Question

i)The dipole moment of HBr is 2.602 ×〖10〗^(-30)C m and its bond length is 141 pm. Calculate its percentage ionic character.

Answer 1

Question #46437, Chemistry, Inorganic Chemistry

i) The dipole moment of HBr is $2.602 \times \ll 10^{3}$ (e·d) C m and its bond length is 141 pm. Calculate its percentage ionic character.

Answer:

m - dipole moment

e - electron charge (1.602·10⁻¹⁹)

d - bond length

X - ionicity connection

$$X = m / (e \cdot d)$$ $$X = (2.602 \cdot 10^{-30}) / 1.602 \cdot 10^{-19} \cdot 141 \cdot 10^{-12} = 0.115$$ $$0.115 \cdot 100\% = 11.5\%$$

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