`NaOH` behaves as a base while `Zn(OH)_(2)` is amphoteric why ?

Explanation

In `Na-O,chi_(O) - chi_(Na) gt chi_(O) - chi_(H)`, therefore it is the `Na - O` bond breaks and releases `OH^(o-)` ions. So, `NaOH` is basic.
But in `Zn-O-H,chi_(O) - chi_(Zn) = chi_(O) - chi_(H)`, therefore there is equal probability that the bond breaks in both ways (releasing both `H_(3)O^(o+)` and `OH^(o-)` in aqueous solution). So, `Zn(OH)_(2)` is amphoteric.