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

Question

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 (A) -3.60 (B) 5.65 (C) 1.55 (D) 3.60

Tardigrade Admin · Accepted Answer

When values of bond energies are in kcal

m o l^{- 1}

then by Pauling's electronegativity scale

χ_{F} - χ_{H} = 0.182 Δ_{H - F}

where,

Δ_{H - F}

(stabilisation energy)

= (BE)_{H F} - (BE)_{H - H} (BE)_{F - F}

= 134.6 - 104.2 \times 36.6

= 72.84

∴ χ_{F} = χ_{H} + 0.182 72.84

= 2.05 + 1.55 = 3.60

Q. Given, Bond energy of H−H bond =104.2kcalmol−1 Bond energy of F−F bond =36.6kcalmol−1 Bond energy of H−F bond =134.6kcalmol−1 Electronegativity of H on Pauling's scale =2.05 Thus, electronegativity of fluorine is

-3.60

5.65

1.55

3.60

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

Q. Given,
Bond energy of $H - H$ bond $= 104.2 k c a l m o l^{- 1}$
Bond energy of $F - F$ bond $= 36.6 k c a l m o l^{- 1}$
Bond energy of $H - F$ bond $= 134.6 k c a l m o l^{- 1}$
Electronegativity of $H$ on Pauling's scale $= 2.05$ Thus, electronegativity of fluorine is