Lewis dot diagram

simple! have you ever heard methane is soluble in water? Probably not. But Ammonia is. Moreover NH3 undergoes hydrogen bonding and thus increasing its solubility. And thus c) is ruled out! and thus 3 options becomes wrong too!

@Prasun Biswas

Simply wrong.

(A)

No. B and C have a higher boiling point due to the intermolecular hydrogen bonding (not seen in CH4).

(B)

No. H2O highest bond is about 104.45°.

In tetrahedral structures our highest bonds are about 109.5°.

Thus, 109.5° > 104.45° is why B) is correct. Saying that a bond in methane has 129.5° and (as it is a carbon atom bonded equivalently to 4 H atoms), that all 4 bonds form the same angle, you violate geometry rules.

(C)

Partially correct.

It is not about the solvents nature.

CH4 has 0 dipole moment, therefore it will not be easily dissolved in water.

Many carboxylic acids, for instance, are miscible in water.

Chọn a,b

Note that only (A) here is a nonpolar molecule, this means $CH_4$ molecules experience dispertion forces intermolecularly. Since (B) is polar, therefore the $NH_3$ molecules experience dipole-dipole which is a stronger intermolecular force than dispertion, hence the $CH_4$ molecules are easier to be broken apart than $NH_3$ so a) is true.

Since (A),(C) both have four domains where (C) has two unshared domains, therefore the unshared electrons exert a larger repulsive force on the adjacent domains, causing the bond angle between $OH$ to decrease, hence b) is also true.

For c) we can use the same reasoning that since (A) is nonpolar the molecules are harder to be pulled apart by $H_2O$ then the polar (B), hence it's false.