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Choose the correct order of second ionisation enthalpies Carbon, Nitrogen, Oxygen and Fluorine.
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The correct answer is:
Oxygen $>$ Fluorine $>$ Nitrogen $>$ Carbon
The corresponding electronic configurations for the given species are:
$\mathrm{C}: 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^2, \mathrm{~N}: 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^3, \mathrm{O}: 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^4$
F : $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^5$.
Since there are three 2 p-orbitals in the 2 p-subshell, it is evident that the loss of a second electron from the valence shell will be most difficult in oxygen as it would disturb the stable half-filled p-subshell.
The removal of second electron will be easiest for carbon as it would give a fully-filled $2 \mathrm{~s}$-subshell.
Thus, the order of second ionization energy will be:Oxygen $>$ Fluorine $>$ Nitrogen $>$ Carbon
$\mathrm{C}: 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^2, \mathrm{~N}: 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^3, \mathrm{O}: 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^4$
F : $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^5$.
Since there are three 2 p-orbitals in the 2 p-subshell, it is evident that the loss of a second electron from the valence shell will be most difficult in oxygen as it would disturb the stable half-filled p-subshell.
The removal of second electron will be easiest for carbon as it would give a fully-filled $2 \mathrm{~s}$-subshell.
Thus, the order of second ionization energy will be:Oxygen $>$ Fluorine $>$ Nitrogen $>$ Carbon
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