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# The correct order of ionisation energy of C, N, O and F is:A.$F{\text{ }} < {\text{ }}N{\text{ }} < {\text{ }}C{\text{ }} < {\text{ }}O$B.$C{\text{ }} < {\text{ }}N{\text{ }} < {\text{ }}O{\text{ }} < {\text{ }}F$C.$C{\text{ }} < {\text{ }}O{\text{ }} < {\text{ }}N{\text{ }} < {\text{ }}F$D.$F{\text{ }} < {\text{ }}O{\text{ }} < {\text{ }}N{\text{ }} < {\text{ }}C$  Hint: In the periodic table, the ionisation energy increases from left to right in a period and decreases down the group as per the trend follows.

- Ionisation energy is the minimum amount of energy required to remove a valence electron from an isolated gaseous atom. As we move across the period, the atomic number of the atom increases and simultaneously the number of electrons increases in the same valence shell. This leads to decrease in the atomic radius dye to increase in effective nuclear charge. -Therefore, the energy required to remove an electron from its valence shell is higher at the right end of the periodic table.
-Ionisation energy is higher for those atoms which have stable electronic configuration i.e. fully and half-filled electronic configuration.
Let us look at the electronic configurations of C, N, O and F
$C{\text{ }} = {\text{ }}1{s^2}2{s^2}2{p^2}$
$N{\text{ }} = {\text{ }}1{s^2}2{s^2}2{p^3}$
$O{\text{ }} = {\text{ }}1{s^2}2{s^2}2{p^4}$
$F{\text{ }} = {\text{ }}1{s^2}2{s^2}2{p^5}$
-As per the trend, the order must be $C{\text{ }} < {\text{ }}N{\text{ }} < {\text{ }}O{\text{ }} < {\text{ }}F$
-But N is half-filled and more stable; and requires more energy to lose an electron as compared to O which has an extra electron to release easily to attain the most stable half-filled configuration.
-That is why the correct order of ionisation energy is $C{\text{ }} < {\text{ }}O{\text{ }} < {\text{ }}N{\text{ }} < {\text{ }}F$.

Hence, the correct option is (C).

Note:
This order is valid only for first ionisation energy. For the second ionisation energy, the order changes to ${\text{C < N < F < O}}$because losing an electron makes O very stable due to half-filled electronic configuration.
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