Dipole moment $\alpha$ diff. in ${e^ - }$ unity is HF > HCl > HBr > HI
A.True
B.False

Hint: The dipole moment depends on the s-character the carbon atom in the molecule and the bond length between the atoms. Greater is the distance between the oppositely charged ions greater will be the dipole moment. The polarity of the compound is related to the electronegativity of the atom.

Complete step by step answer:
Dipole moment is the measure of the extent for the polarity of the compound which is influenced by the electronegativity of the atom present in the compound. When there is the atom attached with the higher electronegativity value, then the dipole moment will be high.
Among all the halogens atoms, fluorine is the most electronegative atom. In the H-F bond, the fluorine will pull the electrons towards itself. As we move down in the group in the periodic table the electronegativity decreases due to the increase in atomic size.
Hence fluorine is the most electronegative element in the halogen group and iodine is the least electronegative element. Therefore, HF shows more polarization than HI.
Hence, HF has a higher dipole moment than other atoms in the halogen group.
The order for the decreasing order of electronegativity is shown below.
HF > HCl > HBr > HI

Therefore, the statement Dipole moment $\alpha$ diff. in ${e^ - }$ unity is HF > HCl > HBr > HI is false.

Note: The dipole moment is a vector quantity, as it possesses both magnitude and direction. As it is a vector quantity the dipole moment can be zero also because the two oppositely charged bond dipoles can cancel out each other.