Chemistry — Class 12 practice questions

Worked examples for Chemistry — Class 12

1. Question 1

   Sodium ethanoate is prepared as

   $\mathrm{CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O}$

   $M_{\mathrm{CH_3COOH}} = 60\ \text{g mol}^{-1}$

   $M_{\mathrm{NaOH}} = 40\ \text{g mol}^{-1}$

   $M_{\mathrm{CH_3COONa}} = 82\ \text{g mol}^{-1}$

   The percentage atom economy of this preparation for sodium ethanoate is

   1. A. $82$Correct answer

      Percentage atom economy $= \frac{M_{\text{desired product}}}{\sum M_{\text{reactants}}} \times 100 = \frac{82}{60 + 40} \times 100 = 82$.

   2. B. $18$

      You put the by-product water in the numerator; atom economy weighs the DESIRED product against the reactants, and water is the waste here.

   3. C. $122$

      You inverted the ratio and divided the reactant mass by the product mass, which can never give a percentage below one hundred for a reaction that makes waste.

   4. D. $137$

      You left sodium hydroxide out of the denominator; every reactant in the balanced equation enters the sum, not only the one carrying the carbon skeleton.

   Explanation

   Atom economy is the mass of the wanted product expressed as a percentage of the total mass of all reactants in the balanced equation, so it measures how much of what goes in actually ends up in the product. It is fixed by the stoichiometry alone and does not depend on how much product is obtained in practice, which is the percent yield. Here one molecule of water leaves as waste, so the figure falls below one hundred.

2. Question 2

   Propanone (acetone) is warmed with zinc amalgam ($\mathrm{Zn-Hg}$) and concentrated $\mathrm{HCl}$. The major organic product is

   1. A. prop-1-ene

      You reduced the ketone to propan-2-ol and then dehydrated it. Clemmensen conditions neither stop at the alcohol nor dehydrate it; dehydration needs conc. $\mathrm{H_2SO_4}$ on the alcohol.

   2. B. propan-2-ol

      You stopped the reduction at the alcohol stage. A hydride reagent such as $\mathrm{NaBH_4}$ or $\mathrm{LiAlH_4}$ gives propan-2-ol; Clemmensen removes the carbonyl oxygen completely.

   3. C. propaneCorrect answer

      Zinc amalgam with concentrated $\mathrm{HCl}$ is the Clemmensen reduction: the carbonyl group is reduced all the way to $\mathrm{-CH_2-}$, so $\mathrm{CH_3-CO-CH_3}$ gives $\mathrm{CH_3-CH_2-CH_3}$.

   4. D. 2-chloropropane

      You treated the concentrated $\mathrm{HCl}$ as a chlorinating reagent for the carbonyl carbon. It only provides the acidic medium for the zinc amalgam; 2-chloropropane would come from propan-2-ol with conc. $\mathrm{HCl}$ and anhydrous $\mathrm{ZnCl_2}$ (Lucas reagent).

   Explanation

   Clemmensen reduction uses zinc amalgam in concentrated hydrochloric acid and converts the carbonyl group of an aldehyde or ketone into a methylene group, so the product is the hydrocarbon with the same carbon skeleton. Reagents that merely add hydride, such as $\mathrm{NaBH_4}$ and $\mathrm{LiAlH_4}$, stop at the alcohol; the complete removal of the oxygen is what distinguishes Clemmensen (and Wolff–Kishner) from them.

3. Question 3

   Cobalt(III) forms octahedral complexes with ligands of both strong and weak field strength.

   Given: atomic number of $\mathrm{Co} = 27$.

   Which of the following complex ions is paramagnetic with four unpaired electrons?

   1. A. $\mathrm{[Co(NH_3)_6]^{3+}}$

      You took ammonia for a weak-field ligand. Ammonia pairs the six $3d$ electrons of $\mathrm{Co^{3+}}$ into three orbitals, leaving two $3d$ orbitals free for $d^2sp^3$, so this inner-orbital complex has no unpaired electron and is diamagnetic.

   2. B. $\mathrm{[CoF_6]^{3-}}$Correct answer

      The metal is $\mathrm{Co^{3+}}$, $3d^{6}$. Fluoride is a weak-field ligand and does not pair those electrons, so four stay unpaired; the metal uses outer $4d$ orbitals, $sp^3d^2$, and the complex is an outer-orbital, high-spin, paramagnetic one.

   3. C. $\mathrm{[Co(CN)_6]^{3-}}$

      Cyanide is the strongest field ligand of the four and pairs the $3d^{6}$ set completely; this is the standard low-spin, diamagnetic cobalt(III) complex, not a paramagnetic one.

   4. D. $\mathrm{[Co(en)_3]^{3+}}$

      Ethylenediamine is bidentate, so three of them give a coordination number of six — that part is true, but the ligand is strong-field and pairs the $3d^{6}$ electrons, leaving the chelate diamagnetic.

   Explanation

   The magnetic behaviour of a complex is settled by the $d$ configuration of the metal ion in its oxidation state and then by what the ligand field does to it. Every sphere here holds cobalt in the $+3$ state, a $3d^{6}$ ion, so the field strength of the ligand is the only thing that separates them: a strong field pairs all six electrons and gives an inner-orbital, diamagnetic complex, while a weak field leaves four unpaired and forces the metal to use outer $d$ orbitals. Fluoride is the one weak-field ligand in the set, which is why cobalt(III) breaks its usual diamagnetic pattern only there.

4. Question 4

   Consider the three acids $\mathrm{CH_3-CH_2-COOH}$ (propanoic acid), $\mathrm{CH_3-CH(Cl)-COOH}$ (2-chloropropanoic acid) and $\mathrm{Cl-CH_2-CH_2-COOH}$ (3-chloropropanoic acid).

   The correct order of their acid strength is

   1. A. 3-chloropropanoic acid $<$ propanoic acid $<$ 2-chloropropanoic acid

      You treated the distant chlorine as releasing electrons and so placed that acid below the unsubstituted acid; a chlorine at any position withdraws electrons, so both chloro acids are stronger than propanoic acid.

   2. B. 2-chloropropanoic acid $<$ 3-chloropropanoic acid $<$ propanoic acid

      You reversed the inductive effect and treated the electron-withdrawing chlorine as weakening the acid; withdrawal of electrons stabilises the carboxylate ion and therefore increases acid strength.

   3. C. propanoic acid $<$ 2-chloropropanoic acid $<$ 3-chloropropanoic acid

      You ignored how far the chlorine stands from the carboxyl group and ordered the two chloro acids by their locants; the inductive effect falls off sharply along the chain, so the 2-chloro acid is the stronger of the two.

   4. D. propanoic acid $<$ 3-chloropropanoic acid $<$ 2-chloropropanoic acidCorrect answer

      Chlorine withdraws electrons and stabilises the carboxylate, so both chloro acids are stronger than propanoic acid; the inductive effect weakens with distance, so the chlorine on the carbon next to $\mathrm{-COOH}$ raises the acidity more than one two carbons away.

   Explanation

   The strength of a carboxylic acid is decided by how well the carboxylate ion left after losing the proton is stabilised. An electron-withdrawing group disperses the negative charge and raises acidity, an electron-releasing alkyl group lowers it, and the effect is transmitted through the sigma bonds, so it grows weaker as the group moves away from the carboxyl group.

5. Question 5

   A reaction of the first order has a half-life of $6.93\ \text{min}$.

   Given: $\log 3 = 0.4771$

   The time in which two-thirds of the reactant is consumed is

   1. A. $11.0\ \text{min}$Correct answer

      $k = \frac{0.693}{6.93} = 0.1\ \text{min}^{-1}$; one-third of the reactant remains, so $t = \frac{2.303}{k}\log 3 = 23.03 \times 0.4771 = 11.0\ \text{min}$.

   2. B. $4.06\ \text{min}$

      You substituted the fraction CONSUMED, taking $\log\frac{3}{2} = 0.1761$; the integrated law needs the concentration remaining, which is one-third of the initial value, so the logarithm is $\log 3$.

   3. C. $25.3\ \text{min}$

      You kept the factor $2.303$ and also took a natural logarithm, $2.303\ln 3$; the $2.303$ is exactly what converts $\ln$ to $\log_{10}$, so using both counts the conversion twice.

   4. D. $0.100\ \text{min}^{-1}$

      You stopped at the rate constant obtained from the half-life; a first-order $k$ carries the unit $\text{min}^{-1}$, while the question asks for a time in $\text{min}$.

   Explanation

   A first-order half-life fixes the rate constant alone, since $t_{1/2} = 0.693/k$ does not involve the initial concentration. The time for any stated extent then follows from $k = \frac{2.303}{t}\log\frac{[\mathrm{A}]_0}{[\mathrm{A}]}$, in which the concentration that appears is the one REMAINING: two-thirds consumed leaves one-third, so the ratio is $3$ and the logarithm is a table value.

6. Question 6

   Arrange aniline, p-nitroaniline, ammonia and methylamine in INCREASING order of basic strength in aqueous solution.

   1. A. aniline $<$ p-nitroaniline $<$ ammonia $<$ methylamine

      You treated the nitro group as electron releasing; it is strongly electron withdrawing, so it reduces the availability of the nitrogen lone pair and makes p-nitroaniline the weakest base of the four.

   2. B. p-nitroaniline $<$ aniline $<$ ammonia $<$ methylamineCorrect answer

      The lone pair of aniline is delocalised into the ring, so aniline is weaker than ammonia; the electron-withdrawing nitro group pulls further electron density away and makes p-nitroaniline weaker still, while the $+\mathrm{I}$ effect of the methyl group makes methylamine the strongest.

   3. C. p-nitroaniline $<$ aniline $<$ methylamine $<$ ammonia

      You placed ammonia above methylamine; the methyl group pushes electron density towards nitrogen by the inductive effect, so methylamine is the stronger base of the two.

   4. D. ammonia $<$ p-nitroaniline $<$ aniline $<$ methylamine

      You ranked by carbon content, assuming any organic amine beats ammonia; both aromatic amines have their lone pair delocalised into the ring and are weaker bases than ammonia.

   Explanation

   Basic strength of an amine depends on how available the lone pair on nitrogen is. An alkyl group releases electron density by the inductive effect and raises it above ammonia, while in an aromatic amine the lone pair is delocalised into the ring and is lowered below ammonia; an electron-withdrawing substituent such as nitro on that ring lowers it further.

7. Question 7

   For the reaction $\mathrm{CH_4(g) + Cl_2(g) \rightarrow CH_3Cl(g) + HCl(g)}$ the mean bond enthalpies at $298\ \text{K}$ are given.

   $\mathrm{C-H} = 414\ \text{kJ mol}^{-1}$

   $\mathrm{Cl-Cl} = 243\ \text{kJ mol}^{-1}$

   $\mathrm{C-Cl} = 331\ \text{kJ mol}^{-1}$

   $\mathrm{H-Cl} = 431\ \text{kJ mol}^{-1}$

   The enthalpy change of the reaction is

   1. A. $-105\ \text{kJ mol}^{-1}$Correct answer

      One $\mathrm{C-H}$ and one $\mathrm{Cl-Cl}$ bond are broken, one $\mathrm{C-Cl}$ and one $\mathrm{H-Cl}$ bond are formed: $\Delta H = (414 + 243) - (331 + 431) = -105\ \text{kJ mol}^{-1}$.

   2. B. $+105\ \text{kJ mol}^{-1}$

      You took bonds formed minus bonds broken. Bond enthalpies enter as broken minus formed, the opposite order to formation enthalpies, so the magnitude is right but the sign is wrong.

   3. C. $+1137\ \text{kJ mol}^{-1}$

      You broke all four $\mathrm{C-H}$ bonds of methane but did not re-form the three that survive in $\mathrm{CH_3Cl}$; only one $\mathrm{C-H}$ bond is actually replaced.

   4. D. $-105\ \text{kJ}$

      Bond enthalpies are molar quantities, so their difference is an enthalpy change per mole of reaction; a bare kilojoule would be the heat for some stated amount, which this question does not give.

   Explanation

   Breaking bonds absorbs energy and forming them releases it, so from bond enthalpies $\Delta_r H = \sum \text{B.E.(bonds broken)} - \sum \text{B.E.(bonds formed)}$ — the reverse of the products-minus-reactants order used with formation enthalpies. Only the bonds that actually change need be counted: the three $\mathrm{C-H}$ bonds that survive in $\mathrm{CH_3Cl}$ appear on both sides and cancel. The chlorination is exothermic because the two bonds made are together stronger than the two broken.

8. Question 8

   A gas is compressed by an external pressure, so that work is done ON the gas. According to the sign convention used by the textbook ($W = -P_{ext}\Delta V$), the signs of $W$ and $\Delta V$ for this compression are

   1. A. $W$ positive, $\Delta V$ negativeCorrect answer

      On compression the final volume is smaller, so $\Delta V = V_2 - V_1 < 0$; with the minus sign in $W = -P_{ext}\Delta V$ the work comes out positive, which is how the book marks energy entering the system.

   2. B. $W$ negative, $\Delta V$ negative

      You dropped the minus sign of $W = -P_{ext}\Delta V$ and took the sign of $W$ straight from $\Delta V$; that is the physics convention, in which work done BY the gas is positive.

   3. C. $W$ positive, $\Delta V$ positive

      A positive $\Delta V$ is an expansion; compression reduces the volume, so $\Delta V$ cannot be positive here.

   4. D. $W$ negative, $\Delta V$ positive

      These are the signs for a reversible or irreversible EXPANSION against an external pressure, not for a compression.

   Explanation

   The textbook defines pressure-volume work as $W = -P_{ext}\Delta V$, so the sign of the work is fixed by the sign of the volume change and the external pressure, which is always positive. A decrease in volume makes $\Delta V$ negative and hence $W$ positive, consistent with the broader convention that energy added to the system is counted positive and energy leaving it negative.