Question

The sets of molecules in which central atom has no lone pair of electrons are:
i. $\text{SnCl}_2, \text{NH}_3, \text{SF}_4$
ii. $\text{HgCl}_2, \text{SO}_3, \text{SF}_6$
iii. $\text{BeCl}_2, \text{BF}_3, \text{PCl}_5$
iv. $\text{ClF}_3, \text{BrF}_5, \text{XeF}_6$

• A. i, iv only
• B. ii, iii only
• C. ii, iii, iv only
• D. i, ii, iii only

Hint:

To quickly count lone pairs on the central atom: $LP = \frac{1}{2} (\text{Valence } e^- - \text{Bonding } e^-)$. The number of bonding electrons is usually the number of atoms attached, but treat double/triple bonds with $\sigma$ and $\pi$ contributions or formal charges carefully. For most simple molecular structures, this formula works directly.

Answer 1

The Correct Option is B

Step 1: Determine the number of valence electrons and lone pairs on the central atom for each molecule.
The number of lone pairs = (Total valence electrons of central atom - number of bonding electrons) / 2.

Step 2: Analyze set i: $\text{SnCl_2, \text{NH}_3, \text{SF}_4$.}
$\text{SnCl}_2$: Sn (Group 14) has 4 valence $e^-$. Forms 2 single bonds. Remaining $e^- = 4-2=2$. Lone pairs = 1. (Has a lone pair)
$\text{NH}_3$: N (Group 15) has 5 valence $e^-$. Forms 3 single bonds. Remaining $e^- = 5-3=2$. Lone pairs = 1. (Has a lone pair)
$\text{SF}_4$: S (Group 16) has 6 valence $e^-$. Forms 4 single bonds. Remaining $e^- = 6-4=2$. Lone pairs = 1. (Has a lone pair)
Set i is incorrect.

Step 3: Analyze set ii: $\text{HgCl_2, \text{SO}_3, \text{SF}_6$.}
$\text{HgCl}_2$: Hg (Group 12) has 2 valence $e^-$. Forms 2 single bonds. Remaining $e^- = 2-2=0$. Lone pairs = 0. (No lone pair)
$\text{SO}_3$: S (Group 16) has 6 valence $e^-$. Forms 3 double bonds (or $\sigma$ and $\pi$ bonds). Remaining $e^- = 6-6=0$. Lone pairs = 0. (No lone pair)
$\text{SF}_6$: S (Group 16) has 6 valence $e^-$. Forms 6 single bonds. Remaining $e^- = 6-6=0$. Lone pairs = 0. (No lone pair)
Set ii is correct.

Step 4: Analyze set iii: $\text{BeCl_2, \text{BF}_3, \text{PCl}_5$.}
$\text{BeCl}_2$: Be (Group 2) has 2 valence $e^-$. Forms 2 single bonds. Remaining $e^- = 2-2=0$. Lone pairs = 0. (No lone pair)
$\text{BF}_3$: B (Group 13) has 3 valence $e^-$. Forms 3 single bonds. Remaining $e^- = 3-3=0$. Lone pairs = 0. (No lone pair)
$\text{PCl}_5$: P (Group 15) has 5 valence $e^-$. Forms 5 single bonds. Remaining $e^- = 5-5=0$. Lone pairs = 0. (No lone pair)
Set iii is correct.

Step 5: Analyze set iv: $\text{ClF_3, \text{BrF}_5, \text{XeF}_6$.}
$\text{ClF}_3$: Cl (Group 17) has 7 valence $e^-$. Forms 3 single bonds. Remaining $e^- = 7-3=4$. Lone pairs = 2. (Has lone pairs)
$\text{BrF}_5$: Br (Group 17) has 7 valence $e^-$. Forms 5 single bonds. Remaining $e^- = 7-5=2$. Lone pairs = 1. (Has a lone pair)
$\text{XeF}_6$: Xe (Group 18) has 8 valence $e^-$. Forms 6 single bonds. Remaining $e^- = 8-6=2$. Lone pairs = 1. (Has a lone pair)
Set iv is incorrect.

Step 6: Conclude the final correct option.
The correct sets are ii and iii.