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A metal $M$ readily forms its sulphate $MSO_{4}$ which is water-soluble. It forms its oxide $MO$ which becomes inert on heating. It forms an insoluble hydroxide $M(OH)_{2}$ which is soluble in $NaOH$ solution. Then $M$ is

Option 1)
$Mg\;$

Option 2)
$\; Ba\;$

Option 3)
$\; Ca\;$

Option 4)
$\; Be$

Answers (1)

best_answer

As we learnt in

Action of oxygen with alkaline earth metals -

Form oxides Ba and Sr on heating with excess $o_{2}$ forms peroxides

- wherein

$2M+O{_2}\overset{\Delta }{\rightarrow}2MO$ (M=Be, Mg, Ca)

$M^{{}'}+O_{2}\overset{\Delta }{\rightarrow}M^{{}'}O{_2}$ ( $M^{{}'}$ =Ba, Sr)

and

Anomalous behavious of Be(OH)2 -

$Be(OH)_{2}$ is amphoteric reacts with caustic alkalis and mineral acids

- wherein

$Be(OH){_2}+2NaOH \rightarrow Na{_2}BeO{_2}+BeCl{_2}+2H{_2}O$

$Be$ readily forms its sulphate $BeSO_{4}$ which is water soluble. $BeO$ is thermally stable. $Be\left ( OH \right )_{2}$ is an insoluble Hydroxide. $Be\left ( OH \right )_{2}$ is soluble in $NaOH$ .

$Be\left ( OH \right )_{2}+2NaOH\rightarrow Na_{2}BeO_{2}+2H_{2}O$

Option 1)

$Mg\;$

This option is incorrect.

Option 2)

$\; Ba\;$

This option is incorrect.

Option 3)

$\; Ca\;$

This option is incorrect.

Option 4)

$\; Be$

This option is correct.

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Plabita

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