration, which leads to \(dsp^2\) hybridization.

**Hint:** Use the electronic configuration of the metal ion to determine which orbitals are involved in hybridization.

### Step 5: Conclusion
Thus, the hybridization of Cu²⁺ in the complex \([Cu(NH_3)_4]^{2+}\) is \(dsp^2\).

**Final Answer:** The hybridization of Cu²⁺ in the complex \([Cu(NH_3)_4]^{2+}\) is \(dsp^2\).