Which of the following statement is true with respect to H_2 O, NH_3 and CH_4 ? A. The central atoms of all the molecules are sp^3 hybridized. B. The H-O-H, H-N-H and H-C-H angles in the above molecules are 104.5^\circ, 107.5^\circ and 109.5^\circ , respectively. C. The increasing order of dipole moment is CH_4D. Both H_2 O and NH_3 are Lewis acids and CH_4 is a Lewis base. E. A solution of NH_3 in H_2 O is basic. In this solution NH_3 and H_2 O act as Lowry-Bronsted acid and base respectively. Choose the correct answer from the options given below:

Chemical Bonding and Molecular Structure PYQ — JEE Main 2025 · 24 Jan · Shift 1 · Q13 with Solution

View written solutionFree

Correct answer: C

Check statement A: hybridization

In $\mathrm{H_2O}$ , the central atom O has 2 bond pairs and 2 lone pairs, i.e. steric number $4$ .
In $\mathrm{NH_3}$ , the central atom N has 3 bond pairs and 1 lone pair, i.e. steric number $4$ .
In $\mathrm{CH_4}$ , the central atom C has 4 bond pairs, i.e. steric number $4$ .

For steric number $4$ , the hybridization is $\mathrm{sp^3}$ .

So, A is true.

Check statement B: bond angles

Known bond angles are:

In $\mathrm{H_2O}$ : $\angle \mathrm{H-O-H} = 104.5^\circ$
In $\mathrm{NH_3}$ : $\angle \mathrm{H-N-H} \approx 107.5^\circ$ (often written $107^\circ$ )
In $\mathrm{CH_4}$ : $\angle \mathrm{H-C-H} = 109.5^\circ$

These match the statement.

So, B is true.

Check statement C: increasing order of dipole moment

Dipole moments:

$\mathrm{CH_4}$ is perfectly tetrahedral and symmetric, so $\mu = 0$ .
$\mathrm{NH_3}$ has a nonzero dipole moment.
$\mathrm{H_2O}$ has a larger dipole moment than $\mathrm{NH_3}$ .

Thus,

\mu(\mathrm{CH_4}) < \mu(\mathrm{NH_3}) < \mu(\mathrm{H_2O})

So, C is true.

Check statement D: Lewis acid/base behavior

$\mathrm{H_2O}$ has lone pairs, so it can donate an electron pair: it behaves as a Lewis base, not Lewis acid in the usual sense here.
$\mathrm{NH_3}$ also has a lone pair, so it is a Lewis base.
$\mathrm{CH_4}$ has no lone pair to donate, so it is not a Lewis base.

Therefore the statement

Both $\mathrm{H_2O}$ and $\mathrm{NH_3}$ are Lewis acids and $\mathrm{CH_4}$ is a Lewis base is false.

So, D is false.

Check statement E: Bronsted-Lowry acid-base roles in aqueous ammonia

In water,

\mathrm{NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-}

Here:

$\mathrm{NH_3}$ 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.

The solution is basic, but the given statement says:

$\mathrm{NH_3}$ and $\mathrm{H_2O}$ act as Lowry-Bronsted acid and base respectively.

That is reversed.

So, E is false.

Select the correct option

True statements: A, B, C only.

Hence the correct option is:

\boxed{\text{C}}

More from Chemical Bonding and Molecular Structure

View written solutionFree

Correct answer: C

Check statement A: hybridization

In $\mathrm{H_2O}$ , the central atom O has 2 bond pairs and 2 lone pairs, i.e. steric number $4$ .
In $\mathrm{NH_3}$ , the central atom N has 3 bond pairs and 1 lone pair, i.e. steric number $4$ .
In $\mathrm{CH_4}$ , the central atom C has 4 bond pairs, i.e. steric number $4$ .

For steric number $4$ , the hybridization is $\mathrm{sp^3}$ .

So, A is true.

Check statement B: bond angles

Known bond angles are:

In $\mathrm{H_2O}$ : $\angle \mathrm{H-O-H} = 104.5^\circ$
In $\mathrm{NH_3}$ : $\angle \mathrm{H-N-H} \approx 107.5^\circ$ (often written $107^\circ$ )
In $\mathrm{CH_4}$ : $\angle \mathrm{H-C-H} = 109.5^\circ$

These match the statement.

So, B is true.

Check statement C: increasing order of dipole moment

Dipole moments:

$\mathrm{CH_4}$ is perfectly tetrahedral and symmetric, so $\mu = 0$ .
$\mathrm{NH_3}$ has a nonzero dipole moment.
$\mathrm{H_2O}$ has a larger dipole moment than $\mathrm{NH_3}$ .

Thus,

\mu(\mathrm{CH_4}) < \mu(\mathrm{NH_3}) < \mu(\mathrm{H_2O})

So, C is true.

Check statement D: Lewis acid/base behavior

$\mathrm{H_2O}$ has lone pairs, so it can donate an electron pair: it behaves as a Lewis base, not Lewis acid in the usual sense here.
$\mathrm{NH_3}$ also has a lone pair, so it is a Lewis base.
$\mathrm{CH_4}$ has no lone pair to donate, so it is not a Lewis base.

Therefore the statement

Both $\mathrm{H_2O}$ and $\mathrm{NH_3}$ are Lewis acids and $\mathrm{CH_4}$ is a Lewis base is false.

So, D is false.

Check statement E: Bronsted-Lowry acid-base roles in aqueous ammonia

In water,

\mathrm{NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-}

Here:

$\mathrm{NH_3}$ 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.

The solution is basic, but the given statement says:

$\mathrm{NH_3}$ and $\mathrm{H_2O}$ act as Lowry-Bronsted acid and base respectively.

That is reversed.

So, E is false.

Select the correct option

True statements: A, B, C only.

Hence the correct option is:

\boxed{\text{C}}

Chemical Bonding and Molecular Structure question

Chemical Bonding and Molecular Structure question

More from Chemical Bonding and Molecular Structure

Chemical Bonding and Molecular Structure question

Chemical Bonding and Molecular Structure question

More from Chemical Bonding and Molecular Structure