Answer:
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)
