Calculate the percentage of ionic character of HF. Given that the dipole moment of HF is 1.91 D and its bonding length is 0.92Å.

Calculate the percentage of ionic character of HF. Given that the dipo

1.	Calculate the percentage of ionic character of HF. Given that the dipole moment of HF is 1.91 D and its bonding length is 0.92Å.
Answer» Solution :If HF is 100% ionic, each atom would CARRY a charge equal to one unit, ie.., `4.8xx10^(-10)` esu. As the bond length of HF is 0.92 Å, itsdipoel moooment for 100% ionic CHARACTER would be `p_("ionic")=qxxd=4.8xx10^(-10)esuxx0.92xx10^(-8)CM` `=4.416xx10^(-18)esucm=4.416D` `[because 10^(-18)esucm=1D]` `therefore%` ionic character`=(mu_("observed"))/(mu_("ionic"))xx100=(1.91xx100)/(4.416)=43.25`