Size of the species $Cl$, $C{l^ - }$ and $C{l^ + }$ decreases as?
a.) $Cl > C{l^ + } > C{l^ - }$
b.) $C{l^ + } > C{l^ - } > Cl$
c.) $C{l^ - } > C{l^ + } > Cl$
d.) $C{l^ - } > Cl > C{l^ + }$
Answer
676.2k+ views
Hint: Try to recall that the size of cation is always smaller than that of its parent atom whereas the size of an anion is always greater than that of its parent atom. Now by using this you can easily find the correct option from the given options.
Complete answer:
It is known to you that the size of an anion is always greater than that of its parent atom. This happens because an anion is formed when a neutral gaseous atom gains one or more electrons. Due to the addition of one or more electrons, there is an increase in repulsions among electrons and the electron cloud of the atom expands. In other words there is an increase in the ionic radius. So, from above we can say that $C{l^ - }$ is larger in size compared to $Cl$i.e.
$C{l^ - }$ > $Cl$.
Also, we know that the radius of cation is always smaller than that of its parent atom. This happens because a cation is formed by the loss of one or more electrons from the neutral gaseous atom and this generally causes removal of the whole shell. As a result the size of cation is smaller than the parent atom from which it is formed. Further due to removal of electrons from the parent atom, the number of electrons in the cation decreases and as a result of this effective nuclear charge increases and hence, size decreases. . So, from above we can say that $C{l^ + }$ is smaller in size compared to $Cl$ i.e. $Cl$ > $C{l^ + }$.
Therefore, from above discussion we can say that option D is the correct option for the given question.
Note: It should be remembered that ions of different elements which have the same number of electrons but different magnitudes of the nuclear charge are called isoelectronic species.
Also, you should remember that the ionic radii of isoelectronic ions decrease with increase in magnitude of effective nuclear charge.
Complete answer:
It is known to you that the size of an anion is always greater than that of its parent atom. This happens because an anion is formed when a neutral gaseous atom gains one or more electrons. Due to the addition of one or more electrons, there is an increase in repulsions among electrons and the electron cloud of the atom expands. In other words there is an increase in the ionic radius. So, from above we can say that $C{l^ - }$ is larger in size compared to $Cl$i.e.
$C{l^ - }$ > $Cl$.
Also, we know that the radius of cation is always smaller than that of its parent atom. This happens because a cation is formed by the loss of one or more electrons from the neutral gaseous atom and this generally causes removal of the whole shell. As a result the size of cation is smaller than the parent atom from which it is formed. Further due to removal of electrons from the parent atom, the number of electrons in the cation decreases and as a result of this effective nuclear charge increases and hence, size decreases. . So, from above we can say that $C{l^ + }$ is smaller in size compared to $Cl$ i.e. $Cl$ > $C{l^ + }$.
Therefore, from above discussion we can say that option D is the correct option for the given question.
Note: It should be remembered that ions of different elements which have the same number of electrons but different magnitudes of the nuclear charge are called isoelectronic species.
Also, you should remember that the ionic radii of isoelectronic ions decrease with increase in magnitude of effective nuclear charge.
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