S-Block Chemistry Flashcards
Group 1 favourite oxidation state:
Group 1 is dominated by the +1 oxidation state, so are really good reducing agents, as they try to lose an electron
Lithium burns in oxygen, show the equation:
4Li (s) + O2 (g) -> 2Li2O (s)
Sodium burns in oxygen, show the equation:
2Na (s) + O2 (g) -> Na2O2 (s)
produces sodium peroxide, peroxide ion is larger, charge is spread out across 2 molecules,
Potassium burns in oxygen, show the equation:
K (s) + O2 (g) -> KO2 (s)
produces a superoxide, size is likely bigger than the peroxide ion, as it has a bigger mass to charge ration
oxidation state of K in KO2 = +1, O = -0.5
What oxides do group 1 make if they are heavier than potassium?
Superoxides, because larger ions make larger oxides
Hydroxide formation equation from sodium:
2Na (s) + 2H2O (l) -> 2NaOH (aq) + H2 (g)
Hydroxide formation equation from sodium peroxide:
Na2O2 (s) + H2O (l) -> 2NaOH (aq) + 1/2O2 (g)
Group 1 hydroxide reaction trend
Reactions are increasing exothermic down the group -> heat ignites the hydrogen produced
The reaction of group 1 metals with water creates vigorous reactions. Explain why rubidium and caesium reactions are dangerous
2Rb (s) + 2H2O (l) -> 2RbOH (aq) + H2 (g)
(1) reaction is extremely exothermic which ignites the H2, which creates a risk of explosion.
Most important group 1 Halide:
NaCl (Soduim chloride)
Lithium halide formation equation:
Li2CO3 (s) + 2HCl (aq) -> 2LiCl (aq) + CO3^2- + 2H+
Lithium carbonate decomposition equation:
Li2CO3 (s) -> Li2O (s) + CO2 (g)
Write a chemicale equation for the decomposition of lithium carbonate under heat. Explain why sodium carbonate doesn’t decompose in the same way.
Li2CO3 (s) -> Li2O (s) + CO2 (g) [under heat conditions - add delta above arrow]
Sodium preferentially forms Na2O2 because the m/c ratio stabilises the peroxide ion:
Na2CO3 (s) -> Na2O2 (s) + CO (g) [under heat conditions - add delta above arrow]
[indicative of the reaction, may not be correct equation]
Group 2 favourite oxidation state:
Group 2 is dominated by the +2 oxidation state, so are really good reducing agents, as they try to lose an electron
Group 2 oxides formula
Generally: 2M (s) + O2 (g) -> 2MO (s)
Heavier metals (e.g Barium): 2BaO (s) + O2 (g) -> BaO2 (s)