According to the Mulliken scale of electronegativity, if the first ionization energy () and the elec — Periodic Table and Periodicity Chemistry Question
Question
According to the Mulliken scale of electronegativity, if the first ionization energy ($IE_1$) and the electron affinity ($EA$) of a hypothetical element are experimentally found to be $13.0 \text{ eV/atom}$ and $3.8 \text{ eV/atom}$ respectively, what is its approximate electronegativity on the Pauling scale?
Answer: B
💡 Solution & Explanation
On the Mulliken scale, electronegativity is the arithmetic mean of IP and EA in eV/atom: $EN_M = \frac{13.0 + 3.8}{2} = 8.4$. The relationship between the Pauling scale ($X_P$) and the Mulliken scale ($X_M$) is given by $X_P = \frac{X_M}{2.8}$. Therefore, Pauling EN $= \frac{8.4}{2.8} = 3.0$.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes