The observed dipole moment of a gaseous diatomic molecule HX is 1.03 D and its bond length is 127 pm. Taking percent ionic character =100xmu_obs/(e d) with e=1.6x10^-19 C and 1 D=3.336x10^-30 C m, calculate the percent ionic character (answer to one decimal place).

JEE Main levelNumerical answerDifficulty 2 of 3cb188

The observed dipole moment of a gaseous diatomic molecule $HX$ is $1.03 D$ and its bond length is $127 pm$ . Taking percent ionic character $= 100 \times μ_{o b s} / (e d)$ with $e = 1.6 \times 1 0^{- 19} C$ and $1 D = 3.336 \times 1 0^{- 30} C m$ , calculate the percent ionic character (answer to one decimal place).

Numerical answer type, answer in %, tolerance 0.1.

For a

100%

ionic bond

μ_{i o ni c} = e d = 1.6 \times 1 0^{- 19} \times 127 \times 1 0^{- 12} C m

, which in debye is 6.091 D. Percent ionic character

= 100 \times 1.03/6.091 = 16.91%

, i.e. about 16.9%.