A covalent molecole `AB_(4)` (not a complex) will have which of the following hybridisation if it is square planar.

Options

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

Explanation

To determine the hybridization of a covalent molecule \( AB_4 \) that has a square planar geometry, we can follow these steps:

### Step 1: Understand the Geometry
The question states that the molecule \( AB_4 \) is square planar. In a square planar geometry, the central atom (A) is surrounded by four atoms (B) in a plane, with 90° angles between them.

### Step 2: Identify the Steric Number
The steric number is calculated as the sum of the number of sigma bonds and lone pairs of electrons around the central atom. For \( AB_4 \):
- There are 4 sigma bonds (one for each B atom).
- We need to determine the number of lone pairs on the central atom.

Since the geometry is square planar, we can infer that there are 2 lone pairs of electrons on the central atom (A) to achieve this shape.

Thus, the steric number \( SN \) is:
\[ SN = \text{Number of sigma bonds} + \text{Number of lone pairs} = 4 + 2 = 6 \]

### Step 3: Determine the Hybridization
The hybridization can be determined from the steric number:
- A steric number of 6 corresponds to \( sp^3d^2 \) hybridization.

### Step 4: Conclusion
Thus, the hybridization of the molecule \( AB_4 \) with a square planar geometry is \( sp^3d^2 \).

### Final Answer
The hybridization of the covalent molecule \( AB_4 \) that is square planar is \( sp^3d^2 \).

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