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Calculate the electronegativity of Flourine, given the following: $E_{H-H} = 104.2 \; kcal\;mol^{-1},\; E_{F-F} = 36.6\; kcal\;mol^{-1},\;E_{H-F} = 134.6 \; kcal\;mol^{-1},\;\chi_H = 2.1$

$\begin{array}{1 1} 3.766 \\ 3.543 \\ 3.98 \\ 3.666 \end{array}$

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Answer: 3.766
According to Pauli's formula for electronegativity: $\chi_{A} - \chi_{B}=\left(\text{kcal/mol}\right)^{-1/2}\sqrt{E_d{{AB}}-\frac{E_d{{AA}}+E_d{{BB}}}{2}}$
$\chi_F - 2.1 = 0.208 \times \large\sqrt { \normalsize 134.6 - \large\frac{104.2+36.6}{2}}$
$\chi_F - 2.1 = 0.208 \times \sqrt{64.2} = 1.666 \rightarrow \chi_F = 3.766$
answered Jul 24, 2014 by balaji.thirumalai
 

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