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Date	May 2016	Marks available	4	Reference code	16M.2.sl.TZ0.1
Level	SL	Paper	2	Time zone	TZ0
Command term	Show, Calculate, Deduce, and Determine	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) Outline whether you expect the bonds in phosphine to be polar or non-polar, giving a brief reason.

(iii) Explain why the phosphine molecule is not planar.

(iv) Phosphine has a much greater molar mass than ammonia. Explain why phosphine has a significantly lower boiling point than ammonia.

[6]

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:

P4 (s) + 3OH⁻ (aq) + 3H₂O (l) → PH₃ (g) + 3H₂PO₂⁻ (aq)

(i) Identify one other element that has allotropes and list two of its allotropes.

Element:

Allotrope 1:

Allotrope 2:

(ii) The first reagent is written as P₄, not 4P. Describe the difference between P₄ and 4P.

(iii) 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.

(iv) State the oxidation state of phosphorus in P₄ and H₂PO₂⁻.

P₄:

H₂PO₂⁻:

(v) 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.

[10]

2.478 g of white phosphorus was used to make phosphine according to the equation:

P₄(s) +3OH⁻(aq)+3H₂O(l) → PH₃(g)+3H₂PO₂⁻(aq)

(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.

[4]

Markscheme

(i)

Accept structures using dots and/or crosses to indicate bonds and/or lone pair.

(ii)
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 (iv).

(iii)
4 electron domains/pairs/negative charge centres «around the central atom»
OR
a lone/non-bonding pair «and three bonding pairs around the central atom»
repulsion between electron domains/pairs/negative charge centres «produces non-planar shape»
OR
«repulsion causes» tetrahedral orientation/pyramidal shape

(iv)
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”.

(i)
Element
carbon/C
OR
oxygen/O/O₂

Allotropes
Award [1] for two of:
diamond
graphite
graphene
C₆₀ / buckminsterfullerene
OR
ozone/O₃ AND «diatomic/molecular» oxygen/O₂

Accept two correctly named allotropes of any other named element (S, Se, Sn, As, etc.).

Accept fullerene, “buckyballs” etc. instead of buckminsterfullerene.

(ii)
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

(iii)

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

(iv)

P₄ : 0
H₂PO₂⁻ : +1

Do not accept 1 or 1+ for H₂PO₂⁻.

(v)

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.