| Date | May 2016 | Marks available | 9 | Reference code | 16M.2.hl.TZ0.1 |
| Level | HL | Paper | 2 | Time zone | TZ0 |
| Command term | Show, Calculate, Determine, Outline, State, and Suggest | Question number | 1 | Adapted from | N/A |
Question
Phosphine (IUPAC name phosphane) is a hydride of phosphorus, with the formula PH3.
(i) Draw a Lewis (electron dot) structure of phosphine.
(ii) State the hybridization of the phosphorus atom in phosphine.
(iii) Deduce, giving your reason, whether phosphine would act as a Lewis acid, a Lewis base, or neither.
(iv) Outline whether you expect the bonds in phosphine to be polar or non-polar, giving a brief reason.
(v) Phosphine has a much greater molar mass than ammonia. Explain why phosphine has a significantly lower boiling point than ammonia.
(vi) Ammonia acts as a weak Brønsted–Lowry base when dissolved in water.
Outline what is meant by the terms “weak” and “Brønsted–Lowry base”.
Weak:
Brønsted–Lowry base:
Phosphine is usually prepared by heating white phosphorus, one of the allotropes of phosphorus, with concentrated aqueous sodium hydroxide. The equation for the reaction is:
(i) The first reagent is written as P4, not 4P. Describe the difference between P4 and 4P.
(ii) The ion H2PO2− is amphiprotic. Outline what is meant by amphiprotic, giving the formulas of both species it is converted to when it behaves in this manner.
(iii) State the oxidation state of phosphorus in P4 and H2PO2−.
P4:
H2PO2−:
(iv) Oxidation is now defined in terms of change of oxidation number. Explore how earlier definitions of oxidation and reduction may have led to conflicting answers for the conversion of P4 to H2PO2− and the way in which the use of oxidation numbers has resolved this.
2.478 g of white phosphorus was used to make phosphine according to the equation:
(i) Calculate the amount, in mol, of white phosphorus used.
(ii) This phosphorus was reacted with 100.0 cm3 of 5.00 mol dm−3 aqueous sodium hydroxide. Deduce, showing your working, which was the limiting reagent.
(iii) Determine the excess amount, in mol, of the other reagent.
(iv) Determine the volume of phosphine, measured in cm3 at standard temperature and pressure, that was produced.
Impurities cause phosphine to ignite spontaneously in air to form an oxide of phosphorus and water.
(i) 200.0 g of air was heated by the energy from the complete combustion of 1.00 mol phosphine. Calculate the temperature rise using section 1 of the data booklet and the data below.
Standard enthalpy of combustion of phosphine, 
Specific heat capacity of air = 1.00Jg−1K−1=1.00kJkg−1K−1
(ii) The oxide formed in the reaction with air contains 43.6% phosphorus by mass. Determine the empirical formula of the oxide, showing your method.
(iii) The molar mass of the oxide is approximately 285 g mol−1. Determine the molecular formula of the oxide.
(iv) State the equation for the reaction of this oxide of phosphorus with water.
(v) Suggest why oxides of phosphorus are not major contributors to acid deposition.
(vi) The levels of sulfur dioxide, a major contributor to acid deposition, can be minimized by either pre-combustion and post-combustion methods. Outline one technique of each method.
Pre-combustion:
Post-combustion:
Markscheme
(i)
Accept structures using dots and/or crosses to indicate bonds and/or lone pair.
(ii)
sp3
Do not allow ECF from a (i).
(iii)
Lewis base AND has a lone pair of electrons «to donate»
(iv)
non-polar AND P and H have the same electronegativity
Accept “similar electronegativities”.
Accept “polar” if there is a reference to a small difference in electronegativity and apply ECF in 1 a (v).
(v)
PH3 has London «dispersion» forces
NH3 forms H-bonds
H-bonds are stronger
OR
London forces are weaker
Accept van der Waals’ forces, dispersion forces and instantaneous dipole – induced dipole forces.
Accept “dipole-dipole forces” as molecule is polar.
H-bonds in NH3 (only) must be mentioned to score [2].
Do not award M2 or M3 if:
• implies covalent bond is the H-bond
• implies covalent bonds break.
Accept “dipole-dipole forces are weaker”.
(vi)
Weak: only partially dissociated/ionized «in dilute aqueous solution»
Brønsted–Lowry base: an acceptor of protons/H+/hydrogen ions
Accept reaction with water is reversible/an equilibrium.
Accept “water is partially dissociated «by the weak base»”.
(i)
P4 is a molecule «comprising 4P atoms» AND 4P is four/separate «P» atoms
OR
P4 represents «4P» atoms bonded together AND 4P represents «4» separate/non-bonded «P» atoms
(ii)
can act as both a «Brønsted–Lowry» acid and a «Brønsted–Lowry» base
OR
can accept and/or donate a hydrogen ion/proton/H+
HPO22– AND H3PO2
(iii)
P4: 0
H2PO2–: +1
Do not accept 1 or 1+ for H2PO2−.
(iv)
oxygen gained, so could be oxidation
hydrogen gained, so could be reduction
OR
negative charge «on product/H2PO2 »/gain of electrons, so could be reduction
oxidation number increases so must be oxidation
Award [1 max] for M1 and M2 if candidate displays knowledge of at least two of these definitions but does not apply them to the reaction.
Do not award M3 for “oxidation number changes”.
(i)
«\(\left\langle {\frac{{2.478}}{{4 \times 30.97}}} \right\rangle \)»= 0.02000 «mol»
(ii)
n(NaOH) = «0.1000 × 5.00 =» 0.500 «mol» AND P4/phosphorus is limiting reagent
Accept n(H2O) = \(\frac{{100}}{{18}}\) = 5.50 AND P4 is limiting reagent.
(iii)
amount in excess «= 0.500 - (3 × 0.02000)» = 0.440 «mol»
(iv)
«22.7 × 1000 × 0.02000» = 454 «cm3»
Accept methods employing pV = nRT, with p as either 100 (454 cm3) or 101.3 kPa (448 cm3). Do not accept answers in dm3.
(i)
temperature rise «=\(\frac{{750 \times 1.00}}{{0.2000 \times 1.00}}\)»=3750«°C/K»
Do not accept −3750.
(ii)
n(P)«=\(\frac{{43.6}}{{30.97}}\)»=1.41 «mol»
n(O) «=\(\frac{{100 - 43.6}}{{16.00}}\)»=3.53 «mol»
«\(\frac{{n\left( {\rm{O}} \right)}}{{n\left( {\rm{P}} \right)}}\)=\(\frac{{3.53}}{{1.41}}\) = 2.50 so empirical formula is» P2O5
Accept other methods where the working is shown.
(iii)
«\(\frac{{285}}{{141.9}}\)=2.00, so molecular formula = 2×P2O5=»P4O10
(iv)
P4O10(s) + 6H2O (l) → 4H3PO4 (aq)
Accept P4O10(s) + 2H2O (l) → 4HPO3 (aq) (initial reaction)
Accept P2O5(s) + 3H2O(l) → 2H3PO4(aq)
Accept equations for P4O6/P2O3 if given in d (iii).
Accept any ionized form of the acids as the products.
(v)
phosphorus not commonly found in fuels
OR
no common pathways for phosphorus oxides to enter the air
OR
amount of phosphorus-containing organic matter undergoing anaerobic decomposition is small
Accept “phosphorus oxides are solids so are not easily distributed in the atmosphere”.
Accept “low levels of phosphorus oxide in the air”.
Do not accept “H3PO4 is a weak acid”.
(vi)
Pre-combustion:
remove sulfur/S/sulfur containing compounds
Post-combustion:
remove it/SO2 by neutralization/reaction with alkali/base
Accept “lime injection fluidised bed combustion” for either, but not both.
