Polarity of molecules — vector sum of bond dipoles: Question bank

Prerequisites worth re-skimming: Electronegativity, VSEPR Theory, Molecular Geometry and Shapes, and the maths in Vectors and the Cosine Rule.

Symbols and formulas used on this page (build them first)

Before any trap, here is every symbol you will meet, in plain words and pictures — nothing is assumed.

Definition The symbols

$μ$ (Greek letter "mu") is the dipole moment: an arrow whose length says how strong the charge separation is and whose direction points from the positive end $δ^{+}$ toward the negative end $δ^{-}$ .
$q_{i}$ is the partial charge on atom number $i$ — how much extra $+$ or $-$ charge sits on that atom because electrons were pulled toward the more electronegative partner.
$r_{i}$ is the position vector of that charge — an arrow from a fixed origin (the central atom) out to where the charge $q_{i}$ sits.
$θ$ (Greek "theta") is the bond angle: the angle between the two bonds at the central atom, i.e. the angle between the two dipole arrows drawn tail-to-tail. Examples: CO₂ has $θ = 180°$ (straight line), water has $θ = 104.5°$ (a bend), BF₃ has $θ = 120°$ between neighbours.
The Debye (D) is the unit chemists use for $μ$ ; $1 D = 3.336 \times 1 0^{- 30} C\cdotm$ . A typical single bond dipole is roughly $0.5$ – $2 D$ .

Formula Where the resultant formulas come from

The total dipole is just charge times position summed over all charges: $μ_{net} = \sum_{i} q_{i} r_{i}$ WHY a plain sum? Because charge is additive, and each $q_{i} r_{i}$ is an ordinary vector, so we add them head-to-tail like any vectors (Vectors and the Cosine Rule). Take two equal bond dipoles of length $μ$ meeting at bond angle $θ$ , placed tail-to-tail at the central atom (see figure). They form two sides of a triangle; the cosine rule gives the third side — the resultant: $μ_{net} = μ_{1}^{2} + μ_{2}^{2} + 2 μ_{1} μ_{2} cos θ$ Setting $μ_{1} = μ_{2} = μ$ and using $1 + cos θ = 2 cos^{2} (θ /2)$ : $μ_{net} = μ 2 + 2 cos θ = 2 μ cos (\frac{θ}{2})$ This is the master shortcut for two equal dipoles. The figure shows both the tail-to-tail arrangement and the cancelling/adding limits.

Figure — Polarity of molecules — vector sum of bond dipoles

True or false — justify

Every polar bond guarantees a polar molecule

False. Bond dipoles are vectors; in symmetric shapes (linear CO₂, planar BF₃, tetrahedral CCl₄) equal arrows cancel to

μ_{net} = 0

despite very polar bonds.

A nonpolar molecule must be built from nonpolar bonds

False. CO₂ and CCl₄ have strongly polar bonds — they are nonpolar only because geometry makes the vector sum zero, not because the bonds are nonpolar.

If two bond dipoles are equal and point in exactly opposite directions, the molecule is nonpolar

True — provided nothing else contributes. Opposite equal vectors give

2 μ cos (180°/2) = 2 μ cos 90° = 0

, so they perfectly cancel.

Any molecule with an odd number of identical polar bonds around a centre must be polar

False. Three identical dipoles at

120°

in a plane (BF₃) sum to zero; oddness alone says nothing — the arrangement decides.

A molecule with zero net dipole cannot dissolve well in water

Mostly true as a rule of thumb. Nonpolar molecules mix poorly with polar water ("like dissolves like", see Solubility — Like Dissolves Like), though other forces can create exceptions.

Doubling every bond dipole magnitude in a molecule doubles the net molecular dipole

True.

μ_{net} = \sum q_{i} r_{i}

is linear, so scaling all bond dipoles by the same factor scales the resultant by that same factor — direction unchanged.

Lone pairs never affect the dipole moment because we only add bond arrows

False. Lone pairs are concentrated charge regions with their own moment; in NH₃ and H₂O they add to the bond moments and are a major reason those molecules are strongly polar.

Two molecules with the same bonds always have the same polarity

False. NH₃ (pyramidal) and BF₃ (planar) both have three polar bonds to a central atom, yet one is polar and one is not — shape from VSEPR Theory is the deciding factor.

A trigonal bipyramidal molecule with five identical polar bonds (e.g. PF₅) is polar

False. The two axial dipoles cancel each other and the three equatorial dipoles (at

120°

in a plane) cancel among themselves, so PF₅ is nonpolar despite five polar bonds.

An octahedral molecule with six identical polar bonds (e.g. SF₆) is polar

False. The six dipoles come in three opposite pairs pointing along

\pm x

\pm y

\pm z

; each pair cancels, giving

μ_{net} = 0

— SF₆ is nonpolar.

Spot the error

"CO₂ is polar because oxygen is more electronegative than carbon."

Error: electronegativity difference makes the bonds polar, but the linear geometry (

θ = 180°

) makes the two opposite C=O dipoles cancel, so the molecule is nonpolar.

"For water I just add 1.5 D + 1.5 D = 3.0 D."

Error: dipoles add as vectors, not numbers. With a typical O–H bond dipole

μ \approx 1.5 D

and water's bond angle

θ = 104.5°

(so half-angle

θ /2 = 52.25°

), use

2 μ cos (θ /2) = 2 (1.5) cos 52.25° \approx 1.84 D

; only at

θ = 0°

would magnitudes simply add.

"BF₃ is polar because each B–F bond has a large dipole."

Error: large bond dipoles do not imply a polar molecule. Three equal dipoles at

120°

in a plane sum to the zero vector, so BF₃ is nonpolar.

"The dipole arrow points from the more electronegative atom to the less electronegative one."

Error (chemistry convention): the crossed-plus arrow points from

δ^{+}

toward

δ^{-}

, i.e. toward the more electronegative atom — the opposite of what was stated.

"NH₃ and NF₃ have the same shape, so they have equal dipole moments."

Error: same pyramidal shape but opposite internal balance. In NF₃ the N–F bond dipoles point away and partly oppose the lone-pair moment, so NF₃ is far less polar than NH₃.

"CH₄ is nonpolar because carbon and hydrogen have identical electronegativity."

Error: C and H differ slightly, so each C–H bond is weakly polar. CH₄ is nonpolar because tetrahedral symmetry cancels the four equal dipoles, not because the bonds are nonpolar.

"Since

μ_{net} = 2 μ cos (θ /2)

, a bigger bond angle always means a bigger dipole."

Error:

cos (θ /2)

decreases as

θ

grows from

0°

180°

, so a bigger bond angle gives a smaller net dipole (reaching zero at

180°

Why questions

Why must we treat dipole moment as a vector rather than a scalar

Because it comes from charge × position (

μ = \sum q_{i} r_{i}

), and position has direction; only vector addition lets opposite arrows cancel and partial ones combine — see Vectors and the Cosine Rule.

Why does replacing one H in CH₄ with Cl (giving CHCl₃) make it polar

Substitution breaks the tetrahedral symmetry — the four bonds are no longer identical, so they no longer cancel, leaving a net dipole along the H–C axis.

Why is the cosine rule the right tool for combining two bond dipoles

Two dipoles placed tail-to-tail form two sides of a triangle at the bond angle

θ

; the cosine rule is precisely the formula that gives the third side (the resultant) from two sides and their included angle.

Why does polarity control boiling point and solubility

Polar molecules attract each other through dipole–dipole forces (a type of Intermolecular Forces), which raise boiling point and make them dissolve in other polar substances.

Why do three vectors at 120° in a plane sum to zero

They are the three symmetry directions of an equilateral triangle; their horizontal components and vertical components each cancel in balanced pairs, leaving no resultant — the geometric reason BF₃ is nonpolar.

Why does using

2 μ cos (θ /2)

require the two bonds to be equal

The compact form comes from setting

μ_{1} = μ_{2} = μ

in the cosine rule; if the bonds differ you must keep the full

μ_{1}^{2} + μ_{2}^{2} + 2 μ_{1} μ_{2} cos θ

instead.

Why can a molecule's centre of positive and negative charge tell you polarity at a glance

If those two centres coincide there is no charge separation and

μ = 0

(nonpolar); if they are offset, that offset is the net dipole (polar).

Edge cases

What is the net dipole of two equal bond dipoles at exactly

θ = 180°

Zero:

2 μ cos 90° = 0

. This is the linear-cancellation case (CO₂), the boundary where a polar-bonded molecule becomes exactly nonpolar.

What is the net dipole of two equal bond dipoles at exactly

θ = 0°

(both parallel)

Maximum:

2 μ cos 0° = 2 μ

. The arrows point the same way, so magnitudes simply add — the only angle where naive addition is correct.

Is a diatomic molecule of two identical atoms (e.g. O₂, N₂) polar

No. Identical atoms have equal electronegativity, so there is no charge separation and the single bond dipole is itself zero — genuinely nonpolar at the bond level.

Can a molecule with only ONE polar bond ever be nonpolar

No. A single dipole has nothing to cancel it, so any molecule with exactly one polar bond (e.g. HCl) is necessarily polar.

What happens to the net dipole formula at the degenerate case

μ = 0

(nonpolar bonds)

Everything collapses to zero:

2 (0) cos (θ /2) = 0

for any angle. Geometry is irrelevant if the bonds carry no dipole to begin with.

If a molecule is bent but its two bonds have unequal dipoles, does the simple

2 μ cos (θ /2)

still work

No — that shortcut assumes equal magnitudes. You must revert to the full cosine rule

μ_{1}^{2} + μ_{2}^{2} + 2 μ_{1} μ_{2} cos θ

with the two different values.

A perfectly square-planar molecule with four identical polar bonds (e.g. XeF₄) — polar or not

Nonpolar. The four equal dipoles point to the corners of a square; opposite pairs cancel exactly, so the vector sum is zero.

Recall Final self-check

A polar-bonded molecule is nonpolar when… ::: its equal bond dipoles are arranged symmetrically (linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral, square planar) so the vector sum is zero.
The single most common trap on this topic is… ::: assuming "polar bonds ⇒ polar molecule" without checking geometry.