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

A bond is polar when the two atoms have different electronegativity — the greedier atom pulls the shared pair and becomes ${\delta }^{-}$. (See Electronegativity.)

• (a) $\text{H–H}$ — same atom, equal pull → nonpolar.
• (b) $\text{H–Cl}$ — Cl is more electronegative → polar, ${\delta }^{-}$ on Cl.
• (c) $\text{Cl–Cl}$ — same atom → nonpolar.
• (d) $\text{C–O}$ — O is more electronegative → polar, ${\delta }^{-}$ on O.

(a) SI unit: coulomb·metre ($\text{C⋅m}$). Chemistry unit: Debye ($\text{D}$), where $1\text{D}=3.336\times 10^{-30}\text{C⋅m}$. (b) The arrow points from positive to negative (${\delta }^{+}\to {\delta }^{-}$). Why: it tracks where the electrons went, so it aims at the electron-rich (negative) end.

WHAT we do: turn the partial charge into coulombs, then multiply by the distance — this is literally $\mu =q\cdot d$. WHY: $q$ was given as a fraction of $e$; the formula needs $q$ in coulombs. $q=0.41\times 1.6\times 10^{-19}=6.56\times 10^{-20}\text{C}$ $\mu =q\cdot d=(6.56\times 10^{-20})(0.917\times 10^{-10})=6.02\times 10^{-30}\text{C⋅m}$ Convert to Debye by dividing by $3.336\times 10^{-30}$: $\mu =\frac{6.02\times 10^{-30}}{3.336\times 10^{-30}}\approx 1.80\text{D}$ This matches the experimental ${\mu }_{\text{HF}}\approx 1.82\text{D}$. ✔

WHAT we do: rearrange $\mu =q\cdot d$ to $q=\mu /d$, then divide by $e$ to express as a fraction. WHY: we know $\mu$ and $d$, so $q$ is the only unknown. First put $\mu$ in $\text{C⋅m}$: $\mu =0.40\times 3.336\times 10^{-30}=1.334\times 10^{-30}\text{C⋅m}$ $q=\frac{\mu }{d}=\frac{1.334\times 10^{-30}}{1.09\times 10^{-10}}=1.224\times 10^{-20}\text{C}$ As a fraction of $e$: $\frac{q}{e}=\frac{1.224\times 10^{-20}}{1.6\times 10^{-19}}\approx 0.077\approx 0.08e$ So each C–H carries only about $8\%$ of an electron's charge — a very weakly polar bond.

WHAT we do: compute the hypothetical dipole if the bond were 100% ionic ($q=e$, full electron transferred), then take the ratio. WHY: $\%$ ionic character compares the real dipole to the maximum possible one over the same bond length (see Ionic vs Covalent character). ${\mu }_{\text{ionic}}=e\cdot d=(1.6\times 10^{-19})(0.917\times 10^{-10})=1.467\times 10^{-29}\text{C⋅m}$ In Debye: ${\mu }_{\text{ionic}}=\frac{1.467\times 10^{-29}}{3.336\times 10^{-30}}=4.40\text{D}$. $\%\text{ ionic}=\frac{{\mu }_{\text{obs}}}{{\mu }_{\text{ionic}}}\times 100=\frac{1.82}{4.40}\times 100\approx 41\%$ HF is $\sim 41\%$ ionic — more ionic than HCl ($\sim 17\%$), consistent with fluorine's larger electronegativity.

Compute ${\mu }_{\text{HI}}$: $q=0.05\times 1.6\times 10^{-19}=8.0\times 10^{-21}\text{C}$ $\mu =(8.0\times 10^{-21})(1.61\times 10^{-10})=1.29\times 10^{-30}\text{C⋅m}=0.39\text{D}$ Ranking: ${\mu }_{\text{HF}}(1.82)>{\mu }_{\text{HCl}}(1.03)>{\mu }_{\text{HI}}(0.39)$. WHY distance did not save HI: $\mu =q\cdot d$ multiplies charge by distance. HI's bond is longest ($d$ largest), but its partial charge $q$ collapses to just $0.05e$ because H and I have nearly equal electronegativity. The tiny $q$ crushes the modest gain in $d$. So here the charge term dominates.

WHAT we do: add the two bond vectors. Only the components along the angle bisector survive; the sideways components point opposite ways and cancel. WHY the cosine of half the angle: put the bisector as the reference line. Each bond leans away from it by $\theta /2$. Its component along the bisector is ${\mu }_b\cos (\theta /2)$ (look at the red projections in the figure). Two of them add: ${\mu }_{\text{net}}=2{\mu }_b\cos \left(\frac{\theta }{2}\right)=2(1.5)\cos (60^{\circ })=2(1.5)(0.5)=1.5\text{D}$ So a $120^{\circ }$ bent ${\text{XY}}_2$ keeps a net dipole of $1.5\text{D}$. (Geometry from VSEPR Theory.)

Using the same formula: ${\mu }_{\text{net}}=2{\mu }_b\cos \left(\frac{180^{\circ }}{2}\right)=2{\mu }_b\cos (90^{\circ })=2{\mu }_b(0)=0$ WHAT this means: at $180^{\circ }$ the two arrows point exactly opposite — they fully cancel. The half-angle $90^{\circ }$ has $\cos =0$, killing the sum. Real example: ${\text{CO}}_2$ (O=C=O), linear, $\mu =0$ despite two polar bonds. The right panel of the figure shows the head-to-tail cancellation.

WHAT we do: place three equal vectors of magnitude ${\mu }_b$ at angles $90^{\circ },210^{\circ },330^{\circ }$ (any $120^{\circ }$-spaced set works). Add their $x$- and $y$-components. $\sum x={\mu }_b\left[\cos 90^{\circ }+\cos 210^{\circ }+\cos 330^{\circ }\right]={\mu }_b\left[0-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right]=0$ $\sum y={\mu }_b\left[\sin 90^{\circ }+\sin 210^{\circ }+\sin 330^{\circ }\right]={\mu }_b\left[1-\frac{1}{2}-\frac{1}{2}\right]=0$ Both components vanish, so ${\mu }_{\text{net}}=0$ — the "Mercedes-logo" cancellation. Symmetry, not weak bonds, kills the dipole.

WHAT we do: invert ${\mu }_{\text{net}}=2{\mu }_b\cos (\theta /2)$ to solve for $\theta$. WHY: we know ${\mu }_{\text{net}}$ and ${\mu }_b$; the only unknown is the geometry. $\cos \left(\frac{\theta }{2}\right)=\frac{{\mu }_{\text{net}}}{2{\mu }_b}=\frac{1.85}{2(1.51)}=\frac{1.85}{3.02}=0.6126$ $\frac{\theta }{2}=\arccos (0.6126)=52.2^{\circ }\Rightarrow \theta =104.5^{\circ }$ The tool — why $\arccos$? $\arccos$ is the question "which angle has this cosine?" It undoes the cosine so we can recover the angle from the dipole ratio. The answer $104.5^{\circ }$ is exactly water's bond angle — we reconstructed molecular shape from a dipole measurement. This is the logic linking $\mu$ to VSEPR Theory.

Rule: a vibration is IR-active only if it changes the molecular dipole $\mu$ (see Infrared Spectroscopy).

• (a) ${\text{CO}}_2$ symmetric stretch: both C=O bonds lengthen together. By the linear cancellation (L4.2), $\mu$ stays $0$ throughout — no change ⇒ IR-inactive for that mode.
• (b) ${\text{H}}_2\text{O}$ symmetric stretch: both O–H lengthen together, but the molecule is bent, so the net dipole along the bisector grows and shrinks — $\mu$ changes ⇒ IR-active. Punchline: geometry decides not only whether $\mu \ne 0$, but whether a vibration shows up in an IR spectrum.

(a) Partial charge. Convert $\mu$ to $\text{C⋅m}$: $\mu =1.08\times 3.336\times 10^{-30}=3.60\times 10^{-30}\text{C⋅m}$. $q=\frac{\mu }{d}=\frac{3.60\times 10^{-30}}{1.13\times 10^{-10}}=3.19\times 10^{-20}\text{C},\frac{q}{e}=\frac{3.19\times 10^{-20}}{1.6\times 10^{-19}}\approx 0.20$ So $q\approx 0.20e$. (b) Percent ionic character. ${\mu }_{\text{ionic}}=e\cdot d=(1.6\times 10^{-19})(1.13\times 10^{-10})=1.808\times 10^{-29}\text{C⋅m}=5.42\text{D}$ $\%\text{ ionic}=\frac{1.08}{5.42}\times 100\approx 19.9\%\approx 20\%$ (c) At $\sim 20\%$ ionic it is slightly more ionic than HCl ($17\%$). (The IR absorption simply confirms $\mu \ne 0$ — a polar diatomic must be IR-active.)