Arrange the following ions in order of their increasing size: `Li^(+), Mg^(2+), K^(+), AI^(3+)`.

Answer

`AI^(3+) lt Mg^(2+) lt Li^(+) lt K^(+)`

Explanation

`K^(+)` has more number of shells than `Mg^(2+)` and `AI^(3+)` and `Mg^(2+)` are isoelectronic but `AI^(3+)` has higher nuclear charge, so `AI^(3+) lt Mg^(2+), Mg^(2+)` and `Li^(+)` have diagonal relationship. But due to `+2` charge in `Mg^(2+)`, the `Mg^(2+)` is smaller than `Li^(+)`. Hence `AI^(3+)` is the smallest one.
`K^(+) = 1.38 Å, Li^(2+) = 0.76 Å,Mg^(2+) = 0.72Å` and `AI^(3+) = 0.535 Å`.