# Dipolar moment between HI and HF [duplicate]

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.

• 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 4 '18 at 13:29
• F is more electronegative than I, how do I evalue if it compensates for the smaller distance? – Mattia Mariantoni Jan 4 '18 at 13:36
• One method is to look up the "rulr" for ionic character versus electronegativity difference, see the plot in chemistry.stackexchange.com/questions/9222/…. Use the curve to figure out where HI might be. Which is the bigger factor now, atomic size or charge separation? – Oscar Lanzi Jan 4 '18 at 14:08
• – Mithoron Jan 4 '18 at 16:18

## 1 Answer

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.

• How exactly is the charge related to the electronegativity? – pentavalentcarbon Jan 4 '18 at 15:02
• Higher the charge per unit area on atomic surface more is the tendency of that atom to attract electrons. (Allred Rochow Electronegativity). – user57147 Jan 4 '18 at 15:33