nctc_08_solutions.pdf

UNCTC (2008) DETAILED SOLUTIONS Q1 CORRECT ANSWER: (D) If the percent abundance of lithium-6 is assumed to be x, then: 6.9409 = 6.015121 x + 7.016003 (1 x)

Solving the above equation gives x = 7.503%

(A) is incorrect; it is the percent abundance of lithium-7. (B) is incorrect; it is the obtained incorrectly from 6.015121/6.9409. (C) is incorrect; it is obtained incorrectly from 6.015121/(6.015121 + 7.016003). (E) is obviously not correct.

Q2 CORRECT ANSWER: (D)

The amount of carbon from 1.471 g of CO2 is 1.471 x (12/44) = 0.4012 g. The amount of oxygen from 0.226 g of H2O is 0.226 x (2/18) = 0.02511 g. The amount of oxygen in the terephthalic acid sample is 0.6943 (0.4012 + 0.02511) = 0.268 g. Dividing the above amounts of carbon, hydrogen, and oxygen by the respective atomic masses of C, H, and O give then the ratio of the number of atoms of the three elements as: 4:3:2. With the information on the molar mass of the compound, the molecular formula of terephthalic acid is confirmed to be C8H6O4.

(A) is incorrect; the number of oxygen atoms in the compound is obtained here from the difference between the expected molar mass of the compound and the total mass of C4H3.

(B) is incorrect; it is probably just a guess. (C) is incorrect; it is the empirical formula. (E) is obviously not correct.

Q3 CORRECT ANSWER: (E)

From equation (1), 3 x 0.002 mol of I2 would be generated from 0.002 mol of IO3. From equation (2), 3 x 0.002 x 2 mol of S2O32 would be needed to react with 3 x 0.002 mol of I2 produced from reaction (1). So, the amount of S2O32 needed to react with 3 x 0.002 mol of I2 is 3 x 0.002 x 2 = 0.012 mol.

(A) is incorrect; it was obtained based on equation (2) using 0.002 as the number of mol of I2, i.e., 0.002 x 2 = 0.004 mol.

(B) (D) are incorrect; students are probably guessing.

Q4 CORRECT ANSWER: (B)

The number of moles of CuCl2 available is 0.038 x 0.5 = 0.019. The number of moles of (NH4)2S available is 0.042 x 0.6 = 0.025. The amount of CuCl2 is limiting, and so the amount of CuS that can be formed is 0.019 x (63.5 + 32.1) = 1.82 g.

(A) is incorrect; it is obtained based on the amount of (NH4)2S available which is in excess.

(C) (E) are incorrect; students are probably guessing.


In quantum mechanics, atomic orbitals are described as wave functions over space, and indexed by the n, l, and ml quantum numbers of the orbital. As electrons cannot be described as solid particles (as a planet or a moth) in this way, a more accurate analogy would be that of a huge atmosphere, the spatially distributed electron, around a tiny planet which is the atomic nucleus. Hence the term "orbit" was substituted with something else: orbital. The principal quantum number n is always a positive integer. Each atom has, in general, many orbitals associated with each value of n; these orbitals together are sometimes called a shell. The azimuthal quantum number is a non-negative integer. Within a shell where n is some integer no, l ranges across all (integer) values satisfying the relation 0 l (no 1) . For instance, the n = 1 shell has only orbitals with l = 0, and the n = 2 shell has only orbitals with l = 0, and l = 1. The set of orbitals associated with a particular value of l are sometimes collectively called a subshell. The magnetic quantum number ml is also always an integer. Within a subshell where l is some integer lo, ml ranges thus: lo ml lo. Subshells are usually identified by their n- and l-values. n is represented by its numerical value, but l is represented by a letter as follows: 0 is represented by 's', 1 by 'p', 2 by 'd', 3 by 'f', and 4 by 'g'. For instance, one may speak of the subshell with n = 2 and l = 0 as a '2s subshell'.

(A) is incorrect; incorrect value for ml.

(B) is incorrect; incorrect values for n, l, and ml.

(C) is incorrect; incorrect value for l.

(E) is incorrect; incorrect value for l.


In the ground state of an atom (the condition in which it is ordinarily found), the electron configuration generally follows the Aufbau principle. According to this principle, electrons enter into states in order of the states' increasing energy; i.e., the first electron goes into the lowest-energy state, the second into the next lowest, and so on. Because electrons have only two possible spin states, an atomic orbital cannot contain more than two electrons (Pauli exclusion principle). According to the above principles (Aufbau and Pauli), it can therefore be concluded that (E) is a ground state electron configuration, and (A) (D) are all not ground state electron configurations. Students making (A) (D) as the choice answer may not understand either or both of the above principles.


The energy needed to remove one or more electrons from a neutral atom to form a positively charged ion is a physical property that influences the chemical behavior of the atom. By definition, the first ionization energy of an element is the energy needed to remove the outermost, or highest energy, electron from a neutral atom in the gas phase. In general, the first ionization energy increases as we go from left to right across a row of the periodic table, because the force of attraction between the nucleus and an electron becomes larger as the number of protons in the nucleus of the atom becomes larger. The second trend is that the first ionization energy decreases as we go down a column of the periodic table. This second trend results from the fact that the principal quantum number of the

orbital holding the outermost electron becomes larger as we go down a column of the periodic table. Although the number of protons in the nucleus also becomes larger, the electrons in smaller shells and subshells tend to screen the outermost electron from some of the force of attraction of the nucleus. Furthermore, the electron being removed when the first ionization energy is measured spends less of its time near the nucleus of the atom, and it therefore takes less energy to remove this electron from the atom. Elements with the highest first ionization energies are therefore to be found in the upper right region of the periodic table.

(A), (B), (C), and (D) are incorrect; students may not fully understand the factors affecting the first ionization energy.

Q8 CORRECT ANSWER: (A)

The structure of cis-HClC=CHCl is

Cl

H

Cl

H Both C-Cl and C-H are polar bonds, but with different magnitudes. From the structure

above, one can see that the vector addition of the polarity of the bonds gives a non-zero value. On the other hand, if the compound is trans-HClC=CHCl, the compound would have zero dipole moments. (B) is incorrect; student may not know what dipole moments are. A dipole moment can

only be formed from molecules with bonds linking atoms with unequal electro-negativities. In this case, the bond is between two identical atoms O2 molecule and hence cannot give rise to dipole moments.

(C) is incorrect; student may be thinking that the molecular geometry of BF3 is pyramidal and hence the molecule will be polar and have a non-zero dipole moment. Students may not be aware that boron is able to form compounds without satisfying the octet rule. The Lewis structure of BF3 has boron forming three covalent bonds. The resultant electron pair structure and molecular geometry is therefore trigonal planar. Although the B-F bond is polar, the trigonal planar arrangement cancels out the polarity, giving the molecule zero dipole moment.

(D) is incorrect; students may not be familiar with shapes of molecules. The central atom in ICl4 has two lone pairs and four single covalent bonds. The resultant electron pair geometry of molecule, ICl4 is octahedral. Since it has two lone pairs, the four Cl atoms forms a square planar geometry. Although the four I-Cl bonds are polar, they are arranged in a square and the vector sum of the dipole moments is zero.

(E) is incorrect; students may not realize that the shape of CO2 is linear, and hence the polarities of the two C=O bonds cancel out each other.

Q9 CORRECT ANSWER: (C)

NaF has a melting point of 993C. To melt an ionic solid, energy must be supplied to break the ionic bonds that hold ion together in a regular array. To melt a molecular compound, it

is the intermolecular bonds that needs to be broken. Ionic bonds are typically stronger than other intermolecular bonds, hence ionic solids typically have higher melting points than molecules. Comparing between ionic solids with ions of same stoichiometry and charge, due to Coulombs law, the smaller the distance is between the anion and cation, the stronger the ionic bond is. In this case, comparing between NaCl and NaF, fluoride ions are smaller than chloride. Hence, NaF has a higher melting point than NaCl.

(A) is incorrect; NaCl has a melting point of 801 C. Student realizes that ionic solids should have the highest melting point, but failed to realize that, given similar charge of the ions, the smaller the size of the ions, the stronger the ionic bond and therefore the higher the melting point.

(B) is incorrect; octane has a melting point of 57 C. Students may mistakenly think that the higher the relative molar mass, the higher the melting point. C8H18.

(D) is incorrect; rubidium Chloride has a melting point of 718 C. Student realizes that ionic solids should have the highest melting point, but failed to realize that, given similar charge of the ions, the smaller the size of the ions, the stronger the ionic bond and therefore the higher the melting point.

(E) is incorrect; hydrogen Chloride has a melting point of -114.2C. Students may not realize that HCl is not an ionic solid, but is a molecular compound


In this compound, the central atom S has one lone pair and four covalent bonds with the atoms H, F, Cl and Br. The VSEPR electron pair geometry is therefore trigonal bipyramidal. With one lone pair, the molecular geometry is a seesaw

S

A

D

BC

The four different atoms attached to S can be either on the equatorial positions or the axial

positions. By simple permutation, there are six ways to divide the four different atoms into the two positions. Since the molecules mirror image is not superimposable onto itself, it is chiral. This multiplies the possible number of isomers by two, bringing the total number of isomers to 12.

(B) is incorrect; student recognizes the concepts that lead to choice A but fail to recognize that the molecules are chiral.

(C) is incorrect; students may be making a random guess.

(D) is incorrect; students may not have recognized that the lone pair geometry of the molecule is trigonal bypyramidal. Instead, students may be thinking that the geometry is tetrahedral.

(E) is incorrect; Students may have thought that the molecular geometry is tetrahedral and did not consider the possibility that the molecule is chiral.


KBr is an ionic compound with a very high boiling point of 1708 K. It is brittle. It does not conduct electricity in its solid state but does so when it is melted.

(A) HCl has a boiling point of 188 K. (B) Al conducts electricity in its solid state. (D) SiF4 is a colorless gas at room temperature; its boiling point is 178 K. (E) I2 has a boiling point of 457.4 K.


The ideal gas equation is given by

PV = nRT

PV = (m/M)RT

density = m/V = PM/RT

M = (dRT)/P

= ( )

mmHg760atmmmHg.124.8

K2.298molKatmL082.0

L125.0g0125.0

= 74.94 g/mol.

Therefore, B6H10 has a molar mass of 74.94 g.

Q13 CORRECT ANSWER: (A) The ideal gas equation is given by PV = nRT

33

/8.46)50273(31.8

106.125 mmolesRTP

Vn =+

== The molar mass of NO2 is 14.0+(216.0) g/mol = 46.0g/mol. Density = (46.846) g/m3 = 2153 g/m3 = 2.15g/L (B), (C), (D), (E) Students may be confused by the unit conversions. Q14 CORRECT ANSWER: (A)

Statement I is not correct: For a non-ideal gas (CH4), at moderately high external pressure and constant temperature, values of PV/(RT) < 1 are due to intermolecular attractions. As pressure rises and volume of sample decreases, average intermolecular distance becomes smaller and molecular attraction between molecules lowers the force of collision of molecules with the wall. Therefore, P decreases and the numerator of PV/(RT) < 1. Statement II is not correct:

At very high pressure and constant temperature, values of PV/(RT) >1 are due predominantly to molecular volume. This is because the free volume between molecules is significantly less than the container volume because of the volume of the molecules themselves. Statement III is correct: At high pressure (and low temperature), non-ideal gas behaviour is most significant. This is close to conditions under which gas liquefies.


Statement I is true since PV = nRT. When T = 0K (absolute zero), V = 0. Statement II us false since PV = nRT, only a plot of P vs. T will be a straight line. Statement III is true. Two cases to consider. When the opposite arm is evacuated, no gas exerts pressure on either mercury surfaces and the mercury levels in the two arms will be the same. When a gas is in the opposite arm, it pushes the mercury level such that the level rises in the closed-arm. Statement IV is true from the postulate that collisions are elastic.


This question tests your understanding of enthalpy. The quantity of heat needs to be calculated based on the number of moles of isooctane. For one mole of isooctane, the corresponding Horxn = (10,992 2) kJ mol1.

C8H18 (l) + 12 O2 (g) 8 CO2 (g) + 9 H2O (l) Horxn = 5,496 kJ mol1

The mass of isooctane = volume density = 1000 mL 0.69 g mL1 = 690 g

The number of moles of isooctane = mass molecular weight = 690 g 114 g mol1 = 6.05 mol

The heat of combustion of 1 L (6.05 mol) isooctane = 5,496 kJ mol1 6.05 mol = 33,251 kJ.

The quantity of heat transferred = 33,251 kJ since heat, a quantity of energy, cannot be negative. The sign refers to the direction of heat transfer.

(A) is incorrect; because it is simply the Horxn of the given reaction without taking into account the number of moles of isooctane.

(B) is incorrect; because the Horxn was not divided by 2 in order to obtain the correct number of moles.

(D) is incorrect; because the quantity of heat cannot be negative.

(E) is incorrect; because it is simply based on dividing the Horxn by the number of moles.


This question tests your understanding of Hess's Law and your ability to balance and manipulate chemical equations.

Equation (i) + Equation (ii) x 0.5 Equation (iii) x 3, gives

N2 (g) + 3 H2 (g) 2 NH3 (g) Ho = 91.8 kJ 2 NH3 (g) + 5/2 O2 (g) 2 NO (g) + 3 H2O (l) Ho = 453.1 kJ 3 H2O (l) 3 H2 (g) + 3/2 O2 (g) Ho = +725.4 kJ

N2 (g) + O2 (g) 2 NO (g) Ho = 91.8 kJ 453.1 kJ +725.4 kJ = 180.5 kJ 1/2 N2 (g) + 1/2 O2 (g) NO (g) Ho = 0.5 x 180.5 kJ = 90.25 kJ = 90 kJ

(B) (E) are incorrect answers that arise from algebraic mistakes that may be made in

manipulating the equations.

Q18 CORRECT ANSWER: (D) The system is the reactants and products, and the surroundings are the pistons, the cooling system, and the rest of the car. Heat is released by the system, so q is negative. Work is done by the system to push the pistons outward, so w is also negative. Therefore, q = -325 J, w = -451 J, E = q + w = -325 J + (-451 J) = -776 J


This question tests your understanding of the concepts of oxidation and reduction. By definition, the oxidant should itself be reduced and the reductant should undergo oxidation. Therefore, Fe(OH)3 contains Fe3+ which undergoes reduction to Fe2+, and I is oxidized to I2 (via I).

(A) is incorrect; because it is the opposite of (B).

(C) is incorrect; because, while H+ can be reduced, Fe(OH)3 does not undergo oxidation.

(D) is incorrect; because Fe2+ does not get reduced and I2 does not get oxidized.

(E) is incorrect; because I does not get reduced and H+ does not get oxidized.


This is a redox reaction. Mn is reduced from an oxidation state of +7 to +2, while S is oxidized from an oxidation state of +4 to +6. The equation is balanced therefore with the coefficients for MnO4 and HSO3 being 2 and 5, respectively.

(A) is incorrect; it is probably a result based just on the number of atoms.

(B) is incorrect; probably just a random guess.

(D) is incorrect; it is a correct coefficient for HSO3 with the corresponding coefficient of 4 for MnO4; but this is not the smallest integers possible. (E) is obviously not correct.


This question tests your understanding of electrochemical potentials and batteries. The reactions should both first be written as reductions. In the case of Cr/Cr3+, the direction of the reaction must be switched so the sign of the potential must also be changed.

Ce3+(aq) + 3e Ce(s) E0 = 2.34 1/3 Cr3+(aq) + e 1/3 Cr(s) E0 = 0.74 V for reduction

In order to obtain the spontaneous direction, the reaction with the more negative half-cell potential must be subtracted from the reaction with the more positive half-cell potential.

E0cell = 0.74 (2.34) = U+1.60 V

The E0cell must be positive in order for the reaction to be spontaneous. This is because the G will be negative (G0 = nFE0cell).

(A) is incorrect; because it is simply the difference in potential between the two written reactions (2.34 0.74) and it is negative.

(B) is incorrect because the sign is incorrect.

(D) is incorrect; it is the answer that would be obtained if the half-cell potential for the Cr/Cr3+ system were divided by the number of electrons (3). The number of electrons transferred in the reaction does not affect the potential.

(E) is incorrect because it is simply the difference in potential between the two written reactions.


The magnitude of the equilibrium constant of a chemical reaction is a function of temperature and change only with change in temperature.

(A) is incorrect; removing products as they are formed will shift the position of equilibrium, but will not change the magnitude of the equilibrium constant.

(B) is incorrect; adding more of a reactant will shift the position of equilibrium, but will not change the magnitude of the equilibrium constant.

(C) is incorrect; adding a catalyst will accelerate a reaction toward equilibrium, but will not change the magnitude of the equilibrium constant.

(D) is incorrect; increasing pressure may change the equilibrium position of a reaction involving gaseous reactants or products, but will not change the magnitude of the equilibrium constant.


From the kinetics point of view, when a chemical system is at equilibrium, the rate of the forward reaction and that of the reverse reaction are equal, the concentrations of the reactants and products, hence, have constant values.

(A) is incorrect; the concentrations of the reactants and those of the products are not necessarily equal when a chemical reaction is at equilibrium.

(C) is incorrect; the forward and reverse reactions are still on when a chemical reaction is at equilibrium.

(D) is incorrect; when a chemical reaction is at equilibrium, the reaction quotient, Q, has a constant value which may be a minimum or a maximum depending on the direction toward which the equilibrium is approached.

(E) is incorrect; when a chemical reaction is at equilibrium, the reaction quotient, Q, has a constant value which may be a minimum or a maximum depending on the direction toward which the equilibrium is approached.


The equilibrium constant of the second equilibrium is a reciprocal of that of the first equilibrium, i.e., 2.5 x 103 = 1/(4.00 x 104)

(A) is incorrect; the equilibrium constant of the second equilibrium is not equal to half of that of the first equilibrium.

(B) is incorrect; the equilibrium constant of the second equilibrium is not equal to the negative of that of the first equilibrium.

(C) is incorrect; the equilibrium constant of the second equilibrium is not equal to half of the reciprocal of that of the first equilibrium.

(E) is incorrect; the reciprocal of 4.00 x 104 is 1/(4.00 x 104), not 4.00 x (1/104).

Q25 CORRECT ANSWER: (D) The solubility constant for Ag2SO4 can be described by [Ag+]2[SO42+] = 1.710-5.

Considering the stoichiometry of the salt (1 mole SO42+ to 2 mole Ag+), and the common ion provided by the 0.1 M sulfuric acid, the equation to solve is

(2x)2(x+0.01) = 1.710-5

(A) is incorrect; student did not include the common ion effect. The resultant equation becomes (2x)2x = 1.710-5.

(B) is incorrect; student erroneously used the approximation that the common ion (SO42+ ) provided by the sulphuric acid is so high that one can ignore the ions provided by Ag2SO4. The resultant equation to solve becomes

(2x)20.01= 1.710-5 (C) is incorrect; student has incomplete understanding of solubility constant. Student may

have used the following expression for the solubility constant of Ag2SO4: [Ag+][SO42+] = 1.710-5, and also ignored the common ion effect.

(E) is incorrect; student understood the common ion effect but may have forgotten to factor in the stoichiometry of the ions that make up the salt. Consequently, student may have end up with this equation to solve: x(x+0.01)= 1.710-5.


Increase in temperature increases the speed and hence the energy of molecular collisions. The rate of chemical reaction is increased then because there are now more collisions with energy exceeding the activation energy required for reaction.

(A) is incorrect; increase in temperature does not necessarily lead to an increase in the number of collisions having the correct orientation of molecules.

(B) is incorrect; the activation energy of a chemical reaction is characteristics of a reaction and does not increase with temperature.

(C) is incorrect; the activation energy of a chemical reaction is characteristics of a reaction and does not decrease with temperature.

(D) is incorrect; temperature does not act as a catalyst for a chemical reaction.

Q27 CORRECT ANSWER: (D) Br2 is a product and Br is a reactant of the reaction, so the rate expressed in terms of change in concentration of Br2 is positive and that in terms of change in concentration of Br is negative. The factor of 0.6 is derived from the stoichiometric coefficients for Br2 and Br (0.6 = 3/5).

(A) is incorrect; H2O is a product of the reaction, so rate in terms of change in concentration of H2O should be positive, not negative.

(B) is incorrect; BrO3 is a reactant of the reaction, so rate in terms of change in concentration of BrO3 should be negative.

(C) is incorrect; the correlating factor is incorrect. (E) is obviously not correct.

Q28 CORRECT ANSWER: (B) From the stoichiometry of the reaction, for every one mole of C produced, there will be two mole of A consumed, therefore the question can be recast as asking how long it will take for reactant A to go from 3 M to 1 M. For a first order reaction, the time for the reactant concentration to become one third its initial concentration, t1/3 is related to the half life, t1/2 as: t/ = ln3/ln2. Therefore the answer is

(A) is incorrect; student may be just making a random guess.

(C) is incorrect; student understands the stoichiometry of the reaction and knows that there should be a 2 mol/L depletion of A. However, the student does not know the relationship between t1/3 and t1/2 and assumes that: t/t = 3/2, thus leading to the answer.

(D) is incorrect; student may not have taken into consideration the stoichiometry of the reaction.

(E) is incorrect; student simply divided 25 by 3. Student shows no understanding of stoichiometry of reaction and the concept of first order reactions.


All 3 oxidation states are stable. This can be deduced by calculating the difference in successive I.E.s as follows: (I,II) = 13.58 6.82 = 6.8eV ( 12eV. Hence Ti in oxidation states II, III and IV are stable.

Q30 CORRECT ANSWER: (D) Calculate V for Li+ (1/0.60 = 1.666). Then calculate values of V for Be2+ 2/0.31 = 6.45, (VLi+-VBe2+) = -4.78 Mg2+ 2/0.65 = 3.08, (VLi+-VMg2+) = -1.41 Ca2+ 2/0.99 = 2.02, (VLi+-VCa2+) = -0.35 Sr2+ 2/1.13 = 1.77, (VLi+-VSr2+) = -0.10 Ba2+ 2/1.35 = 1.48, (VLi+-VBa2+) = +0.19 Thus, out of the series, the closest match for V(Li+) is (V)Sr2+. Hence Sr2+ will resemble Li+ most.


H2CO3 and HCO3- are weak acids. Therefore, the CO32- solution would hydrolyze to become slightly basic (A) Al3+ hydrolyzes in water to give an acidic solution. (B) SO2 reacts with water to give H2SO3, which then reacts with O2 to give H2SO4, a

strong acid. (C) Limestone (CaCO3) is insoluble in water, the supernatant should be close to neutral. (E) Acetic acid is a weak acid, even at 1 mM, the solution is slightly acidic.

Q32 CORRECT ANSWER: (D) For a strong acid, the amount of H+ is a constant before or after dilution. The concentration

of H+ after dilution will be 1/10 of its initial value, so pH will be one unit higher (pH = - log[H+]). So choice (C) is wrong: pH = 5 should be for strong acid. For a weak acid with reasonable approximation, [H+] = (Ka C)1/2, where Ka is the acid dissociation constant. Plugging this to equation pH = -log[H+] gives pH = 4.5 when it is diluted to 10 times of its volume. Choice (A) is wrong and a pH increase of 2 is totally irrelevant when it is diluted to 10 times of its value. Choice (B) is wrong because it is never possible to get a higher H+ concentration (hence a lower pH) after dilution. Choice (E) is irrelevant because the pH should be predictable after the dilution.


Compound I and III will slow down the reaction but not change the amount of hydrogen gas generated. I. The reaction is: Zn + OHAc Zn(OAc)2 + H2. Adding water to the reaction mixture

will lower the [H+], so the reaction is slower, but the total amount of HOAc is not affected.

II. Na2CO3 is a base and reacts with HOAc to give CO2, water, and NaOAc. Obviously, the reaction will slow down with lower [H+], but the amount of H2 generated will also decrease since some HOAc is neutralized by Na2CO3.

III. Addition of NaOAc to HOAc solution makes a buffer and will lower the [H+], so the reaction will be slower, but the amount of HOAc is not affected.

IV. This is analogous to II because of the reaction of ZnO and HOAc. V. Addition of Cu to a mixture of Zn and HOAc will make a primary cell and the

reaction is faster (but the amount of H2 is the same).

Q34 CORRECT ANSWER: (C) CaO will react with water to give Ca(OH)2 and this reaction is highly exothermic. Consequently, the temperature of solution will be higher and the solubility of the Ca(OH)2 will be lower. (A) Solubility is a function of temperature and other compounds present. Here the

temperature is not a constant. (B) The solution is saturated and it cannot dissolve more solute [Ca(OH)2] when the

temperature is kept constant. (D) This choice is irrelevant and the concentration of the solute is predictable. (E) This choice is wrong. If the student chose this, it means that he/she does not know the

reaction of CaO and water.

Q35 CORRECT ANSWER: (A) No H2 gas will be produced at anode since Cu2+ will be preferentially reduced to Cu metal; Br- gets oxidized at cathode before the evolution of O2. Unless in concentrated acid, the equilibrium will unlikely move to the left to give Cl2 gas. (B) 3 Ag + 4 H+ + NO3- (dilute) 3 Ag+ + NO (g) + 2 H2O. (C) MnO2 + 4H+ + 2Cl- Mn2+ + Cl2 (g) + 2 H2O. (D) 2 Al + 2 H2O + 2 OH- 2 AlO2- + 3 H2 (g). (E) Zn + 2 H+ Zn2+ + H2 (g).


At the high temperatures required in extraction and purification processes, Ti will burn in oxygen and nitrogen, forming oxides and nitrides, which remain as contaminants.

(B) The contrary is true. There is increasing demand for the use of this light-weight

material in the aircraft industry in the manufacture of both engines and airframes, in chemical processing and marine equipment.

(C) Ti occurs predominantly as TiO2 and in the mineral ilmenite FeTiO3, which are easily mined in large deposits in several countries.

(D) The ores are in concentrated deposits. (E) The metal has very low density, only ca. 57% that of steel. Alloyed with small

quantities of Al and Sn, has superior mechanical strength, compared to other engineering materials. It is only brittle when contaminated by traces of O, N and C impurities.

Q37 CORRECT ANSWER: (B) The metal ions in complexes (C), (D) and (E) are not transition elements. (A) is not

colourless. Students are not expected to memorize the colour of each compound. However, if they have done the classic KMnO4 redox titration, they should know that compound (B) is colourless in a dilute solution.


Only hydrogen can adopt all three oxidation states. Examples are HCl (+1), H2 (0) and NaH (-1).

(A) Thallium can form +1 and +3 compounds but not -1 derivatives. (B) Although fluorine can form molecular F2 and fluoride, it cannot lose any of its

seven electrons to form +1 compounds. (C), (E) Lithium and cesium are in the same group. They cannot form compounds with

the -1 oxidation state. Both metals have low ionization energies and thus tend to form compounds with the +1 oxidation state only.


Only statement I is correct. I. The major difference between second and third row elements is that the third row

elements may use its empty d-orbitals to accept more than the octet electrons. Thus sulfur may form more than 4 bonds.

II. Silicon can use its d-orbital to accept electrons from the oxygen lone pairs in the other SiO2 unit, thus forming a polymeric compound. It is not due to the atomic weight of silicon. In fact, chlorine has a higher atomic weight than Si but forms gaseous Cl2.

III. This statement is incorrect. If their electronic configurations are allowed, heavier atoms still exist in their mono-atomic forms (noble gases) and diatomic molecules (like I2). Students should find out why S2 and P2 cannot be formed.

IV. O3 is a non-linear molecule. Q40 CORRECT ANSWER: (D)

Au (s) + 4Cl- (aq) + 3 NO3- (aq) + 6 H+ (aq) [AuCl4]- (aq) + 3 NO2 (g) + 3 H2O (l) The statement in (D) is not correct because the final proton concentration should be in between those of the HNO3 and HCl before mixing. Even if we account for the small volume change during the mixing, the proton concentration is unlikely to be significantly higher.

(A) This is a common mistake. As shown above, the oxidation reaction consumes

protons to make water. A high concentration of proton increases the reduction potential of nitrate ions (more oxidative). From a different perspective, a mixture of concentrated NaNO3 and NaCl should not be as corrosive.

(B) Students who choose this option is not clear about the coordination chemistry of Au: Au3+ (aq) + 4 Cl- (aq) AuCl4- (aq)

A high concentration of Cl- drives the equilibrium to the right, effectively reducing the concentration of Au3+, and thus reducing the reduction potential of Au3+/Au.

(C ), (E) Students who choose these options are not familiar with the redox chemistry of nitric acid. HNO3 is an oxidant; NO2 is often released at a high HNO3 concentration and NO2 at lower HNO3 concentrations.


They are all the same image.


The rule here is to follow the rules set out by IUPAC that uses the 3 parts of a prefix, parent and suffix for the naming of compounds. The prefix describes what are the substituents, the parent describes how many carbons and the suffix describes to which family the molecule belongs. Thus, we start by finding the longest continuous chain in the molecule, which is 10 (dec), and use it as the parent name. We also know that the hydrocarbon is saturated and can add the suffix as -ane (for alkanes). Thus, we can deduce the first part of the compounds name as decane. On this basis, we can see only answers (A) and (D) can be correct and answers (B), (C) and (E) are wrong, as they do not follow the IUPAC naming rule. Returning to the IUPAC rule, the numbering of branched molecules starts at the end nearest to the first branch point. The first branch point occurs at C2 in answer (A) whereas it begins at C(5) in answer (D). Again, if the IUPAC rules are to be correctly followed, then answer (A) can only be the correct answer.

Q43 CORRECT ANSWER: (D) The concept here is to recognize that resonance and the inductive effect are important

factors for carbocation stability. In principle, the more a neighboring group can reduce the +ve charge of the carbocation by either re-distributing the +ve charge across more carbon centres or donate bonding electrons or both, the more stable is the carbocation. Carbocation I, which contains both a benzene ring and 2 methyl groups that can donate the bonding electrons toward the +ve carbon makes it the most stable carbocation species. Thus, initially only answers (A) and (D) can be the only possible correct answers. Answers (B) and (E) are incorrect as compound III is a primary carbocation stabilized by hydrogen groups which are poorer than carbon in donating electron density to the +ve charge and is therefore the least stable. Answer (C) is incorrect because compound II is a primary carbocation and there is no conjugation to provide resonance stabilization by the benzene ring, therefore compound II is therefore less stable than compound I. The second compound in answer (A) is compound IV and in answer (D) it is compound V. Compound IV is a secondary carbocation whereas compound V is a tertiary carbocation. Thus, as tertiary carbocations are more stable than secondary carbocation since they contain one more carbon substituent, one might expect compound V to be more stable than Compound IV and the correct answer to be thus answer (D).


The reaction is an alkene hydrobromination, which the addition of a Br and H atom across the C=C bond. Typically, such reactions proceed with bromination at the more substituted carbon centre due to the Markovnikov rule, which states that formation of tertiary carbocation intermediate is more favourable than a secondary carbocation intermediate as the tertiary carbocation intermediate is more stable.

(B) This choice is incorrect as compound II is the product of alkene hydrogenation, which

is not possible as the reagent is HBr and not H2. (C) This choice is incorrect as compound III is the product of bromination at the vicinal

carbon to the alkene functionality.

(D) This choice is incorrect as compound IV is the product of alkene dibromination, which is not possible as the reagent is HBr and not Br2.

(E) This choice is incorrect as compound V, the regioisomer of compound I, is the anti-Markovnikov product. This means the reaction proceed with bromination at the less substituted carbon centre and does not follow the Markovnikov rule. This is not favourable as this mean the reaction proceeds via the formation of a less stable secondary carbocation intermediate over a tertiary carbocation intermediate.


The process is both a substitution reaction and one that occurs via a radical pathway.

(A) This choice is incorrect as the overall outcome of the reaction is the replacing of a H atom by a Cl atom and is therefore a substitution and not an addition.

(B) This choice is incorrect as the conditions employed (h) state the reaction is conducted under photolytic and not thermal conditions.

(C) This choice is incorrect as the conditions employed (h) state that the reaction should occur via a radical pathway. Electrophilic substitution would imply a cationic intermediate which is not possible in radical reactions.

(D) This choice is also incorrect for the same reason as in (C). Nucleophilic substitution would imply a anionic intermediate which is not possible in radical reactions.

Q46 CORRECT ANSWER: (A) CONH2 has meta orienting effect. (B), (C), (D) Substituents are ortho and para directors. (E) The mono-bromonation product induces more ortho or para product than meta product. Q47 CORRECT ANSWER: (D) OH has an activating effect on the ring, while Cl, SO3H and CHO have deactivating effects on the ring. Therefore the molecule of choice (D) undergoes electrophilic substitution on the ring most rapidly.

Q48 CORRECT ANSWER: (A) The alkene is more susceptible to electrophilic attack than the benzene ring as the benzene ring is stabilized by aromaticity. So, answers (B) and (D) are incorrect. H2SO4 is an acid, so the alkene will be protonated to generate a carbocation. This would then be attacked by the benzene ring in an intramolecular Friedel-Crafts reaction to make a second carbocyclic ring. (C) is incorrect because HSO4- is a very weak nucleophile. We can choose between (A) and (E) by thinking about the possible carbocations. (A) would be formed by via a tertiary carbocation and (E) via a primary carbocation. Therefore (A) is correct. Q49 CORRECT ANSWER: (E) Chromium(VI) compounds is one of the convenient reagents for oxidation of secondary alcohols to ketones.

(A) n-BuLi is used as a strong base or nucleophile. If a secondary alcohol is treated with n-BuLi, deprotonation takes place

(B) LiAlH4 is used as a strong reducing reagent or base. If a secondary alcohol is treated with LiAlH4, deprotonation takes place.

(C) This is clearly not correct, (D) SOCl2 is used for replacing hydroxy group with chlorine. If a secondary alcohol is

treated with SOCl2, secondary alkyl chloride is obtained. Q50 CORRECT ANSWER: (C) Proton is electrophilic and firstly interact with an electron lone pair on oxygen atom of hydroxy group, to form oxonium ion.

(A) This choice is clearly not correct, (B) In acidic conditions, an alkene is formed from an oxonium intermediate generated via

protonation of the hydroxy group. (D) This choice is clearly not correct, (E) In acidic conditions, a carbocation is generated by releasing H2O from the oxonium

intermediate.

Q51 CORRECT ANSWER: (D) Addition of water to alkene (Markovnikovs rule) and reduction of ketone are the representative and practical methods to prepare secondary alcohol. (A) Formal Markovnikov addition of H2O to alkene (III) and reductive preparation (IV)

also should be considered. (B) Regioselectivity of hydroboration of alkene is often governed by the steric factor and

therefore hydroboration-oxidation sequence (II) is used for preparation of alkan-1-ol (formal anti-Markovnikov addition).

(C) Oxymercuration of terminal alkene followed by a reductive workup (III) gives formal Markovnikov addition of H2O. Branched alcohols are obtained.

(E) Hydroboration-oxidation sequence (II) is not suitable to synthesize secondary alcohols from terminal alkenes as shown in Choice B.


This is the correct answer because the enolate is involved. (A) This is incorrect because Iodine is not highly electrophilic. (B) Iodide is a good nucleophile, but that is not relevant to the mechanism, (D) This might be important if this is an acid catalysed reaction, but it is not. (E) This would be important if this was a free radical reaction.

Q53 CORRECT ANSWER: (D) Electronegativity of F > Cl > H. More electronegative, more electron-withdrawing and more stable is the anion. Greater stability of anion leads to greater acidity of the parent acid. Thus the correct answer is (D)

Q54 CORRECT ANSWER: (E) The PCl5 reaction results in the formation of compound I if reaction is complete. There could be compounds II and III if the reaction is incomplete. However, compounds I and II are not possible as the acid chloride is readily hydrolysed to the acid upon addition of water. Thus III is a possible product. Compound IV could result from compound II prior to the hydolysis step. Compound V is not a possible product. Thus the answer is (E) Q55 CORRECT ANSWER: (A) Polycarbonate is a polymer with the monomers linked through ester groups. Base catalyzed hydrolysis of esters give the carboxylate group and the alcohol, which is phenol in this case. Thus, the answer is (A). Q56 CORRECT ANSWER: (B) Amides are prepared by reacting acyl chlorides with amines.

(A) Methoxy (-O-CH3) is not good leaving group. (C) Amide is not good nucleophile toward alkyl chloride. (D) Amide is not good nucleophile, and a hydroxy moiety is not good leaving group. (E) Students may be making a wild guess.

Q57 CORRECT ANSWER: (B) Trialkylamines can work as a base.

(A) Phenylamine is a weaker base than alkylamines because the benzene ring pull electrons away from the nitrogen toward the benzene ring.

(C) Nitriles are neutral coupounds. (D) Amides are almost neutral. (E) Nitro compounds are also neutral.

Q58 CORRECT ANSWER (E) In the first step, only alkyl chloride was reacted with KCN to give alkanenitrile. It is noted that aryl chloride does not undergo same type of reaction with KCN. In the second step, resulting nitrile was reduced with lithium aluminun hydride, affording alkyl amine (E). (A) Aryl chloride does not reacted with KCN. (B) Reduction of nitrile gives an methylene amine (-CH2-NH2) moiety, not amine (-NH2). (C) Aryl chloride does not react with KCN. (D) Reduction of nitrile gives an methylene amine (-CH2-NH2) moiety.

Q59 CORRECT ANSWER: (B) Hydrolysis of this amide with dilute alkali gives para-methyl phenylamine and the sodium salt of benzoic acid. Only para-methyl phenylamine can be extracted with diethyl ether.

(A) After hydrolysis with dilute alkali, the sodium salt of benzoic acid was obtained, which cannot be extracted with diethyl ether.

(C) After hydrolysis with dilute alkali, the sodium salt of benzoic acid was obtained. So, benzoic acid cannot be obtained.

(D) Students may be making a wild guess. (E) The sodium salt of benzoic acid cannot be extracted with diethyl ether.

Q60 CORRECT ANSWER: (A) The tripeptide must contain both alanine and serine, and no other amino acids. So, there must be two molecules of either one, but, from the data, we do not know which. (A) satisfies these requirements; (B) is only a dipeptide; (C) also contains glycine; (D) contains only alanine and serine, but one of the linkages is an ester rather than a peptide (amide) bond; (E) contains only alanine and serine, but one of the linkages is through a urea, rather than a peptide.

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