Chemistry - Complete s-block Question with Solution | TestHub

Here statement in option A is not correct as Be and Al both dissolve in NaOH to liberate hydrogen.
$\text{Be}\text{+}\text{2NaOH}\to \underset{\text{Sod}\text{.beryllate}}{{\text{Na}}_{\text{2}}{\text{BeO}}_{\text{2}}}\text{+}{\text{H}}_{\text{2}}\text{(g)}$
$\mathrm{A}\mathrm{l}+\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}+2{\mathrm{H}}_2\mathrm{O}\to 2\mathrm{N}\mathrm{a}\mathrm{A}\mathrm{l}{\mathrm{O}}_2+3{\mathrm{H}}_2\left(\mathrm{g}\right)$
Rest other statements are correct.