When the following five anions are arranged in order of decreasing ionic radius , the correct sequence is :

Options

A. `Se^(2-), I^(-), Br^(-), O^(2-), F^(-)`
B. `I^(-), Se^(2-), Br^(-), F^(-), O^(2-)`
C. `Se^(2-), I^(-), Br^(-), F^(-),O^(2-)`
D. `I^(-), Se^(2-), Br^(-), O^(2-), F^(-)`

Answer

D

Explanation

Across the period ionic size decreases as nuclear charge increases for successive addition of an electron but down the group increases due to increase in the number of atomic shells (effective nuclear charge remains nearly same).
`O^(2-) = 140 pm, Se^(2-) = 198 pm, F^(-) = 133 pm, Br^(-) = 196 pm, I^(-) = 220 pm`.
So, the correct decreasing order of ionic radii. `I^(-) gt Se^(2-) gt Br^(-) gt O^(2-) gt F`.