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 as per the data is

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3.60

5.65

1.55

-2.60

answer is A.

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Detailed Solution

When values of bond energies are in kcal mol-1 then by Pauling's electronegativity scale χF−χH=0.182ΔH−F where, ΔH−F (stabilisation energy) =(BE)HF−(BE)H−H(BE)F−F=134.6−104.2×36.6=72.84∴ χF=χH+0.18272.84=2.05+1.55=3.60

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