A-LevelChemistryChemical periodicity9701/22 · Oct/Nov 2020 · Q317 Marks

PCl5, PCl3 and NCl3 are halides of Group 15 elements.

(a)PCl5 can be formed from the reaction of phosphorus with chlorine. PCl5 has a melting point of 161°C.
(a)(i)Write an equation for the formation of PCl5 from the reaction of phosphorus and chlorine.[1]
(a)(ii)State the type of structure and bonding shown by liquid PCl5.[1]
(b)A small amount of PCl5 is added to excess water. The PCl5 reacts vigorously to form a colourless solution.
(b)(i)Give one other observation you would make when PCl5 reacts with excess water.[1]
(b)(ii)Write the equation for the reaction of PCl5 with excess water.[1]
(b)(iii)Estimate the pH of the resulting solution.[1]
(c)PCl3 is used to convert alcohols to chloroalkanes, such as compound T. [Figure 3c_scheme1] A possible synthesis of T is shown. [Figure 3c_scheme2]
(c)(i)Identify a reagent that could be used in reaction 1.[1]
(c)(ii)T exhibits optical isomerism. Explain what is meant by the term optical isomer and circle any atom(s) in T that give rise to optical isomerism. [Figure 3c(ii)][2]
(c)(iii)T is a minor product in the reaction of compound S with excess HCl. [Figure 3c(iii)] Draw the structure of the major product of the reaction of S with excess HCl.[1]
(d)NCl3 is a yellow liquid that can be used to bleach flour.
(d)(i)Predict the shape of the NCl3 molecule and the Cl-N-Cl bond angle.[2]
(d)(ii)NCl3 reacts with water to form HOCl, a weak Brønsted-Lowry acid. Explain fully what is meant by the term weak Brønsted-Lowry acid.[2]
(d)(iii)NCl3(l) decomposes according to the equation shown. 2NCl3(l) → N2(g) + 3Cl2(g) A sealed container of volume 250 cm³ contains an unreactive gas at a pressure of 1.00 x 105 Pa. 0.241 g of NCl3(l) was injected into the sealed container. The sealed container was heated to make the NCl3(l) decompose fully and then cooled to 20°C. Calculate the final total pressure inside the sealed container at 20 °C after the NCl3(l) has fully decomposed.[4]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.

[1] P4 + 10Cl2 → 4PCl5
[1] simple / molecular AND covalent
[1] steamy / misty fumes
[1] PCl5 + 4H2O → H3PO4 + 5HCl
[1] 0 to 4
[1] LiAlH4 OR lithium tetrahydridoaluminate((III))
[2] M1: molecule with a non super-(im)posable mirror image M2: [diagram of T with chiral carbon circled]
[1] [structure of 2,4-dichloro-2,4-dimethylpentane]
[2] M1: (trigonal) pyramidal M2: 107
[2] M1: proton / H+ donor M2: partially dissociates (in solution)
[4] method 1 M1: show the number of mol gas produced from 0.241 g NCl3 [Mr(NCl3) = 120.5 1 mol NCl3 produces 2 mol gas] n = 2 × 0.241 / 120.5 OR n = 0.0040 (mol gas produced.) M2: correct conversion of T to Kelvin, V to m³ and correct value of R M3: use of equation P = nRT/V with M2 values for n, R, T and V to find pressure of mol gas produced increase in p = nRT/V = 0.0040×8.31×293 / 250×10-6 = 3.90 × 104 M4: = 1.00 × 105 + M3 (Pa) total pressure = 1.00 × 105 + 3.90 × 104 = 1.39 × 105 (Pa) method 2 M1: calculate the produced from 0.241 g NCl3 0.003 (mol) Cl2 AND 0.001 (mol) N2 M2: conversion of T to Kelvin, V to m³ and correct value of R in all PV/RT equations used M3: use of equation PV/RT= n for both calculations or a combined equation with M2 values for R,T and V to find partial pressure for each of the gases ppCl2 = 29217.96 AND ppN2 = 9739.32 M4: Use Ptotal = pp unreacted gas + ppCl2 + ppN2 1 x 105 + 2.92 × 104 + 9.74 × 103 = 138 940