Code: XXX-01
48th International
Chemistry Olympiad
Theoretical Problems
Answer sheets
28 July 2016
Tbilisi, Georgia
48th IChO Theoretical Problems, Official English version 1
, Code: XXX-01
Problem 1 5% of the total
1.1. 1.2. 1.3. 1.4. 1.5. 1.6. 1.7. Sum
3 1 2 2 2 4 3 17
1.1. On which electrode does nitrogen trifluoride form?
Cathode Anode
Write a balanced chemical equation for the electrode half reaction for the
formation of NF3.
NH4+ + 3 F– ⟶ NF3 + 4 H+ + 6 e– or NH4F + 2 HF ⟶ NF3 + 6 H+ + 6 e– or equ.
3p (1p choosing anode, 1p species, 1p coefficients)
1.2. Which of NF3, NHF2 or NH2F compound is expected to condense at the lowest
temperature?
NF3 NHF2 NH2F NF3
1.3. Assign the N-F bond lengths (136, 140, 142 pm) to the molecules.
Molecule NH2F NHF2 NF3 The partial positive charge on N
N-F bond increases in this order, so the attraction
length, pm
also increases between the N and F.
1.4. Write a balanced chemical equation for the formation of the binary nitrogen –
fluorine compound.
2 NHF2 + 2 KF ⟶ N2F2 + 2 KHF2 or 2 NHF2 ⟶ N2F2 + 2 HF
2p (1p N2F2, 1p equation)
1.5. Propose a suitable reagent for the formation of NF4+ and write a balanced
chemical equation for the reaction.
NF3 + F2 + SbF5 ⟶ NF4+ + SbF6− any strong fluoride acceptor (AsF5, BF3)
2p (1p species, 1p coefficients)
48th IChO Theoretical Problems, Official English version 2
, Code: XXX-01
1.6. Write a balanced chemical equation for the hydrolysis of NF4+.
2 NF4+ + 2 H2O ⟶ 2 NF3 + O2 + 2 HF + 2 H+ ⇒ n(O2):n(NF3) = 1:2
2p (1p species, 1p coefficients)
Write a balanced chemical equation for a possible side reaction that can
decrease the theoretically expected O2:NF3 mole ratio.
e.g.: NF4+ + 2 H2O ⟶ NF3 + H2O2 + HF + H+ HOF, O3, OF2 also accepted.
2p (1p species, 1p coefficients)
1.7. Determine the formula of the salt in question.
Your work:
From the NF3:F2 ratio it is clear that the anion also contains fluorine.
Moreover, the starting NF4+: liberated F2 ratio is 1:2, and all fluorine content
is released. With a (NF4)xAFy composition, x:y = 1:4.
8𝑥𝑀(F) 𝑀(A)
= 0.656 ⇒ = 65.7 g/mol
8𝑥𝑀(F)+𝑥𝑀(N)+𝑀(A) 𝑥
With x=1, the atomic mass of A is close to zinc, but not all fluorine would be
released with Zn. If x=2, then the element is xenon, and the formula is
(NF4)2XeF8
3p (1.5p for Zn)
Formula:
48th IChO Theoretical Problems, Official English version 3
48th International
Chemistry Olympiad
Theoretical Problems
Answer sheets
28 July 2016
Tbilisi, Georgia
48th IChO Theoretical Problems, Official English version 1
, Code: XXX-01
Problem 1 5% of the total
1.1. 1.2. 1.3. 1.4. 1.5. 1.6. 1.7. Sum
3 1 2 2 2 4 3 17
1.1. On which electrode does nitrogen trifluoride form?
Cathode Anode
Write a balanced chemical equation for the electrode half reaction for the
formation of NF3.
NH4+ + 3 F– ⟶ NF3 + 4 H+ + 6 e– or NH4F + 2 HF ⟶ NF3 + 6 H+ + 6 e– or equ.
3p (1p choosing anode, 1p species, 1p coefficients)
1.2. Which of NF3, NHF2 or NH2F compound is expected to condense at the lowest
temperature?
NF3 NHF2 NH2F NF3
1.3. Assign the N-F bond lengths (136, 140, 142 pm) to the molecules.
Molecule NH2F NHF2 NF3 The partial positive charge on N
N-F bond increases in this order, so the attraction
length, pm
also increases between the N and F.
1.4. Write a balanced chemical equation for the formation of the binary nitrogen –
fluorine compound.
2 NHF2 + 2 KF ⟶ N2F2 + 2 KHF2 or 2 NHF2 ⟶ N2F2 + 2 HF
2p (1p N2F2, 1p equation)
1.5. Propose a suitable reagent for the formation of NF4+ and write a balanced
chemical equation for the reaction.
NF3 + F2 + SbF5 ⟶ NF4+ + SbF6− any strong fluoride acceptor (AsF5, BF3)
2p (1p species, 1p coefficients)
48th IChO Theoretical Problems, Official English version 2
, Code: XXX-01
1.6. Write a balanced chemical equation for the hydrolysis of NF4+.
2 NF4+ + 2 H2O ⟶ 2 NF3 + O2 + 2 HF + 2 H+ ⇒ n(O2):n(NF3) = 1:2
2p (1p species, 1p coefficients)
Write a balanced chemical equation for a possible side reaction that can
decrease the theoretically expected O2:NF3 mole ratio.
e.g.: NF4+ + 2 H2O ⟶ NF3 + H2O2 + HF + H+ HOF, O3, OF2 also accepted.
2p (1p species, 1p coefficients)
1.7. Determine the formula of the salt in question.
Your work:
From the NF3:F2 ratio it is clear that the anion also contains fluorine.
Moreover, the starting NF4+: liberated F2 ratio is 1:2, and all fluorine content
is released. With a (NF4)xAFy composition, x:y = 1:4.
8𝑥𝑀(F) 𝑀(A)
= 0.656 ⇒ = 65.7 g/mol
8𝑥𝑀(F)+𝑥𝑀(N)+𝑀(A) 𝑥
With x=1, the atomic mass of A is close to zinc, but not all fluorine would be
released with Zn. If x=2, then the element is xenon, and the formula is
(NF4)2XeF8
3p (1.5p for Zn)
Formula:
48th IChO Theoretical Problems, Official English version 3
