Chapter 12: Problem 102
Calculate the electronegativity of fluorine from the following data. \(\mathrm{E}_{\mathrm{H}-\mathrm{H}}=104.2 \mathrm{Kcal} \mathrm{mol}^{-1}\) \(\mathrm{E}_{\mathrm{F}-\mathrm{F}}=36.6 \mathrm{Kcal} \mathrm{mol}^{-1}\) \(\mathrm{E}_{\mathrm{H}-\mathrm{F}}=134.6 \mathrm{Kcal} \mathrm{mol}^{-1}\) \(\mathrm{X}_{\mathrm{H}}=2.1\) (a) \(2.86\) (b) \(3.76\) (c) \(1.86\) (d) \(3.26\)
Short Answer
Step by step solution
Use Pauling's Electronegativity Formula
Convert Bond Energies
Calculate the Difference in Electronegativity
Solve for Electronegativity of Fluorine
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Pauling's Electronegativity Formula
- \(D_{AB}\): Bond energy of the A-B bond
- \(D_{AA}\), \(D_{BB}\): Bond energies of the A-A and B-B bonds
Bond Energy Conversion
- \(1 \text{ Kcal/mol} = 0.0434 \text{ eV}\)
Fluorine Electronegativity
Pauling Scale
- Fluorine, as the benchmark for highest electronegativity, has a value of \(4.0\).
- Hydrogen, often a reference point, holds a value of \(2.1\).
Chemical Bond Energies
- \(H-H\), with an energy of \(104.2 \text{ Kcal/mol}\)
- \(F-F\), with an energy of \(36.6 \text{ Kcal/mol}\)
- \(H-F\), a significant \(134.6 \text{ Kcal/mol}\)