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CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions

Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) to familiarize themselves with the exam format and marking scheme.

CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions

Time allowed : 3 hours
Maximum marks : 70

General Instructions:

  1. All questions are compulsory.
  2. Questions number 1 to 8 are very short answer questions and carry 1 mark each.
  3. Questions number 9 to 18 are short answer questions and carry 2 marks each.
  4. Questions number 19 to 27 are also short answer questions and carry 3 marks each.
  5. Questions number 28 to 30 are long answer questions and carry 5 marks each.
  6. Use log tables if necessary. Use of calculators is not allowed.
    Deleted from Syllabus.

SET I

Question 1.
Why are crystalline solids anisotropic?
Answer:
Crystalline solids show different values of their some properties like electrical conductivity, refractive index, thermal expansion etc. in different directions.

Question 2.
What are emulsions? Name an emulsion in which water is a dispersed phase.
Answer:
Emulsions : An emulsion is a colloidal dispersion in which both the dispersed phase and dispersion medium are liquids.
Water in oil → Butter, cold creams.

Question 3.
What are the collectors used in froth floatation process? Name a substance that can be used as such.
Answer:
Collectors : Pine oils, fatty acids etc. enhance non-wettability of the mineral particles.

Question 4.
Why is F2 a stronger oxidising agent than Cl2?
Answer:
Due to low bond dissociation enthalpy and high electronegativity of Fluorine, it has strong tendency to accept electrons and thus get reduced.
F + e → F
Therefore F2 acts as strong oxidising agent, while Cl2 is weak oxidising agent due to low electronegativity.

Question 5.
Name the alcohol that is used to make the following ester :
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 1
Answer:
Alcohol used : Propan-2ol

Question 6.
Give a test to distinguish between propan-2-one and pentan-3-one.
Answer:
Propan-2-one and pentan-3-one can be distinguished by Iodoform test.
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 2

Question 7.
How does a homopolymer differ from a copolymer?
Answer:
Homopolymers : Polymers whose repeating structural units are derived from only one type of monomer units are called homopolymers.
Example : Polyethene, PVC, teflon.

Copolymers : Polymers whose repeating structural units are derived from two or more types of monomer molecules are copolymers.
Example : Buna-S, Nylon 66

CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions

Question 8.
Define a ‘Peptide linkage’.
Answer:
Peptide linkage : It is an amide linkage formed between -COOH group of one a-amino acid and NH2, group of the other α-amino acid by loss of a molecule of water. -CO – NH – bond is called Peptide linkage.

Question 9.
Set up Neverest equation for the standard dry cell. Using this equation show that the voltage of a dry cell has to decrease with use.
Answer:
Cell reaction of a dry cell can be represented as
Zn + Hg2+ → Zn2+ + Hg (n = 2)

Nernst equation Ecell = \($\mathrm{E}_{\text {cell }}^{\circ}$\) – \(\frac{0.0591}{2}\) log \($\frac{\left[\mathrm{Zn}^{2+}\right]}{\mathrm{Hg}^{2+}}$\)
The voltage of dry cell has to decrease because the concentration of electrolyte decreases in the reaction.

Question 10.
How does a change in temperature affect the rate of a reaction? How can this effect on the rate constant of a reaction be represented quantitatively?
Sol.
The rate constant of a reaction increases with increase of temperature and becomes nearly double for ever, 10°C rise in temperature.
The effect can be represented quantitatively by Arrhenius equation
K = Ae-Ea/RT
where Ea = Activation energy of the reaction
A = Frequency factor

Question 11.
Describe the underlying principle of each of the following processes :
(i) Recovery of silver from the solution obtained by leaching silver ore with a solution of NaCN.
(ii) Electrolytic refining of a crude metal.
Answer:
(i) NaCN acts as a leaching agent or oxidising agent, thus oxidises Ag to Ag+ which then combines with CN~ ions to form respective soluble complex
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 3

(ii) Electrolytic refining : Here the impure metal is made to act as anode and a strip of the same metal in pure form is used as cathode. When they both are put in suitable electrolyte containing soluble salt of same metal, the more basic metal remains in the solution and the less basic ones go to the anode mud.
Example : In refining of Cu
At anode : (oxidation)
Cu → Cu2+ + 2e
At cathode : (reduction)
Cu2+ + 2e → Cu

Or

Describe the principle involved in each of the following processes
(i) Zone refining of a metal
(ii) Vapour phase refining of metals
Answer:
(i) Zone refining of metals : It is based on the principle that the impurities are more soluble in melt than in the solid state of the metal.
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 4

(ii) Vapour phase refining :
Vapour phase refining of inetals Here the metal is converted into its volatile compound and collected elsewhere which then decomposed to give pure metal. The requirements are

the metal should form a volatile compound with an available reagent.
the volatile compound should be easily decomposable so that recovery is easy.

Example :
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 5

Question 12.
Complete the following chemical equations :
(i) SO2 + MnO4 + H2O →
(ii) F2+ (g) + H2O (l) →
Answer:
(i) 5SO2 + 2MnO4 + 2H2O → 5SO42- + 2Mn2+ + 4H+

(ii) Fluorine oxidises H2O to O2 and O3
2F2 (g) + 2H2O (l) → 4H+ (aq) + 4F (aq) + O2 (g)
3F2 (g) + 3H2O (l) → 6H+ (aq) + 6F (aq) + O3 (g)

Question 13.
Assign reasons for the following :
(i) Copper(I) ion is not known to exist in aqueous solutions.
(ti) Both O2 and F2 stabilize high oxidation states of transition metals but the ability of oxygen to do so exceeds that of fluorine.
Answer:
(z) See Q. 23 (i), 2014 Comptt. (II Outside Delhi). [Page 78
(i) Cu2+ (aq) is much more stable than Cu2+ (aq). This is because although second ionization
enthalpy of copper is large but ∆hydH for Cu2+ (aq) is much more negative than that for Cu+ (aq) and hence it more than compensates for the second ionization enthalpy of copper. Therefore, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation as follows:
2Cu+ → Cu2+ + Cu

(ii) Oxygen stabilizes the highest oxidation state even more than fluorine.
Example : Highest fluoride of Mn is MnF4 whereas highest oxide is Mn4O7. It is due to ability of oxygen to form multiple bonds with the metal atoms.

Question 14.
Write the IUPAC names of the following compounds :
(i) CH2 = CHCH4Br
(ii) (CCl3)3 CCl
Answer:
(i) CH2 = CHCH4 – Br
IUPAC name : 3-Bromopropene

(ii) (CCl3)3 – C – Cl
IUPAC name : 2-(Trichloromethyl) – 1, 1, 1, 2, 3, 3, 3-heptachloropropane

Question 15.
What are ambident nucleophiles? Explain with an example.
Answer:
Ambident nucleophile : A nucleophile that can form new bonds at two or more spots in its structure, usually due to resonance contributors.
Example : S = C = N can act as a nucleophile with either the S or N attacking.

Question 16.
(i) Arrange the following compounds in an increasing order of basic strength :
C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH3NH2
(ii) Arrange the following compounds in a decreasing order of pKb values :
C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2
Answer:
(i) Increasing order of basic strength :
C6H5NH2 < C6H5N(CH3) < CH3NH2 < (C2H5)2NH (More +I effect)

(ii) Decreasing order of pKb values :
C6H5NH2 < C6H5NHCH3 < C2H5NH2 < (C2H5)2NH

Question 17.
Give a chemical test to distinguish between each of the following pairs of compounds
(i) Ethylamine and Aniline
(ii) Aniline and Benzylamine
Answer:
(i) Ethylamine and aniline :
By Azodye test: It involves the reaction of any aromatic primary amine with HNO2 (NaNO2 + dil. HCl) at 273-278 K followed by treatment with an alkaline solution of 2-napthol when a brilliant yellow, orange or red coloured dye is obtained.
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 6
Aliphatic 1° amines under these conditions give a brisk evolution of N2 gas with the formation of 1° alcohol i.e., solution remains clear.
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 7

(ii) Distinction between Aniline and Benzylamine :
By Nitrous acid test : Benzylamine reacts with HNO2 to form a diazonium salt which being unstable even at low temperature, decomposes with evolution of N2 gas
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 8
Aniline on the other hand, reacts with HNO2, to form benzenediazonium chloride which is stable at 273-278 K and hence does not decompose to evolve N2 gas.
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 9

Question 18.
Write the names and structures of monomers used for getting the following polymers :
(i) Buna-S
(ii) Nylon-6, 6
Answer:
(i) Buna-S :
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 10

(ii) Nylon-6, 6 : It has two repeating monomers
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 11

CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions

Question 19.
Iron has a body centred cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm-3. Use this information to calculate Avogadro’s number. [3]
(Gram atomic mass of Fe = 55.84 g mol-1).
Answer:
Given :
a = 286.65 pm = 286.65 × 10-10, d = 7.87 g cm-3, M = 56 g mol-1
z = 2, NA = ?

Using formula : d = \(\frac{zM}{a^3 \mathrm{~N}_{A}}\) or NA = \(\\frac{z \mathrm{M}}{a^3 d}\)
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 12

Question 20.
The resistance of 0.01 M NaCl Solution at 25° C is 200 Ω. The cell constant of the conductivity cell used is unity. Calculate the molar, conductivity of the solution.
Solution:
For 0.01 M NaCl solution,
R = 200 Ω, cell constant is unity.
Conductivity (K) = \(\frac{\text { Cell constant }}{\text { Resistance }}\) = \(\frac{1}{200}\) = 0.005 Sm-1
Concentration of solution = 0.01 M = 0.01 mol L-1
= 0.01 × 103 mol m-3 = 10 mol m-3

Molar conductivity = \(\frac{K}{C_m}\) = \(\frac{0.005}{10}\) = 5 × 10-4 Sm2 mol-1

Question 21.
For a decomposition reaction, the values of k at two different temperatures are given below:
k1 = 2.15 × 10-8L/(mol.s) at 650 K
k2 = 2.39 × 10-7L/(mol.s) at 700 K
Calculate the value of Ea for the reaction.
(Log 11.11 = 1.046) (R = 8.314 J K-1 mol-1)
Solution:
Given :
k1 = 2.15 × 10-8 L mol-1 S-1, T1 = 650 K
k2 = 2.39 × 10-7 L mol-1 S-1, T2 = 700 K
R = 8.314 J K-1 mol-1, Ea = ?
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 13

Question 22.
Giving appropriate examples, explain how the two types of processes of adsorption (physisorption and chemisorption) are influenced by the prevailing temperature, the surface area of adsorbent and the activation energy of the process?
Answer:

PhysisorptionChemisorption
1. Temperature: It decreases with increase in temperature.1. It increases with increase in temperature initially after adsorption it is decreasing
2. Surface area of adsorbent: The adsorption increases with increase in surface area of adsorbent.2. It also increases with increase in surface area of adsorbent.
3. Activation energy: It requires very low or no activation energy.3. It requires high activation energy.

Or

Explain clearly how the phenomenon of adsorption finds application in (i) production of vacuum in a vessel
(ii) heterogeneous catalysis
(iii) froth floatation process in metallurgy.
Answer:
(i) Production of vacuum : In Dewar flasks, activated charcoal is placed between the walls of the flask so that any gas which enters into annular space either due to glass imperfection or diffusion through glass is adsorbed and create a vacuum.

(ii) Heterogeneous catalysis : If the catalyst is present in a different phase than that of the reactants, it is called a heterogeneous catalyst and this type of catalysis is called Heterogeneous catalysis.
Example: Manufacture of NH3 from N, and H2 by Haber’s process using iron as catalyst
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 14
Reactants are gaseous whereas catalyst is solid.

(iii) Froth Floatation process : This method is used for removing gangue from suiphide ores.
In this powdered ore is mixed with collectors (e.g. pine oils, fatty acids etc.) and froth stabilisers (e.g. cresols, aniline) which enhance non-wettability of the mineral particles and froth stabilisation respectively. As a result of which ore comes with froth and gangue remain in the solution.
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 15

Question 23.
Give reasons for the following :
(i) Transition metals exhibit a wide range of oxidation states.
(ii) Cobalt(II) is very stable in aqueous solutions but gets easily oxidised in the presence of strong ligands.
(iii) Actinoids exhibit a greater range of oxidation states than lanthanoids.
Answer:
(i) The variability of oxidation state of transition elements is due to incompletely filled d-orbitals as ns, and (n – 1) d electrons have approximate equal energies.
(ii) Co2+ ion is easily oxidised to Co3+ ion in presence of a strong ligand because of its higher crystal field energy which causes pairing of electrons to give inner orbital complexes (d2sp3).
(iii) Because of very small energy gap between 5f, 6d and 7s subshells all their electrons can take part in bonding and shows variable oxidation states.

Question 24.
Write the IUPAC name and draw the structure of each of the following complex entities :
(i) \(\left[\mathrm{Co}\left(\begin{array}{c}
\mathrm{COO} \\
| \\
\mathrm{COO}
\end{array}\right)_3\right]^{3-}\)
(ii) [Cr(CO)6]
(iii) [PtCl3(C2H4)]
(At. nos. Cr = 25, Co = 27, Pt = 78) [3]
Answer:
(i) IUPAC name : Trioxalato chromate (III) ion
(ii) IUPAC name : Hexa carbonyl chromium (o)
(iii) IUPAC name : Trichlorido ethylene platinum (II)

CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions

Question 25.
Explain the following with an example for each :
(i) Kolbe’s reaction
(ii) Reimer-Tiemann reaction
(iii) Williamson ether synthesis [3]
Solution:
(i) Kolbe’s reaction : Phenol reacts with CO2 in presence of sodium hydroxide (NaOH) at 4-7 Atm and 390 – 410 K giving salicytic acid
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 16
(ii) Reimer-Tiemann reaction :
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 17

(iii) Williamson synthesis of an ether : AlkvI Halide reacts with Alkoxide
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 18

Question 26.
What are essential and non-essential amino acids? Give two examples of each. 3
Answer:
Essential amino acids : Amino acids which the body cannot synthesize are called essential amino acids.
Example : Valine, leucine etc. Therefore they must be supplied in diet.

Non-essential amino acids : Amino acids which the body can synthesize are called non¬essential amino acids. Therefore, they may or may not be present in diet.
Example : Glycine, alanine etc.

Question 27.
What is meant by the following terms? Explain with an example for each.
(i) Target molecules as used in medicinal chemistry
(ii) Food preservatives
(iii) Non-ionic detergents [3]
Answer:
(i) Drugs interact with macromolecules like proteins, carbohydrates, lipids etc. and are called as target molecules.
(ii) Food preservatives : They are used to prevent spoilage of food due to microbial growth. Example ; Table salt, vegetable oils, sodium benzoate etc.
(iii) Non-ionic detergents : These are esters of high molecular mass alcohols obtained by reaction between polyethylene glycol and steric acid.

Example :
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 19

Question 28.
(a) What is van’t Hoff factor? What types of values can it have if in forming the solution, the solute molecules undergo
(i) Dissociation?
(ii) Association?

(b) How many mL of a 0.1 M HCl solution are required to react completely with 1 g of a mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both?
(Molar mass : Na2CO3 = 106 g, NaHCO3 = 84 g) [3]
Answer:
(i) Van’t Hoff factor : It is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property
i = \(\frac{\text { Experimental value }}{\text { Calculated value }}\)
If there is dissociation of the solute in the solution, the Van’t Hoff factor T will be greater than one i.e. i > 1.
It means observed colligative property will be greater than calculated value.

(ii) Association : If there is association of solute in the solution, the Van’t Hoff factor T’ will be less than one i.e. i < 1. Thus, observed colligative property will be less than the calculated value.

(b) Calculation of no. of moles of the components in the mixture
Suppose Na2CO3 in the mixture = x g
∴ NaHCO3 in the mixture = (1 – x) g
Molar mass of Na2CO3 = 106 g mol-1
Molar mass of NaHCO3 = 84 g mol-1

∴ Moles of Na2CO3 = \(\frac{x}{106}\)
and Moles of NaHCO3 = \(\frac{1-x}{84}\)
As the mixture contains equimolar amounts of both
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 20
No. of moles of HCl required
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
NaHCO3 + HCl → NaCl + H20 + CO2

1 mole of Na2CO3 requires HCl = 2 moles
∴ 0.00526 mole of Na2C03 requires HCl = 0.00526 × 2
= 0.01052 mole
Similarly, 0.00526 mole of NaHC03 requires HCl = 0.00526
∴ Total HC1 required = 0.01052 + 0.00526 = 0.01578 moles
Thus 0.1 mole of HCl is present in 1000 mL of HCl
∴ 0.01578 mole of HCl present in 1000 mL = \(\frac{1}{2}\) × 0.01578
= 157.8 mL

Or

(a) Define
(i) Mole fraction (ii) Molality (iii) Raoult’s law
(b) Assuming complete dissociation, calculate the expected freezing point of a solution prepared by dissolving 6.00 g of Glauber’s salt, Na2SO-1.10H2O in 0.100 kg of water. (Kf for water = 1.86 K kg mol2, Atomic masses : Na = 23, S = 32, O = 16, H = 1) [3,2]
Answer:
(i) Mole fraction : Mole fraction of a constituent is the fraction obtained by dividing number of moles of that constituent by the total number of moles of all the constituents present in the solution. It is denoted by ‘x’.
Example: x1 = \(\frac{\text { No. of moles of } x_1}{\text { Total no. of moles }}\) = \(\frac{n x_1}{n x_1+n x_2}\)

(ii) Molality: Molality of a solution is defined as the number of moles of the solute dissolved in 1000 grams (1 kg) of the solvent. ¡t is denoted by ‘m’.
zi’ 1000
m = \(\frac{w×1000}{M×W}\)
where r = ‘eight of solute in grams
M = Molecular mass of solute
W = Weight of solvent in grams

(iii) Raoult’s law: For a solution of volatile liquids the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.
P = P°x
Non-ideal solution shows positive and negative deviations from Raoult’s law.
→ is greater than the corresponding ideal solution of same composition. Such behaviour is called positive deviation.
Example : Mixtures of ethanol + cyclohexane
Mixture of acetone + carbon disulphide

→ Negative deviation from Raoult’s law : When the total vapour pressure will be less than corresponding vapour pressure, then it is termed as negative deviation.
Example : Chloroform + Acetone
Chloroform + Diethylether

(b) Since Na2SO4.10H2O is an ionic compound, so undergoes complete dissociation.
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 21

Question 29.
(a) Write the formula and describe the structure of a noble gas species which is isostructural with
(i) IBr2
(ii) BrO3
(b) Assign reasons for the following :
(i) SF6 is kinetically inert.
(ii) NF3 is an exothermic compound whereas NCl3 is not.
(iii) HCl is a stronger acid than HF though flourine is more electronegative than chlorine. 2,3
Answer:
(a) (i) IBr2 : It has 2 bond pairs and 3 lone pairs. Therefore according to VSEPR theory it should have Linear structure.
IBr2 has (7 + 2 × 7 + 1) i.e. 22 valence electrons.
No noble gas has 22 electrons
∴ The isoelectronic species for IBr2 is XeF2
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 22

(ii) IBr3 : It has 3 bond pairs and 1 lone pair of electrons. Therefore according to VSEPR theory it has pyramidal structure.
It has 26 valence electrons i.e. (7+ 3 × 6 + 1 = 26).
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 23
A noble gas having 26 valence electrons is XeO3 (8 + 3 × 6 = 26). Thus XeO3 also has pyramidal structure.
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 24

(b) (i) Because SF6 is showing steric hindrance due to 6 (six) fluorine atoms which make it unable to react further with any other atom.
(ii) Due to smaller size of F as compared to Cl, the N – F bond is much stronger than N-Cl bond while bond dissociation energy of F2 is much lower than that of Cl2. Therefore, energy released during the formationlof NF3 molecule is more than the energy needed to break N2 and F2 molecules into individual atoms. In other words, formation of NF3 is an exothermic reaction.
The energy released during the formation of NCl3 molecule is less than the energy needed to break N2 and Cl2 molecule into individual atoms. Thus formation of NCl is an endothermic reaction.

(iii) Because HCl has large size and less bond strength than HF which makes easier liberation of H+.

Or

+(a) How is ammonia prepared on a large scale? Name the process and mention the optimum conditions for the production of ammonia by this process.
(b) Assign reasons for the following :
(i) H2S is more acidic than H2O.
t(ii) NH3 is more basic than PH3.
(iii) Sulphur has a greater tendency for catenation than oxygen. [2,3]
Answer:
(a) According to Le Chatelier’s principle the favourable conditions for the production of NH3 by Haber’s process are

  • pressure of 200 atmosphere
  • temperature of ~700 K
  • use of catalyst like Fe2O3 with small amount of K2O and AL2O3
    N2 + 3H2 → 2NH3

(b) (i) H2S is more acidic than H2O because bond dissociation enthalpy of H – S bond in
H2S is less than that of H – O bond in H2O.
(ii) Since both P and N contains lone pairs of electrons but due to small size and high electronegativity of Nitrogen in NH3, the electron density is much higher than PH3, therefore it can easily donate electrons and acts as strong Lewis base than PH3.

(iii) The greater catenation tendency of sulphur is due to two reasons :

  • The lone pair of electrons feels more repulsion in O – O bond than S – S bond due to its small size and thus S – S forms strong bond.
  • As the size of atom increases down the group from O to P, the strength of bond increases and therefore catenation tendency also increases.

CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions

Question 30.
(a) Write the IUPAC names of the following compounds :
(i) CH3CO(CH2)4CH3
(ii) Ph – CH = CH – CHO
(b) Describe the following conversions in not more than two steps :
(i) Ethanol to 3-Hydroxybutanal
(ii) Benzoic acid to m-Nitrobenzyl alcohol
(iii) Propanone to Propene [2,3]
Answer:
(a) (i) CH3CO(CH2)4CH3 :
IUPAC name : Heptan-2-one

(ii) Ph – CH = CH – CHO :
IUPAC name : 3-phenylprop-2-en-1-al

(b) (i) Ethanol to 3-hydroxybutanal :
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 25

(ii) Benzoic acid to m-Nitrobenzyl alcohol
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 26

(iii) Propanone to Propene
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 27

Or

(a) Draw the structures of the following compounds :
(i) 4-Chloropentan-2-one
(ii) p-Nitropropiophenone
(b) Give tests to distinguish between the following pairs of compounds
(i) Ethanal and Fropanal
(ii) Phenol and Benzoic acid
(iii) Benzaldehyde and Acetophenone
Answer:
(a) (i) 4-Chloropentane-2-one
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 28

(ii) p-Nitropropiophenone
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 29

(b) (i) Ethanal and Propanal: Ethanal and propanal can be distinguished by iodoform test. Warm each compound with iodine and sodium hydroxide solution in water. Ethanal gives yellow crystal of iodoform while propanal does not respond to iodoform test.
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 30

(ii) Phenol and Benzoic acid: On addition of NaHCO3 to both solutions carbondioxide gas is evolved with benzcic acid while phenol does not form CO2.
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 31

(iii) Benzaldehyde and Acetophenone :
By Iodoform test: Acetophenone being a methyl ketone on treatment with I2/NaOH (NaOI) undergoes iodoform test to give yellow ppt. of iodoform but benzaldehyde does not.
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 32

SET II

Note : Except for the following questions, all the remaining questions have been asked in Set-I.

Question 1.
What is meant by ‘antiferromagnetism’? [1]
Answer:
Antiferromagnetism : These substances possess zero Viet magnetic moment because of presence of equal number of electrons with opposite spins.

Question 2.
Define dialysis. [1]
Answer:
Dialysis : The process of separating the particles of colloids from those of crystalloids by diffusion of the mixture through a parchment or an animal membrane is known as dialysis.

Question 3.
What is the role of CO2 in the extractive metallurgy of aluminium from its ore? [1]
Answer:
The solution of sodium meta-aluminate is neutralized by passing CO2 when hydrated alumina separates out while sodium silicate remains in the solution.

Question 4.
Why is nitrogen gas very unreactive? [1]
Answer:
Nitrogen gas is not particularly reactive due to presence of triple bond between two N-atoms. The bond length is very small and hence the bond dissociation energy is very large. As a result of which nitrogen gas is unreactive.

Question 9.
Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with change in temperature. [2]
Answer:
Conductivity : Conductivity of a solution is defined as the conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross-section. It is represented by K.
Its unit is S cm-1

Molar conductivity : Molar conductivity of a solution at a dilution V is the conductance of all the ions produced from one mole of the electrolyte dissolved in V cm3 of the solution when the electrodes are 1 cm apart and the area of the electrodes is so large that the whole of the solution is contained between them. It is represented by Λm.
Its unit is S cm2 mol-1
Conductivity and molar conductivity of electrolytes increase with increasing temperature.

CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions

Question 10.
Define each of the following :
(i) Specific rate of a reaction
(ii) Energy of activation of a reaction [2]
Answer:
(i) Specific rate of a reaction: Specific rate of reaction is the rate of reaction when the molar concentration of each of the reactants is unity.

(ii) Activation energy of a reaction : The minimum extra amount of energy absorbed by the reactant molecules so that their energy becomes equal to threshold value, is called activation energy.

Question 12.
Complete the following chemical reaction equations :
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 33

Question 14.
Write the structures of the following organic halogen compounds :
(i) 4-tert-Butyl-3-iodoheptane
(ii) 4-Bromo-3-methylpent-2-ene [2]
Solution:
(i) 4-tert-Butyl-3-iodoheptane
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 34

(ii) 4-Bromo-3-methylpent-2-ene
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 35

Question 23.
Assign reasons for the following : [3]
(i) Cu(I) ion is not known to exist in aqueous solutions.
(ii) Transition metals are much harder than the alkali metals.
(iii) From element to element actinoid contraction is greater than the lanthanoid contraction. .
Answer:
(i) Cu2+ (aq) is much more stable than Cu+(aq). This is because although second ionization enthalpy of copper is large but AhydH for Cu2+ (aq) is much more negative than that for Cu+ (aq) and hence it more than compensates for the second ionization enthalpy of copper. Therefore, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation as follows :
2Cu+ → Cu2+ + Cu

(ii) In transitional elements, in addition to metallic bonding there is extra covalent bonding due to presence of unpaired electrons in their ‘d’ orbitals, hence they are much harder.

(iii) The actinoid contraction is greater than lanthanoid contraction due to poorer shielding of 5f electrons as they are extended in space beyond 6s and 6p orbitals whereas 4f orbitals are burried deep inside the atom.

Question 24.
Giving one example in each of the following caSes, discuss briefly the role of coordination compounds in
(i) extraction metallurgy of metals
(ii) analytical chemistry [3]
Answer:
(i) Extraction metallurgy of metals : Cold and silver are extracted from their ores through formation of cyanide complexes [ Ag(CN)2] and [(Au(CN)2] respectively.
Example : Ag2S + 4NaCN ⇔ 2Na[Ag(CN)2l + Na2S

(ii) In analytical chemistry, they are used in qualitative analysis in which basic radicals are determined by converting them into suitable complexes with specific colour.
Example : Ni2+ is determined by DMG (Dimethyl Glyoxime) in the presence of NH4OH and forms a red ppt. of Ni DMG complex.
Similarly cobalt, Fe, Zn are also determined by converting them into complexes.

Question 27.
Answer the following questions :
(i) Why should medicines not be taken without consulting a doctor?
(ii) What is meant by ‘broad spectrum antibiotics’?
(iii) What are the main constituents of Dettol? [3]
Answer:
(i) Because medicines can cause harm to human body if a person does not know its physiological function on body.
(ii) Antibiotics which kill or inhibit a wide range of gram positive and gram negative bacteria, are called broad spectrum antibiotics.
(iii) Dettol is mixture of chloroxylenol and a-terpineol in a suitable solvent.

SET III

Note : Except for the following questions, all the remaining questions have been asked in Set-I and Set-11.

Question 1.
Write a distinguishing feature of a metallic solid compared to an ionic solid. [1]
Answer:
Metallic solid conducts electricity in solid state but ionic solids do so only in molten state or in solution. Metals conduct electricity through electrons while ionic substances conduct through ions. Metallic solids are malleable and ductile while ionic solids are hard and brittle.

Question 2.
What are enzymes? [1]
Answer:
Enzymes are protein molecules which act as catalyst in biochemical reaction.

Question 3.
Name the chief ores of aluminium and zinc. [1]
Answer:
Aluminium ore → Bauxite
Zinc ore → Zinc blende

CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions

Question 4.
Draw the structure of PCl5 (s) molecule.
Solution:
PCl5 (s)
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 36

Question 10.
A reaction is of second order with respect to its reactant. How will its reaction rate be affected if the concentration of the reactant is (i) doubled (ii) reduced to half? [2]
Solution:
Since Rate = K[A]2 For second order reaction
Let [A] = a then Rate = Ka2

(i) If [A] = 2a then Rate = K (2a)2 = 4 Ka2
∴ Rate of reaction becomes 4 times

(i) If [A] = \(\frac{a}{2}\) then Rate = K (\(\frac{a}{2}\))2 = 4 \(\frac{Ka^2}{4}\)
∴ Rate of reaction will be \(\frac{1^th}{4}\)

Question 12.
Complete the following chemical equations : [2]
(i) P4 + SOCl2
(ii) F2 (Excess) + Cl2 \(\xrightarrow{300^{\circ} \mathrm{C}}\)
Solution:
(i) P4 + 8SOCl2 → 4PCl3 + 4SO2 + 2S2Cl2
(ii) 3F2 + Cl2 \(\xrightarrow{300^{\circ} \mathrm{C}}\) 2ClFsub>3 (excess)

Question 13.
Assign reasons for the following :
(i) Transition metals and many of their compounds act as good catalysts.
(ii) Transition metals generally form coloured compounds. [2]
Answer:
(i) The catalytic properties of the transition elements are due to the presence of unpaired electrons in their incomplete d-orbitals and variable oxidation states.
(ii) Because presence of unpaired d electrons, which undergoes d-d transition by absorption of energy from visible region and then the emitted light shows complementary colours.

Question 14.
Write the structures of the following organic halogen compounds :
(i) p-Bromochlorobenzene
(ii) 1-Chloro-4-ethylcyclohexane [2]
Answer:
(i) P-Bromochlorobenzene
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 37

(ii) 1-chloro-4-ethylcyclohexane
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 38

Question 24.
Write down the IUPAC names of the following complexes and also give stereochemistry and magnetic moment of the complexes :
(i) [Co(NH3)5Cl]Cl2
(ii) [CrCl3(py)3]
(iii) K4[Mn(CN)6]
(At. nos. Cr = 24, Mn = 25, Co = 27, py = pyridine)
Solution:
(i) [CO(NH3)5Cl]Cl2
IUPAC name : Pentaammine chlorido cobalt (III) chloride
C.N. of Co = 6
Shape = octahedral
o.s. of Co : x + 0 – 1 = +2
∴ x = +3
Magnetic moment (μ) = \(\sqrt{0(0 + 2)}\) = 0 BM

(ii) [CrCl3(py)3]
IUPAC name : Trichloridotripyridine chromium (III)
C.N. of Cr = 6
Shape = octahedral
o.s. of Cr : x – 3 + 0 = 0
∴ x = +3
n = 3,
∴ μ = \(\sqrt{3 (3 + 2)}\) = \(\sqrt{15}\) = 3.87 BM

(iii) K4[Mn(CN)6]
IUPAC name : Potassium hexacyano manganate (II)
C.N. of Cr = 6
Shape = octahedral
o.s. of Cr : x – 6 = 4
∴ x = +2
n = 1
∴ μ = \(\sqrt{1 (1 + 2)}\) = \(\sqrt{3}\) = 1.73 BM

CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions

Question 25.
How are the following conversions carried out?
(i) Propene → Propan-2-ol
(ii) Ethylmagnesium chloride → Propan-1-ol
(iii) Benzyl chloride → Benzyl alcohol [3]
Answer:
(i) Propene to propan-2-ol
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 39

(ii) Ethylmagnesium chloride to propan-1-ol
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 40

(iii) Benzyl chloride to benzyl alcohol
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions 41

Question 27.
Answer the following :
(i) Why is the use of aspartame limited to cold foods and drinks?
(ii) How do antiseptics differ from disinfectants?
(iii) Why do soaps not work in hard water?
Answer:
(i) Use of aspartame is limited to cold foods and soft drinks because it is unstable at cooking temperature.

(ii)

AntisepticDisinfectants
1. They are chemical substances which prevent the growth of micro-organisms and may even kill them.1. They are chemical substances which kill micro-organisms.
2. They are safe to be applied to the living tissues.2. They are not safe to be applied to the living tissues.
3. They are generally applied on wounds, cuts, ulcers and diseased skin surfaces.
Example : Furacin, soframycin, dettol and savlon.
3. They are used to kill micro-organisms present in the drains, toilets, floors etc.
Example : Phenol (> 1% solution) and chlorine (0.2 to 0.4 ppm).

(iii) Hard water contains insoluble calcium and magnesium chlorides which forms insoluble ppt (scum) with soap and thus cannot be rinsed off easily.


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