Question

Dipole moment of $HX$ is $2.59\times {10}^{-30}$ coulomb-metre. Bond length of $HX$ is $1.39\mathring{A}$. Calculate percentage ionic character of molecule.

Answer 1

Dipole Moment $\mu$ observed=$2.59\times {10}^{-30}$ coulomb-metre.

Bond Length= 1.39$A^o$=$1.39\times {10}^{-10}$ coulomb metre.

Charge=$1.69\times {10}^{-19}$ C.

Ionic character=$\frac{\mu }{BondLength\times Charge}$

=$\frac{2.59\times {10}^{-30}}{1.39\times {10}^{-10}\times 1.69\times {10}^{-19}}$

=$\frac{2.59\times {10}^{-30}}{2.224\times {10}^{-29}}$

=$1.16\times {10}^{-1}$

=$0.116$

Percentage Ionic Character=$0.116\times 100$=11.6%.