Let's analyze the bond angles for the given molecules.

Molecule 1: ${\text{CH}}_4$ (Methane)

In methane, the central carbon atom is ${\text{sp}}^3$ hybridized. It has four bond pairs and no lone pairs. The geometry is tetrahedral, and the ideal bond angle (H-C-H) is $109.5^{\circ }$. So, $x=109.5^{\circ }$.

Molecule 2: ${\text{NH}}_3$ (Ammonia)

In ammonia, the central nitrogen atom is also ${\text{sp}}^3$ hybridized. It has three bond pairs and one lone pair. Due to the greater repulsion exerted by the lone pair on the bond pairs, the bond angle (H-N-H) is slightly reduced from the ideal tetrahedral angle. The bond angle in ammonia is approximately $107^{\circ }$. So, $y=107^{\circ }$.

Comparison:

Comparing the bond angles, we have $x=109.5^{\circ }$ and $y=107^{\circ }$. Therefore, $x>y$.

Option Analysis:

• A) $x>y$: This is correct, as $109.5^{\circ }>107^{\circ }$.
• B) $y>x$: This is incorrect, as $107^{\circ }$ is not greater than $109.5^{\circ }$.
• C) $x=y$: This is incorrect, as the bond angles are different due to the presence of a lone pair in ammonia.
• D) None of these: This is incorrect, as option A correctly describes the relationship.

Correct Answer: (A)