Question

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°, 107.5° and 109.5° respectively. 
(C) The increasing order of dipole moment is CH\(_4\)<NH\(_3\)<H\(_2\)O. 
(D) 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. 
 

• A. A, B, and C only
• B. C, D, and E only
• C. A, D, and E only
• D. A, B, C, and E only

Hint:

In molecules with sp\(^3\) hybridized atoms, the bond angles are typically close to 109.5°. Lone pairs on atoms distort these angles, which is why H\(_2\)O has a bond angle of 104.5° and NH\(_3\) has 107.5°.

Answer 1

The Correct Option is A

To determine which statements are true regarding H\(_2\)O, NH\(_3\), and CH\(_4\), let's analyze each statement one by one: 

1. Statement (A): The central atoms of all the molecules are sp\(^3\) hybridized.
   • H\(_2\)O (Water) - The oxygen atom is sp\(^3\) hybridized, forming a bent shape due to two lone pairs.
   • NH\(_3\) (Ammonia) - The nitrogen atom is sp\(^3\) hybridized, forming a trigonal pyramidal shape due to one lone pair.
   • CH\(_4\) (Methane) - The carbon atom is sp\(^3\) hybridized, forming a tetrahedral shape with no lone pairs.
2. Statement (B): The H–O–H, H–N–H and H–C–H angles in the above molecules are 104.5°, 107.5° and 109.5° respectively.
   • The bond angle in H\(_2\)O is about 104.5° due to the repulsion between lone pairs.
   • The bond angle in NH\(_3\) is about 107.5° due to the presence of one lone pair.
   • The bond angle in CH\(_4\) is 109.5° as it is a perfect tetrahedral shape.
3. Statement (C): The increasing order of dipole moment is CH\(_4\)<NH\(_3\)<H\(_2\)O.
   • CH\(_4\) (Methane) has no net dipole moment as it is symmetrical.
   • NH\(_3\) (Ammonia) has a dipole moment due to its trigonal pyramidal shape.
   • H\(_2\)O (Water) has the highest dipole moment due to its bent shape and strong hydrogen bonding.
4. Statement (D): Both H\(_2\)O and NH\(_3\) are Lewis acids and CH\(_4\) is a Lewis base.
   • Lewis Acid - An electron pair acceptor. H\(_2\)O and NH\(_3\) are not typically Lewis acids; rather, they can donate electrons.
   • Lewis Base - An electron pair donor. CH\(_4\) is not a Lewis base as it does not have available lone pairs to donate.
5. Statement (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.
   • In an aqueous solution, NH\(_3\) acts as a Bronsted-Lowry base as it accepts a proton from H\(_2\)O.
   • H\(_2\)O acts as a Bronsted-Lowry acid as it donates a proton to NH\(_3\).

After evaluating each statement, the true statements are (A), (B), and (C). Therefore, the correct answer is: A, B, and C only.

Answer 2

Let's analyze the given options:

Option A:

The central atoms in all three molecules (H\(_2\)O, NH\(_3\), CH\(_4\)) are sp\(^3\) hybridized. This is true because the oxygen, nitrogen, and carbon atoms form single bonds with surrounding atoms, which requires sp\(^3\) hybridization in each case.

Option B:

The bond angles in H\(_2\)O, NH\(_3\), and CH\(_4\) are 104.5°, 107.5°, and 109.5°, respectively. This is correct: - H\(_2\)O has an angle of 104.5° due to lone pair repulsion. - NH\(_3\) has 107.5°, with one lone pair on nitrogen. - CH\(_4\) has 109.5°, as it is tetrahedral with no lone pairs.

Option C:

The increasing order of dipole moment is CH\(_4\) < NH\(_3\) < H\(_2\)O. This is true because: - CH\(_4\) has no dipole moment due to its symmetrical tetrahedral shape. - NH\(_3\) has a dipole moment due to the lone pair on nitrogen. - H\(_2\)O has the highest dipole moment because of its bent shape and the high electronegativity of oxygen.

Option D:

Both H\(_2\)O and NH\(_3\) are Lewis acids and CH\(_4\) is a Lewis base. This statement is incorrect. H\(_2\)O and NH\(_3\) act as Lewis bases (donors), not acids. CH\(_4\) is a Lewis base, as it has a pair of electrons on the carbon atom that can donate. Therefore, this statement is false.

Option E:

A solution of NH\(_3\) in H\(_2\)O is basic, and in this solution, NH\(_3\) acts as a base and H\(_2\)O acts as an acid. This is true because NH\(_3\) accepts a proton from water to form NH\(_4^+\) and OH\(^-\), making the solution basic.

Conclusion:

The correct answer is \( \boxed{(1)} \), which corresponds to **A, B, and C only** being true.