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Calculate the electronegativity of fluorine from the following data: E_(H-H)=10.4.2 kcal mol^(-1) E_(F-F)= 36.6 kcal mol^(-1) E_(H-P)= 134.6 kcal mol^(-1) Electronegativity of H= 2.05 |
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Answer» Solution :ENERGY for 100% COVALENT bond (H-F)= `sqrt(E_(H-H) XX E_(F-F))` =`sqrt(104.2 xx 36.6)` =61.75 kcal Resonance energy `(Delta)=` acutal bond energy- energy for 100% covalent bond =134.6-61.75 =72.85 kcal Electronegativity difference= `0.18 sqrt(Delta)` Electronegativity of F- electronegativity of `H= 0.18 sqrt(Delta)` Electronegativity of `F= 0.18 xx sqrt(72.85) + 2.05` = 3.586 |
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