DP Chemistry (first assessment 2025)

Question 21N.2.SL.TZ0.3

Date	November 2021	Marks available	[Maximum mark: 9]	Reference code	21N.2.SL.TZ0.3
Level	SL	Paper	2	Time zone	TZ0
Command term	Calculate, Deduce, Explain, Formulate, Justify, Sketch, State	Question number	3	Adapted from	N/A

[Maximum mark: 9]

21N.2.SL.TZ0.3

White phosphorus is an allotrope of phosphorus and exists as P₄.

An equilibrium exists between PCl₃ and PCl₅.

PCl₃(g) + Cl₂(g) $⇌$ PCl₅(g)

(a(i))

Sketch the Lewis (electron dot) structure of the P₄ molecule, containing only single bonds.

[1]

Markscheme

Accept any diagram with each P joined to the other three.

Accept any combination of dots, crosses and lines.

(a(ii))

Write an equation for the reaction of white phosphorus (P₄) with chlorine gas to form phosphorus trichloride (PCl₃).

[1]

Markscheme

P₄(s) + 6Cl₂ (g) → 4PCl₃ (l) ✔

(b(i))

Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl bond angle in PCl₃.

[3]

Markscheme

Electron domain geometry: tetrahedral ✔

Molecular geometry: trigonal pyramidal ✔

Bond angle: 100«°» ✔

Accept any value or range within the range 91−108«°» for M3.

(b(ii))

Explain the polarity of PCl₃.

[1]

Markscheme

polar AND unsymmetrical distribution of charge
OR
polar AND dipoles do not cancel
OR
«polar as» dipoles «add to» give a «partial» positive «charge» at P and a «partial» negative «charge» at the opposite/Cl side of the molecule ✔

Accept “polar AND unsymmetrical molecule”.

(c(i))

Calculate the standard enthalpy change (ΔH^⦵) for the forward reaction in kJ mol⁻¹.

ΔH^⦵_f PCl₃(g) = −306.4 kJ mol⁻¹

ΔH^⦵_f PCl₅(g) = −398.9 kJ mol⁻¹

[1]

Markscheme

«−398.9 kJ mol⁻¹ − (−306.4 kJ mol⁻¹) =» −92.5 «kJ mol⁻¹» ✔

(c(ii))

State the equilibrium constant expression, K_c, for this reaction.

[1]

Markscheme