Answer: [tex]P^{3-}[/tex]> [tex]Cl^{-}[/tex] > [tex]K^{+}[/tex] > [tex]Ca^{2+}[/tex]
Explanation:
Isoelectronic specie are defined as the molecules which have the same number of electrons.
Atomic number of calcium is 20 and thus has 20 electrons.[tex]Ca^{2+}[/tex] is formed by loss of 2 electrons and thus has 18 electrons.
Atomic number of potassium is 19 and thus has 19 electrons. [tex]K^{+}[/tex] is formed by loss of 1 electron and thus has 18 electrons.
Atomic number of chlorine is 17 and thus has 17 electrons. [tex]Cl^{-}[/tex] is formed by gain of 1 electron and thus has 18 electrons.
Atomic number of phosphorous is 15 and thus has 13 electrons.[tex]P^{3-}[/tex] is formed by gain of 3 electrons and thus has 18 electrons.
As the number of electrons are same , the more is the nuclear charge i.e. the number of protons , more will the force of attraction towards the nuclei and thus smaller will be the radii and vice versa.
Thus, order of decreasing ionic radii for the given ions is as follows.
[tex]P^{3-}[/tex]> [tex]Cl^{-}[/tex] > [tex]K^{+}[/tex] > [tex]Ca^{2+}[/tex]
The correct order of decreasing size of the following ions is P3– > Cl– > K+ > Ca2+.
Isoelectronic species can be defined as the molecules which have the same number of electrons.
Since the number of electrons are same , there is an increase in the nuclear charge i.e. the number of protons and decrease in the radii which is depicted in option B.
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