Based on what you know about the unknown metal sample propose a balanced chemical equation for (4 points) a. the formation of the Group 1 metal oxide b. the formation of the Group 2 metal oxide

Group 1.

2Na(s) + O₂(g) → Na₂O(g) ;  2K(s) + O₂(g) → K₂O(g)

4Li (s)+ O₂(g) →   2Li₂O(g) ;  3 Rb(s) + RbO₂(g) → 2Rb₂O(g)

2Fr(s) + O₂(g) → Fr₂O(g) ;  2Cs+ O₂(g)  → Cs₂O(g)

Group 2.

2Mg(s) + O₂(g) → 2MgO(s) ; 2Be(s) + O₂(g) → 2BeO(s)

2Ca(s) + O₂(g) → 2CaO(s) ; 2Sr(s) + O₂(g) → 2SrO(s)

2Ba(s) + O₂(g) → 2BaO(s) ; 2Ra(s) + O₂(g) → 2RaO(s)

Explanation:

Elements from group 1 are: Li, Na, K, Rb, Cs and Fr

When they react to oxygen, they produce the correspond oxide.

2Na(s) + O₂(g) → Na₂O(g)

2K(s) + O₂(g) → K₂O(g)

4Li (s)+ O₂(g) →   2Li₂O(g)

When Rb reacts with the oxygen, it makes the superoxide

Rb(s)+ O₂(g) → RbO₂(g)

The superoxide reduces to Rb₂O by excess of Rb, according to:

3 Rb(s) + RbO₂(g) → 2Rb₂O(g)

2Fr(s) + O₂(g) → Fr₂O(g)

2Cs+ O₂(g)  → Cs₂O(g)

In the presence of excess oxygen, Na and K forms peroxide

2Na(s) + O₂(g) → Na₂O₂(g)

2K(s) + O₂(g) → K₂O₂(g)

K also forms, the peroxide when it burns into the air

K(s) + O₂(g) → KO₂

Group 2. 2Mg(s) + O₂(g) → 2MgO(s)

2Be(s) + O₂(g) → 2BeO(s) It does not react in water, so we can't produce, the Be(OH)₂ but the hidroxide surely, can be decomposed:

2Be(OH)₂ → 2BeO + 2H₂O

2Ca(s) + O₂(g) → 2CaO(s)

2Sr(s) + O₂(g) → 2SrO(s)

2Ba(s) + O₂(g) → 2BaO(s)

2Ra(s) + O₂(g) → 2RaO(s)

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