Shweta thakur Asked a Question
May 14, 2021 1:25 ampts 30 pts
= 2.54 pleCalculate the electronegativity of fluorine atom from the following data EH-4104.2 kcal mol, Ep_ = 36.6 kcal mol-1 Eg-p134.6 kcal moll (electronegativity of H = 2.1) Solution: We know that X-X=0.18 (AE)/2) AE = E -EqH H p-F R/2 = 134.6- [104.2 x 36.611/2 = 72.85 kcal
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  • Dinesh khalmaniya Best Answer
    Yes , shweta we take first EN of high EN atom then EN of less EN atom. Reason is that EN is always positive value
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    Dinesh khalmaniya
    any doubt?
  • Yashasvi yadav thankyou
    Yes
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