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)

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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