Class - XII Chemistry Sample Paper

Instructions:

1. Section-A  Q. no. 1-8 carry 1 mark each.
2. Section-B  Q. no. 9-18 carry 2 marks each.
3. Section-C  Q. no. 19-27 carry 3 marks each.
4. Section-D  Q. no. 28-30 carry 5 marks each.

Section-A

1. What is the total number of atoms per unit cell in a face-centred cubic (fcc) structure?

Ans. The total number of atoms per unit cell in a face-centred cubic (fcc) structure is 4 (8 corner atoms  atom per unit cell + 6 face-centred atoms  atom per unit cell)

2. Why is an alkylamine more basic than ammonia.

Ans. An alkylamine is more basic than ammonia because of inductive effect (+I effect). Alkyl group or   ‘R’ has an electron-releasing effect, which increases electron density over nitrogen atom. This increases   its basicity.

3. Which point defect in crystals of a solid decreases the density of the solid?

Ans. Vacancy defect decreases the density of a substance. Vacancy defect in ionic solids is known as Schottky defect.

4. Name one solid which has both Schottky and Frenkel defects.

Ans. AgBr (Silver Bromide)

5. Why is Bi (V) a stronger oxidant than Sb (V)?

Ans. Bismuth and antimony both belong to the nitrogen family and exhibit the +5 oxidation state. However, on moving down the group, i.e., from antimony to bismuth, the stability of the +5 oxidation  state decreases. This is due to the inert pair effect. Thus, Bi (V) is a stronger oxidant than Sb (V).

6. Give the IUPAC name of the following compound:

Ans. 2-bromo,3-methyl pent-2-en,1-ol.

7. Name a substance that can be used as an antiseptic as well as a disinfectant.

Ans. Phenol can be used as an antiseptic as well as a disinfectant. 0.1% solution of phenol is used as an antiseptic and 1% solution of phenol is used as a disinfectant.

8. What causes Brownian movement in a colloidal solution?

Ans. The particles of a colloidal solution keep moving in a zigzag pattern. This happens because of unbalanced bombardment of the colloidal particles by the molecules of the dispersion medium.

    The particles of the dispersion medium keep striking the particles of the dispersion phase from all directions and with different force. This transfers the kinetic energy from the particles of the dispersion medium to the particles of the dispersion phase in a random manner. As the collisions are random in nature, the movement of the particles is also random.

Section-B

9. The atomic radii increases considerably from N to P but very little increase is observed from As to Bi. why?

Ans. There is a considerable increase in size from N to P as expected but due to the presence of completely filled d- orbitals which have very poor shielding effects, the increases in size is very little from As to Bi.

10. List the reactions of glucose which cannot be explained by its open-chain structure.

Ans. Limitations of the open-chain structure:

Although the open-chain structure of D(+)-glucose explains most of its reactions, it fails to explain the following facts.

(i) D(+)-glucose does not undergo certain reactions of aldehydes. For example, glucose does not from 2, 4-DNP derivative, and does not respond to Schiff’s reagent test.

(ii) Glucose reacts with NH₂OH to form an oxime, but glucose penta-acetate does not.

(iii) Glucose does not form the hydrogen sulphite addition product with NaHSO₃.

(iv) The process of mutarotation in glucose.

11. On what basis are polymers classified ?

Ans. Polymers are classified on the basis of

(a) Source

(b) Structure

(c) Mode of polymerization

(d) Molecular forces

12. Illustrate the following reactions giving a chemical equation for each:

(i) Kolbe’s reaction,
(ii) Williamson synthesis.

Ans. (i) Kolbe’s Reaction

It involves the reaction between sodium phenoxide and carbon dioxide under a pressure of 4 − 7 atmospheres at 398 K to form sodium salicylate which on hydrolysis with minerals acids gives salicylic acid.

(ii) Williamson synthesis

This reaction is used to prepare both symmetrical as well as unsymmetrical ethers. It involves the treatment of alkyl halide with sodium alkoxide or sodium phenoxide.

CH₃I  +  NaOC₂H₅  ----> CH₃OC₂H₅ + NaI
Mrthyl    Sodium            Methoxy 
iodide     ethoxide           ethane

13. Complete the following chemical reaction equations:

(i) P₄ +  NaOH + H₂ O ---->
(ii) I^- + H₂ O + O₃ ---->

Ans.
(i) P_4(s) + 3NaOH_(aq) + 3H₂O_(e) ----> PH₃ + 3NaH₂PO₂

(ii) 2I^-_(aq) + H₂O_(l) + O_3(g) ----> I_2(g) + O_2(g) + 2OH^-_(aq)

14. Which ones in the following pairs of substances undergoes S_N2 substitution reaction faster and why?

(I) 

(II)
 or   

Ans.

undergoes S_N2 substitution reaction faster than . This is because the alkyl group present in benzyl chloride increases its basicity due to +I effect. Stronger the base, lesser is its leaving ability. So, reacts faster.

(ii) Iodide is a weaker base than chloride. Weaker the base, greater is its leaving ability. So,  undergoes S_N2 substitution reaction faster.

15. Explain the following terms:

(i) Invert sugar
(ii) Polypeptides

Ans. (i) The product formed on the hydrolysis of sucrose with dilute acids or enzyme invertase is called invert sugar. These are known so because the sign of rotation changes from dextro (+) to laevo (−).

(ii) Two amino acids combine to form a peptide bond. When the number of combining amino acids is more than ten, the product obtained is known as polypeptide.

16. Calculate the freezing point depression expected for 0.0711 m aqueous solution of Na₂SO₄. If this solution actually freezes at −0.320°C, what would be the value of Van’t Hoff factor? (K_f for water is 1.86 °C mol⁻¹) 

Ans. Molality, m = 0.0711 m

K_f = 1.86ºC mol^-1

∴ Depression in freezing point = K_f × m
                                            = 1.86 × 0.0711
                                            = 0.132^oC

Freezing point = 0ºC − 0.132ºC
                      = −0.132ºC

Now, Van’t Hoff factor,

[Note − Theoretically the value of Vant Hoff factor should be 3 but according to the values given in the question, the value of Vant Hoff factor is coming out to be 2.42. ]

17. The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if the conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10⁻³ S cm⁻¹?  

Ans. Given,

Conductivity, κ = 0.146 × 10⁻³ S cm⁻¹

Resistance, R = 1500 Ω

∴Cell constant = κ × R
                      = 0.146 × 10⁻³ × 1500
                      = 0.219 cm⁻¹

18. Draw the structures of white phosphorus and red phosphorus. Which one of these two types of phosphorus is more reactive and why?

Ans.

     

White phosphorus is more reactive than red phosphorus.

The various P₄ molecules of white phosphorus are held together by weak Vander Waal’s force of attraction, which makes it very reactive. On the other hand, molecules of red phosphorus are joined by covalent bonds to give a polymeric structure, which makes it very stable and less reactive.

Section-C

19. How would you account for the following?

(i) Frenkel defects are not found in alkali metal halides.
(ii) Schottky defects lower the density of related solids.
(iii) Impurity doped silicon is a semiconductor.

Ans. (i) Frenkel defects are shown by ionic solids having large differences in the sizes of ions. Solids such as ZnS, AgCl show these defects due to the small size of Zn²⁺ and Ag⁺ ions, and the large size of anions. Alkali metals are not so small so as to show these defects. Hence, Frenkel defects are not found in alkali metal halides.

(ii) Schottky defects are basically vacancy defects in ionic solids. In these defects, lattice sites become vacant. As a result, the density of the substance decreases.

(iii) Silicon is an intrinsic semi-conductor in which conductivity is very low. To increase its conductivity, silicon is doped with an appropriate amount of suitable impurity. When doped with electron-rich impurities such as P or As,n-type semi-conductor is obtained, and when doped with electron-deficient impurities, p-type semi-conductor is obtained. In n-type semiconductor, negatively charged electron is responsible for increasing conductivity, and inp-type semiconductor, electron hole is responsible for increasing conductivity.

20. Account for the following:

(i) NH₃ is a stronger base than PH₃.
(ii) Sulphur has a greater tendency for catenation than oxygen.
(iii) Bond dissociation energy of F₂ is less than that of Cl₂.

Ans. (i) On moving from nitrogen to phosphorus, i.e., down the group, the atomic size increases. As the size of the central atom increases, the lone pair of electrons occupy a larger volume. Consequently, the electron density, and hence, the basic strength decrease. Thus, NH₃ is a stronger base than PH₃.

(ii) The tendency for catenation depends upon the bond energy. The bond energy of a sulphur molecule is more than that of an oxygen molecule. Thus, the sulphur − sulphur bond strength is higher, and as a result, the tendency of catenation is also higher. Sulphur shows catenation up to eight atoms.

(iii) Bond dissociation energy of fluorine is less than that of chlorine. It is due to the low value of electron affinity of small-sized fluorine. Also, the value of enthalpy of hydration of fluorine is much higher than that of chlorine.

21. Compare the following complexes with respect to their shape, magnetic behaviour and the hybrid orbitals involved:

(i) [CoF₄]²⁻
(ii) [Cr(H₂O)₂ (C₂O₄)₂]⁻
(iii) [Ni(CO)₄]

(Atomic number: Co = 27, Cr = 24, Ni = 28)

Ans. (i) [ CoF₄]²⁻

Let the oxidation state of Co in [CoF₄]²⁻ be x.

x − 4 = −2
∴   x = +2

Due to the unpaired electrons, it is paramagnetic in nature.

(ii) [Cr(H₂O)₂ (C₂O₄)₂]⁻

Let the oxidation state of Cr in [Cr(H₂O)₂ (C₂O₄)2]⁻ be x.

x − 2 × 2 = −1
∴         x = +3

Thus, it has d²sp³ hybridisation with octahedral shape. Due to the presence of unpaired electrons, it is paramagnetic in nature.

(iii) [Ni(CO)₄]

Let the oxidation state of Ni in [Ni(CO)₄] be x.

x + 4(0) = 0
∴       x = 0

22. What are the following substances? Give one example of each type.

(i) Antacid
(ii) Non-ionic detergents
(iii) Antiseptics

Ans. (i) Antacids are stomach acid neutralisers. They raise the pH to reduce acidity in the stomach.

Example − aluminium hydroxide

(ii) Non-ionic detergents produce electrically neutral colloidal particles in solution.

Example − esters of high molecular mass formed by reaction between stearic acid and polyethylene glycol

(iii) Antiseptics are used for destroying microorganisms off the skin of humans and animals.

Example − boric acid

23. What is the difference between multimolecular and macromolecular colloids? Give one example of each. How are associated colloids different from these two types of colloids?

Ans. In multimolecular colloids, the colloidal particles are aggregates of atoms or small molecules having molecular sizes less than 1 nm. Example − gold sol

    Macromolecular colloids are formed by the dissolution of macromolecules whose sizes lie in the colloidal range. Example − starch

    Certain substances behave as strong electrolytes at low concentrations, but at higher concentrations, these substances exhibit colloidal properties because of the formation of aggregates particles. These are called micelles or associated colloids. Example −soap in water

24. Explain the following observations:

(i) Fluorine does not exhibit any positive oxidation state.
(ii) The majority of known noble gas compounds are those of Xenon.
(iii) Phosphorus is much more reactive than nitrogen.

Ans. (i) Electronic configuration of fluorine is 1s² 2s² 2p_x² 2p_y²2p_z¹. It has only one half-filled orbital and there is no d orbital available for the excitation of electrons. Also, it is the most electronegative element. Hence, it shows −1 oxidation state, and no positive oxidation state.

(ii) All noble gases have full s and p outer electron shells, so they do not form chemical compounds easily. However, in heavier members like Xe, the outermost electrons experience a shielding effect from the inner electrons, so they are comparatively easily ionised. Hence, the first ionisation energy is roughly equivalent to that of molecular oxygen. Xe reacts with electronegative elements like fluorine and oxygen to form stable compounds.

(iii) Small-sized nitrogen forms a triple bond with another nitrogen atom and exists as a very stable N₂ molecule. On the other hand, on moving from nitrogen to phosphorus, atomic size increases. Phosphorus is more reactive owing to the bond strain in the P₄ molecule. Also, bond enthalpy is much less for phosphorus molecule.

25. Explain a chemical test to distinguish between primary, secondary and tertiary alcohols.

Ans. Primary tertiary and secondary alcohols can be distinguished by oxidation reaction.Primary alcohols give aldehyde with CrO₃

RCH₂OH----> RCHO (in presence of CrO₃)

Secondary alcohols give ketone with CrO₃

R₂ CHOH ----> R₂ CO

Tertiary alcohol do not get oxidized with CrO₃

R₃ COH ----> no reaction (in presence of Cr O₃ )

26. How do antiseptics differ from disinfectants? Give one example of each type.

Ans. A disinfectant is used for destroying microorganisms off any surface such as floors and non-living objects. Examples of disinfectant are alcohols and aldehydes.

An antiseptic is used for destroying microorganisms off the skin of humans or animals. Examples of antiseptic are boric acid and hydrogen peroxide.

27. Name the following coordination compounds according to IUPAC system of nomenclature:

(i) [Co(NH₃)₄ (H₂O)Cl] Cl₂
(ii) [CrCl₂ (en)₂] Cl, (en = ethane − 1, 2 − diamine)

Ans. (i) Tetraammineaquachlorocobalt(III) chloride

(ii) Dichlorobis(ethylenediamine)chromium(III) chloride

Section-D

28. Describe the following giving a suitable example in each case:

(i) Decarboxylation
(ii) Cannizzaro’s reaction

Ans. (i) Decarboxylation refers to the reaction in which carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with soda-lime.

    Decarboxylation also takes place when aqueous solutions of alkali metal salts of carboxylic acids are electrolysed. This electrolytic process is known as Kolbe’s electrolysis.

(ii) The self oxidation−reduction (disproportionation) reaction of aldehydes having no α-hydrogens, on treatment with concentrated alkalis is known as the Cannizzaro reaction.

    In this reaction, two molecules of aldehydes participate, and one of these is reduced to alcohol while the other is oxidised to carboxylic acid. For example, when ethanal is treated with concentrated potassium hydroxide, ethanol and potassium ethanoate are produced.

29. Explain the following observations:

(i) In general the atomic radii of transition elements decrease with atomic number in a given series.

(ii) The  for copper is positive (+ 0.34 V). It is the only metal in the first series of transition elements showing this type of behaviour.

(iii) The E⁰ value for Mn³⁺ | Mn²⁺ couple is much more positive than for Cr³⁺ | Cr²⁺ or Fe³⁺ | Fe²⁺ couple.

Ans. (i) Among the elements of a particular transition series, as atomic number increases, atomic radius decreases. This is due to the fact that on moving across the period, electrons are added to the same d orbital. Thus, effective nuclear charge increases, which is only partly cancelled by the increased screening effect of electrons. As a result, atomic radius decreases.

(ii) Copper shows a unique behaviour in its Eº value. The electrode potentials are the measure of the value of enthalpy change. Furthermore, enthalpy change in a reaction is the sum of ionisation energy, enthalpy of hydration of metal ion and sublimation of metal. Cu has very low enthalpy of hydration & high ΔHº value, thereby resulting in a positive electrode potential value.

(iii) The change from Mn³⁺ to Mn²⁺ results in the formation of half-filled d⁵configuration, which has extra stability. Eº_M³⁺/_M²⁺ has a high positive value since the third ionisation energy is higher. On the other hand, the low value of Fe is due to the extra stability of Fe³⁺, i.e., d⁵ configuration. Cr²⁺ is reducing as its configuration changes from d⁴ to d³. In d³ configuration, t_2glevel is half-filled. So, Cr³⁺/ Cr²⁺ has a low value.

30. (a) Calculate the emf of the cell
Mg(s) | Mg²⁺ (0.1M) || Cu²⁺ (1 × 10⁻³ M) | Cu (s)
Given: E° Cu²⁺/Cu = +0.34V, E° Mg²⁺/Mg = −2.37 V

(b) Explain with examples the terms weak and strong electrolytes.

Ans. (a) For the given cell,

Now, emf of the cell is given by

        = 2.651 V

(b) Strong electrolytes are substances which dissociate completely (100%) in solution. Strong acids (such as HCl, HNO₃), strong bases (such as NaOH, KOH) and salts that are not weak acids or weak bases (such as NaCl, KBr) are strong electrolytes.

Weak electrolytes are substances which dissociate partially (1% − 10%) in solution. Weak acids (such as CH₃COOH, H₂CO₃) and weak bases (such as NH₃, C₅H₅N) are weak electrolytes.