1. Check the Assertion (A)

Both $\mathrm{NH_3}$ and $\mathrm{NF_3}$ have nitrogen as the central atom with:

• $3$ bond pairs
• $1$ lone pair

So, by VSEPR theory, both molecules have trigonal pyramidal shape.

Thus, the first part of the assertion is true.

Now compare dipole moments:

• In $\mathrm{NH_3}$, nitrogen is more electronegative than hydrogen, so each $\mathrm{N-H}$ bond dipole is directed towards N.
• The lone pair on nitrogen also contributes an orbital dipole in the same general direction as the resultant of the bond dipoles.
• Therefore, the dipole moments add up, giving a larger net dipole moment.

For $\mathrm{NF_3}$:

• Fluorine is more electronegative than nitrogen, so each $\mathrm{N-F}$ bond dipole is directed towards F.
• The resultant of the three $\mathrm{N-F}$ bond dipoles is opposite to the direction of the lone-pair dipole on nitrogen.
• Hence, they partially cancel, making the net dipole moment small.

Therefore, $\mu(\mathrm{NH_3}) > \mu(\mathrm{NF_3})$

So, Assertion (A) is true.

2. Check the Reason (R)

The reason states:

• In $\mathrm{NH_3}$, the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the $\mathrm{N-H}$ bonds.
• Fluorine is the most electronegative element.

This is correct.

Why is mentioning fluorine relevant? Because due to high electronegativity of F, the $\mathrm{N-F}$ bond dipoles in $\mathrm{NF_3}$ are directed away from nitrogen, opposite to the lone pair dipole, leading to cancellation.

So, Reason (R) is true.

3. Does (R) correctly explain (A)?

Yes.

The larger dipole moment of $\mathrm{NH_3}$ compared to $\mathrm{NF_3}$ is explained by the direction of:

• bond dipoles, and
• lone pair dipole.

In $\mathrm{NH_3}$, these reinforce each other. In $\mathrm{NF_3}$, they oppose each other because fluorine is highly electronegative.

Hence, (R) is the correct explanation of (A).

4. Correct option

$\boxed{\text{D}}$

Both (A) and (R) are true and (R) is the correct explanation of (A).