1. Tardigrade
  2. Question
  3. Chemistry
  4. Given, Bond energy of H - H bond =104.2 kcal mol -1 Bond energy of F - F bond =36.6 kcal mol -1 Bond energy of H - F bond =134.6 kcal mol -1 Electronegativity of H on Pauling's scale =2.05 Thus, electronegativity of fluorine is

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Q. Given,
Bond energy of $H - H$ bond $=104.2 \,kcal \,mol ^{-1}$
Bond energy of $F - F$ bond $=36.6 \,kcal \,mol ^{-1}$
Bond energy of $H - F$ bond $=134.6\, kcal\, mol ^{-1}$
Electronegativity of $H$ on Pauling's scale $=2.05$ Thus, electronegativity of fluorine is

Classification of Elements and Periodicity in Properties

Solution:

When values of bond energies are in kcal $mol ^{-1}$ then by Pauling's electronegativity scale

$\chi_{F}-\chi_{H}=0.182 \sqrt{\Delta_{H-F}}$

where, $\Delta_{ H - F }$ (stabilisation energy)

$=(B E)_{H F}-\sqrt{(B E)_{H-H}(B E)_{F-F}}$

$=134.6-\sqrt{104.2 \times 36.6}$

$=72.84$

$ \therefore \chi_{ F } =\chi_{ H }+0.182 \sqrt{72.84} $

$=2.05+1.55=3.60 $