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 the covalent molecule \( AB_4 \) that has a square planar shape, we can follow these steps:

### Step 1: Identify the steric number
The steric number is the sum of the number of bonded atoms and the number of lone pairs around the central atom. For the molecule \( AB_4 \):
- There are 4 bonded atoms (4 B atoms).
- To achieve a square planar shape, we need to have 2 lone pairs.

Thus, the steric number is:
\[ \text{Steric Number} = \text{Number of Bonded Atoms} + \text{Number of Lone Pairs} = 4 + 2 = 6 \]

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

### Step 3: Confirm the geometry
With \( sp^3d^2 \) hybridization, the electron geometry is octahedral. However, when there are 2 lone pairs, they will occupy positions opposite each other to minimize repulsion, resulting in a square planar molecular shape.

### Conclusion
Thus, the hybridization of the covalent molecule \( AB_4 \) that has a square planar shape is \( sp^3d^2 \).

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