Question

Identify the pair of molecules which have the same hybridisation as the hybridisation in Xenon (II) fluoride.

• A. \( \text{XeO}_3 \), \( \text{SF}_4 \)
• B. \( \text{BrF}_5 \), \( \text{PF}_5 \)
• C. \( \text{ClF}_3 \), \( \text{SF}_4 \)
• D. \( \text{PCl}_3 \), \( \text{NH}_3 \)

Hint:

Isoelectronic species have the same number of electrons. To identify them, calculate the total number of electrons for each species in the set.

Answer 1

The Correct Option is C

Step 1: Known Information.
The molecule in question is Xenon (II) fluoride (\( \text{XeF}_2 \)).
We need to determine the hybridization of \( \text{XeF}_2 \) and identify another pair of molecules with the same hybridization.
Step 2: Determine the Hybridization of \( \text{XeF}_2 \).
1. Molecular Formula: \( \text{XeF}_2 \)
2. Central Atom: Xenon (Xe)
3. Valence Electrons of Xe: Xenon has 8 valence electrons.
4. Bonding Electrons: Each fluorine atom forms a single bond with Xe, contributing 2 electrons per bond.
Total bonding electrons = \( 2 \times 2 = 4 \) electrons.
5. Lone Pairs: Remaining electrons are lone pairs.
Lone pairs = \( 8 - 4 = 4 \) electrons (2 lone pairs).
6. Geometry and Hybridization:
\( \text{XeF}_2 \) has a linear geometry.
Linear geometry corresponds to sp hybridization.
Thus, the hybridization of \( \text{XeF}_2 \) is sp.
Step 3: Identify Molecules with sp Hybridization.
To have sp hybridization, a molecule must:
- Have a linear geometry.
- Have 2 regions of electron density around the central atom (2 sigma bonds).
Let's analyze the options: 1. Option 1: \( \text{XeO}_3 \), \( \text{SF}_4 \)
\( \text{XeO}_3 \): Xenon has 3 bonding pairs and 2 lone pairs, resulting in trigonal bipyramidal geometry (sp³d hybridization).
\( \text{SF}_4 \): Sulfur has 4 bonding pairs and 1 lone pair, resulting in seesaw geometry (sp³d hybridization).
Neither molecule has sp hybridization.
2. Option 2: \( \text{BrF}_5 \), \( \text{PF}_5 \)
\( \text{BrF}_5 \): Bromine has 5 bonding pairs and 1 lone pair, resulting in square pyramidal geometry (sp³d² hybridization).
\( \text{PF}_5 \): Phosphorus has 5 bonding pairs, resulting in trigonal bipyramidal geometry (sp³d hybridization).
Neither molecule has sp hybridization.
3. Option 3: \( \text{ClF}_3 \), \( \text{SF}_4 \)
\( \text{ClF}_3 \): Chlorine has 3 bonding pairs and 2 lone pairs, resulting in T-shaped geometry (sp³d hybridization).
\( \text{SF}_4 \): Sulfur has 4 bonding pairs and 1 lone pair, resulting in seesaw geometry (sp³d hybridization).
Neither molecule has sp hybridization.
4. Option 4: \( \text{PCl}_3 \), \( \text{NH}_3 \)
\( \text{PCl}_3 \): Phosphorus has 3 bonding pairs and 1 lone pair, resulting in trigonal pyramidal geometry (sp³ hybridization).
\( \text{NH}_3 \): Nitrogen has 3 bonding pairs and 1 lone pair, resulting in trigonal pyramidal geometry (sp³ hybridization).
Neither molecule has sp hybridization.
Step 4: Correct Answer. The correct pair of molecules with sp hybridization is: \[ \boxed{\text{XeF}_2, \text{CO}_2} \]