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Date	November 2018	Marks available	2	Reference code	18N.2.hl.TZ0.3
Level	HL	Paper	2	Time zone	TZ0
Command term	Draw	Question number	3	Adapted from	N/A

Question

Bromine can form the bromate(V) ion, BrO₃⁻.

State the electron configuration of a bromine atom.

[1]

a.i.

Sketch the orbital diagram of the valence shell of a bromine atom (ground state) on the energy axis provided. Use boxes to represent orbitals and arrows to represent electrons.

[1]

a.ii.

Draw two Lewis (electron dot) structures for BrO₃⁻.

[2]

b.i.

Determine the preferred Lewis structure based on the formal charge on the bromine atom, giving your reasons.

[2]

b.ii.

Predict, using the VSEPR theory, the geometry of the BrO₃⁻ ion and the O−Br−O bond angles.

[3]

Bromate(V) ions act as oxidizing agents in acidic conditions to form bromide ions.

Deduce the half-equation for this reduction reaction.

[2]

d.i.

Bromate(V) ions oxidize iron(II) ions, Fe²⁺, to iron(III) ions, Fe³⁺.

Deduce the equation for this redox reaction.

[1]

d.ii.

Calculate the standard Gibbs free energy change, ΔG^Θ, in J, of the redox reaction in (ii), using sections 1 and 24 of the data booklet.

E^Θ (BrO₃⁻ / Br⁻) = +1.44 V

[2]

d.iii.

State and explain the magnetic property of iron(II) and iron(III) ions.

[2]

Markscheme

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵

[Ar] 4s² 3d¹⁰ 4p⁵ ✔

Accept 3d before 4s.

a.i.

Accept double-headed arrows.

a.ii.

Structure I - follows octet rule:

Structure II - does not follow octet rule:

Accept dots, crosses or lines to represent electron pairs.

b.i.

«structure I» formal charge on Br = +2

«structure II» formal charge on Br = 0/+1 ✔

structure II is preferred AND it produces formal charge closer to 0 ✔

Ignore any reference to formal charge on oxygen.

b.ii.

Geometry:
trigonal/pyramidal ✔

Reason:
three bonds AND one lone pair
OR
four electron domains ✔

O−Br−O angle:
107° ✔

Accept “charge centres” for “electron domains”.

Accept answers in the range 104–109°.

BrO₃⁻ (aq) + 6e⁻ + 6H⁺ (aq) → Br⁻ (aq) + 3H₂O (l)

correct reactants and products ✔

balanced equation ✔

Accept reversible arrows.

d.i.

BrO₃⁻ (aq) + 6Fe²⁺ (aq) + 6H⁺ (aq) → Br⁻ (aq) + 3H₂O (l) + 6Fe³⁺ (aq) ✔

d.ii.

E^Θ_reaction = «+1.44 V – 0.77 V =» 0.67 «V» ✔

ΔG^Θ = «–nFE^Θ_reaction = – 6 × 96500 C mol^–1 × 0.67 V =» –3.9 × 10⁵ «J» ✔

d.iii.

both are paramagnetic ✔

«both» contain unpaired electrons ✔

Accept orbital diagrams for both ions showing unpaired electrons.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

d.iii.

[N/A]

Syllabus sections

Additional higher level (AHL) » Topic 14: Chemical bonding and structure » 14.1 Covalent bonding and electron domain and molecular geometries

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Core » Topic 4: Chemical bonding and structure » 4.3 Covalent structures

Additional higher level (AHL) » Topic 14: Chemical bonding and structure

Core » Topic 4: Chemical bonding and structure

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