Answer
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Hint: The size of isoelectronic atoms or ions is determined by the number of protons. The greater the nuclear charge the smaller will be the radius of isoelectronic atoms or ions. The size of an atom normally decreases across the periodic table and increases down the table.
Complete answer:
The formation of an ion takes place either when one or more electrons are removed from a neutral atom constituting a positive ion (cation) or when an additional electron gets added over a neutral atom constituting a negative ion (anion).
Ionic radius defines the distance between the nucleus to the outermost electron. In case of cations since they are formed by losing electrons the size of cations will be comparatively lower than their parent atom and at the same time in case of anions as they are formed by gaining electrons their size will be greater than the parent atom.
Normally the size of an atom decreases from left to right across the periodic table and size increases from top to bottom. But here since they are ions, we have to consider their ionic values also.
In the case of ${N^{3 - }}$ it has gained 3 electrons additionally whereas ${O^{2 - }}$ has gained 2 electrons additionally \[{F^ - }\] gained one electron additionally but in case of \[N{a^ + }\] it has lost an electron.
When an electron is lost the repulsion between electrons in a shell decreases as there will be fewer electrons present and at the same time effective nuclear charge experienced by the remaining electrons will be more as there will be fewer electrons to shield one another from the nucleus cause shrinking of ions and decrease in size.
But when an electron is gained there will be an increase in electron-electron repulsion and thereby cause an effective decrease in effective nuclear charge resulting in an increase in size.
Hence $N{a^ + }$ will be the smallest among others.
$\therefore $Option (D) is the answer.
Note: Ionic radius defines the distance between the nucleus to the outermost electron. Depending on the ionic charge the ions can be either smaller or larger.
For isoelectronic ions
Ionic size $ \propto $ 1/atomic number
Complete answer:
The formation of an ion takes place either when one or more electrons are removed from a neutral atom constituting a positive ion (cation) or when an additional electron gets added over a neutral atom constituting a negative ion (anion).
Ionic radius defines the distance between the nucleus to the outermost electron. In case of cations since they are formed by losing electrons the size of cations will be comparatively lower than their parent atom and at the same time in case of anions as they are formed by gaining electrons their size will be greater than the parent atom.
Normally the size of an atom decreases from left to right across the periodic table and size increases from top to bottom. But here since they are ions, we have to consider their ionic values also.
In the case of ${N^{3 - }}$ it has gained 3 electrons additionally whereas ${O^{2 - }}$ has gained 2 electrons additionally \[{F^ - }\] gained one electron additionally but in case of \[N{a^ + }\] it has lost an electron.
When an electron is lost the repulsion between electrons in a shell decreases as there will be fewer electrons present and at the same time effective nuclear charge experienced by the remaining electrons will be more as there will be fewer electrons to shield one another from the nucleus cause shrinking of ions and decrease in size.
But when an electron is gained there will be an increase in electron-electron repulsion and thereby cause an effective decrease in effective nuclear charge resulting in an increase in size.
Hence $N{a^ + }$ will be the smallest among others.
$\therefore $Option (D) is the answer.
Note: Ionic radius defines the distance between the nucleus to the outermost electron. Depending on the ionic charge the ions can be either smaller or larger.
For isoelectronic ions
Ionic size $ \propto $ 1/atomic number
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