Answer :
Answer:
K+ < Cl- < S2- < P3-.
Explanation:
The electronic configuration of K is 2,8,8,1. So,K has 19 electrons.Hence,K+ will have 18 electrons.
Similarly,electronic configuration of P is 2,8,5. So,P has 15 electrons.Hence,P^3- will have 18 electrons .
Similarly,electronic configuration of S is 2,8,6. So, S has 16 electrons. Hence,S^2- will have 18 electrons.
Similarly,electronic configuration of Cl is 2,8,7. So,Cl has 17 electrons.Hence,Cl^ 1- will have 18 electrons .
As,all of these species are isoelectronic, so the smallest radius among these will be of that which possesses more amount of +ve charge on it.The reason is that a species having +ve charge on it will be effectively able to attract the outermost valence electrons to the nucleus.
Hence,the order of increasing order of radii is as follows:
K+ < Cl- < S2- < P3-.
Based on decreasing atomic number, the order of increasing order of ionic radii is as follows: K+ < Cl- < S2- < P3-.
What is ionic radius?
The ionic radius of an ion is the distance from the nucleus to the valence electron.
The ionic radius decreases with increasing atomic number for metal ions while it increases with increasing atomic number for non-metallic ions.
For isoelectronic ions, the ionic radius decreases with increasing atomic number.
K+, P3–, S2– and Cl– all have 18 electrons.
K+ has atomic number of 19
P3- has atomic number of 15
S2 has atomic number of 16
Clhas atomic number of 17.
Therefore, the order of increasing order of ionic radii is as follows: K+ < Cl- < S2- < P3-.
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