Foundations — Polarity of bonds — dipole moment μ = q·d

This page assumes nothing. Before you use a symbol you will see it drawn. We go through every letter, sign, and piece of notation the parent note leans on, in an order where each one rests on the one before it.

0. Charge — what the symbols $+$, $-$, $q$, $e$ actually mean

The picture: think of $+$ as "missing electrons here" and $-$ as "extra electrons here."

Why the topic needs $e$: the parent measures partial charges as fractions of $e$ (like $0.17e$ for HCl). Without knowing $e$ in coulombs you cannot turn "$0.17e$" into a real number to multiply.

Look at the figure: the shared electron cloud (blue) is dragged toward the greedy atom, so that atom's side reads ${\delta }^{-}$ and the other reads ${\delta }^{+}$. The tiny letter $\delta$ (delta) is the whole point — it means "a little bit," reminding you the charge is partial.

1. Distance $d$ and the idea of a bond length

Why the topic needs $d$: a dipole is not just about how much charge is separated but how far apart it sits. Two charges nose-to-nose barely separate; the same charges pulled apart make a stronger dipole. $d$ carries the "how far" half of $\mu =q\cdot d$.

2. Vectors — the arrow that carries direction

The figure shows the three rules you must have for the parent page:

• Opposite arrows cancel (left): equal length, pointing $180°$ apart → they add to a zero-length arrow. This is exactly why CO₂ has $\mu =0$.
• Angled arrows add to a shorter arrow along the middle (centre): this is why bent H₂O keeps a net dipole.
• Subtraction (right): ${\vec{r}}_{+}-{\vec{r}}_{-}$ flips ${\vec{r}}_{-}$ and adds it, giving the arrow from the negative charge to the positive charge. Keep this picture — we use it in the next section.

3. From two charges to a sum over many — where $\vec{\mu }={\sum }_i{q}_i{\vec{r}}_i$ comes from

WHY a sum, and how does it grow out of $\mu =q\cdot d$? Watch the two-charge case build itself, then extend.

Step 1 — Write the two-charge sum. Our bond has just two charges: $+q$ at ${\vec{r}}_{+}$ and $-q$ at ${\vec{r}}_{-}$. Applying the sum: $\vec{\mu }={\sum }_i{q}_i{\vec{r}}_i=(+q){\vec{r}}_{+}+(-q){\vec{r}}_{-}.$ Why this step? Each term is "charge times where it sits." The second charge is negative, and here is the key detail: a negative $q_i$ automatically flips its arrow ${\vec{r}}_{-}$ around (that is exactly the "negative of a vector" from the last section). So the sign does the direction-bookkeeping for you — you never have to decide by hand which way to point.

Step 2 — Factor out $q$ using subtraction. $\vec{\mu }=q{\vec{r}}_{+}-q{\vec{r}}_{-}=q({\vec{r}}_{+}-{\vec{r}}_{-}).$ Why this step? Both magnitudes are equal, so $q$ pulls out, leaving a vector subtraction — the very thing we just defined. The bracket is the separation arrow.

Step 3 — Name the separation. Let $\vec{d}={\vec{r}}_{+}-{\vec{r}}_{-}$. By the subtraction rule this arrow runs from the negative charge to the positive charge, with length $d$. So $\boxed{\vec{\mu }=q\vec{d},\mu =q\cdot d.}$

Step 4 — Why keep the general $\sum$ at all? A real molecule has many charges — several nuclei and several partial charges. The formula $\vec{\mu }={\sum }_i{q}_i{\vec{r}}_i$ handles all of them at once: two positive and two negative bond ends in H₂O, three in BF₃, and so on. Each term flips itself correctly by its own sign, and adding them tip-to-tail gives the net molecular arrow. The two-charge $\mu =q\cdot d$ is simply the smallest case of this master sum.

4. The catch: this only behaves nicely for a neutral system

Step 1 — Shift the origin. Move the reference dot by some fixed arrow $\vec{s}$. Every position becomes ${\vec{r}}_i\to {\vec{r}}_i-\vec{s}$. The new moment is ${\vec{\mu }}_{\text{new}}={\sum }_i{q}_i({\vec{r}}_i-\vec{s})={\sum }_i{q}_i{\vec{r}}_i-\left({\sum }_i{q}_i\right)\vec{s}=\vec{\mu }-Q_{\text{total}}\vec{s},$ where $Q_{\text{total}}={\sum }_i{q}_i$ is the total charge.

Step 2 — Read the result in both cases.

• If the system is neutral, $Q_{\text{total}}=0$, so the extra term vanishes: ${\vec{\mu }}_{\text{new}}=\vec{\mu }$. The dipole moment is origin-independent — a real, unambiguous property.
• If the system has net charge ($Q_{\text{total}}\ne 0$), the moment changes when you move the origin. Then $\vec{\mu }$ alone is not a well-defined property; you must also state where the origin is.

Why this matters here: a whole molecule (H₂O, CO₂, HCl) is electrically neutral — every ${\delta }^{+}$ is balanced by an equal ${\delta }^{-}$. That is exactly why the parent page can quote a single number for $\mu$ without ever mentioning an origin. The neutrality of molecules is the quiet reason the formula works.

5. The dipole moment symbol $\mu$ — and the sign convention twist

Why $\mu$ exists as its own symbol: it fuses the two halves — charge $q$ and distance $d$ — into a single measurable score for "how polar." That lets us compare molecules with one number.

6. Units — coulomb·metre and the Debye

To go from C·m to Debye you divide by $3.336\times 10^{-30}$; to go the other way you multiply. That single conversion is the "3-3-3-6" mnemonic on the parent page.

The number line in the figure shows why: raw C·m values crowd near $10^{-30}$ (unreadable), while the Debye axis spreads everyday molecules across a friendly $0$–$5$ range.

7. Electronegativity — the cause behind the charge separation

Why the topic needs it: it explains why $q$ is nonzero at all. Same electronegativity (e.g. Cl–Cl) → equal sharing → $q=0$ → $\mu =0$. The full story lives in Electronegativity.

8. The cosine $\cos$ — how we split an angled arrow

Why the topic needs $\cos$: for H₂O the two O–H dipoles point at half the bond angle away from the symmetry line, so only their $\cos (104.5°/2)$ components add up. That is where the parent's ${\mu }_{\text{net}}=2{\mu }_{\text{OH}}\cos \left(\frac{104.5°}{2}\right)$ comes from. (Full trig lives with the geometry from VSEPR Theory.)

The prerequisite map

Read it top-down: charge ideas and distance feed the formula $\mu =q\cdot d$; vectors, subtraction and cosine feed the molecular dipole (the cancellations); neutrality guarantees the sum is origin-independent; electronegativity is the cause upstream of $q$.

Equipment checklist

Test yourself — cover the right side and answer out loud.

Connections

• Parent topic — where these foundations are put to work.
• Electronegativity — the cause of the partial charge $q$.
• VSEPR Theory — supplies the geometry (angles) the vector sum and $\cos$ need.
• Ionic vs Covalent character — extends the partial-charge idea to percent ionic character.