Chemistry — Class 11 practice questions

Worked examples for Chemistry — Class 11

1. Question 1

   At STP (273 K, 1 atm) the molar volume of a gas is $22.4\ \text{dm}^{3},\text{mol}^{-1}$.

   Given: $N_{A} = 6.022 \times 10^{23}\ \text{mol}^{-1}$

   Oxygen, $\mathrm{O_2}$, and sulfur dioxide, $\mathrm{SO_2}$, are compared at STP. Which of the following statements is correct?

   1. A. Equal volumes of the two gases contain equal numbers of atoms.

      You carried equal numbers of molecules over to equal numbers of atoms without using the atomicity: each $\mathrm{O_2}$ molecule has 2 atoms and each $\mathrm{SO_2}$ molecule has 3, so equal volumes hold atoms in the ratio $2:3$.

   2. B. Equal masses of the two gases occupy equal volumes.

      You treated mass as though it fixed the number of moles. The molar masses differ, so equal masses are different numbers of moles and therefore occupy different volumes at STP.

   3. C. Equal volumes of the two gases contain equal numbers of molecules.Correct answer

      This is Avogadro's law: equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules, whatever the gas. Equal volumes at STP mean equal numbers of moles, hence equal numbers of molecules.

   4. D. $1\ \text{dm}^{3}$ of each gas contains $6.022 \times 10^{23}$ molecules.

      You attached Avogadro's number to one cubic decimetre instead of to one mole; it is $22.4\ \text{dm}^{3}$ at STP, not $1\ \text{dm}^{3}$, that holds $6.022 \times 10^{23}$ molecules.

   Explanation

   Volume at a fixed temperature and pressure measures the number of moles of gas and nothing else, so the same volume of any two gases holds the same number of molecules. The number of atoms, however, needs the atomicity of each molecule, and the mass needs the molar mass, which is why statements phrased in atoms or in masses do not follow from equal volumes. The molar volume links volume to moles only at the stated standard conditions, and only per mole.

2. Question 2

   Benzene is treated with methyl chloride, $\mathrm{CH_3Cl}$, in presence of anhydrous $\mathrm{AlCl_3}$. The major product is

   1. A. chlorobenzene

      Chlorine replaces a ring hydrogen only when the reagent is $\mathrm{Cl_2}$ with anhydrous $\mathrm{AlCl_3}$ (halogenation); here the electrophile generated from $\mathrm{CH_3Cl}$ and $\mathrm{AlCl_3}$ is the methyl group, not chlorine.

   2. B. acetophenone

      This is the Friedel–Crafts acylation product, obtained with acetyl chloride $\mathrm{CH_3COCl}$ and anhydrous $\mathrm{AlCl_3}$; the stem gives an alkyl halide, which delivers an alkyl group and not an acyl group.

   3. C. tolueneCorrect answer

      Anhydrous $\mathrm{AlCl_3}$ abstracts the chloride from $\mathrm{CH_3Cl}$ to give the methyl carbocation, which substitutes one ring hydrogen: this is Friedel–Crafts alkylation, and the product is methylbenzene (toluene).

   4. D. benzyl chloride

      You kept the chlorine on the group that entered the ring. $\mathrm{C_6H_5-CH_2Cl}$ comes from free-radical chlorination of the side chain of toluene with $\mathrm{Cl_2}$ in sunlight, a later step that needs no $\mathrm{AlCl_3}$.

   Explanation

   Benzene is aromatic, so it reacts by electrophilic substitution and not by addition. The Lewis acid anhydrous $\mathrm{AlCl_3}$ polarises the carbon–chlorine bond of the alkyl halide to generate an alkyl carbocation as the electrophile; this attacks the ring and a ring hydrogen is lost, so an alkyl group takes its place. An alkyl halide therefore alkylates the ring, while an acid chloride under the same catalyst acylates it.

3. Question 3

   Butylated hydroxytoluene (BHT) is added in a very small amount to a packet of fried snacks. Its function in the food is to

   1. A. kill the micro-organisms that spoil the food

      Killing micro-organisms is germicidal action, which an antioxidant does not have; the rancidity it prevents is a purely chemical change.

   2. B. delay the oxidation of the unsaturated fats presentCorrect answer

      BHT is an antioxidant; it is consumed by oxygen in place of the unsaturated chains of the fat, so the oxidation that would give volatile aldehydes and acids, and hence the rancid smell, is postponed.

   3. C. prevent the hydrolysis of the fat into free fatty acids

      Hydrolysis is the other route to rancidity, but it is moisture that brings it about and an antioxidant does not stop it; BHT acts only on the oxidation route.

   4. D. neutralise the free fatty acids already formed in the fat

      Neutralising an acid would need an alkali, which would saponify the fat itself; an antioxidant prevents spoilage from beginning instead of removing its products.

   Explanation

   Oils and fats become rancid in two ways: hydrolysis, which liberates free fatty acids, and oxidation of the unsaturated chains to volatile aldehydes and acids that give the stale smell and taste. Antioxidants such as BHT and vitamin E are added in trace amounts because they react with the oxidising species first and so slow the oxidative route down.

4. Question 4

   The allyl carbocation $\mathrm{CH_2=CH-CH_2^{+}}$ is much more stable than the propyl carbocation $\mathrm{CH_3-CH_2-CH_2^{+}}$. The main reason for this is

   1. A. Hyperconjugation by the alpha hydrogen atoms

      Hyperconjugation needs $\mathrm{C-H}$ bonds on an adjacent saturated carbon. In the allyl cation the neighbouring carbon is doubly bonded, so no such bonds are available; it is the propyl cation that has two alpha hydrogen atoms.

   2. B. The $+\mathrm{I}$ effect of the vinyl group

      You gave the vinyl group the electron-releasing sign of an alkyl group. A doubly bonded carbon is electron-withdrawing ($-\mathrm{I}$), so the inductive effect here opposes stability rather than creating it.

   3. C. Delocalisation of the positive charge by resonanceCorrect answer

      The empty $p$ orbital on the charged carbon overlaps the adjacent $\pi$ bond, so the positive charge is shared between the two terminal carbons; spreading the charge lowers the energy.

   4. D. $sp^{3}$ hybridisation of the positively charged carbon

      A carbocation carbon is $sp^{2}$ hybridised and planar with an empty $p$ orbital, in the allyl cation as much as in the propyl cation, so this cannot distinguish the two.

   Explanation

   A carbocation is stabilised by whatever spreads its positive charge. When the charged carbon's empty $p$ orbital sits next to a $\pi$ bond, the charge is delocalised over two carbons by resonance, which is a far larger stabilisation than the $\sigma$-bond delocalisation of hyperconjugation available to a simple primary cation.

5. Question 5

   A freshly obtained precipitate of $\mathrm{Fe(OH)_3}$ is converted into a stable colloidal sol by shaking it with a small quantity of $\mathrm{FeCl_3}$ solution. This process is called

   1. A. Coagulation

      Coagulation is the reverse change — an electrolyte added to a sol neutralises the charge on its particles and settles them as a precipitate.

   2. B. PeptizationCorrect answer

      Breaking up a fresh precipitate into colloidal particles by adding a small amount of a suitable electrolyte, whose ions are adsorbed on the particles, is peptization.

   3. C. Dialysis

      Dialysis is a purification step: excess electrolyte is removed from a sol through a parchment membrane, so it does not make the sol from a precipitate.

   4. D. Emulsification

      Emulsification is the dispersion of one liquid in another liquid with an emulsifier; here a solid precipitate, not a liquid, is being dispersed.

   Explanation

   A fresh precipitate can be redispersed because the added electrolyte supplies ions that are adsorbed on the surface of the precipitate particles; the like charges so acquired make the particles repel one another and break the precipitate down to colloidal size. The electrolyte used for this is called a peptizing agent, and only a small quantity is needed, since a larger amount would instead coagulate the sol formed.

6. Question 6

   Which of the following statements about a pure liquid is correct?

   1. A. Its vapour pressure increases as the temperature rises.Correct answer

      A rise in temperature raises the fraction of molecules having enough kinetic energy to escape from the surface, so more vapour is present at equilibrium and the vapour pressure rises.

   2. B. Its surface tension increases as the temperature rises.

      Surface tension falls with rise in temperature, because the intermolecular attractions that pull a surface molecule inward are weakened by the greater thermal motion.

   3. C. Its viscosity increases as the temperature rises.

      Viscosity falls with rise in temperature: the molecules gain enough energy to slip past one another, which is why oils flow more freely on warming.

   4. D. It boils at a higher temperature on a mountain top than at sea level.

      A liquid boils when its vapour pressure equals the EXTERNAL pressure; atmospheric pressure is lower on a mountain, so the boiling point is lower there, not higher.

   Explanation

   Of the three temperature-dependent properties of a liquid, only the vapour pressure rises on heating; surface tension and viscosity both fall, since both arise from intermolecular attraction that thermal motion opposes. The boiling point is not a property of the liquid alone — it is the temperature at which the vapour pressure matches the pressure above the liquid, so it falls as the external pressure falls.

7. Question 7

   In phosphorus pentachloride, $\mathrm{PCl_5}$, the central phosphorus atom forms five sigma bonds and carries no lone pair. The hybridisation of the phosphorus atom is

   1. A. $sp^{3}$

      Four hybrid orbitals would account for only four sigma bonds; phosphorus here expands its octet to five bonds, which needs a $d$ orbital as well.

   2. B. $sp^{3}d^{2}$

      Six hybrid orbitals are one too many: $sp^{3}d^{2}$ belongs to six sigma bonds and an octahedral shape, as in $\mathrm{SF_6}$.

   3. C. $dsp^{2}$

      This set has only four hybrid orbitals, giving a square planar arrangement; it cannot hold five sigma bonds.

   4. D. $sp^{3}d$Correct answer

      The number of hybrid orbitals equals the number of sigma bonds plus lone pairs, here $5 + 0 = 5$; mixing one $s$, three $p$ and one $d$ orbital gives five $sp^{3}d$ hybrid orbitals, directed to the corners of a trigonal bipyramid.

   Explanation

   Hybridisation is counted from the number of sigma bonds plus lone pairs on the atom, since each of those needs one hybrid orbital; pi bonds use unhybridised orbitals and are never counted. Five such orbitals on phosphorus require one $s$, three $p$ and one $d$ orbital, which is $sp^{3}d$.

8. Question 8

   Diethyl ether, $\mathrm{CH_3-CH_2-O-CH_2-CH_3}$, and methyl propyl ether, $\mathrm{CH_3-O-CH_2-CH_2-CH_3}$, both have the molecular formula $\mathrm{C_4H_{10}O}$. The type of isomerism shown by this pair is

   1. A. metamerismCorrect answer

      Both are ethers, and the four carbon atoms are shared differently between the two alkyl groups attached to the same oxygen (2 and 2 against 1 and 3). Unequal alkyl groups on either side of the same functional group is metamerism.

   2. B. chain isomerism

      Chain isomers differ in the branching of one carbon skeleton, as butan-1-ol and 2-methylpropan-1-ol do. Here neither chain is branched; the carbons are merely distributed differently about the ether oxygen.

   3. C. position isomerism

      Position isomers keep the same skeleton and the same group, moving only its locant, as propan-1-ol and propan-2-ol do. These two molecules do not share one skeleton with the oxygen shifted along it.

   4. D. functional isomerism

      Functional isomers belong to different classes, as diethyl ether and butan-1-ol do. Both compounds here carry the ether group, so the functional group has not changed.

   Explanation

   Structural isomers with the same molecular formula are sorted by what actually differs. When the functional group is the same and the molecule is divided by that group into two alkyl parts, a redistribution of carbon atoms between those two parts is metamerism, the kind shown by ethers, ketones, secondary amines and esters.