Q25 / 200 · Manipal (MET) 2013 · Chemistry

Calculate the electronegativity of chlorine from bond energy of $\mathrm{Cl}-\mathrm{F}$ bond ( $61 \mathrm{kcal} \mathrm{mol}^{-1}$ ) $\mathrm{F}-\mathrm{F}\left(38 \mathrm{kcal} \mathrm{mol}^{-1}\right)$ and $\mathrm{Cl}-\mathrm{Cl}$ bond ( 58 kcal $\mathrm{mol}^{-1}$ ) and electronegativity of fluorine 4.0 eV

  • A
    1.42 eV
  • B
    1.89 eV
  • C
    2.67 eV
  • Correct answer
    D
    3.22 eV

Solution

Based on Pauling's scale $(\mathrm{EN})_{\mathrm{F}}-(\mathrm{EN})_{\mathrm{Cl}}=k \sqrt{\Delta}=0.208 \sqrt{\Delta}$ where, $\Delta$ = resonance energy and $k=$ conversion factor which is 0.208 for converting k cal into eV . $\begin{aligned} \Delta & =(\mathrm{BE})_{\mathrm{Cl}-\mathrm{F}}-\sqrt{(\mathrm{BE})_{\mathrm{Cl}-\mathrm{Cl}}(\mathrm{BE})_{\mathrm{F}-\mathrm{F}}} \\ & =61-\sqrt{58 \times 38} \\ & =61-46.95=14.05 \mathrm{kcal} \\ (\mathrm{EN})_{\mathrm{Cl}} & =(\mathrm{EN})_{\mathrm{F}}-0.208 \sqrt{\Delta} \\ & =4.0-0.208 \sqrt{14.05} \\ & =4.0-0.78=3.22 \mathrm{eV} .\end{aligned}$

Hence, the correct answer is option: 'D'.

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