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Date May 2021 Marks available 3 Reference code 21M.2.hl.TZ2.2
Level HL Paper 2 Time zone TZ2
Command term Suggest and Justify Question number 2 Adapted from N/A

Question

The properties of elements can be predicted from their position in the periodic table.

Explain why Si has a smaller atomic radius than Al.

[2]
a(i).

Explain why the first ionization energy of sulfur is lower than that of phosphorus.

[2]
a(ii).

State the condensed electron configurations for Cr and Cr3+.

[2]
b(i).

Describe metallic bonding and how it contributes to electrical conductivity.

[3]
b(ii).

Deduce, giving a reason, which complex ion [Cr(CN)6]3− or [Cr(OH)6]3− absorbs higher energy light. Use section 15 of the data booklet.

[1]
b(iii).

[Cr(OH)6]3− forms a green solution. Estimate a wavelength of light absorbed by this complex, using section 17 of the data booklet.

[1]
b(iv).

Deduce the Lewis (electron dot) structure and molecular geometry of sulfur tetrafluoride, SF4, and sulfur dichloride, SCl2.

[4]
c.

Suggest, giving reasons, the relative volatilities of SCl2 and H2O.

[3]
d.

Markscheme

nuclear charge/number of protons/Z/Zeff increases «causing a stronger pull on the outer electrons» ✓

same number of shells/«outer» energy level/shielding ✓

a(i).

P has «three» unpaired electrons in 3p sub-level AND S has one full 3p orbital «and two 3p orbitals with unpaired electrons»
OR
P: [Ne]3s23px13py13pz1 AND S: [Ne]3s23px23py13pz1


Accept orbital diagrams for 3p sub-level for M1. Ignore other orbitals or sub-levels.

 

repulsion between paired electrons in sulfur «and therefore easier to remove» ✓


Accept “removing electron from S gives more stable half-filled sub-level" for M2.

a(ii).

Cr:
[Ar] 4s13d5


Cr3+:

[Ar] 3d3

 

Accept “[Ar] 3d54s1”.

Accept “[Ar] 3d34s0”.

Award [1 max] for two correct full electron configurations “1s22s22p63s23p64s13d5 AND 1s22s22p63s23p63d3”.

Award [1 max] for 4s13d5 AND 3d3.

b(i).

electrostatic attraction ✓

between «a lattice of» cations/positive «metal» ions AND «a sea of» delocalized electrons ✓

mobile electrons responsible for conductivity
OR
electrons move when a voltage/potential difference/electric field is applied ✓

 

Do not accept “nuclei” for “cations/positive ions” in M2.

Accept “mobile/free” for “delocalized” electrons in M2.

Accept “electrons move when connected to a cell/battery/power supply” OR “electrons move when connected in a circuit” for M3.

b(ii).

[Cr(CN)6]3− AND CN/ligand causes larger splitting «in d-orbitals compared to OH»
OR
[Cr(CN)6]3− AND CN/ligand associated with a higher Δ/«crystal field» splitting energy/energy difference «in the spectrochemical series compared to OH » ✓

 

Accept “[Cr(CN)6]3− AND «CN» strong field ligand”.

b(iii).

any value or range between 647 and 700 nm ✓

b(iv).

 

SF4/SCl2 structure does not have to be 3-D for mark.

Penalize missing lone pairs of electrons on halogens once only.

Accept any combination of dots, lines or crosses for bonds/lone pairs.

Accept “non-linear” for SCl2 molecular geometry.

Award [1] for two correct electron domain geometries, e.g. trigonal bipyramidal for SF4 and tetrahedral for SCl2.

c.

H2O forms hydrogen bonding «while SCl2 does not» ✓

SCl2 «much» stronger London/dispersion/«instantaneous» induced dipole-induced dipole forces ✓


Alternative 1:

H2O less volatile AND hydrogen bonding stronger «than dipole–dipole and dispersion forces» ✓


Alternative 2:

SCl2 less volatile AND effect of dispersion forces «could be» greater than hydrogen bonding ✓


Ignore reference to Van der Waals.

Accept “SCl2 has «much» larger molar mass/electron density” for M2.

d.

Examiners report

[N/A]
a(i).
[N/A]
a(ii).
[N/A]
b(i).
[N/A]
b(ii).
[N/A]
b(iii).
[N/A]
b(iv).
[N/A]
c.
[N/A]
d.

Syllabus sections

Core » Topic 4: Chemical bonding and structure » 4.4 Intermolecular forces
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