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Which between $\ce{HI}$ and $\ce{HF}$ has a greater dipolar moment? I think it is $\ce{HI}$ because the atomic radius of $\ce{I}$ is greater.

bon
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Mattiatore
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    How much charge separation is between the atoms on each compound? What is the difference between electronegativities of H and I? H and F? – Oscar Lanzi Jan 04 '18 at 13:29
  • F is more electronegative than I, how do I evalue if it compensates for the smaller distance? – Mattiatore Jan 04 '18 at 13:36
  • One method is to look up the "rulr" for ionic character versus electronegativity difference, see the plot in https://chemistry.stackexchange.com/questions/9222/why-are-bonds-ionic-when-the-electronegativity-difference-between-bonded-atoms-i. Use the curve to figure out where HI might be. Which is the bigger factor now, atomic size or charge separation? – Oscar Lanzi Jan 04 '18 at 14:08
  • see https://chemistry.stackexchange.com/questions/31049/dipole-moment-anomaly-in-fluorine-and-chlorine-compounds-what-factors-affect-th – Mithoron Jan 04 '18 at 16:18

1 Answers1

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The dipole moment of $\ce{HF}$ is greater than $\ce{HI}$:

  • $\ce{HF} = \pu{1.91 D}$
  • $\ce{HI} = \pu{0.42 D}$

This is due to high electronegativity of fluorine. Hence from $\mu = \vec{q} \cdot \vec{d}$, the charge of fluorine is larger than iodine, but the bond length changes only a small amount.

pentavalentcarbon
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