JEE Chemistry 2025 Question with Solution

Given: We need to identify which given molecules have square pyramidal geometry.

Find: The correct option containing only molecules with square pyramidal shape.

Using VSEPR theory, square pyramidal geometry corresponds to species of type $AX_5E$, that is, 5 bonding pairs and 1 lone pair on the central atom.

For $\mathrm{BrF_5}$:

There are 5 bond pairs and 1 lone pair on bromine, so its electron geometry is octahedral and its molecular geometry is square pyramidal.

For $\mathrm{XeOF_4}$:

Xenon has one bond with oxygen, four bonds with fluorine, and one lone pair. Thus it also has 5 bonding regions and 1 lone pair, giving square pyramidal geometry.

For $\mathrm{SbF_5}$:

There are 5 bond pairs and no lone pair, so the geometry is trigonal bipyramidal, not square pyramidal.

For $\mathrm{PCl_5}$:

There are 5 bond pairs and no lone pair, so the geometry is again trigonal bipyramidal.

Therefore, the molecules having square pyramidal geometry are $\mathrm{BrF_5}$ and $\mathrm{XeOF_4}$.

The correct option is A.

Given: The molecules listed are $\mathrm{BrF_5}$, $\mathrm{XeOF_4}$, $\mathrm{SbF_5}$, and $\mathrm{PCl_5}$.

Find: Which of these show square pyramidal geometry.

1. A square pyramidal molecule is generally of type $AX_5E$.
2. This means the central atom must have:

• 5 bonding pairs
• 1 lone pair
• total steric number $= 6$

3. Steric number $6$ gives octahedral electron-pair geometry.
4. With one lone pair in an octahedral arrangement, the molecular shape becomes square pyramidal.

Now examine each molecule:

For $\mathrm{BrF_5}$:

• Bromine has 7 valence electrons.
• It forms 5 bonds with fluorine.
• One lone pair remains.
• So it is of type $AX_5E$. Hence, $\mathrm{BrF_5}$ is square pyramidal.

For $\mathrm{XeOF_4}$:

• Xenon has 8 valence electrons.
• It forms one bond with oxygen and four bonds with fluorine.
• One lone pair remains on xenon.
• Effective arrangement is again $AX_5E$ with steric number $6$. Hence, $\mathrm{XeOF_4}$ is square pyramidal.

For $\mathrm{SbF_5}$:

• Antimony forms 5 bonds and has no lone pair.
• Type is $AX_5$. Hence, its geometry is trigonal bipyramidal.

For $\mathrm{PCl_5}$:

• Phosphorus forms 5 bonds and has no lone pair.
• Type is $AX_5$. Hence, its geometry is trigonal bipyramidal.

So only $\mathrm{BrF_5}$ and $\mathrm{XeOF_4}$ satisfy the condition.

Therefore, the correct option is A.