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STD: XII
QUARTERLY EXAMINATION – 2024
CHEMISTRY
Total Marks: 70
Time: 3.00 Hours

I) Answer all the questions: 15 × 1 = 15

  1. Wolframite ore is separated from tinstone by the process of
    a) Smelting
    b) Calcination
    c) Roasting
    d) Electromagnetic separation
  2. Which of these is not a monomer for a high molecular mass silicone polymer?
    a) Me₃SiCl
    b) PhSiCl₃
    c) MeSiCl₃
    d) Me₂SiCl₂
  3. P₂O₃ reacts with cold water to give
    a) H₃PO₃
    b) H₄P₂O₃
    c) HPO₃
    d) H₃PO₄
  4. In acidic medium, potassium permanganate oxidises oxalic acid to
    a) Oxalate
    b) Carbon dioxide
    c) Acetate
    d) Acetic acid
  5. Potassium has a bcc structure with nearest neighbour distance 4.52 Å. Its atomic weight is 39. Its density will be
    a) 915 kg m⁻³
    b) 2142 kg m⁻³
    c) 452 kg m⁻³
    d) 390 kg m⁻³
  6. If the initial concentration of the reactant is doubled, the time for the half reaction is also doubled. Then the order of the reaction is
    a) Zero
    b) one
    c) Fraction
    d) None
  7. For the reaction 2A + B → 3C + D which of the following does not express the reaction rate?
    a) d[D]/dt
    b) −d[A]/2dt
    c) d[C]/3dt
    d) −d[B]/dt
  8. Equal volumes of three acid solutions of pH 1, 2 and 3 are mixed in a vessel. What will be the H⁺ ion concentration in the mixture?
    a) 3.7 × 10⁻²
    b) 10⁻⁶
    c) 0.111
    d) None of these
  9. Assertion: Tertiary alcohols undergo dehydration more readily than primary alcohol.
    Reason: Tertiary alcohols are less acidic than primary alcohols.

a) Both assertion and reason are true and reason is the correct explanation of assertion.
b) Both assertion and reason are true and reason is not the correct explanation of assertion.
c) Assertion is true but reason is false.
d) Both assertion and reason are false.

  1. HO–CH₂–CH₂–OH on heating with periodic acid gives
    a) methanol
    b) ethanol
    c) Methanal
    d) CO₂
  2. The IUPAC name of the given compound is
    a) but-3-enoic acid
    b) but-1-ene-4-oic acid
    c) but-2-ene-1-oic acid
    d) but-3-ene-1-oic acid
  3. Which one of the following pairs is not correctly Matched?

Reducing agent Name of the reaction

a) Zn/Hg/Conc HCl Clemmensen reduction
b) LiAlH₄ Wolf-kishner’s reduction
c) Pd/BaSO₄ Rosenmund’s reduction
d) SnCl₂/Conc HCl Stephen’s reduction

  1. The oxidation state of chlorine in Cl₂O₇ is
    a) +6
    b) +7
    c) +4
    d) +5
  2. Reason for Lanthanoid contraction is
    a) Increasing nuclear charge
    b) decreasing nuclear charge
    c) Imperfect shielding effect of 4f orbitals
    d) both (a) & (c)
  3. In diborane, the number of electrons that account for banana bonds is
    a) six
    b) two
    c) four
    d) three

II) VERY SHORT ANSWER

  1. Give the basic requirement for vapour phase refining.

The metal is treated with a suitable reagent to form a volatile compound. Then the volatile compound is decomposed to give the pure metal at high temperature.

  1. CO is a reducing agent, justify with an example.

Carbon monoxide acts as a strong reducing agent. Carbon monoxide thus has a relatively high tendency to be oxidised to form carbon dioxide.

3CO + Fe₂O₃ → 2Fe + 3CO₂

  1. What are interhalogen compounds? Give examples.

Each halogen combines with other halogens to form a series of compounds called inter halogen compounds.

Example: ClF₃, IF₇

  1. Why d-block elements form complexes?

i. Transition metal ions are small and highly charged.
ii. They have vacant low energy orbitals to accept an electron pair donated by other group.

Examples: [Fe(CN)₆]⁴⁻, [Co(NH₃)₆]³⁺, etc.

  1. Define unit cell.

A basic repeating structural unit of a crystalline solid is called a unit cell.

  1. Identify the order for the following reaction.

a) Rusting of iron – First order reaction
b) ₉₂U²³⁸ Radioactive disintegration – First order reaction

  1. What are the limitations of Arrhenius concept?

i. Arrhenius theory does not explain the behaviour of acids and bases in non aqueous solvents such as acetone, Tetrahydrofuran etc.

ii. This theory does not account for the basicity of the substances like ammonia (NH₃) which do not possess hydroxyl group.

  1. Benzoin Condensation
    • C6H5-C+H-C-C6H5
  1. Alcohols having higher boiling point than aldehydes, alkanes and ethers – Why?

Because of intermolecular Hydrogen bonding between alcohol, They are having higher boiling point than others.

III) SHORT ANSWER

  1. Give the limitations of Ellingham diagram.
  2. It gives information about the thermodynamic feasibility of a reaction.
  3. It does not tell anything about the rate of the reaction.
  4. More over it does not give any idea about the possibility of other reactions that might be taking place.
  5. The interpretation of ΔG is based on the assumption that the reactants are in equilibrium with the product which is not always true.
  6. Explain McAfee process.

Aluminium chloride is obtained by heating a mixture of alumina and coke in a current of chlorine.

2Al₂O₃ + 3C + 6Cl₂ → 4AlCl₃ + 3CO₂

  1. Uses of potassium permanganate:
  2. It is used as a strong oxidizing agent.
  3. It is used for the treatment of various skin infections and fungal infections of the foot.
  4. It is used in water treatment industries to remove iron and hydrogen sulphide from well water.
  5. It is used as a Bayer’s reagent for detecting unsaturation in an organic compound.
  6. It is used in quantitative analysis for the estimation of ferrous salts, oxalates, hydrogen peroxide and iodides.
  7. Write a note on Frenkel defect.

Frenkel defect arises due to the dislocation of ions from its crystal lattice. The ion which is missing from the lattice point occupies an interstitial position. This defect occurs when cation and anion differ in size. Unlike Schottky defect, this defect does not affect the density of the crystal.

Ex: AgBr

  1. Derive integrated rate law for a zero order reaction.

A → Product

Rate = k[A]⁰

∴ [A]⁰ = 1

−d[A]/dt = k

−d[A] = kdt

Integrating between the limits of concentration [A₀] to [A] and time t = 0 to t = t.

−∫[A₀]^[A] d[A] = k∫₀ᵗ dt

−([A] − [A₀]) = k(t)₀ᵗ

[A₀] − [A] = kt

k = ([A₀] − [A]) / t

  1. Lewis acids and Lewis bases

Lewis acids:

  1. Electron deficient molecules
    Ex: BF₃, AlCl₃
  2. All metal ions (or) atoms
    Ex: Fe²⁺, Fe³⁺, Cr³⁺
  3. Molecules that contain a polar double bond
    Ex: SO₂, CO₂
  4. Molecules in which the central atom can expand its octet due to the availability of empty d-orbitals
    Ex: SiF₄, SF₄, FeCl₃ etc.
  5. Carbonium ion: (CH₃)₃C⁺

Lewis bases:

  1. Molecules with one (or) more lone pairs of electrons.
    Ex: NH₃, H₂O, R–OH
  2. All anions.
    Ex: F⁻, Cl⁻, CN⁻
  3. Molecules that contain carbon – carbon multiple bond.
    Ex: CH₂=CH₂, CH≡CH
  4. All metal oxides.
    Ex: CaO, MgO, Na₂O etc.
  5. Carbanion: CH₃⁻
  6. Preparation of Trinitroglycerine (TNG)?

CH₂–OH
|
CH–OH + 3 HONO₂
|
CH₂–OH

        Conc. H₂SO₄
        −3H₂O

→

CH₂–O–NO₂
|
CH–O–NO₂
|
CH₂–O–NO₂

Propan-1,2,3-triol
glycerol

1,2,3-trinitroxy propane

  1. Tests for carboxylic acid group?

i) In aqueous solution carboxylic acid turn blue litmus red.

ii) Carboxylic acids give effervescence with sodium bicarbonate due to the evolution of CO₂.

iii) When carboxylic acid is warmed with alcohol and Con. H₂SO₄ it forms an ester, which is detected by its fruity odour.

  1. Sulphurous Acid H₂SO₃

HO–S(=O)–OH

Marshall’s Acid (H₂S₂O₈)

HO–S(=O)₂–O–O–S(=O)₂–OH

IV. DETAIL

  1. (a) Explain froth floatation method.

This is used to concentrate sulphide ores such as galena (PbS), Zinc blende (ZnS) etc.

Metallic ore particles preferentially wetted by oil can be separated from gangue. Crushed ore is mixed with water and a frothing agent like pine oil or eucalyptus oil.

A small amount of sodium ethyl xanthate is added as a collector. A froth is formed by blowing air through the mixture. The collector molecules attach to the ore particles and make them water repellent. As a result ore particles wetted by the oil rise to the surface along with the froth. The froth is skimmed off and dried to recover the concentrated ore.

Gangue particles preferentially wetted by water settle at the bottom. When sulphide ore contains other metal sulphides as impurities, depressing agents such as sodium cyanide, sodium carbonate etc. are used to selectively prevent other from coming to the froth.

For example, when impurities such as ZnS is present in Galena (PbS), Sodium cyanide NaCN is added to depress the flotation property of ZnS by forming a layer of zinc complex Na₂[Zn(CN)₄] on the surface of ZnS.

  1. (b) (i) What is Inorganic benzene? How it is obtained?

When treated with excess ammonia at low temperatures diborane gives diboranediammoniate. On heating at higher temperatures it gives borazole.

3B₂H₆ + 6NH₃
→ 3B₂H₆·2NH₃ (or) 3[BH₃(NH₃)₂][BH₄]

3B₂H₆ + 2NH₃
High temp / Closed vessel
→ 2B₃N₃H₆ (Borazole or Borazine – Inorganic benzene)

(ii) Uses of silicones.

i) Silicones are used for high temperature oil baths.

ii) They are used for making water proofing clothes.

iii) They are used as insulting material in electrical motor and other electrical appliances.

iv) They are mixed with paints and enamels to make them resistant towards high temperature, sunlight, dampness and chemicals.

  1. (a) How bleaching powder is prepared?

It is prepared by passing chlorine gas through dry slaked lime (calcium hydroxide).

Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O

(b) Holmes signal:

Phosphine is used for producing smoke screen.

In a ship, a container with a mixture of calcium carbide and calcium phosphide, liberates phosphine and acetylene when thrown into the sea. The liberated Phosphine catches fire and ignites acetylene. These burning gases serves as a signal to the approaching ships. This is known as Holmes signal.

  1. (b) Explain the preparation of potassium dichromate.

Extraction of Potassium dichromate from its ore:

  1. Ore: Chromite – FeO.Cr₂O₃
  2. Conversion of chrome iron ore to sodium chromate

4FeO.Cr₂O₃ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂↑

  1. Conversion of Na₂CrO₄ to Na₂Cr₂O₇

2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ + Na₂SO₄ + H₂O

  1. Conversion of sodium dichromate into potassium dichromate

Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl

  1. (a) Percentage efficiency of packing in Simple Cubic Crystal:

Packing efficiency =
(total volume occupied by spheres in a unit cell / volume of the unit cell) × 100

Volume of cube = a × a × a = a³

Radius of the sphere from figure,
a = 2r
r = a/2

Volume of the sphere with radius r:

= 4/3 πr³

= 4/3 π(a/2)³

= πa³/6

The number of spheres belongs to a unit cell in SC arrangement is 1.

Packing efficiency =
(1 × πa³/6 / a³) × 100

= 52.31%

  1. (b) Define half-life of a reaction?

It is defined as the time required for the reactant concentration to reach half its initial value.

Half-life of a first order reaction.

k = 2.303/t log [A₀]/[A]

If t = t₁/₂, then [A] = [A₀]/2

k = 2.303/t₁/₂ log [A₀]/[A₀/2]

k = 2.303/t₁/₂ log 2

k = 2.303 × 0.3010 / t₁/₂

k = 0.6932/t₁/₂

t₁/₂ = 0.6932/k

For a first order reaction, the half-life is a constant i.e., it does not depend on the initial concentration.

  1. (a) Henderson – Hasselbalch equation

In an acidic buffer solution,

[H₃O⁺] = Kₐ [acid]eq / [base]eq

Due to common ion effect,

[acid]eq = [acid]
[base]eq = [salt]

[H₃O⁺] = Kₐ [acid]/[salt]

Taking logarithm on both sides,

log [H₃O⁺] = log Kₐ + log [acid]/[salt]

reverse the sign on both sides,

−log [H₃O⁺] = −log Kₐ − log [acid]/[salt]

  1. (b)

An organic compound (A) of Molecular formula (C₂H₆O) is Ethanol
= CH₃CH₂–OH

CH₃CH₂–OH
Conc. H₂SO₄
→ CH₂=CH₂

CH₂=CH₂ + H₂O
Cold alkaline KMnO₄
[O]
→

CH₂–CH₂
| |
OH OH

ethene ethane-1,2-diol

A – CH₃CH₂–OH Ethanol
B – CH₂=CH₂ Ethylene
C – CH₂–CH₂ Glycol
| |
OH OH

  1. (a) Explain Saytzeff’s rule?

During intramolecular dehydration, if there is a possibility to form a carbon – carbon double bond at different locations, the preferred location is the one that gives the more (highly) substituted alkene i.e., the stable alkene.

Example:

CH₃–C(CH₃)₂–CH(OH)–CH₃
Conc. H₂SO₄
→

Minor product: less substituted alkene

Major product: more substituted alkene

  1. (b) Explain Cannizzaro reaction mechanism.

Cannizzaro reaction

C₆H₅CHO + C₆H₅CHO
50% NaOH
→ C₆H₅CH₂OH + C₆H₅COONa

Benzaldehyde Benzyl alcohol
+
Sodium Benzoate

Step – 1:

C₆H₅–CHO + OH⁻
fast
→ C₆H₅–C(OH)(O⁻)–H

Step – 2:

C₆H₅–C(OH)(O⁻)–H + C₆H₅–CHO
slow
→ C₆H₅–COOH + C₆H₅CH₂O⁻

Step – 3:

C₆H₅–COOH + C₆H₅CH₂O⁻
Proton exchange
→ C₆H₅–COO⁻ + C₆H₅CH₂OH

Benzoate Benzyl alcohol

ANSWER KEY – 2024

  1. d) Electromagnetic separation
  2. a) Me₃SiCl
  3. a) H₃PO₃
  4. b) Carbon dioxide
  5. a) 915 kg m⁻³
  6. a) Zero
  7. c) −d[C]/3dt
  8. a) 3.7 × 10⁻²
  9. b) Both assertion and reason are true and reason is not the correct explanation of assertion
  10. c) Methanal
  11. a) but-3-enoic acid
  12. b) LiAlH₄ – Wolf-kishner reduction
  13. b) +7
  14. d) both (a) & (c)
  15. c) four

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