Blogger Widgets APEX INSTITUTE : Class - XII Chemistry Sample Paper

Tuesday, September 16, 2014

Class - XII Chemistry Sample Paper

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.
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.
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 ---->
(i) P4(s) + 3NaOH(aq) + 3H2O(e) ----> PH3 + 3NaH2PO2
(ii) 2I-(aq) + H2O(l) + O3(g) ----> I2(g) + O2(g) + 2OH-(aq)
14. Which ones in the following pairs of substances undergoes SN2 substitution reaction faster and why?
undergoes SN2 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 SN2 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? (Kf for water is 1.86 °C mol−1
Ans. Molality, m = 0.0711 m
Kf = 1.86ºC mol-1
∴ Depression in freezing point = Kf × m
                                            = 1.86 × 0.0711
                                            = 0.132oC
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?
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.
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 2px2 2py22pz1. 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
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. EºM3+/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, t2glevel 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.

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