The set representing the correct order of ionic radius is : a)`Na^(+) gt Mg^(2+) gt Al^(3+) gt Be^(2+) gt Li^(+)` b)`Na^(+) gt Li^(+) gt Mg^(2+) gt Al^(3+) gt Be^(2+)` c)`Na^(+) gt Mg^(2+) gt Al^(3+) gt Li^(+) gt Be^(2+)` d)`Na^(+) gt Mg^(2+) gt Li^(+) gt Be^(2+) gt Al^(3+)`

Options

A. `Na^(+) gt Li^(+) gt Mg^(2+) gt Be^(2+)`
B. `Li^(+) gt Na^(+) gt Mg^(2+) gt Be^(2-)`
C. `Mg^(2+) gt Be^(2+) gt Li^(+) gt Na^(+)`
D. `Li^(+) gt Be^(2+) gt Na^(+) gt Mg^(2+)`

Answer

A

Explanation

Down the group, ionic radii increases with increasing atomic number because of the increase in the number of shells. But across the period, the ionic radii decreases due to increase on effective nuclear charge as electrons are added in the same shell. `Li^(+)` and `Mg^(2+)` are diagonally related but `Mg^(2+)` having higher charge is smaller than `Li^(+)`, so correct order is `Na^(+) gt Li^(+) gt Mg^(2+) gt Be^(2+)`.
`Be^(2+) = 0.31 Å`
`Mg^(2+) = 0.72 Å`
`Li^(+) = 0.76 Å`
`Na^(+) = 1.02 Å`