Estimate the dipole moments of all five possible HX compunds, where X is a halogen, using...

Estimate the dipole moments of all five possible HX compunds, where X is a halogen, using only the Pauling electronegativities and showing your answers in both Coulomb-meters and Debye. What is responsible for the trend in polarities of these bonds that your calculations predict? (i.e. what causes the differences in the halogen electronegativities?) Follow-up question: How does this trend differ from that of van der waals interactions involving the halogens?

Expert Solution

(1.91 D) / (3x10^29 D/C*m) = 6.37x10^-30 C*m

(0.92 A) (1x10*-10 m/A) = 9.2x10^-11 m

Electronegativies found in Wikipedia:
Electronegativity F = 3.98
Electronegativity H = 2.20
Difference = 3.98 - 2.20 = 2.20
From the table in the reference below, this corresponds to 55% ionic character.

Similarly..calculate the Dipole moment of other Hydrogen Halides

HF	1.91
HCl	1.08
HBr	0.80
HI	0.42 Dipole moment in Debye

Dipole moment means how polar the molecule is, and since the hydrogen halides are diatomic and linear, this translates directly into how polar the bond is.

The polarity of a bond depends on the electronegativity difference between the 2 atoms. The general trend in electronegativity is that it decreases down a column. This is because, while elements in the same column have the same effective nuclear charge (+7 in the case of halogens), they have larger electron clouds as you move down the periodic table. This means the valence (bonded) electrons are farther from the nucleus of Br than they are from the nucleus of F. Therefore, the attraction they feel for the nucleus is less for the larger atoms.

queen_honey_blossom answered 2 years ago

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