Application Note: Dipole Moment in NEET Problems

When μ = 0 (Non-polar molecules)

A molecule has μ = 0 when ALL bond dipoles cancel by vector symmetry:

Geometry	Example	Condition for μ = 0
Linear	$CO_{2}$ , $BeCl_{2}$ , $CS_{2}$	Central atom has no lone pairs
Trigonal planar	$BF_{3}$ , $BCl_{3}$ , $SO_{3}$	No lone pairs on central atom
Tetrahedral	$CCl_{4}$ , $CF_{4}$ , $CH_{4}$	All 4 substituents identical
Octahedral	$SF_{6}$ , $PF_{6}^{-}$	All 6 substituents identical
Square planar	$XeF_{4}$	All 4 substituents identical

When μ ≠ 0 (Polar molecules)

Geometry	Example	Reason
Bent	$H_{2}O$ (1.85 D), $SO_{2}$ (1.63 D)	Lone pairs on central atom break symmetry
Trigonal pyramidal	$NH_{3}$ (1.47 D), $NF_{3}$ (0.24 D), $PCl_{3}$ (0.97 D)	Lone pair on N/P
T-shaped	$ClF_{3}$	2 lone pairs on Cl
Seesaw	$SF_{4}$	1 lone pair on S
Linear with different groups	HCl (1.08 D), HF (1.91 D)	Different atoms

Dipole Moment Comparisons (NEET Traps)

$NF_{3}$ vs $BF_{3}$ : $NF_{3}$ has μ ≠ 0 (trigonal pyramidal); $BF_{3}$ has μ = 0 (trigonal planar)
$CHCl_{3}$ vs $CCl_{4}$ : $CHCl_{3}$ has μ ≠ 0 (asymmetric); $CCl_{4}$ has μ = 0 (symmetric)
$CO_{2}$ vs $SO_{2}$ : $CO_{2}$ has μ = 0 (linear); $SO_{2}$ has μ ≠ 0 (bent)
$O_{2}$ vs CO: $O_{2}$ has μ = 0 (homonuclear); CO has μ ≠ 0 (heteronuclear)
cis-CHCl=CHCl vs trans-CHCl=CHCl: cis has μ ≠ 0; trans has μ = 0

Dipole Moment and Boiling Point

Higher μ → stronger dipole-dipole forces → higher boiling point (all else equal): $\text{bp order: H}_2\text{O} > \text{H}_2\text{S} > \text{H}_2\text{Se}$ ( $H_{2}O$ anomaly: also H-bonding)