Select the species in which the precipitate formed on adding NaOH further dissolves in excess of NaOH.

$\text{ (a) }{\mathrm{ZnCl}}_2+2\mathrm{NaOH}\to {⟶\mathrm{Zn}\left({\mathrm{OH}}_2\right)\downarrow }_{\mathrm{white} \mathrm{ppt}}+2\mathrm{NaCl}$

$\mathrm{Zn}(\mathrm{OH}{)}_2+2\mathrm{NaOH}⟶{{\mathrm{Na}}_2{\mathrm{ZnO}}_2}_{(\mathrm{soluble} \mathrm{sodium} \mathrm{zincate})}+2{\mathrm{H}}_2\mathrm{O}$

$\text{ (b) Al}{\mathrm{Cl}}_3+3\mathrm{NaOH}\to {⟶\mathrm{Zn}\left({\mathrm{OH}}_2\right)\downarrow }_{\mathrm{white} \mathrm{ppt}}+3\mathrm{NaCl}$

$\mathrm{Al}(\mathrm{OH}{)}_3+\mathrm{NaOH}⟶{{\mathrm{NaAIO}}_2}_{(\mathrm{sodium} \mathrm{metaaluminate})}+2{\mathrm{H}}_2\mathrm{O}$

Note :${\mathrm{NaAIO}}_2$ is also written as $\mathrm{Na}\left[\mathrm{Al}(\mathrm{OH}{)}_4\right]$

$\text{ (c) }{\mathrm{FeCl}}_3+3\mathrm{NaOH}⟶\underset{\text{ Brown ppt }}{\mathrm{Fe}(\mathrm{OH}{)}_3\downarrow }+3\mathrm{NaCl}$

$\mathrm{Fe}(\mathrm{OH}{)}_3$ is insoluble in excess of $\mathrm{NaOH}\text{. }$

$\text{ (d) }{\mathrm{CrCl}}_3+3\mathrm{NaOH}⟶\underset{\text{ Green ppt }}{\mathrm{Cr}(\mathrm{OH}{)}_3\downarrow }+3\mathrm{NaCl}$

 $\mathrm{Cr}(\mathrm{OH}{)}_3$ is insoluble in excess $\mathrm{NaOH}\text{. }$

Note :$\mathrm{Cr}(\mathrm{OH}{)}_3$ can be made soluble in $\mathrm{NaOH}$ if ${\mathrm{H}}_2{\mathrm{O}}_2$

$\mathrm{Cr}(\mathrm{OH}{)}_3+2\mathrm{NaOH}+{\mathrm{H}}_2{\mathrm{O}}_2⟶\underset{\begin{array}{c}\text{ Soluble }\\ \text{ (yellow) }\end{array}}{{\mathrm{Na}}_2{\mathrm{CrO}}_4}+{\mathrm{H}}_2\mathrm{O}$

This is redox reaction.