Noesis
2.3.6 · D5Chemical Bonding

Question bank — Polarity of bonds — dipole moment μ = q·d

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True or false — justify

True or false: A molecule made entirely of polar bonds is always a polar molecule.
False. Bond dipoles are vectors; symmetric shapes (CO₂, BF₃, CCl₄) let them cancel to a net — see the middle panel of the figure above and VSEPR Theory.
True or false: If a molecule has , none of its bonds can be polar.
False. Each C=O bond in CO₂ is strongly polar; the molecule's is zero only because the two equal, opposite bond vectors cancel (left of the CO₂/BF₃ panel).
True or false: The in equals one full electron charge .
False. is a partial charge, typically –. Full is used only to compute the hypothetical 100% ionic for percent-ionic-character.
True or false: A larger electronegativity difference always produces a larger dipole moment.
False. also depends on ==bond length == (and on geometry and lone pairs), so a bigger EN alone doesn't guarantee bigger . See Electronegativity.
True or false: The dipole arrow (chemistry convention) points from the negative end to the positive end.
False. By chemistry convention it points from ==positive () to negative ()==, i.e. in the direction the electrons drifted, arrow-head at the greedy atom (panel a of the figure).
True or false: Two identical atoms in a bond (like H–H or Cl–Cl) can still have a nonzero bond dipole.
False. With equal electronegativity the pair is shared equally, so and therefore no matter the bond length.
True or false: Doubling the bond length while keeping the same partial charge doubles the dipole moment.
True. is linear in , so at fixed the moment scales directly with the separation distance.
True or false: A molecule with a permanent dipole moment can still be IR-inactive for one of its vibrations.
True. A vibration is IR-active only if it changes ; a symmetric stretch that keeps constant stays IR-silent. See Infrared Spectroscopy.

Spot the error

Spot the error: "HCl has D, so its charge separation must be one full electron over the bond length."
The error is assuming . The observed divided by gives only ~17% ionic character, so , a fraction of an electron.
Spot the error: "BF₃ has three polar B–F bonds pointing outward, so it must be strongly polar."
Three equal vectors at 120° (a symmetric "star") sum to zero, so despite each bond being polar — right panel of the figure above.
Spot the error: "Water is polar because oxygen is electronegative — the bent shape is irrelevant."
Shape is essential: if H₂O were linear, the two O–H dipoles would cancel like CO₂. The 104.5° bend is exactly why the components along the bisector survive and add (panel c of the figure).
Spot the error: " gives , so HCl's dipole moment is about 10 Debye."
Units confusion: is in C·m. Dividing by C·m/D gives ~3 D-scale values, and HCl is actually ~1.03 D — the raw C·m number is not the Debye value.
Spot the error: "To find a molecule's dipole, just add up the magnitudes of all its bond dipoles."
You must add them as vectors head-to-tail using the geometry, not add magnitudes. Scalar addition ignores direction and wrongly predicts every molecule is polar.
Spot the error: "CCl₄ and CHCl₃ are both symmetric, so both have ."
Only CCl₄ (four identical C–Cl at tetrahedral angles) cancels. In CHCl₃ one C–H replaces a C–Cl, breaking the symmetry, so the bond dipoles no longer cancel and .

Why questions

Why do chemists use the Debye instead of C·m?
Because raw values sit around – C·m (ugly); the Debye is scaled so most molecules land in the readable – D range.
Why does the net dipole of water use rather than ?
The bisector splits the angle in half; each O–H dipole makes half the bond angle with the bisector, so only its component adds along that axis (the perpendicular parts cancel) — see panel c of the figure above.
Why can a molecule with polar bonds have a lower than a molecule with less-polar bonds?
Because geometry can partially cancel the strong bond dipoles, while a less symmetric molecule with weaker bonds may leave more of its (smaller) dipoles uncancelled.
Why do we compare the observed dipole to (not to some fixed number) when finding percent ionic character?
Because the "100% ionic" reference uses a full electron transferred across the actual bond length, so it must use that molecule's own ; a fixed number would ignore how far the charges sit apart.
Why do high- molecules like H₂O and HF have unusually strong intermolecular attractions?
Large permanent dipoles produce strong dipole–dipole (and hydrogen-bonding) interactions, since these forces scale with . See Intermolecular forces and Hydrogen bonding.

Edge cases

Edge case: What is for a bond between two atoms of identical electronegativity?
Exactly zero — equal sharing means , so regardless of how long the bond is.
Edge case: Can a monatomic species (like a lone Ne atom) have a dipole moment?
No — with a single symmetric atom there is no charge separation ( is meaningless), so .
Edge case: A linear triatomic X–Y–Z where the two end atoms differ (e.g. OCS, O=C=S) — does it cancel like CO₂?
No. The two bond dipoles are unequal in magnitude, so even though they're 180° apart they don't fully cancel, leaving a small net .
Edge case: If the partial charge but the bond length stays finite, what happens to ?
: with no charge imbalance there is nothing for the distance to multiply, so the bond becomes nonpolar in the limit.
Edge case: Trans-difluoroethene vs cis-difluoroethene — same bonds, why different ?
In the trans isomer the two C–F dipoles point opposite and cancel (); in the cis isomer they point to the same side and add to a nonzero net dipole — pure geometry deciding polarity (see figure above).
Recall One-line summary of every trap

Reveal ::: needs a partial , a real , and above all vector addition over the molecular shape — symmetry cancels, asymmetry survives.

Connections

  • VSEPR Theory — the geometry that decides cancellation.
  • Electronegativity — sets the partial charge .
  • Ionic vs Covalent character — percent ionic character context.
  • Infrared Spectroscopy — only -changing vibrations absorb.
  • Intermolecular forces — dipole–dipole strength scales with .
  • Hydrogen bonding — dominant in high- molecules.