1. Tardigrade
  2. Question
  3. Chemistry
  4. Calculate the electronegativity of fluorine from the following data: EH -H = 104.2 kcal mol-1 EF-F= 36.6 kcal mol-1 EH-F = 134.6 kcal mol-1 XH = 2.1 Express your answer to the nearest possible integral.

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Q. Calculate the electronegativity of fluorine from the 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$^{-1}$
$X_{H} = 2.1$
Express your answer to the nearest possible integral.

Classification of Elements and Periodicity in Properties

Solution:

Let $X_{H}$ and $X_{F}$ be the electronegativity of $H$ and $F$, then
$X_{H} \sim X_{F}=0.208\left[E_{H-F}\left(E_{H-H}\times E_{F-F}\right)^{1/2}\right]^{1/2}$
$X_{H} \sim X_{F} =0.208\left[134.6-\left(104.2\times36.6\right)^{1/2}\right]^{1/2} $
$X_{H} \sim X_{F} =1.78$ and $X_{H} < X_{F} $
since, $X_{H}=2.1$
$X_{F} =2.1 +1.78=3.88=4$