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Calculate the electronegativity of fluorine from following data : F_(H-H) =104. 2 k cal mol ^(-1) E _(F-F) =36.6 kcal mol ^(-1) E _(H-F) = 134. 6 kcal mol ^(-1) Electronegativity ofH is 2.05 |
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Answer» Solution :On Paulling SCALE: `x _(F) -x_(H) =0.182sqrt(Delta _(H)-F)` (using B.E. in kcal `mol ^(-1))` `Delta _(H-F) = E_(H-F) - SQRT( E_(H-H) xx E _(F -F))` `=13.45 - sqrt( 104.2 xx36.6) = 72.84` kcal From (i) `x_(F) -x_(H) = 0.182 sqrt(72.84)+ 1.5534` `x_(F)=x_(H) + 1.4434=2.05+ 1.5534=3.6034` |
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