
The species Ar, ${{K}^{+}}$ and $C{{a}^{2+}}$ contain the same number of electrons. In which order do their radii increase?
(a)- $C{{a}^{2+}}$ < ${{K}^{+}}$ < $Ar$
(b)- ${{K}^{+}}$ < $Ar$ < $C{{a}^{2+}}$
(c)- $Ar$ < ${{K}^{+}}$ < $C{{a}^{2+}}$
(d)- $C{{a}^{2+}}$ < $Ar$ < ${{K}^{+}}$
Answer
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Hint: In isoelectronic species, as the nuclear charge increases, the force of attraction by the nucleus increases on the electrons also increases due to which the radii decrease.
Complete answer:
Isoelectronic species or ions are the ions of different elements that have the same number of electrons but differ in the magnitude of the nuclear charge.
Besides ions, a neutral atom may also have the same number of electrons hence it is also an isoelectronic species.
Example, sulphide ion (${{S}^{2-}}$ ), chloride ion ($C{{l}^{-}}$), argon (Ar), and potassium ion (${{K}^{+}}$ ) are isoelectronic species.
Variation of size among isoelectronic species: As the nuclear charge increases, the force of attraction by the nucleus on the electrons also increases. As a result, the ionic radius decreases.
The nuclear charge is calculated by the number of electrons + charge on the ion.
In Ar,${{K}^{+}}$ and$C{{a}^{2+}}$ the number of electrons is 18.
The nuclear charge of Ar is 18 because there is no charge on the ion.
The nuclear charge of the ${{K}^{+}}$ion is 19 because potassium ion has a +1 charge.
The nuclear charge of the $C{{a}^{2+}}$ion is 20 because calcium ion has a +2 charge.
So, the calcium ion has the most charge, hence it will be the smallest and Argon has the least charge so it will be the largest.
The size range will be: $C{{a}^{2+}}$ < ${{K}^{+}}$ < $Ar$
So, the correct answer is “Option A”.
Note: With the help of calculation the nuclear charge we can find the radii range of a large number of ions. Example, nitrogen ion (${{N}^{3-}}$ ), oxygen ion (${{O}^{2-}}$ ), fluorine ion (${{F}^{-}}$ ), sodium ion ($N{{a}^{+}}$), magnesium ion ($M{{g}^{2+}}$ ) and aluminium ion ($A{{l}^{3+}}$ ). They have 10 electrons. They have nuclear charge +7, +8, +9, +11, +12 and +13. Hence, the radii will be:
$A{{l}^{3+}}$ < $M{{g}^{2+}}$ < $N{{a}^{+}}$ < ${{F}^{-}}$ < ${{O}^{2-}}$ < ${{N}^{3-}}$
Complete answer:
Isoelectronic species or ions are the ions of different elements that have the same number of electrons but differ in the magnitude of the nuclear charge.
Besides ions, a neutral atom may also have the same number of electrons hence it is also an isoelectronic species.
Example, sulphide ion (${{S}^{2-}}$ ), chloride ion ($C{{l}^{-}}$), argon (Ar), and potassium ion (${{K}^{+}}$ ) are isoelectronic species.
Variation of size among isoelectronic species: As the nuclear charge increases, the force of attraction by the nucleus on the electrons also increases. As a result, the ionic radius decreases.
The nuclear charge is calculated by the number of electrons + charge on the ion.
In Ar,${{K}^{+}}$ and$C{{a}^{2+}}$ the number of electrons is 18.
The nuclear charge of Ar is 18 because there is no charge on the ion.
The nuclear charge of the ${{K}^{+}}$ion is 19 because potassium ion has a +1 charge.
The nuclear charge of the $C{{a}^{2+}}$ion is 20 because calcium ion has a +2 charge.
So, the calcium ion has the most charge, hence it will be the smallest and Argon has the least charge so it will be the largest.
The size range will be: $C{{a}^{2+}}$ < ${{K}^{+}}$ < $Ar$
So, the correct answer is “Option A”.
Note: With the help of calculation the nuclear charge we can find the radii range of a large number of ions. Example, nitrogen ion (${{N}^{3-}}$ ), oxygen ion (${{O}^{2-}}$ ), fluorine ion (${{F}^{-}}$ ), sodium ion ($N{{a}^{+}}$), magnesium ion ($M{{g}^{2+}}$ ) and aluminium ion ($A{{l}^{3+}}$ ). They have 10 electrons. They have nuclear charge +7, +8, +9, +11, +12 and +13. Hence, the radii will be:
$A{{l}^{3+}}$ < $M{{g}^{2+}}$ < $N{{a}^{+}}$ < ${{F}^{-}}$ < ${{O}^{2-}}$ < ${{N}^{3-}}$
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