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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5.65

1.55

-2.60

3.60

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

$χ_{F} - χ_{H} = 0 .182 \sqrt{Δ_{H} - F}$

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

$\begin{array}{l} = (BE)_{H F} - \sqrt{(BE)_{H - H} (BE)_{F - F}} \\ = 134.6 - \sqrt{104.2 \times 36.6} \\ = 72.84 \end{array}$

$\begin{aligned} ∴ χ_{F} = χ_{H} + 0.182 \sqrt{72.84} \\ = 2.05 + 1.55 = 3.60 \end{aligned}$

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