Evaluation of Octet Rule and Lone Pairs in Molecules

To determine the correctness of the given statements, we analyze each statement individually:

Analysis of Statement I:

The Octet rule states that atoms tend to adjust their valence electrons to achieve an octet (8 valence electrons). An expanded octet occurs when an atom has more than 8 valence electrons surrounding it.

Let's calculate the total number of electrons in the valence shell of the sulfur ($\text{S}$) atom in each given molecule:

1. $\text{SO}_2$: The central $\text{S}$ atom forms two double bonds with two oxygen atoms and retains $1$ lone pair. $\text{Total valence electrons around S} = 2 \times 2 \text{ (bonding pairs)} + 2 \text{ (lone pair electrons)} = 10\text{ e}^-$ (Does not obey the Octet rule — Expanded Octet)

2. $\text{SO}_3$: The central $\text{S}$ atom forms three double bonds with three oxygen atoms. $\text{Total valence electrons around S} = 3 \times 2 \text{ (bonding pairs)} = 12\text{ e}^-$ (Does not obey the Octet rule — Expanded Octet)

3. $\text{SF}_4$: The central $\text{S}$ atom forms four single bonds with four fluorine atoms and has $1$ lone pair. $\text{Total valence electrons around S} = 4 \times 1 \text{ (bonding pairs)} + 2 \text{ (lone pair electrons)} = 10\text{ e}^-$ (Does not obey the Octet rule — Expanded Octet)

4. $\text{SF}_6$: The central $\text{S}$ atom forms six single bonds with six fluorine atoms. $\text{Total valence electrons around S} = 6 \times 1 \text{ (bonding pairs)} = 12\text{ e}^-$ (Does not obey the Octet rule — Expanded Octet)

5. $\text{H}_2\text{S}$: The central $\text{S}$ atom forms two single bonds with two hydrogen atoms and has $2$ lone pairs. $\text{Total valence electrons around S} = 2 \times 1 \text{ (bonding pairs)} + 4 \text{ (lone pair electrons)} = 8\text{ e}^-$ (Obeys the Octet rule)

Thus, there are 4 compounds ($\text{SO}_2, \text{SO}_3, \text{SF}_4, \text{SF}_6$) in which sulfur does not obey the octet rule. Statement I claims that this number is $3$, which is incorrect. Therefore, Statement I is false.

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Analysis of Statement II:

To find the number of lone pairs on the central atom of each molecule, we use the formula: $\text{Lone pairs (LP)} = \frac{V - B}{2}$ where $V$ is the number of valence electrons of the central atom and $B$ is the number of bonding electrons participating in single bonds.

• $\text{H}_2\text{O}$: Oxygen (Central) has $V = 6$, $B = 2 \implies \text{LP} = \frac{6 - 2}{2} = 2$
• $\text{ClF}_3$: Chlorine (Central) has $V = 7$, $B = 3 \implies \text{LP} = \frac{7 - 3}{2} = 2$
• $\text{SF}_4$: Sulfur (Central) has $V = 6$, $B = 4 \implies \text{LP} = \frac{6 - 4}{2} = 1$
• $\text{NH}_3$: Nitrogen (Central) has $V = 5$, $B = 3 \implies \text{LP} = \frac{5 - 3}{2} = 1$
• $\text{BrF}_5$: Bromine (Central) has $V = 7$, $B = 5 \implies \text{LP} = \frac{7 - 5}{2} = 1$
• $\text{XeF}_4$: Xenon (Central) has $V = 8$, $B = 4 \implies \text{LP} = \frac{8 - 4}{2} = 2$

Now, let's evaluate each set to check if all molecules in the set have exactly one lone pair:

1. $[\text{H}_2\text{O} (2\text{ LP}), \text{ClF}_3 (2\text{ LP}), \text{SF}_4 (1\text{ LP})]$: Incorrect
2. $[\text{NH}_3 (1\text{ LP}), \text{BrF}_5 (1\text{ LP}), \text{SF}_4 (1\text{ LP})]$: Correct (All molecules have 1 lone pair)
3. $[\text{BrF}_5 (1\text{ LP}), \text{ClF}_3 (2\text{ LP}), \text{XeF}_4 (2\text{ LP})]$: Incorrect
4. $[\text{XeF}_4 (2\text{ LP}), \text{ClF}_3 (2\text{ LP}), \text{H}_2\text{O} (2\text{ LP})]$: Incorrect (All molecules have 2 lone pairs)

The number of sets in which all molecules have one lone pair on the central atom is precisely 1 (Set 2). Therefore, Statement II is true.

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Conclusion:

• Statement I is false
• Statement II is true

Hence, the correct option is D.