Which of the following metal in solution forms a precipitate with `NaOH` which is not soluble in an excess of the base?

Options

A. Fe
B. Sn
C. Pb
D. Zn

Answer

A

Explanation

`Fe^(2+) + 2OH^(-) to Fe(OH)_(2)` (Dark green ppt)
`Fe^(3+) + 3OH^(-) to Fe(OH)_(3)` (Brown ppt)
`Sn^(2+) + 2OH^(-) to Sn(OH)_(2)` (White ppt)
`Pb^(2+) + 2OH^(-) to Pb(OH)_(2)` (White ppt)
`Zn^(2+) + 2OH^(-) to Zn(OH)_(2)` (White ppt)
`Al^(3+) + 3OH^(-) to AI(OH)_(3)` (White ppt)
Precipitates of `Fe^(2+)` and `Fe^(+)` are insoluble in excess of NaOH whereas ppt of Sn,Pb,Zn,Al readily dissolves in excess NaOH and forms clear solution.