Exercises — Polarity of molecules — vector sum of bond dipoles

• (a) $C{O}_2$ — linear, two equal C=O arrows point exactly opposite ⇒ cancel ⇒ nonpolar.
• (b) $H_{2}O$ — bent ($104.5°$), the two O–H arrows both lean toward O and add ⇒ polar.
• (c) $B{F}_3$ — trigonal planar, three equal arrows $120°$ apart form a balanced star ⇒ nonpolar.
• (d) $CC{l}_4$ — tetrahedral, four equal arrows to symmetric corners cancel ⇒ nonpolar.
• (e) $CHC{l}_3$ — one bond swapped (C–H not C–Cl), symmetry broken ⇒ polar. Rule of thumb: equal arrows + symmetric shape (linear, trigonal planar, tetrahedral) = cancel.

${\delta }^{-}$ sits on the more electronegative atom, Cl. In the chemistry convention the arrow runs from ${\delta }^{+}$ to ${\delta }^{-}$, so it points from H toward Cl. (Physics points the other way — do not mix conventions.)

WHAT: we place both arrows tail-to-tail and add them (figure). WHY the formula: equal magnitudes, so use ${\mu }_{\text{net}}=2\mu \cos (\theta /2)$. ${\mu }_{\text{net}}=2(1.5)\cos \left(\frac{120°}{2}\right)=3.0\cos 60°=3.0(0.5)=1.5\text{D}.$ What it looks like: the resultant (pink arrow) bisects the angle. Neat coincidence: at $120°$ the net equals a single bond dipole.

Equal dipoles again, so ${\mu }_{\text{net}}=2(1.5)\cos \left(\frac{104.5°}{2}\right)=3.0\cos (52.25°)=3.0(0.6122)\approx 1.84\text{D}.$ Why it matches: using ${\mu }_{\text{OH}}\approx 1.5\text{D}$ reproduces the experimental $1.85\text{D}$ almost exactly, confirming the vector-sum picture. This polarity is why water dissolves salts — see Solubility — Like Dissolves Like.

Invert ${\mu }_{\text{net}}=2\mu \cos (\theta /2)$: ${\mu }_{\text{SO}}=\frac{{\mu }_{\text{net}}}{2\cos (\theta /2)}=\frac{1.63}{2\cos (59.5°)}=\frac{1.63}{2(0.5075)}=\frac{1.63}{1.015}\approx 1.61\text{D}.$ WHY divide: the formula is a machine turning bond dipole into net dipole; running it backward gives the bond dipole from the measured net.

Put the three arrows at angles $90°,210°,330°$ (each $120°$ apart). Sum the $x$ and $y$ parts. $\sum x=\cos 90°+\cos 210°+\cos 330°=0+(-0.8660)+0.8660=0,$ $\sum y=\sin 90°+\sin 210°+\sin 330°=1+(-0.5)+(-0.5)=0.$ Both components vanish, so ${\vec{\mu }}_{\text{net}}=\vec{0}$. What it looks like: the balanced star in the figure — every arrow is exactly opposed by the combined pull of the other two. This is why $B{F}_3$ is nonpolar.

Reasoning: the net dipole points along the direction where the electron-pulling (Cl) is bunched versus the electron-releasing side. In $C{H}_{2}C{l}_2$ two C–Cl arrows on one side face two weaker C–H arrows on the other, giving a strong resultant. In $CHC{l}_3$ three C–Cl arrows are spread around the axis; their sideways parts cancel and only the axial remainder survives, opposed partly by the lone C–H. The result: $C{H}_{2}C{l}_2$ ($1.60\text{D}$) $>$ $CHC{l}_3$ ($1.04\text{D}$). Lesson: more Cl does not mean more polar — direction and cancellation decide.

Now the magnitudes differ, so the shortcut $2\mu \cos (\theta /2)$ cannot be used — we need the full law of cosines: ${\mu }_{\text{net}}=\sqrt{{\mu }_1^2+{\mu }_2^2+2{\mu }_1{\mu }_2\cos \theta }.$ $=\sqrt{1.0^2+1.6^2+2(1.0)(1.6)\cos 100°}.$ $\cos 100°=-0.1736$, so the cross term is $2(1.6)(-0.1736)=-0.5556$. ${\mu }_{\text{net}}=\sqrt{1.0+2.56-0.5556}=\sqrt{3.0044}\approx 1.73\text{D}.$ What it looks like: the parallelogram in the figure — the diagonal is the net arrow, shorter than $1.0+1.6$ because the obtuse angle makes the arrows partly oppose.

For three equal arrows to sum to zero they must be symmetric in a plane at $120°$ (trigonal planar, like $B{F}_3$). Any pyramidal folding leaves a resultant along the axis (like $N{H}_3$). So the condition is: equal bonds and a planar $120°$ arrangement. WHY: only that arrangement makes both component sums zero (proved in L3.1). Geometry — from VSEPR Theory — is the deciding factor, not the bonds.

Add the four vectors component by component: $x:1+1-1-1=0,y:1-1+1-1=0,z:1-1-1+1=0.$ All three components are zero, so ${\vec{\mu }}_{\text{net}}=\vec{0}$. WHY it works: each coordinate has two $+1$'s and two $-1$'s — perfect balance. This is the algebra behind $C{H}_4$ and $CC{l}_4$ being nonpolar. Bond angle here is $\arccos (-\frac{1}{3})\approx 109.47°$.

The three Cl-directions sum to $(1,-1,-1)+(-1,1,-1)+(-1,-1,1)=(-1,-1,-1)$, which is antiparallel to the H-direction $(1,1,1)$. So every contribution lies on the C–H axis — the net is one-dimensional. Each tetrahedral vector has length $\sqrt{3}$; a unit arrow of magnitude $\mu$ along $(1,1,1)$ contributes $\mu \cdot \frac{(1,1,1)}{\sqrt{3}}$. The three Cl arrows give resultant ${\mu }_{Cl}\cdot \frac{(-1,-1,-1)}{\sqrt{3}}$ but there are three of them each of length... let's do it cleanly with the axial-projection idea: three C–Cl arrows each make angle $109.47°$ with the H-axis, so each projects $\cos (180°-109.47°)=\cos 70.53°=\frac{1}{3}$ of itself onto the H-axis direction. ${\mu }_{\text{net}}=|3{\mu }_{Cl}\cdot \frac{1}{3}-{\mu }_H|=|{\mu }_{Cl}-{\mu }_H|.$ Wait — sign: the three Cl axial projections point opposite to H, and add to $3{\mu }_{Cl}\cdot \frac{1}{3}={\mu }_{Cl}$. So ${\mu }_{\text{net}}={\mu }_{Cl}+{\mu }_H\text{or}{\mu }_{Cl}-{\mu }_H\text{ depending on relative arrow sense.}$ Taking the C–H and C–Cl arrows pointing the same way along the axis (both toward the more electronegative end sum), ${\mu }_{\text{net}}={\mu }_{Cl}+{\mu }_H=1.5+0.4=1.9\text{D}$ (rough model; the real value $1.04\text{D}$ is smaller because real C–H and C–Cl dipoles and projections differ). Key mastery point: by symmetry the net is guaranteed to lie exactly along the C–H axis — the three sideways components of the Cl arrows cancel among themselves.

No. A dipole needs a charge separation, which needs an electronegativity difference — i.e. at least one polar bond. If every bond is nonpolar, every ${\vec{\mu }}_i=\vec{0}$, so their sum is $\vec{0}$. Polarity requires (i) at least one polar bond and (ii) a geometry that does not cancel them. Lone-pair asymmetry (as in Intermolecular Forces discussions of $H_{2}O$, $N{H}_3$) can also supply a net moment, but that still originates from uneven charge distribution.

• Equal dipoles at $\theta$? ::: ${\mu }_{\text{net}}=2\mu \cos (\theta /2)$.
• Unequal dipoles at $\theta$? ::: ${\mu }_{\text{net}}=\sqrt{{\mu }_1^2+{\mu }_2^2+2{\mu }_1{\mu }_2\cos \theta }$.
• Three equal in-plane arrows at $120°$? ::: sum to zero (nonpolar).
• Four tetrahedral equal arrows? ::: sum to zero (nonpolar).
• Polar molecule with only nonpolar bonds? ::: impossible.
• Why $C{H}_{2}C{l}_2$ more polar than $CHC{l}_3$? ::: less internal cancellation of the C–Cl arrows.