DP Chemistry (last assessment 2024)

Test builder »

Question 19M.3.sl.TZ1.3

Select a Test
Date May 2019 Marks available [Maximum mark: 7] Reference code 19M.3.sl.TZ1.3
Level sl Paper 3 Time zone TZ1
Command term Calculate, Explain, Identify, Suggest Question number 3 Adapted from N/A
3.
[Maximum mark: 7]
19M.3.sl.TZ1.3

Lithium has many uses.

(a(i))

Identify the type of bonding in lithium hydride, using sections 8 and 29 of the data booklet.

[1]

Markscheme

ionic  [✔]

Examiners report

Many students correctly used the data tables and chart to deduce the bonding is ionic.

(a(ii))

Explain why lithium is paramagnetic while lithium hydride is diamagnetic by referring to electron configurations.

[2]

Markscheme

lithium has an unpaired electron  [✔]

all electrons in lithium hydride are paired  [✔]

 

Note: Award [1 max] for correct electron configurations of Li AND Li+ AND H- without discussion of pairing.

Examiners report

Only a few students were aware of the connection between electron pairing and magnetic properties.

The emission spectra obtained by ICP-OES for a mixture containing the isotope 6Li (Li-6) and naturally occurring lithium (Li (N)) is shown.

(b(i))

Suggest why ICP-OES does not give good quantitative results for distinguishing 6Li from naturally occurring lithium.

[1]

Markscheme

emission spectra of both «6Li and natural Li» give same colour/produce same «range of» wavelengths
OR
they have same electron transitions/same nuclear charge  [✔]

 

Note: Accept “the spectra are almost identical”.

Examiners report

Surprisingly few candidates referred to the fact that the two spectra shown are almost identical.

(b(ii))

Suggest a better method.

[1]

Markscheme

ICP-MS  [✔]

 

Note: Accept “MS/mass spectrometry”.

Examiners report

Again relatively few students were able to suggest ICP-MS as a more appropriate alternative.

(c)

Lithium is obtained by electrolysis of molten lithium chloride. Calculate the time, in seconds, taken to deposit 0.694 g Li using a current of 2.00 A.

Q (charge) = I (current) × t (time)

[2]

Markscheme

n«=  m M r = 0.694 6.94 » = 0.100 «mol»  [✔]

«t =  0 .100 mol × 96500  C mo l − 1 2 .00 C  s − 1 =»
4830 «s»  [✔]

 

Note: Accept “4820” OR “4825 «s»”.

Award [2] for correct final answer.

Examiners report

About half the candidates who attempted this were able to calculate the amount of lithium being formed and hence the time required.

Syllabus sections