Arrange the following in order of decreasing radii?
a. `F^(ɵ), O^(2-), N^(3-), S^(2-)`, b. `P, Si, N, C`, c. `I^(ɵ), I^(o+), I`

Explanation

a. Since `F^(ɵ), N^(3-), O^(2-)` are isoelectronic species (with 10 electrons). So, size of anions decreases as nuclear charge increases (`Z` for `F^(ɵ), O^(2-), N^(3-) = 9, 8` and `7` respectively) So, decreasing order of size among isoelectronic anions are as:
`N^(3-) gt O^(2-) gt F^(ɵ)`
Since S belongs to the 3rd period while `F, O, N` all belong to the 2nd period. Therefore, size of `S^(2-)` is largest.
Thus overall order of decreasing size is
`S^(2-) gt N^(3-) gt O^(2-) F^(ɵ)`
b. C and N belong to the 2nd period whereas `Si` and `P` belong to the 3rd period. Elements in the 3rd period have higher atomic sizes than those of the 2nd period due to increase in shell. Thus, size of `Si` and `P` are higher than those of C and N respectively.
Moreover, along the period atomic size decrease from the left to the right due to the increased nuclear charge. Thus, size of `C gt N` and `Si gt P`. Thus, overall decreasing order of atomic sizes is
`Si gt P gt C gt N`
c. Decreasing order of radii is
`I^(ɵ) gt I gt I^(o+)`
because the size of anion is always greater than while that of an cation is always smaller than the parent atom.