DP Chemistry (first assessment 2025)

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Question 22M.2.HL.TZ1.6

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Date May 2022 Marks available [Maximum mark: 11] Reference code 22M.2.HL.TZ1.6
Level HL Paper 2 Time zone TZ1
Command term Deduce, Draw, Explain, Show, State Question number 6 Adapted from N/A
6.
[Maximum mark: 11]
22M.2.HL.TZ1.6

Nitric acid is usually produced by the oxidation of ammonia.

A mixture of nitric acid and sulfuric acid can be used to convert benzene to nitrobenzene, C6H5NO2.

(a(i))

Draw arrows in the boxes to represent the electron configuration of a nitrogen atom.

[1]

Markscheme


Accept all 2p electrons pointing downwards.

Accept half arrows instead of full arrows.

Examiners report

Drawing arrows in the boxes to represent the electron configuration of a nitrogen atom was done extremely well.

(a(ii))

Deduce a Lewis (electron dot) structure of the nitric acid molecule, HNO3, that obeys the octet rule, showing any non-zero formal charges on the atoms.

[2]

Markscheme

bonds and non-bonding pairs correct ✔

formal charges correct ✔

 

Accept dots, crosses or lines to represent electron pairs.

Do not accept resonance structures with delocalised bonds/electrons.

Accept + and – sign respectively.

Do not accept a bond between nitrogen and hydrogen.

For an incorrect Lewis structure, allow ECF for non-zero formal charges.

Examiners report

Drawing the Lewis structure of HNO3 was performed extremely poorly with structures that included H bonded to N, no double bond or a combination of single, double and even a triple bond or incorrect structures with dotted lines to reflect resonance. Many did not calculate non-zero formal charges.

(a(iii))

Explain the relative lengths of the three bonds between N and O in nitric acid.

[3]

Markscheme

Any three of:

two N-O same length/order ✔
delocalization/resonance ✔

N-OH longer «than N-O»
OR
N-OH bond order 1 AND N-O bond order 1½ ✔

 

Award [2 max] if bond strength, rather than bond length discussed.

Accept N-O between single and double bond AND N-OH single bond.

Examiners report

Poorly done; some explained relative bond strengths between N and O in HNO3, not relative lengths; others included generic answers such as triple bond is shortest, double bond is longer, single longest.

(a(iv))

State a technique used to determine the length of the bonds between N and O in solid HNO3.

[1]

Markscheme

X-ray crystallography ✔

Examiners report

A majority could not state the technique to determine length of bonds; answers included NMR, IR, and such instead of X-ray crystallography.

(b(i))

Write an equation for the reaction between the acids to produce the electrophile, NO2+.

[1]

Markscheme

HNO3 + 2H2SO4  NO2+ + H3O+ + 2HSO4-

 

Accept “HNO3 + H2SO4  NO2+ + H2O + HSO4-”.

Accept “HNO3 + H2SO4  H2NO3+ + HSO4-AND “H2NO3+  NO2+ + H2O”.

Accept single arrows instead of equilibrium signs.

Examiners report

Many had difficulties writing the balanced equation(s) for the formation of the nitronium ion.

(b(ii))

Draw the structural formula of the carbocation intermediate produced when this electrophile attacks benzene.

[1]

Markscheme

 

Accept any of the five structures.

Do not accept structures missing the positive charge.

Examiners report

Again, many had difficulty drawing the structural formula of the carbocation intermediate produced in the reaction.

(b(iii))

Deduce the number of signals that you would expect in the 1H NMR spectrum of nitrobenzene and the relative areas of these.

[2]

Markscheme

Number of signals: three/3 ✔

Relative areas: 2 : 2 : 1 ✔

Examiners report

Deducing the number of signals in the 1H NMR spectrum of nitrobenzene, which depend on the number of different hydrogen environments, was done poorly. Also, instead of relative areas, the common answer included chemical shift (ppm) values.

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