Question

In which one of the following compounds of xenon, highest number of lone pairs of electrons is present on xenon?

• A. XeF$_6$
• B. XeF$_4$
• C. XeOF$_4$
• D. XeO$_3$
• E. XeF$_2$

Hint:

For noble gases: \[ \text{Lone pairs} = \frac{8 - \text{number of bonds}}{2} \]

Answer 1

Answer: (E)

Concept: To determine the number of lone pairs on xenon, we must:
• Calculate total valence electrons of Xe
• Subtract electrons used in bonding
• Remaining electrons form lone pairs Xenon belongs to noble gases: \[ \text{Valence electrons of Xe} = 8 \]

Step 1: Analyze XeF$_2$
• Xe forms 2 bonds with fluorine atoms
• Each bond uses 1 electron from Xe
• Total electrons used in bonding = 2 Remaining electrons: \[ 8 - 2 = 6 \text{ electrons} \] Number of lone pairs: \[ \frac{6}{2} = 3 \text{ lone pairs} \]

Step 2: Analyze XeF$_4$
• Xe forms 4 bonds
• Electrons used = 4 Remaining electrons: \[ 8 - 4 = 4 \] Lone pairs: \[ \frac{4}{2} = 2 \]

Step 3: Analyze XeF$_6$
• Xe forms 6 bonds
• Electrons used = 6 Remaining: \[ 8 - 6 = 2 \] Lone pairs: \[ 1 \]

Step 4: Analyze XeOF$_4$
• Xe bonded to 4 F and 1 O
• Total bonds = 5 Remaining electrons: \[ 8 - 5 = 3 \] This leads to approximately 1 lone pair (after proper distribution)

Step 5: Analyze XeO$_3$
• Xe forms 3 bonds with oxygen
• Electrons used = 3 Remaining: \[ 8 - 3 = 5 \Rightarrow \text{~2 lone pairs (approx)} \] Final Comparison Table: {c c} Compound & Lone Pairs on Xe
XeF$_2$ & 3
XeF$_4$ & 2
XeF$_6$ & 1
XeOF$_4$ & 1
XeO$_3$ & 1–2
Conclusion: Maximum lone pairs are present in XeF$_2$ \[ \boxed{\text{XeF}_2} \]