Question

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

• A. $S{e}^{2-},I^{-},B{r}^{-},O^{2-},F^{-}$
• B. $I^{-},S{e}^{2-},O^{2-},B{r}^{-},F^{-}$
• C. $S{e}^{2-},I^{-},B{r}^{-},F^{-},O^{2-}$
• D. $I^{-},S{e}^{2-},B{r}^{-},O^{2-},F^{-}$

Answer 1

The correct option is D $I^{-},S{e}^{2-},B{r}^{-},O^{2-},F^{-}$

Across the period, ionic size decreases as nuclear change increases for successive addition of an electron but down the group size increases due to increase in the number of atomic shells (effective nuclear charge remains nearly same).
So, the correct decreasing order of ionic radii.
$I^{-}>S{e}^{2-}>B{r}^{-}>O^{2-}>F^{-}$