2.2 · Shapes & polarity

Molecular polarity

Polar bonds do not always make a polar molecule.

By the end you should be able to:

Decide whether a molecule is polar from its shape and bond dipoles

Key idea

The molecular dipole is a vector sum

A molecule is polar when it has a net dipole moment. Each polar bond contributes a bond dipole pointing toward its $\delta^-$ atom, and the molecular dipole is the vector sum of all of them, so it depends on both bond polarity and shape.

Polar bonds arranged so that they cancel: nonpolar molecule ( $\ce{CO2}$ ).
Polar bonds that do not cancel: polar molecule ( $\ce{H2O}$ ).
Only nonpolar bonds: nonpolar or nearly so ( $\ce{H2}$ , $\ce{Cl2}$ , hydrocarbons). Ozone is the classic exception: all its bonds are O–O, but its bent shape and uneven electron distribution give it a small dipole.

Key idea

Shapes that cancel bond dipoles

With identical outer atoms, these shapes cancel all bond dipoles, so the molecule is nonpolar:

Shape	Class	Nonpolar example
linear	$\mathrm{AX_2}$	$\ce{CO2}$
trigonal planar	$\mathrm{AX_3}$	$\ce{BF3}$
tetrahedral	$\mathrm{AX_4}$	$\ce{CCl4}$
trigonal bipyramidal	$\mathrm{AX_5}$	$\ce{PCl5}$
octahedral	$\mathrm{AX_6}$	$\ce{SF6}$
linear	$\mathrm{AX_2E_3}$	$\ce{XeF2}$
square planar	$\mathrm{AX_4E_2}$	$\ce{XeF4}$

Bent, trigonal pyramidal, seesaw, T-shaped and square pyramidal shapes never cancel, so with polar bonds they are always polar ( $\ce{H2O}$ , $\ce{NH3}$ , $\ce{SF4}$ , $\ce{ClF3}$ , $\ce{BrF5}$ ).

Method

Deciding whether a molecule is polar

Draw the Lewis structure and find the VSEPR molecular geometry.
Mark the polar bonds (ΔEN of 0.4 or more).
Is the shape one of the symmetric ones and are all outer atoms identical? If yes, the molecule is nonpolar.
Otherwise sketch the bond dipoles in 3D and add them. If they do not cancel, the molecule is polar, with its negative end on the side of the more electronegative atoms (and lone pairs).

Common mistake

Polarity traps

Wrong: " $\ce{CO2}$ has polar bonds, so it is polar." Right: its two C=O dipoles are equal and point in opposite directions; $\ce{CO2}$ is nonpolar.

Wrong: " $\ce{CH2Cl2}$ is tetrahedral, so it is nonpolar." Right: a tetrahedron cancels only when all four substituents are identical; $\ce{CH3Cl}$ , $\ce{CH2Cl2}$ and $\ce{CHCl3}$ are all polar.

Wrong: drawing $\ce{CH2Cl2}$ flat with the two Cl atoms opposite each other and concluding the dipoles cancel. Right: in 3D every pair of positions on a tetrahedron is 109.5° apart; no two bonds point in opposite directions.

Wrong: "any lone pair on the central atom makes a molecule polar." Right: in $\ce{XeF2}$ and $\ce{XeF4}$ the lone pairs are arranged symmetrically, and both are nonpolar.

Worked example

Worked example: similar formulas, different polarity

Pair	Shapes	Polarity
$\ce{CO2}$ vs $\ce{SO2}$	linear vs bent ( $\mathrm{AX_2E}$ )	nonpolar vs polar
$\ce{BF3}$ vs $\ce{NF3}$	trigonal planar vs trigonal pyramidal	nonpolar vs polar
$\ce{CCl4}$ vs $\ce{CH2Cl2}$	tetrahedral, identical vs mixed outer atoms	nonpolar vs polar
$\ce{XeF4}$ vs $\ce{SF4}$	square planar vs seesaw	nonpolar vs polar

Every bond in this table is polar; only the shape, or the mix of outer atoms, decides.

Worked example

Worked example: the dipole of CH₂Cl₂

Lewis structure: C bonded to 2 H and 2 Cl, no lone pairs on C; $\mathrm{AX_4}$ , tetrahedral.
Bond polarity: C–Cl has ΔEN = 3.16 − 2.55 = 0.61 (polar, $\delta^-$ on Cl); C–H has ΔEN = 0.35 (essentially nonpolar).
The two C–Cl dipoles point toward two corners of the tetrahedron that are 109.5° apart. Their sum points along the bisector of the Cl–C–Cl angle, and the nearly nonpolar C–H bonds cannot cancel it.

Answer: $\ce{CH2Cl2}$ is polar, with its negative end between the two Cl atoms.

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