A. The atomic number that is proton number of Na is 10 and electrons' count is also 10. The proton number of Na+ is 10 and electrons' count is 9. The attraction will be more, the size of Na+ will be small and Na has the larger radius.
(b)The proton number of I is 35 and electrons count is also 35. The atomic number (proton number) of
I− is 5 and electrons' number is 36. Then, the size of I− will be large but I−has the larger radius.
C. The proton number of S is 16 and electrons' count is also 16. The atomic number of S2− is 16 and electrons' number is 18. the size of S2− will be large.
So, S2− has the larger radius.
D. The proton number of Al is 13 and electrons' count is also 13. The proton number of Al3+ is 13 and electrons' number is 10.
Conclusively, In Al3+ , same effective number charge acting on less number of electrons when compared to Al, making attraction will be more and so the size of
Al3+ will be small.
Therefore, Al has the larger radius
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