Visual walkthrough — Polarity of molecules — vector sum of bond dipoles

We only need two ideas from elsewhere: what an arrow that has both a length and a direction (a vector) is, and how two of them combine — see Vectors and the Cosine Rule. Everything else we draw as we go.

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Step 1 — Draw a single bond as one arrow

WHAT we did: replaced a whole bond by a single arrow of length $\mu$. WHY: because the only thing about the bond that affects polarity is how much charge is separated ($\mu$) and which way ($\to$). An arrow stores exactly those two facts and nothing else. PICTURE: the arrow below points from the pale ${\delta }^{+}$ atom into the dark ${\delta }^{-}$ atom.

$\underset{\text{the arrow}}{\underbrace{\vec{\mu }}}=\underset{\text{charge pulled aside}}{\underbrace{q}}\cdot \underset{\begin{array}{c}\text{step from }{\delta }^{+}\\ \text{to }{\delta }^{-}\end{array}}{\underbrace{\vec{d}}}$

Here $q$ is how much charge got separated and $\vec{d}$ is the little step across the bond. Multiplied together they give the arrow $\vec{\mu }$. Its plain length is $\mu =qd$.

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Step 2 — Put two bonds on one central atom

WHAT: we drew two equal arrows sharing a starting point, opened by angle $\theta$. WHY: dipole moments come from charges, and charges just add. So the molecule's total arrow is whatever these two arrows add up to — nothing more mysterious than that. PICTURE: two magenta arrows fanning out from one navy dot; the wedge between them is labelled $\theta$.

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Step 3 — Add arrows head-to-tail (the parallelogram)

WHAT: we turned "two arrows from a point" into a closed four-sided figure whose diagonal is the answer. WHY: sliding is allowed because a vector only remembers length and direction, not where it starts. The parallelogram is just a bookkeeping trick that makes the diagonal (our answer) something we can measure with geometry. PICTURE: the two solid magenta arrows, their two dashed copies, and the violet diagonal ${\vec{\mu }}_{\text{net}}$.

The diagonal is longest when the arrows point the same way and shrinks to nothing when they point opposite. We now put a number on that diagonal.

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Step 4 — Why we reach for the cosine rule

WHAT: we identified the right tool for "two sides, one angle → third side". WHY: plain addition ($\mu +\mu$) only works when the arrows are parallel. For any other angle the diagonal is shorter, and the cosine rule is precisely the correction for that. PICTURE: the triangle isolated, its two sides marked $\mu$, the interior angle marked, the wanted side marked ${\mu }_{\text{net}}$.

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Step 5 — Plug in equal bonds and simplify

WHAT: we specialised the general formula to two equal bonds. WHY: most real symmetric cases (water, $\mathrm{C}{\mathrm{O}}_2$, $\mathrm{S}{\mathrm{O}}_2$…) have two identical bonds, so this simpler formula covers them. PICTURE: same triangle now with both sides literally equal — an isosceles triangle. The diagonal splits it into two mirror halves; that mirror line is the key to the next step.

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Step 6 — The half-angle trick turns  ​ into a clean cosine

WHAT: we cleaned the square root into a single cosine of the half bond angle. WHY the half angle: the diagonal of the isosceles triangle lies exactly along the bisector of $\theta$. Each arrow leans away from that bisector by $\frac{\theta }{2}$, and only its component along the bisector survives (the sideways parts cancel by mirror symmetry). Projecting an arrow of length $\mu$ onto the bisector gives $\mu \cos (\frac{\theta }{2})$; two of them give $2\mu \cos (\frac{\theta }{2})$. That is the geometric meaning of the algebra. PICTURE: the bisector drawn in orange, each arrow's shadow onto it marked $\mu \cos (\frac{\theta }{2})$, the two side-parts shown cancelling.

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Step 7 — Walk every case of $\theta$

PICTURE: ${\mu }_{\text{net}}$ plotted against $\theta$, with real molecules dropped onto the curve.

$\theta$	$\frac{\theta }{2}$	$\cos (\frac{\theta }{2})$	${\mu }_{\text{net}}$	Meaning
$0^{\circ }$	$0^{\circ }$	$1$	$2\mu$	arrows parallel → maximum, fully add
$90^{\circ }$	$45^{\circ }$	$0.707$	$1.41\mu$	right angle → partial
$104.5^{\circ }$	$52.25^{\circ }$	$0.612$	$1.22\mu$	water → clearly polar
$120^{\circ }$	$60^{\circ }$	$0.5$	$\mu$	bent like $\mathrm{S}{\mathrm{O}}_2$
$180^{\circ }$	$90^{\circ }$	$0$	$0$	arrows opposite → cancel, nonpolar

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The one-picture summary

One figure, the whole story: two equal bond arrows open by $\theta$, their sideways parts cancelling on the mirror line, their forward shadows $\mu \cos (\frac{\theta }{2})$ adding to give ${\mu }_{\text{net}}=2\mu \cos (\frac{\theta }{2})$ — big when $\theta$ is small, zero when $\theta =180^{\circ }$.

This links straight to VSEPR Theory and Molecular Geometry and Shapes (which set $\theta$), and the resulting polarity drives Intermolecular Forces and Solubility — Like Dissolves Like.

Recall Feynman: tell the whole walkthrough to a friend

Every bond is an arrow: it starts on the slightly-plus atom and points to the slightly-minus one, and its length says how strong it is. Stick two identical arrows out of the middle atom at the bond angle $\theta$. To find the molecule's arrow, add them. Adding arrows means sliding one to the tip of the other and drawing the diagonal — that makes a triangle with two known sides and a known angle, and the one tool for that job is the cosine rule. Do the algebra, use the half-angle identity, and it collapses to a beautifully simple thing: $2\mu \cos (\frac{\theta }{2})$. The picture behind that formula: only the part of each arrow along the mirror line survives; the sideways parts kill each other. Each survivor is $\mu \cos (\frac{\theta }{2})$, and there are two, so $2\mu \cos (\frac{\theta }{2})$. When the bonds point straight apart ($\theta =180^{\circ }$) the survivors vanish and the molecule is nonpolar — that's $\mathrm{C}{\mathrm{O}}_2$. When they're bent ($104.5^{\circ }$) the survivors add up and you get a polar molecule — that's water.

Recall Quick checks

Two equal dipoles at $60^{\circ }$, each $\mu =1\text{D}$ — net? ::: $2(1)\cos 30^{\circ }=1.73\text{D}$. Why does $\theta =180^{\circ }$ give zero? ::: $\cos (90^{\circ })=0$; opposite arrows, shadows cancel. Which surviving part of each arrow adds up? ::: The component along the bisector, $\mu \cos (\frac{\theta }{2})$. Wider bond angle → more or less polar? ::: Less; $\cos (\frac{\theta }{2})$ shrinks as $\theta$ grows.