Question
The periodic table provides information about electron configuration, and physical and chemical properties of elements.
(a) Bismuth has atomic number 83. Deduce two pieces of information about the electron configuration of bismuth from its position on the periodic table.
(b) Outline why aluminium is malleable.
(c) An 11.98g block of pure aluminium was heated. Calculate the heat energy absorbed, in J, to increase its temperature from 18.0°C to 40.0°C. The specific heat capacity of aluminium is 0.902J \(g^{−1} K^{−1}\)
(d) Argon has three naturally occurring isotopes, \(^{36}Ar\), \(^{38}Ar\) and \(^{40}Ar\).
(i) Identify the technique used to determine the relative proportions of the isotopes of argon.
The isotopic composition of a sample of argon is 0.34% of \(^{36}Ar\), 0.06% of \(^{38}Ar\) and 99.6% of \(^{40}Ar\).
(ii) Calculate the relative atomic mass of this sample, giving your answer to two
decimal places.
(e) State the full electron configuration of the cobalt(II) ion, \(Co^{2+}\)
Answer/Explanation
Answer:
(a) Any two of:
«group 15 so Bi has» 5 valence electrons
«period 6 so Bi has» 6 «occupied» electron shells/energy levels
«in p-block so» p orbitals are highest occupied
occupied d/f orbitals
has unpaired electrons
has incomplete shell(s)/subshell(s)
(b) «layers of» cations slide over each other without disrupting bonding
OR
attraction between metal ions and delocalized electrons/metallic bonding is not
disrupted by changing position of metal ions
OR
metallic bonds are non-directional
OR
changing the shape does not disrupt the bonding
(c) «heat energy = 11.98 g x 0.902 J \(g^{−1} K^{−1}\) x 22.0 K =» 238 «J»
(d) (i) mass spectrometry
OR
mass spectroscopy
OR
mass spectrum
OR
MS
(ii) (0.0034 x 36) + (0.0006 x 38) + (0.996 x 40)
39.99
(e) \(1s^2\) \(2s^2\) \(2p^6\) \(3s^2\) \(3p^6\) \(3d^7\)
Question
Magnesium is a group 2 metal which exists as a number of isotopes and forms many compounds.
State the nuclear symbol notation, \({}_Z^AX\), for magnesium-26.
Mass spectroscopic analysis of a sample of magnesium gave the following results:
Calculate the relative atomic mass, Ar, of this sample of magnesium to two decimal places.
Magnesium burns in air to form a white compound, magnesium oxide. Formulate an equation for the reaction of magnesium oxide with water.
Describe the trend in acid-base properties of the oxides of period 3, sodium to chlorine.
In addition to magnesium oxide, magnesium forms another compound when burned in air. Suggest the formula of this compound
Describe the structure and bonding in solid magnesium oxide.
Magnesium chloride can be electrolysed.
Deduce the half-equations for the reactions at each electrode when molten magnesium chloride is electrolysed, showing the state symbols of the products. The melting points of magnesium and magnesium chloride are 922 K and 987 K respectively.
Anode (positive electrode):
Cathode (negative electrode):
Answer/Explanation
Markscheme
\({}_{12}^{26}{\rm{Mg}}\)
«Ar =»\(\frac{{24 \times 78.60 + 25 \times 10.11 + 26 \times 11.29}}{{100}}\)
«= 24.3269 =» 24.33
Award [2] for correct final answer.
Do not accept data booklet value (24.31).
MgO(s) + H2O(l) → Mg(OH)2(s)
OR
MgO(s) + H2O(l) → Mg2+(aq) + 2OH–(aq)
Accept \( \rightleftharpoons \).
from basic to acidic
through amphoteric
Accept “alkali/alkaline” for “basic”.
Accept “oxides of Na and Mg: basic AND oxide of Al: amphoteric” for M1.
Accept “oxides of non-metals/Si to Cl acidic” for M2.
Do not accept just “become more acidic”
Mg3N2
Accept MgO2, Mg(OH)2, Mg(NOx)2, MgCO3.
«3-D/giant» regularly repeating arrangement «of ions»
OR
lattice «of ions»
Accept “giant” for M1, unless “giant covalent” stated.
electrostatic attraction between oppositely charged ions
OR
electrostatic attraction between Mg2+ and O2– ions
Do not accept “ionic” without description.
Anode (positive electrode):
2Cl– → Cl2(g) + 2e–
Cathode (negative electrode):
Mg2+ + 2e– → Mg(l)
Penalize missing/incorrect state symbols at Cl2 and Mg once only.
Award [1 max] if equations are at wrong electrodes.
Accept Mg (g).
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.
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.
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.
Answer/Explanation
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)
H2PO2− : +1
OR
negative charge «on product/H2PO2− » /gain of electrons so could be reduction
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.