Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2019 (Series BVM/1) to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2019 (Series BVM/1) with Solutions
Maximum Marks: 70
General Instructions
- All questions are compulsory.
- Section A : Q. no. 1 to 5 are very short-answer questions and carry 1 mark each.
- Section B : Q. no. 6 to 12 are short-answer questions and carry 2 marks each.
- Section C: Q. no. 13 to 24 are also short-answer questions and carry 3 marks each.
- Section D : Q. no. 25 to 27 are long answer questions and carry 5 marks each.
- There is no overall choice. However, an internal choice has been provided in two questions of 1 mark, two questions of 2 marks, four questions of 3 marks and all the three questions of 5 marks weightage. You have to attempt only one. of the choice in such questions.
- Use log tables if necessary. Use of calculators is not allowed.
Set I Code No. 56/1/1
Secton – A
Question 1.
Out of NaCl and AgCl, which one shows Frenkel defect and why? [1]
Answer:
Not in Syllabus.
Question 2.
Arrange the following in increasing order of boiling points:
(CH
3
)
3
N, C
2
H
5
OH,C
2
H
5
NH
2
[1]
Answer:
(CH
3
)
3
N < C
2
H
5
NH
2
< C
2
H
5
OH
Question 3.
Why are medicines more effective in colloidal state? [1]
Answer:
Medicines are more effective in colloidal state because they have larger surface area and hence the absorption, assimilation and digestion of medicine becomes easier.
Or
What is the difference between an emulsion and a gel?
Answer:
Emulsion is a colloidal solution in which both dispersed phase and dispersion medium are liquids. For e.g., milk, cold-cream etc., while Gel is a colloidal solution in which dispersed phase is liquid and dispersion medium is solid. For e.g., cheese, butter, etc.
Question 4.
Define ambidient nucleophile with an example. [1]
Answer:
Ambidient nucleophile is a group of atoms in anions which contain two nucleophilic centres for donation of e
–
pair. For e.g.
–
CN (Cyanide) and
–
NC (isocyanide).
Question 5.
What is the basic structural difference between glucose and fructose? [1]
Answer:
Glucose contains one aldehydic (-CHO) group and one primary alcoholic (-CH
2
OH) group and four secondary alcoholic (-CHOH) groups. Glucose is an aldehexose. Fructose is a ketohexose which contains two primary alcoholic groups at C-1 and C-6 position, one keto group at C-2 position and rest are 3 secondary -CHOH group.
Or
Write the products obtained after hydrolysis of lactose.
Answer:
On hydrolysis, lactose gives β-D-galactose and β-D-glucose
Section – ‘B’
Question 6.
Write balanced chemical equations for the following processes:
(i) XeF
2
undergoes hydrolysis.
(ii) MnO
2
is heated with conc. HCl.
Answer:
Or
Arrange the following in order of property indicated for each set:
(i) H
2
O,H
2
S,H
2
Se,H
2
Te — increasing acidic character
(ii) HF,HCl,HBr,HI — decreasing bond enthalpy
Answer:
(i) Increasing acidic character. H
2
O < H
2
S < H
2
Se < H
2
Te
(ii) Decreasing bond enthalpy. HF > HCl > HBr > HI
Question 7.
State Raoult’s law for a solution containing volatile components. Write two characteristics of the solution which obeys Raoult’s law at all concentrations. [2]
Answer:
Raoult’s Law: For a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly porportional to its mole fraction present in solution.
P = P
0
x
Ideal solutions obey Raoult’s law at all concentrations. Their characteristics are as follows:
(a) There will be no change in volume on mixing the two components i.e. ∆V
mix
= 0
(b) There will be no change in enthalpy on mixing i.e. ∆H
mix
= 0
Question 8.
For a reaction [2]
the proposed mechanism is as given below:
(1) H
2
O
2
+ I
–
→ H
2
O + IO
–
(slow)
(2) H
2
O
2
+ IO
–
→ H
2
O + I
–
+ O
2
(fast)
(i) Write rate law for the reaction.
(ii) Write the overall order of reaction.
(iii) Out of steps (1) and (2), which one is the rate determining step?
Answer:
(i) Rate law for above equation is: k[H
2
O
2
] [I
–
]
(ii) Order w.r.t. H
2
O
2
is 1 and w.r.t. I
–
is 1, so overall order of reaction is 2.
(iii)
Step (1) is slow step, so it is the rate determining step.
Question 9.
When MnOz is fused with KOH in the presence of KNOs as an oxidizing agent, it gives a dark green compound
(A). Compound (A) disproportionates in acidic solution to give purple compound
(B). An alkaline solution of compound (B) oxidises KI to compound (C) whereas an acidified solution of compound (B) oxidises KI to (D). Identify (A), (B), (C), and (D). [2]
Answer:
Question 10.
Write IUPAC name of the complex [Pt(en)
2
Cl
2
]. Draw structures of geometrical isomers for this complex. [2]
Answer:
[Pt (en)
2
Cl
2
]
IUPAC name : Dichloridobis (ethane-1, 2-diamine) platinum (II)
Structures of geometrical isomers:
Or
Using IUPAC norms write the formulae for the following:
(i) Hexaammine cobalt (III) sulphate
(ii) Potassium trioxalatochromate (III)
Answer:
(i) Hexaa;nmine cobalt (III) sulphate. [Co(NH
3
)
6
]
2
(SO
4
)
3
(ii) Potassium trioxalatochromate (III). K
3
[Cr(C
2
O
4
)
3
]
Question 11.
Out of [C0F
6
]
3-
and [Co(en)
3
]
3+
, which one complex is [2]
(i) paramagnetic
(ii) more stable
(iii) inner orbital complex and
(iv) high spin complex
(Atomic no. of Co = 27)
Answer:
(i) [CoF
6
]
3-
is paramagnetic due to presence of four unpaired electrons.
(ii) [Co(en)
3
]
3-
is more stable due to chelation.
(iii) [Co(en)
3
]
3+
is inner orbital complex.
(iv) [CoF
6
]
3-
is a high spin complex.
Question 12.
Write structures of compounds A and 3 in each of the following reactions: [2]
Answer:
Section – C
Question 13.
The decomposition of NH3 on platinum surface is a zero order reaction. If rate constant (k) is 4 × 10
-3
Ms
-1
, how long will it take to reduce the initial concentrataion of NH
3
from 0.1 M to 0.064 M. [3]
Answer:
Given: K = 4 × 10
-3
Ms
-1
[A]
0
= 0.1M [A] = 0.064M t = ?
Using formula
Question 14.
(i) What is the role of activated charcoal in gas mask?
(ii) A colloidal sol is prepared by the given method in the figure. What is the charge on hydrated ferric oxide colloidal particles formed in the test tube? How is the sol represented?
(iii) How does chemisorption vary with temperature?
Answer:
(i) Activated charcoal adsorbs poisonous gases in mines.
(ii) The formed hydrated ferric oxide has negative charge which is represented as Fe
2
O
3
..xH
2
O/OH
–
(iii) Chemisorption initially increases with increase in temperature and then decreases.
Question 15.
An element crystallizes in fee lattice with a cell edge of 300 pm. The density of the element is 10.8 g cm
-3
. Calculate the number of atoms in 108 g of the element. [3]
Answer:
Not in Syllabus.
Question 16.
A 4% solution (w/w) of sucrose (M = 342 g mol
-1
) in water has a freezing point of 271.15K. Calculate the freezing point of 5% glucose (M = 180 g mol
-1
) in water. [3] (Given: Freezing point of pure water = 273.15 K)
Answer:
Given: Molar mass of sucrose = 342 g mol
-1
Molar mass of glucose = 180 g mol
-1
∆T
f
of 4% of sucrose = 271.15K
4% solution of sucrose in water means 4g of sucrose present in 96g (100 – 4) of water.
4% sucrose solution means, w = 4g
Mass of water, W = 96g
Molarity (m) = \(\frac{w \times 1000}{\mathrm{~W} \times \mathrm{M}}\) = \(\frac{4 \times 1000}{96 \times 342}\) = 0.1218 mol kg
-1
∆T
f
for sucrose = 273.15 – 271.15K = 2K
∆T
f
= K
f
× M
= 16.42 × 0.2923 = 4.80 (approx)
∴ Freezing point of 5% glucose = (273.15 – 4.80)K = 268.35K
Question 17.
(a) Name the method of refining which is
(i) used to obtain semiconductor of high purity,
(ii) used to obtain low boiling metal.
(b) Write chemical reactions taking place in the extraction of copper from Cu
2
S. [3]
Answer:
(a) (i) Zone refining is used to obtain semiconductor (Ge, Si) of high purity.
(ii) Distillation is used to obtain low boiling metal eg. Zn, Cd, Hg, etc.
(b) Extraction of copper from Cu
2
S
2Cu
2
S + 3O
2
→ 2Cu
2
O + 2SO
2
2Cu
2
O + Cu
2
S → 6Cu + SO
2
Question 18.
Give reasons for the following: [3]
(i) Transition elements and their compounds act as catalysts.
(ii) E° value for (Mn
2+
| Mn) is negative whereas for (Cu
2+
| Cu) is positive.
(iii) Actinoids show irregularities in their electronic configuration.
Answer;
(i) The catalytic properties of the transition element as due to the presence of unpaired electrons in their incomplete d-orbitals and variable oxidation states.
(ii) E° value for Mn
2+
| Mn is negative due to greater stability of half filled d-subshell (d
5
) in Mn
2+
whereas E° value for Cu
2+
| Cu is positive due to its high enthalpy of atomization and low enthalpy of hydration.
(iii) This happens because the energy difference between 5f, 6d and 7s subshells of the actinoids is very small and hence electrons can be accomodated in any of them.
Question 19.
Write the structures of monomers used for getting the following polymers: [3]
(i) Nylon-6,6
(ii) Glyptal
(iii) Buna-S
Answer:
(i)
(ii) Monomers of Glyptal are: Ethylene glycol
(iii) Buna-S:
Or
(i) Is
a homopolymer or copolymer? Give reason.
Answer:
(ii) Write the monomers of the following polymer:
(iii) What is the role of Sulphur in vulcanization of rubber?
Answer:
(i) Not in Syallbus.
(ii) Monomers of melamine formaldehyde polymer are melamine and formaldehyde
(iii) In vulcanization of rubber, sulphur introduces sulphur bridges or cross-links between polymer chains thereby imparting more tensile strength, elasticity and resistance to abrasion.
Question 20.
(i) What type of drug is used in sleeping pills?
(ii) What type of detergents are used in toothpastes?
(iii) Why the use of alitame as artificial sweetener is not recommended? [3]
Answer:
(i) Tranquillizers are used in sleeping pills.
(ii) Anionic detergents are used in toothpastes.
(iii) Alitame is not recommended as an artificial sweetener because it is a high potency sweetener whose sweetness is difficult to control in food.
Or
Define the following terms with a suitable example in each:
(i) Broad-spectrum antibiotics
(ii) Disinfectants
(iii) Cationic detergents
Answer:
(i) Antibiotics which kill or inhibit a wide range of Gram – positive and Gram – negative bacteria are said to be Broad-spectrum antibiotics.
(ii) Disinfectants. Disinfectants kill or prevent growth of microbes and are applied on
inanimate/non living objects. Example: Phenol.
(iii) Cationic detergents.
i) Cationic detergents. They are quaternary ammonium salts of amines with acetates, chlorides or bromides as anions and the çationic part possess a long hydrocarbon chain, and a positive charge on nitrogen atom. Therefore they are called cationic detergents.
Example. Cetyltrimethyl ammonium bromide
Question 21.
(i) Out of (CH
3
)
3
C —Br and (CH
3
)
3
C — I, which one is more reactive towards S
N
1 and why?
(ii) Write the product formed when p-nitrochlorobenzene is heated with aqueous NaOH at 443 K followed by acidification.
(iii) Why dextro and laevo—rotatory isomers of Butan-2-ol are difficult to separate by fractional distillation? [3]
Answer:
(i) (CH
3
)
3
C—I is more reactive towards S
N
1 because Iodine is bigger in size and less electronegative than Bromine, so a better leaving gr1up.
(ii)
(iii) Dextro and laevo-rotatorv isomers of Butan-2-ol are difficult to seperate by fractional distillation because they are enantioniers of each other and both have same boiling point.
Question 22.
An aromatic compound ‘A’ on heating with Br
2
and KOH forms a compound ‘B’ of molecular formula C
6
H
7
N which on reacting with CHCl
3
and alcoholic KOH produces a foul smelling compound ‘C’. Write the structures and JUPAC names of compounds A, B and C. [3]
Answer:
Question 23.
Complete the following reactions:
Answer:
(i)
Or
Write chemical equations for the following reactions.
(i) Propanone is treated with dilute Ba(OH)
2
.
(ii) Acetophenone is treated with Zn(Hg)/Conc. HCl
(iii) Benzoyl chloride is hydrogenated in presence of Pd/BaSO
4
.
Answer:
(i) This is a self-aldol Condensation reaction:
(ii) Acetophenone is treated with Zn(Hg)/Conc. HCl.
This reaction is called Clemmensen Reduction.
(iii) Benzoyl chloride hydrogenated in the presence of Pd/BaSO
4
.
This reaction is called Rosenmund Reaction.
Question 24.
Differentiate between the following:
(i) Amylose and Amylopectin
(ii) Peptide linkage and Glycosidic linkage
(iii) Fibrous proteins and Globular proteins [3]
Answer:
(i) Amylose is a linear polymer of α-D-glucose in which C
1
of one glucose unit is attached to C
4
of the other through α-glycosidic linkage while Amylopectin is a highly branched polymer containing 20-25 glucose units which are joined together through α-glvcosidic linkages between C
1
of one and C
4
of the other which is further linked to C
6
of other glucose.
(ii) Peptide linkage: A peptide linkage is an amide linkage formed between —COOH group of one α-amino acid and NH
2
group of the other α-amino acid by loss of a molecule of water. The -CO-NH— bonds formed is called peptide linkage.
Glucosidic linkage. The two mohosaccharide unitsare joined together through an etheral or oxide linkage formed by loss of a molecule of water. Such a linkage between two monosaccharide units through oxygen atom is called glycosidic linkage.
(iii)
| Globular Proteins | Fibrous Proteins |
| (i) Globular proteins have almost spheroidal shape due to folding of the polypeptide chain. | Polypeptide chains of fibrous proteins consist of thread like molecules which tend to lie side by side to form fibres. |
| (ii) Globular proteins are soluble in water. | Fibrous proteins are insoluble in water. |
| (iii) Globular proteins are sensitive to small changes of temperature and pH. Therefore they undrgo denaturation on heating or on treatment with acids/bases. | Fibrous proteins are stable to moderate changes of temperature and pH. |
|
(iv) They possess biological activity that’s why they act as enzymes.
Example: Maltase, invertase etc. harmones (insulin) antibodies, transport agents (haemoglobin), etc. |
They do not have any biological activity but serve as chief structural material of animal tissues.
Example: Keratin in skin, hair, nails and wool, collagen in tendons, fibroin in silk etc. |
Or
Write chemical reactions to show that open structure of D-glucose contains the following:
(i) Straight chain
(ii) Five alcohol groups
(iii) Aldehyde as carbonyl group
Answer:
(i) Reaction of open structure of D-glucose contains a straight chain.
(ii) Reaction showing five alcohol groups (Acetylation of glucose)
(iii) Reaction showing aldehvde as carboxy group in glucose
Section – ‘D’
Question 25.
E°
cell
for the given redox reaction is 2.71V
Mg
(s)
+ Cu
2+
(0.01 M)
→ Mg
2+
(0001 M)
+ Cu
(s)
Calculate Eceii for the reaction. Write the direction of flow of current when an external opposite potential applied is 5
(i) less than 2.71 V and
(ii) greater than 2.71 V
Answer:
Given: E°
cell
= 2.71V
Cell Reaction
Mg
(s)
+ Cu
2+
(0.01 M)
→ Mg
2+
(0.001 M)
+ Cu
(s)
Applying nernst equation
(i) Current flows from copper to magnesium (i.e., from cathode to anode) when external opposite potential is less than 2.71V.
(ii) Current flows from magnesium to copper (i.e., from anode to cathode) when external opposite potential becomes greater than 2.71V.
Or
(a) A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeSO
4
and ZnSO
4
until 2.8 g of Fe deposited at the cathode of cell X. How long did the current flow? Calculate the mass of Zn deposited at the cathode of cell Y.
(Molar mass : Fe = 56 g mol
-1
, Zn = 65.3 g mol
-1
, 1F = 96500 C mol
-1
)
(b) In the plot of molar conductivity (Λ
m
) vs square root of concentration (c
1/2
), following curves are obtained for two electrolytes A and B:
Answer the following:
(i) Predict the nature of electrolytes A and B.
(ii) What happens on extrapolation of Λ
m
to concentration approaching zero for electrolytes A and B?
Answer:
(a) For deposition of Fe, reaction is
Fe
2+
+ 2e
–
→ Fe
Thus, 1 mole i.e., 56g Fe is deposited by 2F = 96500C × 2
∴ 2.8 g Fe will be deposited by \(\frac{96500}{56}\) × 2 × 2.8 = 9650C
Now, Q = I × t
∴ t = \(\frac{Q}{I}\) = \(\frac{9650}{2}\) = 4,825 sec.
For deposition of Zn, the reaction is: Zn
2+
+ 2e
–
→ Zn
Thus, 2F = 2 × 96500C deposit Zn = 1 mol = 65.3g
∴ 9650C will deposit Zn = \(\frac{65.3}{2 \times 96500}\) × 9650 = 3.265 g
(b)
(i) Electrolyte A is strong electrolyte while B is a weak electrolyte.
(ii) For strong electrolyte A, the plot becomes linear near high dilutions and thus can be extrapolated to zero concentrations to get Am while for the weak electrolyte Λ
m
increases steeply on dilution and extrapolation to zero concentration is not possible.
(a) How do you convert the following:
(i) Phenol to Anisole
(ii) Ethanol to Propan-2-ol
(b) Write mechanism of the following reaction:
(c) Why phenol undergoes electrophilic substitution more easily than benzene? [5]
(a) (i) Phenol to Anisole
(ii) Ethanol to Propan-2-ol
(b)
(c) Phenol undergoes electrophilic substitution more easily than benzene because of the strong activating effect of the -OH group attached to the benzene ring. The involvement of lone pair of oxygen in delocalisation makes the benzene ring electron rich.
Or
(a) Account for the following:
(i) o-nitrophenol is more steam volatile than p-nitrophenol.
(ii) t-butyl chloride on heating with sodium methoxide gives 2-methylpropene instead of t-butyl methyl ether.
(b) Write the reaction involved in the following:
(i) Reimer-Tiemann reaction
(ii) Friedal-Crafts Alkylation of Phenol
(c) Give simple chemical test to distinguish between Ethanol and Phenol.
Answer:
(a) (i) o-nitrophenol is more steam volatile due to intramolecular hydrogen bonding present in it. While p-nitrophenol is less volatile due to intermolecular hydrogen bonding.
(ii) t-butyl chloride on heating with sodium methoxide gives 2-methylpropene instead of t-butyl methyl ether because t-butyl chloride is 3° and CH3ONa is a strong base which favours elimination reaction.
(b) (i)
(ii)
(c) Chemical tests to distinguish between Ethanol and Phenol:
(i) Add neutral FeCl
3
to both the compounds, phenol will give violet colouration while ethanol does not.
(ii) Ethanol on reacting with I
2
in NaOH gives yellow ppt. of iodo form whereas phenol does not respond to this test.
Question 27.
(a) Give reasons for the following:
(i) Sulphur in vapour state shows paramagnetic behaviour.
\(\dagger\)
(ii) N-N bond is weaker than P-P bond.
(iii) Ozone is thermodynamically less stable than oxygen.
\(\dagger\)(b) Write the name of gas released when Cu is added to
(i) dilute HNO
3
and
(ii) cone. HNO
3
[5]
Answer:
(a) (i) In vapour state sulphur partly exists as S
2
molecule which has two unpaired
electrons in the antibonding π* orbitals like O
2
and hence exhibits paramagnetism.
(ii) Not in Syllabus.
(iii) ozone is thermodynamically very unstable because:
- The decomposition of ozone into oxygen is exothermic in nature. (∆H = -ve)
- There is also increase in entropy which in turn makes ∆G —ve and reaction spontaneous.
(b) Not in Syllabus.
Or
\(\dagger\)(a) (i) Write the disproportionation reaction of H
3
P0
3
.
(ii) Draw the structure of XeF
4
.
(b) Account for the following:
(i) Although Fluorine has less negative electron gain enthalpy yet F
2
is strong oxidizing agent.
(ii) Acidic character decreases from N
2
O
3
to Bi
2
O
3
in group 15.
(c) Write a chemical reaction to test sulphur dioxide gas. Write chemical equation involved.
Answer:
(a) (i) Not in Syllabus.
(ii) Structure of XeF
4
.
(b) (i) Because of small size of fluorine atom and strong electron-electron repulsions in its compact 2p orbitals.
(ii) Not in Syllabus.
(c) Sulphur dioxide (SO
2
) gas can decolourise oxidified KMnO
4
solution 5SO
2
+ 2Mn\(\mathrm{O}_4^{-}\) + 2H
2
O → 5S\(\mathrm{O}_4^{2-}\) + 4H
+
+ 2Mn
2+
SET II Code No. 56/1/2
Question 2.
Arrange the following in increasing order of base strength in gas phase:
(C
2
H
5
)
3
N, C
2
H
5
NH
2
, (C
2
H
5
)
2
NH
Answer:
Basic strength in gas phase: C
2
H
5
NH
2
< (C
2
H
5
)
2
NH < (C
2
H
5
)
3
N
Question 3.
Why conductivity of silicon increases on doping with phosphorus? [1]
Answer:
Conductivity of silicon increases on doping with phosphorous because of the movement of negatively charged extra electron of phosphorus in the lattice which contributes to increased conductivity in the formed semiconductor.
Question 5.
Write IUPAC name of the given compound:
Answer:
4-chlorobenzene sulphonic acid.
Question 8.
Write two differences between an ideal solution and a non-ideal solution. [2]
Answer:
| Ideal Solution | Non Ideal Solution |
| (i) They obey Raoult’s law over the entire range of concentration. | (i) They do not obey Raoult’s law over the entire range of concentration. |
| (ii) Neither the heat is evolved or absorbed during dissolution. | (ii) Heat is evolved or absorbed during dissolution. |
|
(iii) ∆
mix
H = 0
∆ mix V = 0 |
(iii) ∆
mix
H is not equal to 0.
∆ mix V is not equal to 0. |
Question 10.
Write IUPAC name of the complex [Cr(NH3)4Cl2]+]. Draw structures of geometrical isomers for this complex. [2]
Answer:
[Cr(NH
3
)
4
Cl
2
]
+
Name: Tetra amminedichlorido chromium (III) ion
Or
Using IUPAC norms write the formulae for the following:
(i) Pentaamminenitrito-O-cobalt(III) chloride.
(ii) Potassium tetracyanidonickelate (II)
Answer:
(i) Pentaamrnine nitrito-O-cobalt (III) chloride. [CO(NH
3
)
5
(ONO)]Cl
2
(ii) Potassium tetracyanidonickelate (II). K
2
[Ni(CN)
4
]
Question 11.
Out of [CoF
6
]
3-
and [Co(C
2
O
4
)
3
]
3-
, which one complex is [2]
(i) diamagnetic
(ii) more stable
(iii) outer orbital complex
(iv) low spin complex?
(Atomic no. of Co = 27)
Answer:
[COF
6
]
3-
Hexafluorido Cobalt (III) ion
CO
3+
= [Ar] 3d
5
4s
0
4p
0
and
[Co(C
2
O
4
)
3
]
3-
Tris Oxalato Cobalt (III) ion
CO
3+
= [Ar] 3d
5
4s° 4p°
(i) [CO(C
2
O
4
)
3
]
3-
is diamagnetic
(ii) [CO(C
2
O
4
)
3
]
3-
is more stable
(iii) [COF
6
]
3-
forms an outer orbital complex
(iv) [CO(C
2
O
4
)
3
]
3-
forms low spin complex
Question 17.
(i) Write the role of ‘CO’ in the purification of nickel. [3]
(ii) What is the role of silica in the extraction of copper?
(iii) What type of metals are generally extracted by electrolytic method?
Answer:
(i) In the Mond’s process, CO is passed over metallic nickel to produce a volatile complex which decomposes on further heating to give pure nickel.
(ii) Silica acts as a flux and helps to remove impurities (FeO) by forming a slag.
(iii) More reactive metals having large negative electrode potential. Copper, Gold, Silver, Lead, Nickel, Chromium, Zinc, Aluminium etc. are extracted by electrolytic method.
Question 18.
Give reasons for the following: [3]
(i) Transition metals form alloys,
(ii) MN
2
O
3
is basic whereas MN
2
O
7
is acidic.
(iii) Eu
2+
is a strong reducing agent.
Answer:
(i) Transition metals form alloys because their atomic radii in any series are not much different from each other and they can easily replace each other in the crystal lattice to form alloys.
(ii) Mn
2
O
3
is basic while Mn
2
O
7
is acidic because the basic nature decreases as the oxidation state or number of oxygen atoms increases i.e. Mn
2
O
3
(+3) and Mn
2
O
7
(+7).
(iii) Eu
2+
is a strong reducing agent because the most stable oxidation state of lanthanides is +3 so to acquire +3 oxidation state, Eu
2+
. loses an electron to become Eu
3+
.
Question 20.
(i) Why bithional is added in soap?
(ii) Why magnesium hydroxide is a better antacid than sodium bicarbonate?
(iii) Why soaps are biodegradable whereas detergents are non-biodegradable? [3]
Answer:
(i) Bithional acts as deodorant in soaps, hence it works as an antiseptic agent and reduces the odours produced by bacterial decomposition of organic matter on the skin.
(ii) Magnesium hydroxide is a better antacid than sodium bicarbonate because being soluble, it does not increase the pH above neutrality whereas sodium bicarbonate can make the stomach alkaline and trigger the production of even more acid.
(iii) Soaps are sodium potassium salts of long chain fatty acids and being soluble in water can easily be decomposed by microorganisms while detergents are NH4 or sulphate salts which can’t be decomposed by microorganisms easily hence are non- biodegradable.
Or
Define the following terms with a suitable example in each:
(i) Antibiotics
(ii) Artificial sweeteners
(iii) Analgesics
Answer:
(i) Antibiotics: Those chemical substances which are produced completely or partially by chemical synthesis in low concentration and either kill or inhibit the growth of micro-organisms by intervening in their metabolic processes, are known as antibiotics.
Example: Tetracycline, Vancomycin
(ii) Artificial sweeteners: Those chemical substances which are sweet in taste but do not add any calories to our body are called artificial sweetening agents. These are excreted as such through urine.
Example: Saccharin, aspartame etc.
(iii) Analgesics: Analgesics reduce or abolish pain without causing impairment of consciousness, mental confusion, in coordination or paralysis or some other disturbance of nervous system.
They are of two types:
(a) Non-narcotic analgesics Example: Aspirin
(b) Narcotic analgesics Example: Morphine
Question 21.
Write the structure of main products when benzene diazonium chloride reacts with the following reagents: [3]
(i) CuCN
(ii) CH
3
CH
2
OH
(iii) KI
Answer:
Reaction of benzene diazonium chloride with:
SET III Code No. 56/1/3
Question 1.
Arrange the following in decreasing order of solubility in water: [1]
(CH
3
)
3
N, (CH
3
)
2
NH, CH
3
NH
2
Answer:
CH
3
NH
2
> (CH
3
)
2</sub > NH > (CH
3
)
3
N
Question 2.
What type of stoichiometric defect is shown by ZnS and why? [1]
Answer:
Not in Syllabus.
Question 3.
Write one stereochemical difference between S
N
1 and S
N
2 reactions. [1]
Answer:
S
N
1 reactions lead to the formation of racemic mixture whereas S
N
2 reartions lead to inversion in configuration of product.
Question 7.
State Henry’s law and write its two applications. [2]
Answer:
Henry’s Law: “The solubility of a gas in a liquid at a particular temperature is directly proportional to the pressure of the gas in equilibrium with the liquid at that temperature.”
Applications of Henry’s Law:
- In the production of carbonated beverages which are pjepared under high pressure.
- Deep sea divers depend upon compressed air for their oxygen supply.
Question 11.
Write the hybridization and magnetic character of following complexes:
(i) [Fe(H
2
O)
6
]
3+
(ii) [Fe (CO)
5
]
(Atomic no. of Fe = 26)
Answer:
(i) [Fe(H
2
O)
6
]
3+
: The element Fe is in +3 oxidation state. As H
2
O is a weak field ligand, so electron pairing is not possible in this case.
(ii) [Fe(CO)
5
]: Here CO is a strong field ligand
26
Fe = [Ar] 3d
6
4s
2
4p
0
Therefore, Hybridisation – dsp
3
;
Magnetic Character. Since all electrons are pairs, so diamagnetic in nature.
Question 12.
Write structures of main compounds A and B in each of the following reactions:
Answer:
Question 17.
How will you convert the following:
(i) Impure Nickel to pure Nickel
(ii) Zinc blende to Zinc metal
(iii) [Ag(CN)
2
]
–
to Ag
Answer:
(i) Pure Nickel from impure Nickel can be obtained by Mond’s process in which nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetra-carbonyl which is decomposed at higher temperature to obtain pure metal.
(ii) Zinc blende to zinc metal → By Roasting. ZnS is roasted to give ZnO which is heated with reducing agent coke to give Zn.
(iii) The metal is recovered from the complex by reduction or displacement method by using a more electropositive zinc metal.
2[Ag(CN)
2
]
–
+ Zn → 2Ag + [Zn(CN)
4
]
2-
Question 18.
Give reasons for the following: [3]
(i) The transition metals generally form coloured compounds.
(ii) E° value for (Mn
3+
| Mn
2+
) is highly positive than that for (Cr
3+
| Cr
2+
) couple.
(iii) The chemistry of actinoids elements is not so smooth as that of the lanthanoids.
Answer:
(ii) The large positive E° value for Mn
3+
/Mn
2+
shows that Mn
2+
is much more stable than Mn
+3
due to stable half filled configuration (3d
5
). Therefore the 3
rd
ionisation energy of Mn will be very high and Mn
3+
is unstable and can be easily reduced to Mn
2+
. E° value for Cr
3+
| Cr
2+
is positive but small i.e. Cr
3+
can also be reduced to Cr
2+
but less easily. Thus Cr
3+
is more stable than Mn
3+
.
(iii) The chemistry of actinoids elements is not so smooth as that of lanthanoids because they show large number of oxidation states like + 3, + 4, + 5, + 6, + 7 due to very less difference in energy of 5f, 6d and 7s orbitals.
Question 22.
Write equations of the following reactions: [3]
(i) Acetylation of aniline
(ii) Coupling reaction
(iii) Carbylamine reaction
Answer:
(i) Acetylation of aniline
(ii) Coupling reaction: Arenediazoniurn salts react with highly reactive aromatic compounds such as phenols and amines to form brightly coloured azo compounds. Ar – N = N – Ar. This reaction is known as coupling reaction.
‘J , ,
(iii) Carbylamine reaction: Aliphatic and aromatic primary amines on heating with
chloroform and ethanolic KOH form isocyanides or carbylamines which are foul
smelling substances. This reaction is known as carhylamines reaction
R – NH2 + CHCI3 + 3KOH R – NC + 3KC1 + 3H20
Q.24. Define the following with a suitable example in each: 3
(1) Oligosacchariðes (ii) Denaturation of protein (iii) Vitamins
Ans. (i) Oligosaccharides: Those carbohydrates which on hydrolysis give 2-10 molecules of
monosaccharides are called oligosaccharides. Example: sucrose, maltose.
(ii) Denaturation of protein: When a protein in its native form, is subjected to physical
change like change in temperature or chemical change like change in pH, the
hydrogen bonds are disturbed. Due to this, globules unfold and helix gets uncoiled
and protein loses its biological activity. Example: Boiling of egg.
(iii) Vitamins: Organic compounds required in the diet in small amounts to perform
specific biological functions for normal maintenance of optimum growth and health
of the organism. Example: Vitamin A.
Or
Write the reactions involved when D-glucose is treated with the following reagents:
(i) Br
2
water
(ii) H
2
N-OH
(iii) (CH
3
CO)
2
O
Answer:
(i)
(ii) D-glucose reacts with hydroxylamine to form oxime.
(ii) D-glucose reacts with hydroxylamine to form oxime.
