JEE Chemistry 2025 Question with Solution

Given: Statements about $\mathrm{H_2O}$, $\mathrm{NH_3}$, and $\mathrm{CH_4}$ are to be checked.

Find: Which set of statements is true.

Evaluate each statement one by one.

1. Statement (A): The central atoms of all the molecules are $sp^3$ hybridized.

• In $\mathrm{H_2O}$, oxygen is $sp^3$ hybridized.
• In $\mathrm{NH_3}$, nitrogen is $sp^3$ hybridized.
• In $\mathrm{CH_4}$, carbon is $sp^3$ hybridized.

So, (A) is true.

2. Statement (B): The bond angles are $104.5^\circ$, $107.5^\circ$, and $109.5^\circ$ respectively.

• $\mathrm{H_2O}$ has bond angle about $104.5^\circ$ because of two lone pairs.
• $\mathrm{NH_3}$ has bond angle about $107.5^\circ$ because of one lone pair.
• $\mathrm{CH_4}$ has bond angle $109.5^\circ$ for tetrahedral geometry.

So, (B) is true.

3. Statement (C): Increasing order of dipole moment is $\mathrm{CH_4 < NH_3 < H_2O}$.

• $\mathrm{CH_4}$ is symmetrical, so net dipole moment is zero.
• $\mathrm{NH_3}$ has a non-zero dipole moment.
• $\mathrm{H_2O}$ has a higher dipole moment due to bent geometry.

So, (C) is true.

4. Statement (D): Both $\mathrm{H_2O}$ and $\mathrm{NH_3}$ are Lewis acids and $\mathrm{CH_4}$ is a Lewis base.

• $\mathrm{H_2O}$ and $\mathrm{NH_3}$ usually act as Lewis bases because they donate electron pairs.
• $\mathrm{CH_4}$ is not a Lewis base because it does not have a lone pair available for donation.

So, (D) is false.

5. Statement (E): In aqueous solution of $\mathrm{NH_3}$, $\mathrm{NH_3}$ and $\mathrm{H_2O}$ act as Lowry-Bronsted acid and base respectively.

• In water, $\mathrm{NH_3}$ accepts a proton, so it acts as a Bronsted-Lowry base.
• $\mathrm{H_2O}$ donates a proton, so it acts as a Bronsted-Lowry acid.

Therefore the stated acid-base roles are reversed, so (E) is false.

Hence the true statements are (A), (B), and (C) only.

The correct option is A.

Given: Five statements about hybridization, bond angle, dipole moment, and acid-base behavior of $\mathrm{H_2O}$, $\mathrm{NH_3}$, and $\mathrm{CH_4}$.

Find: Which combination of statements is correct.

The hint indicates that for $sp^3$ hybridized central atoms, the ideal bond angle is close to $109.5^\circ$, but lone pairs reduce the angle.

Thus:

• $\mathrm{CH_4}$ is tetrahedral with angle $109.5^\circ$.
• $\mathrm{NH_3}$ is trigonal pyramidal with angle about $107.5^\circ$.
• $\mathrm{H_2O}$ is bent with angle about $104.5^\circ$.

This confirms (A) and (B).

For dipole moment:

• $\mathrm{CH_4}$ has zero net dipole moment due to symmetry.
• $\mathrm{NH_3}$ is polar.
• $\mathrm{H_2O}$ is more polar than $\mathrm{NH_3}$.

So the order $\mathrm{CH_4 < NH_3 < H_2O}$ is correct, confirming (C).

For (D), the solution states $\mathrm{H_2O}$ and $\mathrm{NH_3}$ are not Lewis acids; they more commonly act as Lewis bases. Also $\mathrm{CH_4}$ does not have a lone pair to donate, so it is not a Lewis base. Hence (D) is false.

For (E), in aqueous ammonia:

Here $\mathrm{NH_3}$ accepts a proton and acts as a base, while $\mathrm{H_2O}$ donates a proton and acts as an acid. Therefore the statement as written is false.

So only (A), (B), and (C) are true.

The correct option is A.