Dipolar moment between HI and HF
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.