1、Arrangegaseoushudrogenhalidesinorderofdecreasingdipolemomentsandrationalizeyourorder.2、Thedipolemomentsforthemethylhalidesare:CH3F,1.82D;CH3Cl,1.94D;CH3Br,1.79D;CH3I,1.63DExplain.

Question

1、Arrange gaseous hudrogen halides in order of decreasing dipole moments and rationalize your order.
2、The dipole moments for the methyl halides are:
CH3F,1.82D; CH3Cl,1.94 D; CH3Br,1.79 D; CH3I,1.63 D
Explain.

Answer 1

1. HF>HCl>HBr>HI
dipole moment is separation of charge between two atoms of a chemical bond related to the electronegativity of halogens and fluorine has the highest electronegativity, chlorine next, bromine third and iodine the last while hydrogen has very low.
2. The order of dipole moments generally follow the decrease of the electronegativity of halogens with an exception. The apparent smaller dipole moment of CH3F than CH3Cl can be explained by the shorter bond C-F than C-Cl even though F has higher electronegativity than Cl.
dipole moment mu = delta * d (delta is separation of charge, partial charge; d is distance of the partial charge between two atoms)