Even an ext confusing is the electron affinity can likewise be identified as the ionization power of the singly fee anion type of one atom, i.e. The ionization energy of $\ceCl-$ is same in magnitude to the electron affinity that $\ceCl$. Therefore, it appears that the fads for ionization energy and electron affinity should always match, which is no the case with fluorine and also chlorine.
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edited Nov 30 "17 at 7:59
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To quote chemguide:
The an initial ionisation power is the power required to remove the most loosely held electron from one mole of gaseous atoms to create 1 mole the gaseous ion each with a fee of 1+.
That means, the ionization energy of fluorine is the power of the adhering to reaction:
$$\ceF -> F+ + e-$$
The ionization energy of chlorine is the power of the complying with reaction:
$$\ceCl -> Cl+ + e-$$
Since the many loosely hosted electron in fluorine is closer to the nucleus than chlorine, it takes an ext energy to remove the electron native fluorine than in chlorine.
The ionization energy of chloride ion ($\ceCl-$) is the power of the following reaction:
$$\ceCl- -> Cl + e-$$
The electron affinity of $\ceCl$ is the energy of the adhering to reaction:
$$\ceCl + e- -> Cl-$$
As a result, your energies are equal in magnitude but opposite in sign.
Why walk chlorine have actually a greater electron affinity than fluorine?
To quote this answer:
Fluorine, though higher than chlorine in the regular table, has a very tiny atomic size. This makes the fluoride anion so formed unstable (highly reactive) due to a an extremely high charge/mass ratio. Also, fluorine has actually no d-orbitals, which boundaries its atomic size. As a result, fluorine has actually an electron affinity less than that of chlorine.
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As deserve to be checked out in the second photo, this impact seems to be true for period 2 elements.