Question

If dipole moment of AB molecule is 1.6 × 10−29 Cm and distance of A − B bond is 1.6 Å, then percentage ionic character of the compound will be​

Answer 1

Answer:

62.5% Answer

Explanation:

The diapole moment of 100% ionic character will be:

μ_ionic = q×d = 1.6×10^−19 C × 1.6× 10^-10 m

=2.56×10^-29 C m

The percentage ionic character is =

$\frac{dipole \: moment}{dipole \: moment \: of \: 100\% \: ionic \: character} \times 100$

$\frac{1.6 \times 10 {}^{ - 29} \times 100 }{2.56 \times 10 {}^{ - 29} }$

% Ionic Character = 62.5%