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The \[{\text{I}}{{\text{P}}_{\text{1}}}\] , \[{\text{I}}{{\text{P}}_2}\] , \[{\text{I}}{{\text{P}}_3}\] , \[{\text{I}}{{\text{P}}_4}\] and \[{\text{I}}{{\text{P}}_5}\] of an element are 7.1, 14.3, 34.5, 46.8, 162.2 eV respectively. The element is likely to be:
(A) ${\text{Na}}$
(B) ${\text{Si}}$
(C) ${\text{F}}$
(D) ${\text{Ca}}$

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Last updated date: 13th Jun 2024
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Answer
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Hint: (i) In chemistry, the term ionization energy refers to the amount of energy which is required to remove an electron from an isolated atom which is in gaseous state in its ground state. When expressed in electron volts, it is called ionization potential. It is represented by the symbol IE or IP.
(ii) The smaller the ionization energy or ionization potential, the easier it is for the neutral atom to change itself into a positive ion. Whenever a stable electronic configuration is achieved during ionization, the IE for further ionization increases exceptionally.

Complete step by step answer:
(i) The ionization potential is always positive. The ionization potential to remove the first electron from the isolated gaseous atom is called the first ionization potential, the ionization potential to remove the second electron from the isolated gaseous atom is called the second ionization potential, the ionization potential to remove the third electron from the isolated gaseous atom is called the third ionization potential and so on. They are represented by the symbols \[{\text{I}}{{\text{P}}_{\text{1}}}\] , \[{\text{I}}{{\text{P}}_2}\] , \[{\text{I}}{{\text{P}}_3}\] and so on.
(ii) The second ionization energy of an atom will be higher than the first ionization energy as after the removal of the first electron, the atom becomes positively charged and hence the effective nuclear charge per electron increases. As a result, the remaining electrons are held more tightly to the nucleus and it becomes difficult to remove the second electron.
(iii) Generally, the ionization energy of an atom is higher than the previous ionization energy. But if stable configuration is achieved, then the increase is highly exceptional.
In the given question, the fifth IP increases exceptionally which means after removal of the fourth electron, a stable configuration is achieved.
* For sodium, the electronic configuration is ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{1}}}$ and so after removal of first electron, fully filled stable inert ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}$ configuration is achieved which indicates that the second IP is exceptionally higher than first IP. But this doesn’t match the given data. So, option A is wrong.
* For silicon, the electronic configuration is ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{2}}}$ and here we can see that the removal of the fourth electron gives the stable inert gas configuration of ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}$ which means the fifth electron will be harder to remove and will require very very high IP. This matches the given data and so option B is right.
* For fluorine, the electronic configuration is ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^5}$ and here, the removal of four electrons gives the electronic configuration ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^1}$ which indicates that the removal of fifth electron is also easy and doesn’t require very high energy. So, option C is also wrong.
* For calcium, the electronic configuration is ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^{\text{2}}}$ and here, the removal of four electrons gives the electronic configuration ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^4}$ which indicates that the removal of fifth electron is also easy and doesn’t require very high energy. So, option D is also wrong.

So, the correct answer is B.

Note: Some of the factors affecting ionization energy are:
* Size of atom: Smaller the size of the atom, more tightly held electrons and so greater IE.
* Nuclear charge: Greater the nuclear charge, more energy needed to pull electrons and so greater IE.
* Electronic configuration: Elements having half filled or fully filled valence orbitals have higher IE values than expected normally from their position in the periodic table.