Arrange the following in order of increasing radii ? (i) I, `I^(+) , I^(-) (ii) C, N, Si, P (iii) O^(2), N^(3-) , S^(2-) ,F^(-)`

Answer

`(i) I^(+) lt I lt I^(-) (ii) N lt C lt P lt Si (iii) F^(-) lt O^(2-) lt N^(3-) lt S^(2-) `

Explanation

(i) Size of a cation is always smaller while that of an anion is always bigger than the neutral atom i.e., `I^(+) lt I lt I^(-)`
(ii) C and N lie in 2nd period while Si and P lie below them in the 3rd period .Since elements in the 3th period have higher atomic size than those in the 2nd period therefore, atomic radii of Si and P are higher than those of C and N respectively .Since atomic radii decrease from left to right in a period dut to higher nuclear charge therfore C has higher atomic radius than N and Si has higher atomic radius than P. Thus the overall order of increasing atomic radii is : `N lt C lt P lt Si`
(iii) Among isoelectronic ions the size of anions increase as the nuclear charge decreases. `F^(-) lt O^(2-) lt N^(3-).` Since S belong to third period while `F, O, N` all belong to the second period therefore ionic radius of `S^(2-) lt N^(3-) lt S^(2-)`