Question

The observed dipole-moment of HCl molecule is 1.03 D. If \[\ce{H - Cl}\] bond distance is 1.275 Å and electronic charge is 4.8 × 10^-10 esu. What is the percent polarity of HCl?

Options

• 1.275 Å × 1.03

• `(4.8 xx 10^-10 xx 1.275 xx 10^-8)/1.03`

• `(1.03 xx 100 xx 10^-18)/(4.8 xx 10^-10 xx 1.275 xx 10^-8)`

• `(4.8 xx 10^-10)/1.03 xx 10`

Answer 1

`(1.03 xx 100 xx 10^-18)/(4.8 xx 10^-10 xx 1.275 xx 10^-8)`

Explanation:

`"m"_"calculated" = "e" xx "d"`

= `4.8 xx 10^-10 xx 1.275 xx 10^-8`

d = `1.25 xx 10^-8` cm

1 D = `1 xx 10^-18` esu - cm

`mu_"observed" = 1.03` D

= `1.03 xx 10^-18` esu-cm

% = `mu_"observed"/mu_"cal" xx 100`

`= (1.03 xx 10^-18 xx 100)/(4.8 xx 10^-10 xx 1.275 xx 10^-8)`