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.

Question

Among the species O$_2^+$, N$_2^-$, N$_2^{2-}$ and O$_2^-$ which have same bond order as well as paramagnetic in nature.

Answer 1

Analysis of statements regarding H$_2$O, NH$_3$, and CH$_4$:

Statement (A): Central atoms in H$_2$O, NH$_3$, and CH$_4$ are sp$^3$ hybridized.
- H$_2$O: Oxygen is sp$^3$ hybridized, resulting in a bent geometry due to two lone pairs.
- NH$_3$: Nitrogen is sp$^3$ hybridized, resulting in a trigonal pyramidal geometry due to one lone pair.
- CH$_4$: Carbon is sp$^3$ hybridized, resulting in a tetrahedral geometry with no lone pairs.
Statement (B): Bond angles H–O–H, H–N–H, and H–C–H are approximately 104.5°, 107.5°, and 109.5°, respectively.
- H$_2$O bond angle is ~104.5° due to lone pair-lone pair repulsion.
- NH$_3$ bond angle is ~107.5° due to the influence of one lone pair.
- CH$_4$ bond angle is 109.5° consistent with a perfect tetrahedral structure.
Statement (C): Dipole moments increase in the order CH$_4$<NH$_3$<H$_2$O.
- CH$_4$: Exhibits no net dipole moment due to its symmetrical tetrahedral structure.
- NH$_3$: Possesses a dipole moment owing to its trigonal pyramidal shape.
- H$_2$O: Has the largest dipole moment due to its bent shape and strong hydrogen bonding interactions.
Statement (D): 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; they function as electron pair donors.
- Lewis Base: An electron pair donor. CH$_4$ is not a Lewis base as it lacks available lone pairs for donation.
Statement (E): An aqueous solution of NH$_3$ is basic, with NH$_3$ acting as a Lowry-Bronsted acid and H$_2$O as a Lowry-Bronsted base.
- In water, NH$_3$ acts as a Bronsted-Lowry base by accepting a proton from H$_2$O.
- H$_2$O acts as a Bronsted-Lowry acid by donating a proton to NH$_3$.

Following the evaluation of each statement, statements (A), (B), and (C) are determined to be true. Thus, the correct conclusion is: A, B, and C only.

Show Hint

The Correct Option is A

Solution and Explanation

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