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Calculate the electronegativity of fluor...

Calculate the electronegativity of fluorine from following data :
`E_(H - H) = 104.2` kcal `mol^(-1)`
`E_(F - F) = 36.6` kcal `mol^(-1)`
`E_(H -F) = 134.6` kcal `mol^(-1)`
Electronegativity of H is 2.05.

Text Solution

Verified by Experts

The correct Answer is:
3.82

Suppose the electronegativity of H and F are
reprsented as `chi_(H) and chi_(F)` respectively . Appling
Pauling's formula
`|chi_(H)- chi_(F) |=0.208sqrt(Delta) `
`Delta= BE (H-F) - sqrt(BE(H-H)xxBE(F-F))`
`= 134.6 - sqrt(104.2xx36.6)`
`= 134.6 - 61.8` kcal mol `^(-1)`
72. 8 kcal mol `^(-1)`
`therefore |chi_(H)- chi_(F) |=0.208sqrt(72.8) = 1.77`
As `chi_(F)gt chi_(H),chi_(F) = 1.77 + chi_(H) = 1.77+2.1 = 3.82`
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Knowledge Check

  • Calculate the electronegativity of fluorine using the following data. E_(H-H) = 104.2 kcal "mol"^(-1) , E_(F-F) = 36.6 kcal "mol"^(-1) , E_(H-F) = 144.6 kcal "mol"^(-1) and the electronegativity of H = 2.1.

    A
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    B
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    C
    4
    D
    4.2
  • The electronegativity of carbon from the following data is nearly E_(H - H) = 104.2 "kcal" mol^(-1) , E_(C-C) = 83.1 kcal mol^(-1) , E_(C-H) = 98.8 kcal mol^(-1) , X_(H) = 2.1

    A
    `3.0`
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    D
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