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Identify the correct order of second ionization energy of $\mathrm{C, N, O \: and\: F.}$
Option: 1
$\mathrm{O > F > N > C}$

Option: 2
$\mathrm{O > N > F > C}$

Option: 3
$\mathrm{C > N > O > F}$

Option: 4
$\mathrm{F > O > N > C}$

Answers (1)

best_answer

The electronic configuration of $\mathrm{C, N, O \: and\: F \: }$ are :

$\mathrm{C -[He] 2s^{2} 2p^{2}}$

$\mathrm{N - [He] 2s^{2} 2p^{3}}$

$\mathrm{O -[He] 2s^{2} 2p^{4}}$

$\mathrm{F - [He] 2s^{2} 2p^{5}}$

After removing an electron, the electronic configuration shall be

$\mathrm{C^{+} :[He] 2s^{2} 2p^{1}}$

$\mathrm{N^{+} :[He] 2s^{2} 2p^{2}}$

$\mathrm{ O^{+} :[He] 2s^{2} 2p^{3}}$

$\mathrm{F^{+} :[He] 2s^{2} 2p^{4}}$

If you observe their electronic configuration, you shall find that after removal of one
more electron -
Carbon shall have no electron in p-subshell. Hence it will easily give away the second
electron which has a very low second IP.
Nitrogen will have only one electron in the outermost p subshell.
Oxygen will have two electrons in p subshell which makes it very unstable. So it will
have high $\mathrm{IE.}$
Fluorine will occupy stable half-filled configuration.
So $\mathrm{O>F>N>C}$ is the correct order of $\mathrm{IE.}$

Hence Option 1 is correct.

Posted by

vinayak

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