The radii of $F,{F^ - },{O^ - }\,and\,{O^{2 - }}$is in the order:
A. ${O^{2 - }} > {O^ - } > {F^ - } > F$
B. ${O^{2 - }} > F > {F^ - } > {O^ - }$
C. ${F^ - } > {O^{2 - }} > F > {O^ - }$
D. ${O^{2 - }} > {O^ - } > F > {F^ - }$
Answer
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Hint: Radii of any elements is calculated by the distance of the electron from the nucleus. This radius depends upon the force of attraction exerted by the nucleus on the electron. Lower be the nuclear force of attraction less will be the atomic radii.
Complete step by step solution
First, we will see the comparison of atomic radii of an atom and its anion. In case of anion there will be an increment in the number of electrons by $1$ than its parent atom. While both contain the same number of protons. Since in case of anion there will be a greater number of electrons so there will be more nuclear force of attraction applied on electrons and in case of parent atom there will be no such increment in force of attraction. So, radii will be more in case of anion then its parent atom.
Let us see the electronic configuration of $F,{F^ - },{O^ - }\,and\,{O^{2 - }}$
The electronic configuration of $F$ is $1{s^2}2{s^2}2{p^5}$.
The electronic configuration of ${F^ - }$ is $1{s^2}2{s^2}2{p^6}$
The electronic configuration of ${O^ - }$ is $1{s^2}2{s^2}2{p^5}$
The electronic configuration of ${O^{2 - }}$ is $1{s^2}2{s^2}2{p^6}$
Both oxygen and fluorine belong to the second period with atomic number $8\,and\,9$ respectively. It is clear from the electronic configuration and we also know that in a period respectively. Its electrons are added in the same orbit. And the number of protons also increases in the nucleus, due to which the nuclear charge increases and the valence electrons come closer to the nucleus.
So, by taking both these factors the order of radii will be ${O^{2 - }} > {O^ - } > {F^ - } > F$.
Hence, option A is correct.
Note:
The species having the same number of electrons are also called an isoelectronic species. For the isoelectronic species, atoms are isoelectronic with a high atomic number having small atomic radii.
Complete step by step solution
First, we will see the comparison of atomic radii of an atom and its anion. In case of anion there will be an increment in the number of electrons by $1$ than its parent atom. While both contain the same number of protons. Since in case of anion there will be a greater number of electrons so there will be more nuclear force of attraction applied on electrons and in case of parent atom there will be no such increment in force of attraction. So, radii will be more in case of anion then its parent atom.
Let us see the electronic configuration of $F,{F^ - },{O^ - }\,and\,{O^{2 - }}$
The electronic configuration of $F$ is $1{s^2}2{s^2}2{p^5}$.
The electronic configuration of ${F^ - }$ is $1{s^2}2{s^2}2{p^6}$
The electronic configuration of ${O^ - }$ is $1{s^2}2{s^2}2{p^5}$
The electronic configuration of ${O^{2 - }}$ is $1{s^2}2{s^2}2{p^6}$
Both oxygen and fluorine belong to the second period with atomic number $8\,and\,9$ respectively. It is clear from the electronic configuration and we also know that in a period respectively. Its electrons are added in the same orbit. And the number of protons also increases in the nucleus, due to which the nuclear charge increases and the valence electrons come closer to the nucleus.
So, by taking both these factors the order of radii will be ${O^{2 - }} > {O^ - } > {F^ - } > F$.
Hence, option A is correct.
Note:
The species having the same number of electrons are also called an isoelectronic species. For the isoelectronic species, atoms are isoelectronic with a high atomic number having small atomic radii.
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