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 metal M is:

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Solution

The correct option is A Be
Berilllium form Berillium sulfate ( $B e S O_{4}$ ) which is water soluble
Berilllium form Berillium hydroxide ( $B e (O H)_{2}$ ) which is water insoluble
Berilllium form Berillium oxide ( $B e O$ ) which becomes inert on heating.
$B e (O H)_{2}$ being amphoteric reacts with NaOH.

Hence option A is correct.

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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. The hydroxide is soluble in NaOH. The metal M is:

The correct option is A BeBerilllium form Berillium sulfate (BeSO4) which is water solubleBerilllium form Berillium hydroxide (Be(OH)2) which is water insolubleBerilllium form Berillium oxide (BeO) which becomes inert on heating.Be(OH)2 being amphoteric reacts with NaOH.Hence option A is correct.

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 metal M is: