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Date May 2017 Marks available 1 Reference code 17M.2.sl.TZ2.4
Level SL Paper 2 Time zone TZ2
Command term Deduce Question number 4 Adapted from N/A

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)

State and explain the difference in bond strength between the nitrogen atoms in a hydrazine and nitrogen molecule.

[2]
a.i.

State why hydrazine has a higher boiling point than dinitrogen tetraoxide.

[1]
a.ii.

Determine the oxidation state of nitrogen in the two reactants.

[1]
a.iii.

Deduce, giving a reason, which species is the reducing agent.

[1]
a.iv.

Deduce the Lewis (electron dot) structures of ozone.

[2]
b.

Markscheme

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]

b.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
a.iii.
[N/A]
a.iv.
[N/A]
b.

Syllabus sections

Core » Topic 9: Redox processes » 9.1 Oxidation and reduction
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