1. We need to count how many of the given molecules have non-zero net dipole moment.

2. Rule:

   • A molecule is polar if the vector sum of all bond dipole moments is non-zero.
   • Even if bonds are polar, a symmetric geometry can make the net dipole moment zero.

3. Check each molecule one by one.

(i) $NF_3$

• Shape: trigonal pyramidal (because N has one lone pair)
• The $N-F$ bond dipoles do not cancel completely due to non-planar asymmetric shape.
• Hence, $NF_3$ is polar.

$\Rightarrow$ Net dipole moment $\neq 0$

(ii) $BF_3$

• Shape: trigonal planar
• Three identical $B-F$ bond dipoles are symmetrically arranged at $120^\circ$.
• They cancel out.

$\Rightarrow$ Net dipole moment $=0$

(iii) $BeF_2$

• Shape: linear
• Two identical $Be-F$ bond dipoles are opposite and cancel.

$\Rightarrow$ Net dipole moment $=0$

(iv) $CHCl_3$

• Shape: tetrahedral
• Bonds are not all identical: three $C-Cl$ and one $C-H$.
• Because of unequal bond dipoles, cancellation is not complete.

$\Rightarrow$ Net dipole moment $\neq 0$

(v) $H_2S$

• Shape: bent (due to two lone pairs on S)
• The two $S-H$ bond dipoles do not cancel.

$\Rightarrow$ Net dipole moment $\neq 0$

(vi) $SiF_4$

• Shape: tetrahedral
• Four identical $Si-F$ bonds arranged symmetrically.
• Dipoles cancel.

$\Rightarrow$ Net dipole moment $=0$

(vii) $CCl_4$

• Shape: tetrahedral
• Four identical $C-Cl$ bond dipoles cancel due to symmetry.

$\Rightarrow$ Net dipole moment $=0$

(viii) $PF_5$

• Shape: trigonal bipyramidal
• Symmetric arrangement of identical $P-F$ bonds leads to cancellation.

$\Rightarrow$ Net dipole moment $=0$

4. Polar molecules are:

• $NF_3$
• $CHCl_3$
• $H_2S$

So, the number of molecules having net resultant dipole moment is $3$

5. Comparison with stored correct answer:

• Stored correct answer = $3$
• Derived answer = $3$

They match.