GCE a Level H2 Chemistry Nov 2008 Paper 1 Suggested Solutions
H2 Chemistry Nov 2010 Paper 1 Suggested Solutions
1.B
Recall that across a period, atomic radius decreases. Down a group, atomic radius increases. Combine both trends, the element at lowest left-hand corner of Periodic Table should give largest radius.
2.B
4Na+ 3CO2 2Na2CO3 + C
3.B
4.D
Co-ordinate (special type of covalent), covalent and ionic bonds are relatively very strong bonds compared to vdw. If the lizard’s toe pads are stuck to glass surface by co-ordinate, covalent or ionic bonds, it can forget about leaving!
5.B
PH3 is similar in shape to NH3.
6.C
The oppositely charged ions must be in contact with each other, ie. alternate arrangement.
7.A
The more significant (stronger) the IMF, the gas deviates more from ideal gas behaviour (pV is less than expected). NH3 has hydrogen bonding, much stronger IMF than vdw in ethene, methane and nitrogen.
8.A
Atomisation involves energy absorbed to break bonds (form gaseous atoms).
9.D
Recall: catalyst increases rate constant. Catalyst lowers the EA but does not affect the enthalpy change of reaction.
10.A
When steamcondenses at 100 ºC, the equilibrium H2O (g) ⇌H2O (l) occurs.
Thus ΔG = 0. ΔS = = = -118 J K-1 mol-1
Since 54 g of water involved, ΔS = -118= -354 J K-1
11.B
To act as reducing agent, itself can be oxidised. So we choose between Br- and Fe2+. Fe2+ has higher tendency to be oxidized (=> stronger RA) since Fe3+ has lower tendency to be reduced to Fe2+ (less positive Eº)
12.D
Write balanced overall redox equation.
2 MnO4- + 5SO2 + 2H2O 2Mn2+ + 5SO42- + 4H+
As more and more SO2 is bubbled and reacted with available MnO4-, [H+] increases, pH decreases.
13.C
N2O4(g) ⇌ 2NO2(g)
initial/molesx 0
change/moles-0.5x +x
eqm/moles0.5x x
PN2O4 = PNO2 = 1 -
Kp =
14.B
Mg2+
Al3+(largest)
Fe3+
SiCl4 will hydrolyse to form SiO2, an insoluble solid.
15.C
Recall question.
16.B
Recall question.
17.C
Eº / V
Fe3+ + 3e- Fe-0.44
Fe2+ + 2e- Fe-0.04
Cl2 + 2e- 2Cl-+1.36
Br2 + 2e- 2Br-+1.07
l2 + 2e- 2l-+0.54
Feasibility of forming FeCl2: Eºcell = +1.36 –(-0.04) > 0
Feasibility of forming FeCl3: Eºcell = +1.36 –(-0.44) > 0 (more +ve than above)
Feasibility of forming FeBr2: Eºcell = +1.07 –(-0.04) > 0
Feasibility of forming FeBr3: Eºcell = +1.07 –(-0.44) > 0 (more +ve)
Feasibility of forming Fel2: Eºcell = +0.54 –(-0.04) > 0
18.C
Recall question.
Exhibiting more than one oxidation state is a property of transition metals and its compounds, and not a definition.
19.B
Carbon atom 2: sp hybridized (triple bond), atom 3: sp3 hybridised (all single bonds)
20.C
21.D
Option A: Not readily achieved. Free radical substitution is very random.
Option B: No reaction. Recall C-F bonds are very hard to cleave (inert).
Option C: Recall –CH3 is 2,4-directing.
22.B
Option A: For electrophilic addition of HX, HX must be in gaseous state.
Option B: Correct. From product, it is elimination of HBr, and that is done in alcoholic solvent.
Option C: Wrong state of AlCl3. AlCl3 will hydrolyse in water!
Option D: Cl2 should be in gaseous state.
23.B
Option A: It will give immediate white ppt with AgNO3(aq) due to hydrolysis.
Option B: Correct. Only after nucleophilic substitution is Cl- free.
Option C: Halogenoarenes do not readily undergo nucleophilic substitution.
Option D: Same as option A. (in addition to AgCl, benzoic acid also as ppt.)
24.D
P is C6H5CH2CN. Then it is reduction of –CN to form –CH2NH2.
25.D
For option D, after alkaline hydrolysis, forms CI3CO2- and CH3OH. Clear that both compounds will not give CHI3 with alkaline I2(aq).
26.A
Treating with NaBH4, -CHO reduces to –CH(H)OH, -CO- reduces to –CH(OH)-. Regardless of aldehyde or ketone, 2 hydrogen atoms are incorporated for each group. Each X, Y and Z only has one such group.
Note: NaBH4 does NOT convert C=C to saturated form!
27.B
Recall only phenols and carboxylic acids are weakly acidic, not alcohols.
28.C
3-methylbutyl ethanoate: derived from 3-methylbutan-1-ol and ethanoic acid (or ethanoyl chloride).
29.C
The no. of carbon atoms should remain the same after hydrolysis.
30.D
Recall question.
31.A
We can let x = mass of MgCO3 and y = mass of BaCO3.
1From titration results, we can find total amount of carbonates reacting with HCl. Another equation can be formed and solved simultaneously.
2Again, from volume of gases, we can find total amount of carbonates.
3From here, we can find the amount of BaCO3.
32.B
1Correct.2 N atoms behave as ligands, and Mg2+ is like a TM ion, forming dative covalent bonds between them.
2Correct. See the solid lines denoting single bonds (and hence σ bonds) between N and Mg.
3Wrong. The structure is planar. sp3 hybridisation => tetrahedral
33.A
1This is the process of mixing, and entropy of system will increase.
2Since solvent molecules change from liquid to gaseous state, entropy of system also increases.
3Also a process of mixing. Or another way to look at it: increase in dispersion of the solute in its solution.
34.B
1Correct. As [HBr] increases, rate decreases.
2Correct. The stoichiometry of overall reaction happens to match the respective orders in rate equation.
3Cannot tell from the complex rate equation.
35.C
1Incorrect. Low solubility means it cannot be readily absorbed or dissolved, and so not easy to incorporate into human metabolism.
2Correct. With a considerable long half-life, the level of radioactivity can still do some damage after a number of years.
3Correct. Since they are in same group – can share similar chemical reactions – in this case it can replace calcium compounds in human metabolism.
36.B
1Correct. This is formed on mixing sodium ethanoate and silver nitrate.
2Correct. The fact that silver bromide appears after adding KBr to above.
3Wrong. Precipitation reactions are not redox reactions.
37.C
1Wrong. It should be electrophilic substitution.
2Correct. Nucleophile is OH-.
3Correct. Nucleophile is NH3.
38.A
1Correct. A step involves heterolytic fission of Br2 to form Br-.
2Correct.Removal of Br as Br- during elimination.
3Correct.After substituting for Br, also leaves as Br-.
39.A
1Correct. Due to ketone group.
2Correct. Due to phenolic –OH group.
3Correct. Due to phenolic –OH group.
40.C
1Wrong. No such groups CH3CH(OH)- or CH3CO-.
2Correct. There is the amide group (-NHCO-).
3Correct. Presence of primary and secondary alcohol groups.