A diatomic molecule possesses an experimental dipole moment of and an inter-nuclear bond distance of — Chemical Bonding Chemistry Question
Question
A diatomic molecule $AB$ possesses an experimental dipole moment of $1.2 \text{ D}$ and an inter-nuclear bond distance of $1.0 \times 10^{-8} \text{ cm}$. Assuming the magnitude of electronic charge is $4.8 \times 10^{-10} \text{ esu}$, what is the approximate percentage ionic character of the $A-B$ bond?
Answer: C
💡 Solution & Explanation
Theoretical dipole moment for a 100% ionic bond ($\mu_{theoretical}$) = $q \times d$ = $(4.8 \times 10^{-10} \text{ esu}) \times (1.0 \times 10^{-8} \text{ cm}) = 4.8 \times 10^{-18} \text{ esu cm} = 4.8 \text{ D}$. The percentage ionic character is $(\mu_{observed} / \mu_{theoretical}) \times 100 = (1.2 / 4.8) \times 100 = 25\%$.
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