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A metal M readily forms water soluble sulphate $M S O_{4}$ water insoluble hydroxide $M {(O H)}_{2}$ and oxide MO which becomes inert on heating. The hydroxide is soluble in NaOH. The M is

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M is Be which forms water soluble $B e S O_{4}$ water insoluble $B e {(O H)}_{2}$ and BeO. $B e {(O H)}_{2}$ is soluble in $N a O H$ giving sodium beryllate $N a_{2} B e O_{2}$

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