Question
Define the term isotopes.
A sample of silicon contains three isotopes.
Calculate the relative atomic mass of silicon using this data.
Describe the structure and bonding in silicon dioxide and carbon dioxide.
Draw the Lewis structure of NH3, state its shape and deduce and explain the H–N–H bond angle in \({\text{N}}{{\text{H}}_{\text{3}}}\).
The graph below shows the boiling points of the hydrides of group 5. Discuss the variation in the boiling points.
Explain, using diagrams, why CO and \({\text{N}}{{\text{O}}_{\text{2}}}\) are polar molecules but \({\text{C}}{{\text{O}}_{\text{2}}}\) is a non-polar molecule.
Answer/Explanation
Markscheme
atoms of the same element with the same atomic number/Z/same number of protons, but different mass numbers/A/different number of neutrons;
\((0.9223 \times 28) + (0.0468 \times 29) + (0.0309 \times 30)\);
28.1/28.11;
Working must be shown to get [2], do not accept 28.09 on its own (given in the data booklet).
Silicon dioxide
single covalent (bonds);
network/giant covalent/ macromolecular / repeating tetrahedral units;
Carbon dioxide
double covalent (bonds);
(simple / discrete) molecular;
Marks may be obtained from suitable structural representations of SiO2 and CO2.
;
Allow crosses or dots for lone-pair.
trigonal/triangular pyramidal;
(\( \sim \))107° / less than 109.5°;
Do not allow ECF.
LP-BP repulsion \( > \) BP-BP repulsion / one lone pair and three bond pairs / lone pairs/non-bonding pairs repel more than bonding-pairs;
Do not accept repulsion between atoms.
boiling points increase going down the group (from \({\text{P}}{{\text{H}}_{\text{3}}}\) to \({\text{As}}{{\text{H}}_{\text{3}}}\) to \({\text{Sb}}{{\text{H}}_{\text{3}}}\));
\({M_{\text{r}}}\)/number of electrons/molecular size increases down the group;
Accept electron cloud increases down the group for the second marking point.
greater dispersion/London/van der Waal’s forces;
\({\text{N}}{{\text{H}}_{\text{3}}}\)/ammonia has a higher boiling point than expected due to the hydrogen bonding between the molecules;
Do not accept hydrogen bonding alone.
CO:
Award [1] for showing the net dipole moment, or explaining it in words (unsymmetrical distribution of charge).
\(N{O_2}\):
Award [1] for correct representation of the bent shape and [1] for showing the net dipole moment, or explaining it in words (unsymmetrical distribution of charge).
\(C{O_2}\):
Award [1] for correct representation of the linear shape and [1] for showing the two equal but opposite dipoles or explaining it in words (symmetrical distribution of charge).
For all three molecules, allow either arrow or arrow with bar for representation of dipole moment.
Allow correct partial charges instead of the representation of the vector dipole moment.
Ignore incorrect bonds.
Lone pairs not needed.
Examiners report
In general the definition of isotopes was correct in (a) (i), but there are still some candidates who stated “isotopes are elements” and not “atoms of the same element”.
Nearly everybody gave the correct answer of 28.1 for the relative atomic mass of silicon in (ii).
Part (a) (iii) proved to be very difficult for the candidates. There was a lot of confusion about the two molecules; some candidates stated that they had the same double bond. Not many candidates mentioned the giant covalent structure for the silicon dioxide or the simple molecular structure for the carbon dioxide.
In (b) (i) the majority of candidates drew the Lewis structure of the ammonia molecule correctly showing the lone pair of electrons and the correct shape and angle and (ii) was well answered by most candidates.
They realised that \({\text{N}}{{\text{H}}_{\text{3}}}\) had a higher boiling point than \({\text{P}}{{\text{H}}_{\text{3}}}\) because of the intermolecular hydrogen bonding present in \({\text{N}}{{\text{H}}_{\text{3}}}\).
For (c) most answers given here showed diagrams of the three molecules, including distribution of charges, bonding and shapes. Some candidates gave very good answers showing a good understanding of the polarity of molecules.
Question
Rubidium contains two stable isotopes, \(^{{\text{85}}}{\text{Rb}}\) and \(^{{\text{87}}}{\text{Rb}}\). The relative atomic mass of rubidium is given in Table 5 of the Data Booklet.
Calculate the percentage of each isotope in pure rubidium. State your answers to three significant figures.
State the number of electrons and the number of neutrons present in an atom of \(^{{\text{87}}}{\text{Rb}}\).
Number of electrons:
Number of neutrons:
Answer/Explanation
Markscheme
(let \(x = {\text{fraction of}}{{\text{ }}^{{\text{85}}}}{\text{Rb}}\))
\(\frac{{(x \times 85) + [(100 – x) \times 87]}}{{100}} = 85.47\);
\(^{{\text{85}}}{\text{Rb}} = 76.5\% \) and \(^{{\text{87}}}{\text{Rb}} = 23.5\% \);
Award [2] for correct final answer.
37 (electrons);
50 (neutrons);
Examiners report
This question was answered very well by those that knew the correct mathematical technique; however some candidates did not have any idea how to tackle this problem.
The vast majority of candidates could correctly state the number of electrons and neutrons present in Rubidium- 87.
Question
Chlorine occurs in Group 7, the halogens.
Two stable isotopes of chlorine are \(^{{\text{35}}}{\text{Cl}}\) and \(^{{\text{37}}}{\text{Cl}}\) with mass numbers 35 and 37 respectively.
Chlorine has an electronegativity value of 3.2 on the Pauling scale.
Chloroethene, H2C=CHCl, the monomer used in the polymerization reaction in the manufacture of the polymer poly(chloroethene), PVC, can be synthesized in the following two-stage reaction pathway.
\[\begin{array}{*{20}{l}} {{\text{Stage 1:}}}&{{{\text{C}}_2}{{\text{H}}_4}{\text{(g)}} + {\text{C}}{{\text{l}}_2}{\text{(g)}} \to {\text{ClC}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{Cl(g)}}} \\ {{\text{Stage 2:}}}&{{\text{ClC}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{Cl(g)}} + {\text{HC=CHCl(g)}} + {\text{HCl(g)}}} \end{array}\]
Define the term isotopes of an element.
Calculate the number of protons, neutrons and electrons in the isotopes 35Cl and 37Cl.
Using the mass numbers of the two isotopes and the relative atomic mass of chlorine from Table 5 of the Data Booklet, determine the percentage abundance of each isotope
Percentage abundance 35Cl:
Percentage abundance 37Cl:
Define the term electronegativity.
Using Table 7 of the Data Booklet, explain the trends in electronegativity values of the Group 7 elements from F to I.
State the balanced chemical equation for the reaction of potassium bromide, KBr(aq), with chlorine, Cl2(aq).
Describe the colour change likely to be observed in this reaction.
Determine the enthalpy change, \(\Delta H\), in \({\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\), for stage 1 using average bond enthalpy data from Table 10 of the Data Booklet.
State whether the reaction given in stage 1 is exothermic or endothermic.
Draw the structure of poly(chloroethene) showing two repeating units.
Suggest why monomers are often gases or volatile liquids whereas polymers are solids.
Answer/Explanation
Markscheme
atoms of same element / atoms with same number of protons/atomic number/Z;
Do not allow elements instead of atoms in second alternative.
(but) different numbers of neutrons/mass number/A;
Allow [1 max] for 17 p, 17 e for both if n’s are omitted or incorrect.
Allow [1 max] for 35Cl: 18 n and 37Cl: 20 n if p’s and e’s are omitted.
\(({\text{for}}{{\text{ }}^{{\text{35}}}}{\text{Cl}}:x\% ){\text{ }}35x + 3700 – 37x = 3545\);
Allow other alternative mathematical arrangements.
\(^{{\text{35}}}{\text{Cl}} = 77.5\% \) and \(^{{\text{37}}}{\text{Cl}} = 22.5\% \);
Award [1 max] for correct percentages if no correct working is shown.
ability of atom/nucleus to attract bonding/shared pair of electrons / attraction of nucleus for bonding/shared pair of electrons / OWTTE;
Do not allow element instead of atom/nucleus.
increasing atomic radii (down the group) / OWTTE;
so reduced attraction (for the bonding electrons) / OWTTE;
screening/shielding effect of inner electrons / OWTTE;
Allow more energy levels/electron shells for M1.
Do not accept decrease in nuclear charge.
\({\text{2KBr(aq)}} + {\text{C}}{{\text{l}}_2}{\text{(aq)}} \to {\text{2KCl(aq)}} + {\text{B}}{{\text{r}}_2}{\text{(aq)}}\);
Ignore state symbols.
Allow ionic equation.
colourless/pale yellow/green to yellow/orange/brown;
Start and end colours must both be mentioned.
Bonds breaking:
1 \( \times \) (C=C) \( + \) 4 \( \times \) (C–H) \( + \) 1 \( \times \) (Cl–Cl)
\( = (1)(612) + (4)(413) + (1)(243)/ = ( + )2507{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}{\text{)}}\);
Bonds forming:
1 \( \times \) (C–C) \( + \) 4 \( \times \) (C–H) \( + \) 2 \( \times \) (Cl–Cl)
\( = (1)(347) + (4)(413) + (2)(346)/ = – 2691{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}{\text{)}}\);
Enthalpy change:
\((2507 – 2691 = ){\text{ }} – 184{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}{\text{)}}\);
OR
Bonds breaking:
1 \( \times \) (C=C) \( + \) 1 \( \times \) (Cl–Cl)
\( = (1)(612) + (1)(243)/ = ( + )855{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}{\text{)}}\);
Bonds forming:
1 \( \times \) (C–C) \( + \) 2 \( \times \) (C–Cl)
\( = (1)(347) + (2)(346)/ = – 1039{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}{\text{)}}\);
Enthalpy change:
\((855 – 1039 = ){\text{ }} – 184{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}{\text{)}}\);
Award [3] for correct final answer.
exothermic;
Do not award mark unless based on some value for part (iii).
representation of PVC showing two repeating units;
For example,
Brackets not necessary but continuation bonds must be given.
No penalty if chlorines are not on same side.
No penalty if chlorines are on two middle C atoms or on two end C atoms.
monomers are smaller molecules / monomers have smaller mass / smaller surface area than polymers;
weaker/fewer intermolecular/London/dispersion/van der Waals’ forces (of attraction);
Allow reverse argument.
Allow abbreviation for London/dispersion as FDL or for van der Waals’ as vdW.
Award zero if reference is made to breaking of bonds.
Examiners report
This was by far the most popular choice of question in Section B. Again, part a) (i) proved challenging as many candidates failed to refer to atoms in their definition and scored only 1 mark out of 2.
In a) (ii) most candidates could state the numbers of protons, neutrons and electrons in the isotopes of chlorine. Those who got this wrong gave answers which indicated a complete lack of understanding of atomic structure.
In a) (iii) some candidates remembered the percentage abundance of chlorine isotopes but could not do the calculation.
Part b) (i) required another definition. Again, many candidates lost marks for inarticulate responses.
The explanation in b) (ii) of trends in electronegativity values was reasonably well done, with most candidates scoring at least one mark out of two.
However, writing a balanced equation in b) (iii) was poorly done with many candidates not knowing the formula of KCl, and not knowing what products would be formed. This is clearly on the syllabus in 3.3.1.
Almost no-one knew the colours of aqueous chlorine and aqueous bromine in b) (iv).
In part c) (ii) the calculation of \(\Delta H\) using bond enthalpies was done well. Some candidates failed to use the C=C bond enthalpy value and some did not recall that bond breaking is endothermic and bond formation exothermic.
Nearly everyone scored a mark in c) (iii) as follow-through marks were awarded.
Drawing two repeating units of poly(chloroethene) presented difficulties in c) (iv). Some candidates tried to draw the monomers joined through the chlorine atoms.
In c) (v) most candidates scored at least one out of two for explaining why monomers have a much lower melting point than polymers.