A metal $M$ readily forms water soluble sulphate $M S O_{4}$ , water insoluble hydroxide $M (O H)_{2}$ and oxide $M O$ which becomes inert on heating. If the hydroxide is soluble in $N a O H$ , then $M$ is:

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Solution

The correct option is B Be
Beryllium forms highly soluble (in water) beryllium sulfate $(B e S O_{4})$ . It forms water insoluble beryllium hydroxide $(B e (O H)_{2})$ .
On heating, beryllium oxide becomes inert to heating due to increase in the degree of polymerization..
Beryllium hydroxide dissolves in sodium hydroxide solution to form sodium beryllate $(N a_{2} B e O_{2})$ .
$B e (O H)_{2} + 2 N a O H \to N a_{2} B e O_{2} + H_{2} O$

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A metal M readily forms water soluble sulphate MSO4, water insoluble hydroxide M(OH)2 and oxide MO which becomes inert on heating. If the hydroxide is soluble in NaOH, then M is:

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