The dipole moment of hydrogen chloride is 1.03 D. Its percent ionic character is about 17%. Its bond distance in pm is about

h+ and cl- bear a unit charge of 4.8*10^-10
theoritical value of u=4.8*10^-10 *d
wher d is the bond distance
percentage of ionic character=exp/theoritical*100=17
bond distance=1.27A
thanks and regards
sunil kr
askIITian faculty

Approved

Here, we are given the percent ionic character = 17% =0.17, so if r is the bond distance,

The dipole moment of HCL is (0.17e)r = (0.17 x 1.6 x 10^–19 C) r

= (1.03 x 3.33 x 10^–30 C m)

r = 1.03 x 3.3356 x 10^–30 Cm / 0.17 x 1.6 x 10^–19 C

Thus, we have bond distance= 1.27 x 10^–10 m = 127 pm

Thanks & Regards,

Vasantha Sivaraj,

askIITians faculty

Approved

correct answer and good explanation Thank you Mam

Given :-
Observed dipole moment = 1.03D
% ionic character = 17%
Now we know that,
mu = e * d
= 1.6 * 10^-19 * d/3.335 * 10<sup>-30

mu = 0.48 * 1011 * d(m)

% ionic character = Observed dipole moment/ calculated dipole moment *100

17 = 1.03 * 100/</sup> 0.48 *10¹¹* d
d = 1.03 * 100/0.48 * 10¹¹ * 17
d = 0.127 * 10 ^-9
or, d = 127 pm
Thanks & Regards
Aarti Gupta
askiitians Faculty