Chemistry Q53 – JEE-MAIN 2026

Home Chemistry Chemical BondingQuestion #227

Given below are two statements : \ \textbf{Statement I:} The number of compounds among

S O_{2}

S O_{3}

S F_{4}

S F_{6}

and

H_{2} S

in which sulphur does not obey the Octet rule is 3. \ \textbf{Statement II:} Among

[H_{2} O, C I F_{3}, S F_{4}]

[N H_{3}, B r F_{5}, S F_{4}]

[B r F_{5}, C I F_{3}, X e F_{4}]

and

[X e F_{4}, C I F_{3}, H_{2} O]

, the number of sets in which all the molecules have one lone pair of electrons on the central atom is 1. \ In the light of the above statements, choose the correct answer from the options given below :

Clue 1 of 3: Octet Rule and Expanded Octet

Recall that elements in Period 3 and beyond can accommodate more than 8 electrons in their valence shell, leading to an expanded octet.

🧠 Full Solution Path

Step 1: Evaluate Statement I (Octet Rule)✦ Active

We need to identify compounds where Sulfur (S) does not obey the Octet rule. Sulfur is a Period 3 element and can expand its octet.

- $S O_{2}$ : S has 10 valence electrons (2 double bonds, 1 lone pair) in the most stable Lewis structure. Does not obey Octet rule.

- $S O_{3}$ : S has 12 valence electrons (3 double bonds). Does not obey Octet rule.

- $S F_{4}$ : S has 10 valence electrons (4 single bonds, 1 lone pair). Does not obey Octet rule.

- $S F_{6}$ : S has 12 valence electrons (6 single bonds). Does not obey Octet rule.

- $H_{2} S$ : S has 8 valence electrons (2 single bonds, 2 lone pairs). Obeys Octet rule.

The number of compounds where S does not obey the Octet rule is 4 ( $S O_{2}, S O_{3}, S F_{4}, S F_{6}$ ). Statement I claims this number is 3. Therefore, Statement I is false.

💡 Teacher's Secret Hint

Remember that elements in Period 3 and beyond can have an expanded octet, meaning more than 8 valence electrons.

Step 2: Evaluate Statement II (Lone Pairs on Central Atom)○ Expand

We need to determine the number of lone pairs on the central atom for each molecule:

- $H_{2} O$ : Central atom O (6 valence e⁻). 2 bonds to H. Remaining $6 - 2 = 4$ e⁻. So, 2 lone pairs.

- $C l F_{3}$ : Central atom Cl (7 valence e⁻). 3 bonds to F. Remaining $7 - 3 = 4$ e⁻. So, 2 lone pairs.

- $S F_{4}$ : Central atom S (6 valence e⁻). 4 bonds to F. Remaining $6 - 4 = 2$ e⁻. So, 1 lone pair.

- $N H_{3}$ : Central atom N (5 valence e⁻). 3 bonds to H. Remaining $5 - 3 = 2$ e⁻. So, 1 lone pair.

- $B r F_{5}$ : Central atom Br (7 valence e⁻). 5 bonds to F. Remaining $7 - 5 = 2$ e⁻. So, 1 lone pair.

- $X e F_{4}$ : Central atom Xe (8 valence e⁻). 4 bonds to F. Remaining $8 - 4 = 4$ e⁻. So, 2 lone pairs.

Now, let's check the given sets for molecules having exactly one lone pair on the central atom:

- Set 1: $[H_{2} O (2 L P), C I F_{3} (2 L P), S F_{4} (1 L P)]$ . Not all have 1 LP.

- Set 2: $[N H_{3} (1 L P), B r F_{5} (1 L P), S F_{4} (1 L P)]$ . All have 1 LP. This set qualifies.

- Set 3: $[B r F_{5} (1 L P), C I F_{3} (2 L P), X e F_{4} (2 L P)]$ . Not all have 1 LP.

- Set 4: $[X e F_{4} (2 L P), C I F_{3} (2 L P), H_{2} O (2 L P)]$ . Not all have 1 LP.

Only 1 set (Set 2) contains molecules where all central atoms have one lone pair. Statement II claims this number is 1. Therefore, Statement II is true.

💡 Teacher's Secret Hint

For calculating lone pairs, remember to count only the valence electrons of the central atom and subtract the electrons used in forming bonds with surrounding atoms. The remaining electrons are lone pair electrons.

Step 3: Conclusion○ Expand

Statement I is false.

Statement II is true.

Based on this, the correct option is 'Statement I is false but Statement II is true'.

💡 Teacher's Secret Hint

Carefully match your findings for each statement with the given options.

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Chemistry Question 53 – JEE-MAIN 2026

Clue 1 of 3: Octet Rule and Expanded Octet

🧠 Full Solution Path