Which of the following statement is true with respect to  $$H_2O,\ NH_3$$ 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 $$CH_4 < NH_3 < H_2O.$$ $$D.$$ Both $$H_2O$$   and $$NH_3$$ are Lewis acids and  $$CH_4$$ is a Lewis base.  $$E.$$ A solution of  $$NH_3$$ in $$H_2O$$  is basic. In this solution  $$NH_3$$ and $$H_2O$$ act as Lowry-Bronsted acid and base respectively. Choose the correct answer from the options given below:

We analyze each statement about $$H_2O$$, $$NH_3$$, and $$CH_4$$:

A. Central atoms of all molecules are $$sp^3$$ hybridized.

$$H_2O$$: O has 2 bond pairs + 2 lone pairs = $$sp^3$$. $$NH_3$$: N has 3 bond pairs + 1 lone pair = $$sp^3$$. $$CH_4$$: C has 4 bond pairs = $$sp^3$$. Correct.

B. Bond angles are $$104.5°$$, $$107.5°$$, and $$109.5°$$ respectively.

H-O-H = $$104.5°$$, H-N-H = $$107°$$ (approximately $$107.5°$$), H-C-H = $$109.5°$$. Correct.

C. Increasing order of dipole moment: $$CH_4 < NH_3 < H_2O$$.

$$CH_4$$ has zero dipole moment (symmetric). $$NH_3$$ has $$\mu = 1.47$$ D. $$H_2O$$ has $$\mu = 1.85$$ D. So $$CH_4 < NH_3 < H_2O$$. Correct.

D. Both $$H_2O$$ and $$NH_3$$ are Lewis acids and $$CH_4$$ is a Lewis base.

$$H_2O$$ and $$NH_3$$ have lone pairs and act as Lewis BASES (not acids). $$CH_4$$ is neither a good Lewis acid nor base. Incorrect.

E. A solution of $$NH_3$$ in $$H_2O$$ is basic. $$NH_3$$ and $$H_2O$$ act as Lowry-Bronsted acid and base respectively.

In the reaction $$NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$$, $$NH_3$$ acts as a Bronsted BASE (accepts $$H^+$$) and $$H_2O$$ acts as a Bronsted ACID (donates $$H^+$$). The statement says $$NH_3$$ is the acid and $$H_2O$$ is the base, which is reversed. Incorrect.

Correct statements: A, B, and C only.

The correct answer is Option A: A, B and C only.