The dipole moment of `HBr` is ` 1.6 xx 10^(-30) cm` and interatomic spacing is `1 Å`. The `%` ionic character of `HBr` is

Options

A. 7
B. 10
C. 15
D. 27

Answer

B

Explanation

Charge of electron=`1.6xx10^(-19)`C
Dipole moment HBr=`1.6xx10^(-30)` cm
Interiodic spacing =1 Ã… =`1xx10^(-10)` m
Considering HBr to be ionic , dipole moment
=`1.6xx10^(-19) C xx 10^(-10)m`
`=1.6xx10^(-29)` Cm
Therefore, percentage ionic character of HBr =`(1.6xx10^(-30))/(1.6xx10^(-29))xx100=10%`