Chemistry - Complete s-block Question with Solution | TestHub

$2\mathrm{Al}+2\mathrm{NaOH}+2{\mathrm{H}}_2\mathrm{O}⟶\underset{\begin{array}{c}\text{ sodum meta }\\ \text{ aluminata }\end{array}}{{\mathrm{NaAlO}}_2}+3{\mathrm{H}}_2$ Sodiummetaaluminate, thus formed, is soluble in water and changes into the complex ${\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_2(\mathrm{OH}{)}_4\right]}^{-}$, in which coordination number of $\mathrm{Al}$ is 6 .