Given below are two statements:
Statement I : Elements 'X' and 'Y' are the most and least electronegative elements, respectively among N, As, Sb and P. The nature of the oxides $$X_{2}O_{3}$$ and $$y_{2}O_{3}$$ is acidic and
amphoteric, respectively.
Statement II:$$BCl_{3}$$ is covalent in nature and gets hydrolysed in water. It produces $$[B(OH)_{4}]^{-}$$ and $$[B(H_{2}O)_{6}]^{3+}$$ in aqueous medium.
In the Light of the above statements, choose the correct answer from the options given below:

We need to evaluate two statements about p-block elements.

Statement I Analysis:

Among N, As, Sb, and P, the electronegativity order is: $$N > P > As > Sb$$.

So $$X = N$$ (most electronegative) and $$Y = Sb$$ (least electronegative).

$$X_2O_3 = N_2O_3$$: Nitrogen is a nonmetal, so $$N_2O_3$$ is an acidic oxide. $$\checkmark$$

$$Y_2O_3 = Sb_2O_3$$: Antimony is a metalloid, and $$Sb_2O_3$$ is amphoteric (it reacts with both acids and bases). $$\checkmark$$

Statement I is TRUE.

Statement II Analysis:

$$BCl_3$$ is indeed covalent and gets hydrolysed in water. $$\checkmark$$

However, in aqueous medium:

$$BCl_3 + 3H_2O \rightarrow B(OH)_3 + 3HCl$$

$$B(OH)_3 + H_2O \rightleftharpoons [B(OH)_4]^{-} + H^{+}$$

Boron does form $$[B(OH)_4]^-$$. $$\checkmark$$

However, boron does NOT form $$[B(H_2O)_6]^{3+}$$. Boron has a maximum coordination number of 4 (not 6) because it has no d-orbitals available, and being a very small atom with high charge density, it cannot accommodate six water molecules. $$\times$$

Statement II is FALSE.

Since Statement I is true and Statement II is false, the correct answer is Option D.