I have seen following definitions in my textbook:

Electron gain enthalapy may be defined as enthalpy change taking place when an isolated gaseous atom of the element accepts an electron to form a monovalent gaseous anion.

Electronegativity may be defined as the tendency of an atom in molecule to attract towards itself the shared pair of an electron.

I know that fluorine and Oxygen have low ${\Delta H_{eg}}$ than Chlorine and Sulpher respectively due to the fact when an electron is added to these atoms, the electron goes to the relatively compact second energy level which results in significant repulsion from other electrons present in the shell.

But I have seen that Fluorine and Oxygen have very high electronegativity.

(i)I can't understand how can an atom attract shared pair of electron very well when it is unwilling to add elctron in its valence shell?


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    $\begingroup$ Very unreactive? Flourine still has the second highest electron affinity in the periodic system. Flourine and oxygen are by far smaller than chlorine and sulfur, so the repulsion is higher. Etc. A very similar question is why HI is the strongest of the HHal acids, and HF the weakest. Stability of the ion has no direct relation to electronegativity in a covalent bond. It's a different electronic configuration, you cannot compare the two just like that. $\endgroup$ – Karl Oct 6 '18 at 9:14
  • $\begingroup$ @Karl : thanks for the answer, but the (i) question is still unclear to me. $\endgroup$ – pranjal verma Oct 6 '18 at 9:52
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    $\begingroup$ The one electron is added to an atom orbital. The other phenomenon is the pull on a pair of electrons that are already in a molecular orbital. It's incomparabale. And both oxygen and fluorine have a negative electron affinity, it's just slightly weaker than of sulphur and chlorine. ?! $\endgroup$ – Karl Oct 6 '18 at 12:35

The reaction between atomic hydrogen and atomic halogen can be considered as a two-step reaction: single electron transfer furnished by the formation of bond between halogen anion and hydrogen cation. Even if the real mechanism is different, it is of no importance as long as we consider thermodynamics.

The second step, from the point of view of pure electrostatics, should release more energy in the case of fluorine than in the case of chlorine because the shorter bond is formed.

  • $\begingroup$ Thanks for the answer but I more interested in understanding (i) question as the (ii) question is simply a part of that. $\endgroup$ – pranjal verma Oct 6 '18 at 9:11

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