Correct Answer - B
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%`