DETAILED ANSWERS NCTC 2007

1. Correct Answer C. Since the mole ratio of NO to O2 is 2:1, box IV represents the reactants (ratio of the number of NO molecules to O2 molecules is 6:3 or 2:1). The products are represented by box I.

Choices A and D. The molar ratio of NO to O2 is 1:1.

Choice B. Box II does not contain O2,

Choice E is clearly not correct.

2. Correct Answer E. Empirical formula is the simplest type of formula that shows the relative number of atoms of each element in a compound. In this case, it is CH2O.

Choice A. This choice is two times the correct empirical formula, but students may consider it the empirical formula since it is the simplest formula given.

Choice B. This choice gives the atom number ratio derived from the percentage composition, and it is fundamentally incorrect.

Choice C. This choice is two times the formula for choice A. Students choosing C as the answer have basically the incorrect concept about empirical formula.

Choice D. This is two times the formula for choice B. It is basically incorrect as with choice B.

3. Correct Answer A. Number of moles of Na2S2O3 = (27.54 L/1000)×(0.1153 mol

L-1) = 3.175 × 10-3 mols. Number of moles of I2 that reacted with 3.175 × 10-3 mols of Na2S2O3 = 1.588 × 10-3 mols. Initial number of moles of of I2 = (53.20 L/1000) ×(0.1030 mol L-1) = 5.480 × 10-

3 mols. Therefore, number of moles of I2 that reacted with the vitamin = (5.480-1.588) × 10-3 mols = 3.892 × 10-3 mols. Number of moles of C6H8O6 = 3.892 × 10-3 mols. Molar mass of C6H8O6 = 176 g mol-1

Therefore, mass of ascorbic acid = 3.892 × 10-3 mols × 176 g mol-1 = 0.685 g

Choice B. Students may have assumed that the number of moles of vitamin = number of moles of initial iodine = 5.480 × 10-3 mols. Choice C. Students may have assumed that the no. of moles of moles of I2 that reacts with Na2S2O3 is the same as the no. of moles of I2 that reacts with the

vitamin. The use of the no. of moles of C6H8O6 = 1.588 × 10-3 mols will lead to choice C. Choice D. Students may have assumed that the no. of moles of I2 that reacted with Na2S2O3 is 3.175 × 10-3 mols resulting in the no. of moles of I2 reacting with the vitamin = 2.305× 10-3 mols. Choice E . Error similar to choice D but with no. of moles of C6H8O6 = 2(2.305× 10-3 mols).

4. Correct Answer A. This question requires the student to calculate the grams of

SO2 produced first using molecular weight of S and SO2 and also the mole ratios of the two from the equation for the oxidation reaction. Once this value is calculated the students needs to calculate the grams of sulphuric acid produced using the percentages provided in the question for the amount of SO2 released and for the amount that gets converted to H2SO4.

Choice B. The student has not converted weight of bituminous coal to grams from kilograms before using it for calculating grams of SO2 and using that subsequently to calculate grams of H2SO4. Choice C. The student has not taken into consideration that the only 18.0% of sulfur is produced from 1.00kg of coal. Choice D. Student has wrongly assumed that the amount of SO2 produced is proportional to the amount of H2SO4 produced in the atmosphere. The student has not taken into account the molecular weight of H2SO4 and also the percentage conversion factors that needs to be included in the calculations. Choice E. Student is probably making a wild guess.

5. Correct Answer C. A shell of principal quantum number n, contains subshells with quantum numbers l = 0, 1, 2…n-1. Each subshell with angular quantum number l contains orbitals with magnetic quantum numbers ml = l, …, +1, 0, -1, …, -l. The table below list all the orbitals contained in the n = 4 shell

n l ml

4 0 0

4 1 -1, 0, 1

4 2 -2, -1, 0, 1, 2

4 3 -3, -2, -1, 0, 1, 2, 3

From the table above, the shell n = 4 contains a total of 16 orbitals. Since, according to Pauli’s exclusion principle, each orbital can take up to two electrons, the maximum number of electrons that can occupy the n = 4 shell is 16 × 2 = 32. Choice A. Students may be thinking that the n = 4 shell contains four orbitals that can accommodate up to 8 electrons Choice B. Students may have counted correctly that number of orbitals in the n = 4 shell is 16 but did not take into account that each orbital can accommodate two electrons. Choice D. Students may be making a wild guess. Choice E. Student may have totaled up the number of electrons in all the shells n = 1, 2, 3 and 4.

6. Correct answer D. Boron – IE1 to IE3 represent removal of three electrons from

second level (2s2 and 2p1) at which point atom becomes isoelctronic with He. IE4 and IE5 represent removal of electrons from 1s orbital, which require much greater amounts of energy. Choice A. Beryllium cannot have IE5. The student’s knowledge of electron configuration and/or ionization energy is wrong. Choice B. Oxygen – There should not be a drastic increase in IE from IE3 to IE4 for O since electrons are being removed from the same 2p level. Student’s concept of energies related to electrons occupying the various sublevels and its implication on ionization energy is wrong. Choice C. Silicon is not in the second period. The student is not familiar with the concept of periods and groups. Choice E. Carbon – You do not expect a drastic change in IE values for C when you remove the first and second electrons from a 2s orbital (IE3 to IE4).

7. Correct Answer A. The second ionization energies (IE) of Na and Ne are

4560kJ/mol and 3960kJ/mol, respectively. The second ionization of sodium removes an electron from a filled shell while the second ionization of neon it removes an electron from an incomplete shell. The stability of a filled shell is the main reason for sodium having a greater second IE than neon. Furthermore, sodium has a higher effective nuclear charge than neon. Choice B. This choice will be correct if the question ask for the comparisons for the first IE. Student may be thinking that the trend for the second IE is similar to the first IEs.

Choice C. Sodium and lithium are from the same group. Lithium lies in a lower period than sodium. All IEs of lithium are higher than those of sodium. Choice D. Like choice B, this choice will be correct if the question ask for the comparisons for the first IE. Student may be thinking that the trend for the second IE is similar to the first IEs. Choice E. Student may not have realized that the second ionization for sodium removes a core electron, while it only removes a valence electron from magnesium.

8. Correct Answer A. There are two lone pairs on the double bonded oxygen atom, and one lone pair each on both the nitrogen atoms. The CN triple bond contains two π bonds and the CN and CO double bonds contain one π bond each.

Choice B. Students may have failed to count the lone pairs on either the nitrogen atoms or on the oxygen atom. Choice C. Students may have failed to count the lone pairs on either the nitrogen atoms or on the oxygen atom as well as thinking mistakenly that there is only one π bond in the CN triple bond. Choice D. Students may be thinking mistakenly that there is only one π bond in the CN triple bond. Choice E. Students may not know that a triple bond contains two π bonds.

9. Correct Answer B. This question tests the students on their understanding of

VSEPR theory. According to VSEPR theory, a AXY2 type molecule can have the shapes triangular planar, triangular pyramidal and T-shaped, depending on whether the central atom A’s hybridization is sp2, sp3 or sp3d, respectively. For a T-shaped AXY2 type molecule, it is possible for structural isomers to occur:

AX

YY

AY

XY

On the other hand, triangular planar and triangular pyramidal AXY2 molecules have no isomers. Therefore, the central atom on the molecule in question must be sp3d hybridized, and iodine is the correct answer. Choice A. BHF2 will be trigonal planar in shape. Students may not know that it is not possible for a trigonal planar molecule to have two isomers. Flipping the molecule around gives back the same molecule.

Choice C. NHF2 will be trigonal pyramidal in shape. Students may mistakenly think that NHF2 can have optical isomerism. However, for a AXY2 type molecule having a trigonal pyramidal shape, its mirror image can be superimposed on itself, hence there is no optical isomerism. For trigonal pyramidal molecules, optical isomerism happens for AXYZ type molecules. Choice D. Students may not be well versed in drawing the Lewis dot diagrams for molecules. OHF2 does not exist, since the Lewis dot diagram of OHF2 will violate the octet rule which is impossible for a second period element. Choice E. Similar error as in choice C.

10. Correct Answer A. The P in PCl5 is sp3d hybridized and the I in IO4- is sp3

hybridized. Choice B. Both central atoms are sp3 hybridized. These two molecules are basic examples of molecules with sp3 hybridized central atoms. Choice C. Both central atoms here are sp2 hybridized. In getting this wrong answer, students may not know that boron, with three valence electron conveniently uses its 2s and two of its 2p orbitals to form 3 sp2 hybrid orbitals. Choice D. Both central atoms are sp3d2 hybridized. Students may not be familiar with hybridization involving d orbitals. Choice E. There is no reason to choose this answer, as both N and P are from the same group and hence, in this case, should have similar hybridization.

11. Correct Answer C. BF3 is the only molecule that has a symmetrical geometry in which all the dipole moments cancel each other. Choice A. There are two O-H bonds and two electron lone pairs associated with the central oxygen atom. The resultant dipole moment arising from the two O-H bonds is at opposite direction to that generated from the two lone pairs. However, students need to know that these two opposite dipole moments do not have the same magnitude and therefore cannot cancel each other. As a result, H2O is a polar molecule. Choice B. The same explanations as to why choice B is wrong, apply to Me2S as well. This is because sulfur is just below oxygen in the periodic table. Choice D and Choice E. Similarly, in the cases of PF3 and NH3, each central atom carries one electron lone pair and three substituents. The resultant dipole moment of the three P-F and N-H bonds are of opposite direction to the dipole moment of the electron lone pairs in their central atom, but of different magnitudes.

12. Correct Answer C. ONLY statement III is correct

Students meay not be aware that, unlike dioxygen, two sulfur atoms do not use their 3p orbitals to form a pi-bond. This is due to the inner core repulsion arising from their filled 2p orbitals. However, student should be aware that sulfur does form “double bonds” by using its empty d orbitals to accept electrons from other pi-orbitals to form (dative) pi-bondings. One common example is sulfur dioxide. Indeed SO2 is a gas. Therefore statements I and II are incorrect. Only statement III is correct.

13. Correct Answer C. The diatomic molecules Cl2, Br2 and I2 are not polar owing to their symmetry. Since there are no covalent, ionic or hydrogen bonding, the dominant interaction is van der Waals force, operative between any pair of molecules. Choice A. Student may not know that covalent bonds exist between the two halogen atoms but not inter-molecularly. Choice B. Students may not know that the molecules are not ionic. Choice D. Molecules do not contain H’s, so one needs to be very confused about the concepts to choose this answer. Choice E. Students are making a random guess

14. Correct Answer A. Statement I is correct. Raising the temperature while maintaining constant volume causes the N2 molecules to move at higher average speed and higher energy. They would therefore hit the walls of the container harder and more frequently, causing a higher pressure. Statement II is wrong. With more molecules and at the same temperature, the molecules in the new sample still have the same average energy. However, you would still have more molecules per unit time hitting the walls, and pressure would be higher. Statement III is wrong. Temperature is constant, so average energy remains constant. If volume is decreased, the molecules must hit the walls more frequently. Therefore, pressure should increase. Choices B and D. Students wrongly assume that decreasing the volume will reduce the number of collisions will arrive at these wrong choices. Choices C and E. Statement II is clearly not correct. Students choosing choices C and E may not completely understand the kinetic-molecular theory.

15. Correct Answer D. Students need to know how to apply the ideal gas law.

Initial when bubble is at the bottom of the lake:P1 = 6.3 atm, V1 = 2.1 mL T1 = 8oC = 281 K Final when bubble is at water’s surface: P2 = 1.0 atm, V2 = ? mL T2 = 25oC = 298 K Assume that the amount of air in the bubble remains constant (i.e., n1 = n2), then, (P1V1)/T1 = (P2V2)/T2V2 = V1 × (P1/P2) × (T2/T1) = 2.1 mL × (6.3 atm/1.0 atm) × (298 K/281 K) = 14 mL Choice A. Students may have used oC as the units for temperature instead of K. Thus getting the wrong answer: V2 = V1 × (P1/P2) × (T2/T1) = 2.1 mL × (6.3 atm/1.0 atm) × (25oC/8oC) = 41.3 mL. Choice B. Students may have used a wrong formula, V2 = V1 × (P2/P1) × (T2/T1) = 2.1 mL × (1.0 atm/6.3 atm) × (298 K/281 K) = 0.35 mL Choice C. Students may have wrongly assumed that the volume is twice the initial volume, 2(2.1) mL = 4.2 mL. Choice E. Students assume that the final volume is the same as the initial volume, 2.1 mL.

16. Correct Answer B. Students need to know how to apply the ideal gas law and the

concept of molarity. P = 1.104×105 Pa, V = 45.5 L = 45.5 ×10-3 m3 and T = 22.0 oC = 295 K Number of moles of N2 required = PV/RT = (1.104×105 Pa)( 45.5 ×10-3 m3) /(8.314JK-1mol-1)(295K) = 2.05 mols Mass of NaN3 = (2/3) (2.05mol) (65 gmol-1) = 88.8 g Choices A and E. Students wrongly used the centigrade scale instead of the absolute temperature scale in their calculations. Choice C. Student did not take into account the stoichiometry of the reaction and hence calculated the mass of NaN3 = (2.05mol) (65 gmol-1) = 133.3 g Choice D. Student made a mistake with the stoichiometry of the reaction and calculated the mass of NaN3 = (3/2) (2.05mol) (65 gmol-1) = 199.9 g

17. Correct Answer D. The question is about deviations from the ideal gas law. An ideal gas assumes that the molecules have zero volume and that there is no intermolecular interaction. In this question, the molecules are assumes to have finite volume. The ‘free’ volume that the molecules are in is therefore lesser, and hence the effective pressure is higher compared to the ideal gas pressure. The repulsive nature of the intermolecular also adds to a higher pressure than the ideal gas pressure. However, at low enough pressure, the real gas should bear the ideal gas pressure. The ideal gas pressure at the three volumes, according to equation PV = nRT are 1.16×107Pa, 1.16×105Pa, and 1.16×103Pa, respectively. The real gas pressure should be proportionately higher for the smaller volume, and tend towards ideality at the largest volume. Choice A. Students may be thinking that the repulsive interaction contributes to lowering the pressure, resulting in the wrong conclusion that at some intermediate pressure, when the interaction dominates, the pressure may have dipped below that of the ideal gas pressure. Choice B. Students are not aware that there should be a trend towards ideal gas pressure, with decreasing pressure. Choice C. Students may not be aware that in these conditions, the real gas pressures must all be greater than the ideal gas pressure Choice E. Students may be thinking that the finite molecular volume and repulsive interaction are lowering instead of increasing the effective pressure.

18. Correct Answer D.

The equation is: 4NH3 (g) + 5O2 (g) → 4NO (g) + 6H2O (g). ΔHo = ∑ n ΔHo

f(products) - ∑ m ΔHof(reactants)

= [4 ΔHof(NO) + 6 ΔHo

f(H2O)] – [4ΔHof(NH3) + 5ΔHo

f(O2)] = [4(90.3) + 6(-241.8)] kJ – [4(-45.9) +5(0)] kJ = -906 kJ Choice A. Students may be confused with the signs of the standard enthalpies and simply used the absolute values of the standard enthalpy values [4(90.3) + 6(241.8)] kJ – [4(45.9) +5(0)] kJ = 1628.4 kJ Choice B. Students only considered the enthalpies of formation for the products. Using 4(90.3) + 6(241.8) kJ = 332.1 kJ Choice C. Students neglected to consider the stoichiometry of the reaction. i.e. they ignored n and m, and calculated: [(90.3) + (-241.8)] kJ – [(-45.9) +(0)] kJ = -105.6 kJ Choice E. Student may be making a random guess

19. Correct Answer B. Entropy is a measure of disorder.

Process I results in an increase in entropy. A molecule (glucose) breaks into smaller molecules (C2H5OH and CO2). Moreover, this results in a gas being released. Process II results in a decrease in entropy. In this reaction, the moles of gas decrease (by 3 mol), which would decrease the entropy. Process III results in an increase in entropy. In this reaction, 2 moles of gas are obtained from 1 mole of gas.

20. Correct Answer D. For HCl, the density is 1.02 g mL-1, and its mass = 1.02 g mL-1 × 50.0 mL = 51.0 g. For NaOH, the density is 1.04 g mL-1, and its mass = 1.04 g mL-1 × 50.0 mL = 52.0 g. Mass of the final solution is thus the sum, 103.0 g. The reaction changes the system’s temperature by ΔT = (32.2 – 25.5) oC = 6.7 oC Heat produced by the reaction = mass × specific heat × ΔT =103.0 g × 4.18 J g-1 oC-1 × 6.7 oC = 2.9 × 103 J. No. of moles of HCl = 0.0500 mol. Therefore, energy evolved per mole of HCl = (2.9 × 103 J)/(0.0500 mol HCl) = 58 kJ mol-1. Thus the enthalpy change per mole of acid is -58 kJ mol-1 (negative because reaction is exothermic). Choices A and B. Students may not have considered the number of moles of acid. Choice C. Students made a mistake with the sign. Choice E. Students calculated the energy evolved per mole of HCl by mistakenly using (2.9 × 103 J)×(0.0500 mol HCl) = 0.145 kJ mol-1.

21. Correct Answer E. The second reaction establishes that Ag+ is a more effective electron acceptor than H+; the first reaction demonstrates that H+ is more effective than Cd2+. Finally, the third equation shows that Cd2+ is more effective than Zn2+. Thus, the order of oxidizing strength is Ag+ > H+ > Cd2+ > Zn2+. Students who chose the other choices may not be clear about the differences between oxidizing and reducing agents or may not be able to read redox reactions.

22. Correct Answer C. No. of moles of L-dopa = (It)/(nF) = Q/(nF) = (42.0 × 10-6 C)/(96485×2 C mol-1) = 2.176 × 10-10 mol Therefore, the mass of L-dopa = (2.176 × 10-10 mol) (197.2 g mol-1) = 4.29 × 10-8 g = 42.9 ng Students must know how to convert units: 1 g = 103 mg = 106 μg = 109 ng Choice A. Student may make this wrong choice if 1 electron per molecule is considered instead of 2 electrons.

Choice B. Student may make this wrong choice if 4 electrons per molecule are considered instead of 2 electrons. Choice D. Similar case as Choice A. Furthermore, student got the unit conversion wrong. Choice E. Student got the unit conversion wrong.

23. Correct Answer B. The standard electrode potential for Ag+ + e- → Ag is Eo = +0.80 V, while the standard potential for the electrode potential for Cl2 + 2e- → 2Cl- is Eo = +1.36 V. The standard cell potential for the overall reaction (i) 2Ag (s) + Cl2 (g) → 2Ag+ (aq) + 2 Cl- (aq) is Eo

cell = (1.36-0.80)V = 0.56 V, and (ii) 2Ag+ (aq) + 2Cl- (aq) → 2Ag (s) + Cl2 (g) is Eo

cell = (0.80-1.36)V = -0.54 V. Statement I and II are correct since the overall reaction (i) is a spontaneous reaction while reaction (ii) is not, the reduction half-reaction is given by Cl2 (g) + 2e- → 2Cl- (aq), and the standard cell potential is 0.56 V. Statement III is wrong. The electrons in the external circuit flow from the Ag electrode to the positive (site of Cl2 reduction) electrode. Therefore, the negative ions move in the salt bridge from the Cl2/Cl- half-cell to the Ag/Ag+ half-cell.

24. Correct Answer C. This question tests the student’s knowledge of acid and base

equilibria. This is a typical strong acid weak base titration. We need to obtain [H+] from the acid equilibria equation

]NH[]NH][H[

106.54

310+

+− =×=aK

NH4+ can be viewed as a weak acid. At equivalence point, the number of moles of

NH4+ can be estimated as the number of moles of HCl added.

mol10365.3mol/L005.0L0673.0 4NH4

−×=×=+n

Taking into account of both the volumes of the titrant and the analyte, the concentration [NH4

+] is

M34

4 1065.4L)005.00673.0(

mol10365.3]NH[ −−

+ ×=+×

=

Stoichiometrically, NH4+ dissociates to give equimolar of H+ and NH3. i.e. [H+] =

[NH3] Solving for the Ka equation yields the answer. Choice A. Student may realize that at the equivalence point between a strong acid and weak acid titration, the resultant pH should be acidic. However, student may have difficulty solving the equation and is just making a guess. Choice B. Student may have forgot to take into account both volumes of titrant and the analyte while doing the calculations.

Choice D. Student may not understand the acid base equilibria and mistakenly assume that all acid and base titration end up with pH 7 at equivalence point. Choice E. Student does not understand concept, and is just taking the negative of the logarithm of the Ka constant.

25. Correct Answer B. The concept of solution equilibria and common ion effect is tested in this question. To obtain the answer, students first need to calculate the Ksp of PbCl2. To do so, students need to convert 4.51g PbCl2 to the molar quantity.

1--1

Lmol0162.0g/mol)5.3522.207(

gL51.4⋅=

×+=M

The Ksp can then be calculated by recognizing that the concentrations of the Pb2+ and Cl- ions are [Pb2+] = 0.0162mol/L and [Cl-] = 2 × 0.0162 = 0.0324mol/L, respectively, The resultant Ksp is [Pb2+][Cl-]2 = 0.0162×(0.0324)2 = 1.7×10-5. To calculate how much PbCl2 can dissolve in 0.1M Cl- solution, one needs to solve the equation

25

22

)21.0(107.1

][Cl][Pb

xx

K sp

+=×

=−

−+

where x is the concentration of PbCl2. The answers provided in the multiple choices can be converted to moles, and only choice B can solve the above equation. Choice A. Students did not take into account of the stoichoimetry of the salt. i.e. they may have used [Pb2+] = [Cl-] = 0.0162mol/L and arrived at the wrong Ksp value. Choice C. Students may have wrongly assume that the amount of Cl- contributed by PbCl2 while dissolving in 0.1M NaCl solution is too little and can be ignored. i.e. students tried to solve the equation

25

22

)1.0(107.1

][Cl][Pb

x

K sp

=×

=−

−+

Choice D. Student may be randomly guessing Choice E. Students may be using the incorrect expression for the solubility constant in this case

][Cl][Pb2 −+=spK

26. Correct Answer D. Using the ideal gas law, students can deduce the total number of moles of CO and CO2 gases in the 3L container. From that, student can then deduce the equilibrium constant. The equilibrium constant can then be used to calculate the number of moles of CO and CO2 in the 6L container, and ultimately the final pressure.

Choice A. Student may be just using the original amount of CO2 to calculate the final pressure in the 6L container. Choice B. Student may be making a wild guess. Choice C. Student did not take into account equilibrium and mistakenly assumes that the pressure in the 6L container must be half of that of the 3L container. Choice E. Student mistakenly used the pressure in the 3L container 3.66×105 Pa as the equilibrium constant.

27. Correct Answer D. The question tests the student’s concept on reaction rate law

and the application of the ideal gas law. The rate law for this reaction is ]][[ BAkrate =

For an ideal gas, doubling the pressure at constant temperature leads to a doubling of concentration (moles per unit volume) for both A and B. Therefore, according to the rate law, the rate should increase by four times. Choice A. The student may have mistakenly thought that increasing pressure results in a lowering of concentration Choice B. The student may not recognize from the ideal gas law, the connection between increase in pressure at constant temperature and increase in concentration. Choice C. The student may have gotten a wrong expression for the rate law Choice E. The student may have used the chemical reaction equation to derive the wrong rate law as

][][ 2 BAkrate = Rate law in general does not follow the chemical reaction equation.

28. Correct Answer C. Since it is a first order reaction with respect to X, the amount

of X decreases according to an exponential form. The time taken for X to decrease to one eighth its original amount divided by the time taken for X to decrease to

one third its original amount, is3ln8ln .

Choice A. Student may have mistakenly assumed that it takes 8/3 times the time needed for X to deplete to 1/8 its original amount as compared to 1/3 its original amount. Choice B. Student may be mistakenly calculating the half life of the reaction to be

14432× sec = 96 sec. The student then arrives at the wrong answer 3×96 sec = 288

sec as the answer by reasoning that, to reach one eighth the amount, you need three half lives. Choice D. Student may be making a wild guess. Choice E. Student is assuming that the amount of X remaining is a linear function of time. This only happens if the reaction is zero order in X.

29. Correct Answer D. This question test the students understanding and application

of the Arrhenius equation )exp( RTEAk a−=

The information provided should lead student to formulate two simultaneous equations

)400exp(2)400exp()300exp()300exp(

2211

2211

REAREAREAREA

aa

aa

−=−−=−

Solving the simultaneous equation will lead to the answer Choice A. Students may not be familiar with the Arrhenius equation and attempt to solve the problem by applying Arrhenius equation without the preexponential term A on the relative rates of the two reactions at 400K.

)400exp(2)400exp( 21 RERE aa −=− Choice B. Similar error as choice A, compounded with a careless mistake in positive/negative sign Choice C. Students may have made a careless mistake. However, students can spot this kind of careless mistake by considering the following fact: In comparing two reactions, upon increasing the temperature, the reaction rate of the one with higher activation energy will be increased more than the one with the lower activation energy. In this case rate of reaction 1 increase more than rate of reaction 2 when the temperature is raised from 300K to 400K, hence we can conclude that (Ea1-Ea2) is positive. Choice E. Student may be making a wild guess.

30. Correct Answer A. Al2O3 react with both acid and base.

Al2O3 + 6H+ →2 Al3+ + 3 H2O Al2O3 + 2 OH- → 2AlO2- + H2O

Choice B. Li2O only reacts with acid: Li2O + 2 H+ → 2 Li+ +H2O Choice C: SO3 only reacts with base: SO3 + 2 OH- → (SO4)2- + H2O Choice D: CO2 only reacts with base: CO2 + 2 OH- → (CO3)2- + H2O

Choice E: This choice is wrong because choice A is correct.

31. Correct Answer B. Heating of water cause CO2 solubility to decrease, and CaCO3 to precipitate as limescale. Choice A. This reaction is the reverse reaction of B. Student may not know which compound (Ca(HCO3)2 or CaCO3) is more soluble. Choice C. Student may choose this if they are not familiar with solubility of compounds and think table salt is related to this process. Choice D. This reaction is the industrial preparation of CaO and is not involved in the observed phenomena. Choice E. Addition of CaO can be used to remove hardness of water, but that is not the process occurring when heating water.

32. Correct Answer E. BaSO4 is most insoluble. The insolubility of BaSO4 is useful in the gravimetric analysis of sulfate: one can add a solution of barium chloride to a solution containing sulfate ions. The appearance of a white precipitate, which is barium sulfate, indicates that sulfate anions are present.

Choice A and B. The student is expected to know that the both MgSO4 and CaSO4 are soluble since these compounds contribute to the permanent hardness in water. Also, gypsum is slightly soluble and it accounts partially for the permanent hardness in water.

Choice C. CaCO3 is insoluble but its Ksp cannot be very low since CO2 can dissolve limestone. The student may not be aware of the concept of solubility product and its implications. Choice D. Ba(NO3)2 is very soluble. Student may not be aware that nitrates are very soluble.

33. Correct Answer C. This question involves the knowledge of the reactivity series

for metals, where Ca is more reactive than Mg. However both of these two metals are more reactive than H2. This means that even if elemental Mg were to be formed in aqueous solution, it would have reacted with water to give H2 gas. Therefore, elemental Mg will not form.

Choice A and B. The student may not know that alkaline metals can react with water to give H2. This reaction also produces Ca(OH)2, which is insoluble and therefore precipitates. Choice D. Ca(OH)2 reacts with Mg(HCO3) to give MgCO3 and CaCO3 as precipitates.

Choice E. Students may not be aware that the reaction of Ca with water is exothermic, so the solution will indeed get hot.

34. Correct Answer E. The student needs to know the reactivity series whereby F2 > Cl2 > Br2. Since Br2 is less reactive than Cl2, displacement reaction does not occur. Choice A. Students may not be aware that this is the laboratory method for production of Cl2, MnO2 can oxidize concentrated HCl to Cl2 gas.

Choice B. Students may not be aware that this is the method for the industrial production of Cl2, where Cl- ions are oxidized at the electrode to Cl2 gas

Choice C. KMnO4 is a stronger oxidant than MnO2. Hence, like in the reaction described in choice A, KMnO4 can oxidize concentrated HCl to Cl2 gas. Choice D. Answer D is the reverse of the equilibrium reaction in the question. This reverse reaction, of course, produces Cl2.

35. Correct Answer C. The white precipitate is AgCl, it dissolves upon addition of

NH3. This is because the coordination of NH3 to Ag produces the species [Ag(NH3)2]+ which is soluble and drives the solubility equilibrium to the dissolution of AgCl. Choice A. There is no H+ or OH- involved in these reactions. Therefore answer A is not correct. Choice B. The solubility of AgCl does not change with NH3. Student may not be aware that solubility constant cannot change (if temperature is held constant). Choice D. Student did not account for the possibility of NH3 undergoing coordination to Ag.

Choice E. Student may be making a random guess

36. Correct Answer A. Transition metals with d8 configuration can be either square planar or tetrahedral. Tetrahedral structures are favored sterically with bond angles being 109.5 degree. Square planar structures are more hindered with four ligands and the metal in the same plane and the steric interaction is bigger with 90 degree bond angles. However, a square planar structure has larger crystal field stabilization energy and this structure is electronically favored. The structure of a d8 metal is controlled by these two factors.

Choice B. Students may not know that ligands are more crowded in square planar structures.

Choice C. Students may not know that the structure is controlled by both electronic and steric effects.

Choice D. Students may not know the structure is controlled by both electronic and steric effects.

Choice E. Students may not know that the preferred structure has the lowest energy.

37. Correct Answer B. BF3 is trigonal planar with the boron atom being sp2 hybridized. The angle is 120 deg. Although NH3, H2O and CH4 are all sp3 hybridized, VSERP tells us that more lone pairs will give rise to a smaller bond angle due to the lone pair-lone pair repulsion. H2O has two lone pairs so the angle is the smallest.

Choice A. Students don’t know that the presence of lone pairs will reduce the bond angles.

Choice C. Students don’t know that BF3 has sp2 hybridization and the angel is 120 degree.

Choice D. Students don’t know that the more lone pairs the smaller the bond angles.

Choice E. Students don’t know that the presence of lone pairs will reduce the bond angles.

38. Correct Answer C. Neutral Cu atom has 3d104s1 configuration due to the Hund's rule and due to the fact that orbitals with filled or half filled electrons are most stable. In CuI, the Cu is in the oxidation state of +1. One electron is therefore lost form the neutral Cu atom. The 4s electron is lost to give the 3d104s0 stable configuration.

Choice A. Student does not know the oxidation state of copper in CuI.

Choice B. Student did not find the correct oxidation state of the copper.

Choice D. Student does not know the electron configuration of Cu(0).

Choice E. Student does not know that the electron in 4s orbital is removed first.

39. Correct Answer C. Pt is famous for the anti cancer reagent cis-PtCl2(NH3)2, the cis-platin. It was accidentally discovered in electrolysis.

Choice A. Some Sn complexes show mild antitumour activity, but none of them are famous as antitumour drugs

Choice B. Some Ni complexes show mild antitumour activity, but none of them are famous as antitumour drugs

Choice D. Some Ir complexes show mild antitumour activity but none of them are famous as antitumour drugs

Choice E. Some F containing complexes do exhibit mild antitumour activity, but none of them are famous as antitumour drugs

40. Correct Answer C. The three isomers are 1-butene, Z-2-butene and E-2-butene

Choice A. The student may not understand isomerism. Choice B. The student may have disregarded the possibility of geometrical isomers Choice D. The student may be thinking that the 1-butene has two isomers Choice E. Was the student thinking about enantiomers? – but these molecules are planar.

41. Correct Answer A. This question test the students application of nomenclature

and the application of CIP (Cahn-Ingold-Prelog) rules R is correct; the lone pair of electrons takes the lowest priority Choice B. Incorrect application of the CIP rules Choice C. The student may not realize that the sulfur atom is tetrahedral Choice D. Students may not realize that the CIP system (R/S) can be applied to any atom, not just carbon. Choice E. The DL descriptions are usually only used for sugars and amino acids

42. Correct Answer D. The molecules are enantiomers and should therefore exhibit

same physical and chemical properties. The only difference is their behavior towards plane polarized light where the light is rotated in equal but opposite directions by the isomers. Choices A, B, C and E are incorrect because they are regioisomers. Choice A. The compounds differ in that one is a straight-chain alcohol whereas the other one is a branched-chain alcohol.

Choice B. Similarly, one compound is a straight-chain hydrocarbon whereas the other is a branched-chain hydrocarbon. In both choices A and B, the differences in packing in the solid state (straight-chain molecules fit together and therefore pack better) and the associated stereoelectronic effects, is sufficient to effect differences in their physical properties, such as boiling point. Choices C and E. A similar rationale can be applied, the regioisomers of xylene in choice C and cis and trans geometrical isomers in choice E.

43. Correct Answer D. From the formula, students should deduce that the compound

has one degree of saturation, and therefore conclude that the compound has either a double bond or has all single bonds but exists as a ring structure. Since the compound does not decolourise bromine (i.e. no bromination reaction), it cannot contain double bonds. That leaves these possible ring compounds: three dimethyl cyclopropanes, one ethyl cyclopropane, methyl cyclobutane and cyclopentane Choice A. The student is probably only thinking of cyclopentane and is not aware of the other possible isomers with a ring structure Choice B. The student may be considering cyclopropanes, cyclobutanes and cyclopentanes, but not considering all possible isomers Choice C. The student has underestimated the number of isomers Choice E. the student may have included pentenes, but these would decolourise bromine water

44. Correct Answer D. This compound does not have any highly polarized bonds or

available lone pair of electrons. Choice A to C. These are compounds are nucleophiles because they have lone pairs of electrons available. Choice E. This molecule has a highly polar C-Li bond.

45. Correct Answer A. Although unsaturated, the benzene ring is resonance

stabilized and hence does not readily undergo electrophilic addition. If compound X underwent electrophilic addition, the resulting product would be a non-aromatic compound, thus resonance stabilization provided by aromatization would be lost, making the reaction energetically unfavorable.

Choice B. X can undergo electrophilic substitution at the benzene ring. Electrophilic substitution can proceed as re-aromatization can occur as the H atom of the intermediate can be eliminated to regenerate the benzene ring. In fact, the re-aromatization drives the reaction as product stability is provided by resonance in the product.

Choice C. X can undergo free radical substitution at –CH2–. This reaction is possible because formation of a radical is favored at the benzylic –CH2– centre since the resulting radical species can be resonance stabilized around the benzene ring. Choice D. X can undergo nucleophilic substitution at –CH2Cl. Primary benzyl halides can undergo SN2 substitution reactions. Choice E. X can undergo reduction at –NO2. The concept here is that chemoselective reduction of –NO2 in the presence of a reducing reagent can be achieved.

46. Correct Answer C. Elimination reaction occurs because NaNH2 is a strong base

Choice A. Nucleophilic substitution of vinyl halides is not possible Choice B. Nucleophilic addition to an unactivated alkene is unlikely. Choice D and E. These are reduction reactions, but NaNH2 is not a reducing agent

47. Correct Answer A. The relatively weak C-Cl bond allows Cl to be displaced.

Choice B. Students may not be aware that strength of the C-Br bond and the incorrect geometry for substitution resist the CN- nucleophile Choice C. The strong C-F bond is resistant to CN- nucleophiles Choice D. Student might be only thinking that Cl and Br are good leaving groups Choice E. A wild guess by the student?

48. Correct Answer E. The lone pair of electrons on the N atom delocalizes into the

benzene ring and this renders the benzene ring highly activated towards electrophilic substitution at the ortho and para positions. The two ortho positions are substituted by methyl groups thus bromination can only occur at the para position of the benzene ring.

Choice A. This choice is incorrect. The student has probably made a wild guess, 2,6-dimethylphenylamine is not capable of acting as an oxidant in this case. Choice B. The lone pair of electrons on the N atom delocalizes into the benzene ring and this renders the benzene ring highly activated toward electrophilic substitution at the ortho and para positions. Also electrophilic substitution at this position is not favorable since NH2 is not a good leaving group.

Choice C. The two ortho positions are substituted by methyl groups thus no H atoms are present and bromination can only occur at the para position of the benzene ring. Even so, cleavage of a C-C bond under the stated conditions is not energetically favorable. Choice D. Students may be making a wild guess. These compounds do not form complexes.

49. Correct Answer B. This gives the lowest yield as the use of this to synthesize an

amine from ammonia will lead to a mixture of secondary amines to quaternary ammonium salts.

Choice A. This is the best starting material as only a single reduction step to amine I is needed Choice C. Cyanide addition followed by reduction will give the product Choice D. Requires chlorination, cyanide addition then reduction. This will give the product cleanly. Choice E. Cyanide addition followed by reduction will give the product

50. Correct Answer D. There is always competition between substitution and

elimination for a secondary alkyl halide.

Choice A and B. Students mistakenly think that only elimination products result. Choice C. Students mistakenly think that only substitution products result. Choice E. Students may not be aware of this reaction?

51. Correct Answer A. The reaction requires generating an alkoxide ion for the

nucleophilic displacement of Br

Choice B. The ethoxide ion would encounter a crowded alkyl halide and would prefer to carry out elimination by first extracting a H. Choice C. Na2CO3 is too weak a base to generate the alkoxide ion Choice D. Na2CO3 is too weak a base to generate the alkoxide ion Choice E. Students may be making a wild guess.

52. Correct Answer E. Two factors influence the degree of acidity of a molecule.

These are the strength of the bond being broken and the second is the stability of the ions being formed. Thus, CH3CH3 is considered not acidic as the energy

required to break the C-H bond is too high and the ion formed is not stabilized. CH3CH2OH is so weakly acidic that can hardly be counted as acidic at all. If the hydrogen-oxygen bond breaks to release a hydrogen cation, an intensely negatively charged ethoxide ion would be formed, which will be highly attractive towards hydrogen ions and re-protonation would instantly occur. In CH3CO2

- ion, the delocalized system is distorted towards the two oxygen atoms, therefore essentially sharing the negative charge between them. Since each oxygen atom only has about half the charge it would have if there wasn't any delocalization, neither oxygen is going to be as attractive towards hydrogen ions as it would otherwise be. That means that the ethanoate ion won't take up a hydrogen ion as easily as it might otherwise. If it stays ionized, the formation of the hydrogen ions means that it is acidic. In the C6H5O- ion, the single oxygen atom is still the most electronegative atom present, and the delocalized system will be heavily distorted towards it. That still leaves the oxygen atom with most of its negative charge. What delocalization there is makes the phenoxide ion more stable than it would otherwise be, and so phenol is acidic to an extent. However, the delocalization hasn't shared the charge around very effectively. There is still a lot of negative charge around the oxygen to which hydrogen ions will be attracted - and so the phenol will readily re-form. Phenol is therefore only very weakly acidic and the correct answer is thus answer E.

Choice A. CH3CO2H is more acidic than C6H5OH.

Choice B. The acidity of one molecule of C6H5OH is the same as another molecule of C6H5OH.

Choice C. CH3CH3 is considered not acidic at all.

Choice D. CH3CH2OH is so weakly acidic that it can hardly be counted as acidic, and in any case CH3CO2H and C6H5OH are more acidic.

53. Correct Answer D. The phenol would be deprotonated by NaOH(aq) and

extracted out. The alcohol can undergo dehydration to give C1=C2 or C2=C3 double bonds. C2=C3 alkenes has 2 geometric isomers.

The other answers are probably just wild guesses.

54. Correct Answer B. The transformation is a rearrangement of the molecule

because the product is an isomer of the starting material.

Choices A and E. It is obvious that no atoms are added or eliminated from the reaction. Choices C and D. Aromatization is an oxidative process since the number of hybridized sp2 increases carbons and no sp3 carbons remain on the ring structure.

The carbonyl carbon functionality, on the other hand, is reduced to its alcohol oxidation state, thus overall there is neither oxidation nor reduction.

55. Correct Answer A. Students should recognize that this is an oxidation reaction.

CrO3, H2SO4, also known as Jones Reagent, is an oxidant and would readily convert alcohols to their corresponding carboxylic acids. Choice B. KOH is a base and not able to oxidize an alcohol Choice C. Ozone will cleave the double bond. Choice D. HBr will hydrobrominate the two carbon-carbon double bonds. Choice E. NaSO4 is just a drying agent.

56. Correct Answer A. The reactivity follows the δ+ character of the carbonyl

carbon in the carboxylic acid derivatives. The δ+ character of the carbonyl carbon would be expected to be most positive in acyl chlorides due to the electron-withdrawing chloride atom. A comparison of the electron-withdrawing capabilities of a chloride atom versus that of all the other compounds would tell us that the chloride atom is the only one that is electron-withdrawing. Choice B. The nitrogen in carboxylic amides is electron-donating which would decrease the δ+ character of the carbonyl carbon. Choice C. The oxygen of the ester in carboxylic ester is electron-donating which would decrease the δ+ character of the carbonyl carbon. Choice D. The connecting ethoxy group of the ester in carboxylic ester is electron-donating which would decrease the δ+ character of the carbonyl carbon. Choice E is incorrect; the connecting oxygen of the ester in acyl anhydride is electron-donating (this is the mesomeric effect) which would decrease the δ+ character of the carbonyl carbon. However for ethoxy group of the ester in carboxylic ester is more electron-donating since there is also the inductive effect of the alkyl group (ie mesomeric effect + inductive effect)

57. Correct Answer A. The recognition here is that LiAlH4 can reduce carboxylic

acids and aldehydes but since aldehydes are much more reactive than carboxylic acids, the reduction of aldehydes should be much quicker. In fact, it should be recognized that simple deprotection of the carboxylic acid OH moiety is the reason for the significant difference in kinetic rates.

Choice B. Students may not know that aldehydes are much more reactive than carboxylic acids. An aldehyde is more reactive since they the OH group reduces the δ+ character of the carbonyl carbon more than a H atom.

Choice C. Students may not be aware that aldehydes are more reactive than carboxylic acids for such reactions. Choice D. Students may not be aware that LiAlH4 can reduce both carboxylic acids and aldehydes.

Choice E. Students may not be aware that LiAlH4 can reduce both carboxylic acids and aldehydes

58. Correct Answer E. A methyl group will be ortho/para directing. The meta isomer cannot be produced in this way.

Choices A and B. These might be effective on more activated substrates. Choice C. This is the worst choice because it is not a source of electrophilic nitrogen. Choice D would give a mixture of ortho and para nitration since the methyl group will be ortho and para directing.

59. Correct Answer A. the overall concept is acid-base interactions. The NH2 group

in serine is the only basic group present, thus the only group that is expected to be protonated with hydrochloric acid, the proton source.

Choice B. The carboxylic acid group is an acid and would not be deprotonated by hydrochloric acid.

Choice C. The NH2 group is more likely to be protonated with hydrochloric acid than deprotonated.

Choice D. The carboxylic acid group is an acid and would be not protonated by hydrochloric acid, as this would form an unstable acetal.

Choice E. formation of the zwitterion for serine might be expected at a pH close to 7. The presence of hydrochloric acid means that the pH would be less than pH 7. Thus formation of the zwitterion would be unlikely.

60. Correct Answer A. The carboxylic acid group and the amino group in alanine

approximately compensate for each other so that a solution of alanine is nearly neutral.

Choice B. This statement is true because when ordinary laboratory methods are used to make alanine you are as likely to make one isomer as the other.

Choice C and D are correct statements and are covered by the experiments on amino acids. Choice E. In a neutral environment, the amino acid is thought to be deprotonated at the carboxylic group and protonated at the amino group thus giving no overall charge, this form is known as a zwitterion.