Chemistry Q59 – JEE-MAIN 2026

Home Chemistry Chemical BondingQuestion #228

Given below are two statements : Statement I:

F_{2} O < H_{2} O < C l_{2} O

is the correct trend in terms of bond angle. Statement II:

S i F_{4}

S n F_{4}

and

P b F_{4}

are ionic in nature. In the light of the above statements, choose the correct answer from the options given below :

Clue 1 of 3: VSEPR Theory and Electronegativity

The bond angle in molecules with a central atom and lone pairs is influenced by the repulsion between electron pairs and the electronegativity of the surrounding atoms.

🧠 Full Solution Path

Step 1: Evaluate Statement I (Bond Angles)✦ Active

All three molecules ( $F_{2} O$ , $H_{2} O$ , $C l_{2} O$ ) have a central oxygen atom with two bond pairs and two lone pairs, leading to a bent geometry. The bond angle is influenced by the electronegativity of the surrounding atoms. In $F_{2} O$ , fluorine is more electronegative than oxygen, pulling electron density away from oxygen. This reduces bond pair-bond pair repulsion and increases lone pair-lone pair repulsion, resulting in a smaller bond angle (approx. ${103.2}^{\circ}$ ). In $H_{2} O$ , the bond angle is approx. ${104.5}^{\circ}$ . In $C l_{2} O$ , chlorine is less electronegative than oxygen, and the larger size of chlorine atoms also contributes to increased bond pair-bond pair repulsion, leading to a larger bond angle (approx. ${110.9}^{\circ}$ ). Thus, the trend $F_{2} O < H_{2} O < C l_{2} O$ ( ${103.2}^{\circ} < {104.5}^{\circ} < {110.9}^{\circ}$ ) is correct.

Statement I is TRUE.

Step 2: Evaluate Statement II (Ionic Nature)○ Expand

The ionic or covalent character of Group 14 fluorides ( $S i F_{4}$ , $S n F_{4}$ , $P b F_{4}$ ) can be assessed using Fajan's rules. $S i F_{4}$ : Silicon is a non-metal, and $S i^{4 +}$ is a small, highly charged cation. This favors covalent character, making $S i F_{4}$ a covalent molecule. $S n F_{4}$ : Tin is larger than silicon. While it has some ionic character, it is still largely covalent. $P b F_{4}$ : Lead is the largest among them, and $P b^{4 +}$ is a larger cation. This increases the ionic character significantly, and $P b F_{4}$ is considered to have substantial ionic character. However, the statement claims all three "are ionic in nature." Since $S i F_{4}$ is clearly covalent, the statement is false.

Statement II is FALSE.

Step 3: Conclude based on both statements○ Expand

Since Statement I is true and Statement II is false, the correct option is 'Statement I is true but Statement II is false'.

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Chemistry Question 59 – JEE-MAIN 2026

Clue 1 of 3: VSEPR Theory and Electronegativity

🧠 Full Solution Path