1. We need to count molecules having zero dipole moment among: $\mathrm{CH_4},\ \mathrm{BF_3},\ \mathrm{H_2O},\ \mathrm{HF},\ \mathrm{NH_3},\ \mathrm{CO_2},\ \mathrm{SO_2}$

2. Rule: A molecule has zero dipole moment if the bond dipoles cancel due to a symmetric geometry.

3. Now check each molecule one by one.

(i) $\mathrm{CH_4}$

• Geometry: tetrahedral
• All four $\mathrm{C-H}$ bonds are symmetrically arranged.
• Bond dipoles cancel.

$\mu = 0$

So, $\mathrm{CH_4}$ has zero dipole moment.

(ii) $\mathrm{BF_3}$

• Geometry: trigonal planar
• Three identical $\mathrm{B-F}$ bonds arranged symmetrically at $120^\circ$.
• Dipoles cancel.

$\mu = 0$

So, $\mathrm{BF_3}$ has zero dipole moment.

(iii) $\mathrm{H_2O}$

• Geometry: bent (due to two lone pairs on O)
• $\mathrm{O-H}$ bond dipoles do not cancel.

$\mu \neq 0$

So, $\mathrm{H_2O}$ does not have zero dipole moment.

(iv) $\mathrm{HF}$

• Diatomic molecule with different atoms.
• Bond is polar, so molecule is polar.

$\mu \neq 0$

So, $\mathrm{HF}$ does not have zero dipole moment.

(v) $\mathrm{NH_3}$

• Geometry: trigonal pyramidal (one lone pair on N)
• Dipoles do not cancel.

$\mu \neq 0$

So, $\mathrm{NH_3}$ does not have zero dipole moment.

(vi) $\mathrm{CO_2}$

• Geometry: linear
• Two identical $\mathrm{C=O}$ bond dipoles are opposite and equal.
• They cancel.

$\mu = 0$

So, $\mathrm{CO_2}$ has zero dipole moment.

(vii) $\mathrm{SO_2}$

• Geometry: bent (due to lone pair on S)
• Dipoles do not cancel.

$\mu \neq 0$

So, $\mathrm{SO_2}$ does not have zero dipole moment.

4. Molecules with zero dipole moment: $\mathrm{CH_4},\ \mathrm{BF_3},\ \mathrm{CO_2}$

5. Total number: $3$

Comparison with stored correct answer

• Derived answer = $3$
• Stored correct answer = $3$
• They match.