IB DP Chemistry Topic 9.1 Oxidation and reduction SL Paper 2

Question

Bonds can be formed in many ways.

The landing module for the Apollo mission used rocket fuel made from a mixture of hydrazine, N₂H₄, and dinitrogen tetraoxide, N₂O₄.

N₂H₄(l) + N₂O₄(l) → 3N₂(g) + 4H₂O(g)

a.i.State and explain the difference in bond strength between the nitrogen atoms in a hydrazine and nitrogen molecule.[2]

a.ii.State why hydrazine has a higher boiling point than dinitrogen tetraoxide.[1]

a.iii.Determine the oxidation state of nitrogen in the two reactants.

[1]

a.iv.Deduce, giving a reason, which species is the reducing agent.[1]

b.Deduce the Lewis (electron dot) structures of ozone.[2]

▶️Answer/Explanation

triple bond in nitrogen «molecule» AND single bond in hydrazine

triple bond stronger than single bond
OR
more shared «pairs of» electrons make bond stronger/attract nuclei more

Accept bond enthalpy values from data booklet (158 and 945 kJ\(\,\)mol^–1).

[2 marks]

a.i.

hydrogen bonding «between molecules, dinitrogen tetraoxide does not»

[1 mark]

a.ii.

N₂H₄: –2 AND N₂O₄: +4

[1 mark]

a.iii.

N₂H₄ AND oxidized/oxidation state increases
OR
N₂H₄ AND loses hydrogen
OR
N₂H₄ AND reduces/removes oxygen from N₂O₄

Accept “N₂H₄ AND gives electrons «to N₂O₄»”.

[1 mark]

a.iv.

Accept any combination of lines, dots or crosses to represent electrons.

Do not penalize missing lone pairs if already done in 3b.

Do not accept structure that represents 1.5 bonds.

[2 marks]