varun
Last Activity: 7 Years ago
HELLO,
first we will assume that this molecule is 100% ionic
the dipole moment is U=Q.r
U=1.602x10^-19 x 127.5x10^-12m
=2.04259 x10^-29cm
but actual dipole moment of HCL =3.57X10^-30 cm
therefore,% of ionic character of HCL
=actual dipole moment/calculated dipole moment X 100
=3.57X10^-30/2.04259X10^-29 X 100
=17.478%
so the answer is 17.478%
THANK YOU
