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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 State and explain 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]
c.

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, ΔGΘ, in J, of the redox reaction in (ii), using sections 1 and 24 of the data booklet.

EΘ (BrO3 / Br) = +1.44 V

 

[2]
d.iii.

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

 

[2]
e.

Markscheme

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5

OR

[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

OR

«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°.

c.

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.

EΘreaction = «+1.44 V – 0.77 V =» 0.67 «V» ✔

ΔGΘ = «–nFEΘreaction = – 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.

 

e.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.
[N/A]
d.i.
[N/A]
d.ii.
[N/A]
d.iii.
[N/A]
e.

Syllabus sections

Additional higher level (AHL) » Topic 13: The periodic table—the transition metals » 13.1 First-row d-block elements
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Additional higher level (AHL) » Topic 13: The periodic table—the transition metals
Additional higher level (AHL)

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