Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2014 Comptt (Delhi) to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Delhi) with Solutions
Time allowed : 3 hours
Maximum marks : 70
General Instructions:
- All questions are compulsory.
- Questions number 1 to 8 are very short answer questions and carry 1 mark each.
- Questions number 9 to 18 are short answer questions and carry 2 marks each.
- Questions number 19 to 27 are also short answer questions and carry 3 marks each.
- Questions number 28 to 30 are long answer questions and carry 5 marks each.
-
Use log tables if necessary. Use of calculators is not allowed.
Deleted from Syllabus.
SET I
Question 1.
How many atoms per unit cell (z) are present in bcc unit cell?
Answer:
Number of atoms in a unit cell of a body centred cubic structure :
Contribution by 8 atoms at the corners = \(\frac{1}{2}\) × 8 = 1
Contribution by the atom present within the body = 1
∴ Total number of atoms present in the unit cell = 1 + 1 = 2 atoms
Question 2.
What is the difference between lyophobic sol and lyophilic sol?
Answer:
Lyophobic sols : Substances like metals, their suiphides, etc., when simply mixed with the dispersion medium do not form the colloidal sol. Their colloidal sols can only be prepared by specific methods. They are not much hydrated and are irreversible in nature. They are also called extrinsic colloids.
Example: AS
2
S
3
sol.
Lyophilic sols : Liquid loving colloids in which there is affinity between dispersed phase and dispersion medium.
Example: Starch sol, Gum sol, Gelatin sol
Question 3.
Draw the structure of XeF
2
.
Answer:
structure of XeF
2
Q.4.
What is the function of SiO
2
in the metallurgy of copper?
Answer:
During roasting, copper pyrites are converted into a mixture of FeO and Cu
2
O.
To remove basic FeO, an acidic flux silica is added during smelting. Now FeO combines with SiO
2
(silica) to form ferrous silicate (FeSiO
3
) slag which floats over molten matte.
Question 5.
Why do transition elements show variable oxidation states? [1]
The variability of oxidation state of transition elements is due to incompletely filled d-orbitals and presence of unpaired electrons, i.e. ns and (n – 1) d electrons have approximate equal energies.
Question 6.
Draw the structure of 2-bromopentane.
Answer:
Question 7.
Write the IUPAC name of the following compound:
Answer:
Question 8.
Out of CH
3
NH
2
and (CH
3
)
3
N, which one has higher boiling point?
Answer:
CH
3
-NH
2
has higher boiling point than (CH
3
)
3
N.
Question 9.
How is the vapour pressure of a solvent affected when a non-volatile solute is dissolved in it? [2]
Answer:
The vapour pressure of a solvent decreases when a non-volatile solute is dissolved in it because some solvent molecules are replaced by the molecules of solute.
Question 10.
(a) For a reaction, A + B Product, the rate law is given by, Rate = fc[A]
1
[B]
2
. What is the order of the reaction?
(b) Write the unit of rate constant ‘fc’ for the first order reaction. [2]
Answer:
(a) For a reaction, A + B
Rate = k [A]
1
[B]
2
This is the third order of reaction.
(b) Unit of rate constant for first order reaction is S
-1
Question 11.
Define the following terms:
(i) Roasting
(ii) Calcination
Answer:
(i) Roasting : The process of heating strongly the concentrated sulphide ore in the presence of air is called roasting :
(ii) Calcination : The process of heating strongly the concentrated ore in the absence of air is called calcination :
Question 12.
Draw the structure of each of the following:
(i) H
2
SO
4
(ii) Solid PCl
5
[2]
Answer:
(i) H
2
SO
4
:
(ii) PCl
5
[s]:
Question 13.
Assign a reason for each of the following observations: [2]
(i) The transition metals (with the exception of Zn, Cd and Hg) are hard and have high melting and boiling points.
(ii) The ionization enthalpies (first and second) in the first series of the transition ele¬ments are found to vary irregularly.
Answer:
(i) Because of stronger metallic bonding and high enthalpies of atomization.
(ii) Due to irregulaties in the electronic configuration there is irregularities in the enthalpies of atomisation. Hence there is irregular variation in I.E.
Or
What is lanthanoid contraction? Write a consequence of lanthanoid contraction.
Answer:
Lanthanoid contraction : The overall decrease in atomic and ionic radii with increasing atomic number is known as lanthanoid contraction.
Cause : As we move along the lanthanoid series, the effective nuclear charge increases on addition of electrons and the electrons added in/-subshell causes imperfect shielding which is unable to the counterbalance the effect of the increased nuclear charge. Hence the contraction in size occurs.
Consequences :
(i) Due to small change in atomic radii, the chemical properties of lanthanoids are very similar due to which separation of lanthanoids becomek very difficult.
(ii) There is similarity in size of elements belonging to same group of second and third transition series.
Example : Zr and H/ are known as chemical twins due to their similar radii.
Question 14.
How would you account for the following?
(i) The highest oxidation state of a transition metal is usually exhibited in its oxide.
(ii) The oxidising power of the following three oxoions in the series follows the order:
Answer:
(i) The highest oxidation state of a metal is exhibited in its oxide because oxygen has the ability to form multiple bonds.
(ii) This is due to the increasing stability of the lower species to which they are reduced.
Question 15.
Write the equations involved in the following reactions:
(i) Williamson ether synthesis
(ii) Kolbe’s reaction
Answer:
(i) Williamson ether synthesis: Alkvl Halide reacts with Alkoxide
(ii) Kolbe’s reaction : Phenoxide ion generated by treating phenol with sodium hydroxide is more reactive than phenol and undergoes electrophilic substitution with carbon dioxide. Ortho hydroxybenzoic acid is formed as the main reaction product.
Question 16.
How are the following conversions carried out?
(i) Propene to Propan-2-ol
(ii) Ethyl chloride to Ethanal
Solution:
(i) Propene to propan-2-ol
(ii) Ethyl chloride to ethanal
Question 17.
Answer the following questions:
(i) Why are vitamin A and vitamin C essential for us?
(ii) What is the difference between a nucleoside and a nucleotide? [2]
Answer:
(i) Because deficiency of vitamin A and vitamin C causes night blindness and scurvy respectively.
(ii) Nucleoside : A nucleoside contains only two basic components of nucleic acids i.e. a pentose sugar and a nitrogenous base. During their formation 1-position of the pyrimidine or 9-position of the purine moitey is linked to C
1
of the sugar (ribose or deoxyribose) by a ß-linkage
Question 18.
Enumerate the reactions of glucose which cannot be explained by its open chain structures. [2]
Answer:
Limitations of the open chain structure of glucose
(i) Glucose does not form NaHSO
3
addition product. Despite having aldehyde ammonia group, it does not give test 2,4-DNP and does not respond to Schiff’s reagent test.
(ii) Glucose penta acetate does not react with NH
2
OH due to absence of aldehydic group.
Question 19.
The density of copper is 8.95 g cm
-3
. It has a face centred cubic structure. What is the radius of copper atom?
(Atomic mass Cu = 63.5 g mol
-1
, N
A
= 6.02 × 10
23
mol
-1
) [3]
Solution:
Question 20.
Some ethylene glycol, HOCH
2
CH
2
OH, is added to your car’s cooling system along with 5 kg of water. If the freezing point of water-glycol solution is -15.0°C, what is the boiling point of the solution? [3]
(K
b
= 0.52 K kg mol
-1
and = 1.86 K kg mol
-1
for water)
Solution:
Question 21.
Hydrogen peroxide, H
2
O
2
(aq) decomposes to H
2
O (l) and O
2
(g) in a reaction that is first order in H
2
O
2
and has a rate constant k = 1.06 × 10
-3
min
-1
. [3]
(i) How long will it take for 15% of a sample of H
2
O
2
to decompose?
(ii) How long will it take for 85% of the sample to decompose?
Solution:
(i) Given
k = 1.06 × 10
-3
min
-1
For first order reaction
Question 22.
Define the following terms: [3]
(i) Adsorption
(ii) Peptization
(iii) Sol
Answer:
(i) Adsorption : It is a surface phenomenon which occurs only at the surface of the adsorbent.
(ii) Peptization: It is a process of converting a fresh precipitate into colloidal particles by shaking it with the dispersion medium in the presence of a small amount of a suitable electrolyte.
(iii) Sol : The colloids in which a solid is dispersed in the liquid.
or, In a colloidal sol, the dispersed phase is a solid and the dispersion medium is a liquid.
Question 23.
Write down the IUPAC name for each of the following complexes:
(i) [CO(NH
3
)
5
Cl]Cl
2
(ii) K
3
[Fe(CN)
6
]
(iii) [NiCl
4
]
2
[3]
Solution:
(i) [CO(NH
3
)
5
Cl]Cl
2
:
IUPAC name : Pentaammine chlorido cobalt (III) chloride
(ii) K
3
[Fe(CN)
6
] :
IUPAC name : Potassium hexacyanoferrate (III)
(iii) [NiCl
4
]
2
:
IUPAC name : Tetra chloridonickelate (II)
Or
Draw the structures of optical isomers of each of the following complex ions:
[Cr(C
2
O
4
)
3
]
3-
, [PtCl
2
(en)
2
]
2+
, [Cr(NH
3
)
2
Cl
2
(en)]
+
Solution:
Optical isomers of [Cr(C
2
O
4
)
3
]
3-
:
Name : Trioxalatochromate (III) ion
Optical isomers of [PtCl
2
(en)
2
]
2+
Optical isomers of [Cr(NH
3
)
2
Cl
2
(en)]
+
Question 24.
(a) Which compound in each of the following pairs will react faster in S
N
2 reaction with -OH group?
(i) CH
3
Br or CH
2
I
(ii) (CH
3
)
3
CCl or CH
3
Cl
(b) Write the product of the following reactions:
(i) CH
3
– Cl + KCN → ?
(ii)
Answer:
(a) (i) As I
–
ion is a good leaving group than Br
–
, therefore reacts faster than CH
3
Br in S
N
2 reactions with -OH.
(ii) As 3° alkyl halides have more steric hindrance than 1° alkyl halide therefore CH 3 Cl undergoes S N 2 reaction faster than 3° alkyl halide.
Question 25.
Account for the following:
(i) Aniline does not give Friedel-Crafts reaction.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) pK
b
of methylamine is less than that of aniline. [3]
Answer:
(i) Aniline being a Lewis base reacts with Lewis acid AlCl
3
to form a salt.
As a result, N of aniline acquires positive charge and hence it acts as a strong deactivating group for electrophilic substitution reaction. Consequently, aniline does not undergo Freidel Craft reaction.
(ii) Ethylamine is soluble in water due to its capability to form H-bonds with water while aniline is insoluble in water due to larger hydrocarbon part which tends to retard the formation of H-bonds.
(iii) In aniline due to resonance lone pair of electron of nitrogen atom is delocalised due to which it is weaker base than methyl amine.
Hence Aniline has high pK
b
molecule i.e., methylamine has less pK
b
, molecule.
Question 26.
Write the names of the monomers of the following polymers:
(i) Polythene
(ii) Polyvinyl chloride
(iii) Bakelite [3]
Answer:
(i) Polyethene
(ii) PVC(Polyvinyl chloride)
(in) Bakelite → Phenol and formaldehyde → Condensation
Question 27.
Explain the following terms with a suitable example for each:
(i) Disinfectants
(ii) Antacids
(iii) Food preservatives
Answer:
(i) These are the chemical substances which are used for killing or preventing the growth of micro-organisms but they are not safe for living tissues.
(ii) Antacids : Antiseptics are the chemicals which either kill or prevent growth of microbes on living tissues, e.g., Penicillin.
(iii) Food preservatives : They are used to prevent spoilage of food due to microbial growth.
Example : Table salt, vegetable oils, sodium benzoate etc.
Question 28.
(a) Define the terms conductivity and molar conductivity for the solution of an electrolyte. Comment on their variation with temperature.
(b) The measured resistance of a conductance cell was 100 ohms. Calculate (i) the specific conductance and (ii) the molar conductance of the solution.
(KCl = 74.5 g mol
-1
and cell constant = 1.25 cm
-1
)
Answer:
(a) Conductivity : Reciprocal of resistivity is called conductivity
k = \(\frac{1}{R}\) × \(\frac{l}{A}\)
Molar conductivity : It is defined as the conductivity of solution containing 1 mole solute dissolved per litre when placed between two electrodes of unit area separated by 1 cm.
Molar conductivity and conductivity of solution increase on increasing the temperature.
(b) Given : R = 100 Q
Cell constant = 1.25 cm
-1
Molarity = 74.5 g mol
-1
(i) Specific conductance, K = \(\frac{1}{R}\) × Cell constant = \(\frac{1}{100}\) × 1.25
= 0.0125 Ω
-1
cm
-1
(ii) Molar conductance (Λ
m
) = \(\)
= \(\frac{\mathrm{K} \times 1000}{\text { Molarity }}\) = 0.167 Ω
-1
cm
2
mol
-1
Or
(a) Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO
3
with platinum electrodes.
(ii) An aqueous solution of H
2
SO
4
with platinum electrodes.
(b) Estimate the minimum potential difference needed to reduce Al
2
O
3
at 500°C. The Gibbs energy change for the decomposition reaction
Al
2
O
3
→ Al + O
2
is 960 kJ. (F = 96500 C mol
-1
)
Solution:
Question 29.
(a) Complete the following chemical equations:
(i) P
4
+ NaOH + H
2
O →
(it) XeF
4
+ O
2
F
2
→
(b) How would you account for the following situations?
(i) The acidic strength of these compounds increases in the following order :
PH
3
< H
2
S < HCl
(ii) The oxidising power of oxoacids of chlorine follows the order :
HClO
4
< HClO
3
< HClO
2
< HClO
(iii) In vapour state sulphur exhibits paramagnetic behaviour. 2,3
Solution:
(a)
(b) (i) As the electronegativity increases in the same period from left to right so their electronegativity are in the increasing order, P < S < Cl.
In the same way the acid strength is also in the increasing order i.e. PH
3
< H
3
S < HCl.
(ii)
Acidic strength of oxoacids of tire same halogen increases with increase in oxidation number of the halogen because of the relative stability of the anions left after removal of proton. Thus as the number of oxygen atoms in the anion increases, the dispersal of the negative charge through pπ-pπ back bonding also increases and hence stability and acidic strength increases.
(iii) In vapour state sulphur partly exists as S 2 molecule which has two unpaired electrons in the antibonding π* orbitals and hence exhibits paramagnetism.
Or
(a) Using VSEPR theory predict the probable structures of the following :
(i) N
2
O
3
(ii) BrF
3
(b) Arrange the following groups of substances in the order of the property indicated against each group :
(i) NH
3
, PH
3
, AsH
3
, SbH
3
– increasing order of boiling points.
(ii) O, S, Se, Te – increasing order pi electron gain enthalpy with negative sign.
(iii) F
2
, Cl
2
, Br
2
, I
2
– increasing order of bond dissociation enthalpy. [2,3]
Answer:
(a)
(b)
(i) NH
3
, PH
3
, AsH
3
, SbH
3
(i) PH
3
< AsH
3
< NH
3
< SbH
3
(ii) O < Te < Se < S
(iii) The increasing order of bond dissociation enthalpy is :
Cl
2
> Br
2
> F
2
> I
2
Question 30.
(a) Describe :
(i) Aldol condensation (ii) Cannizzaro reaction
(b) Describe a chemical test to distinguish between
(i) Ethanal and Propanal (ii) Benzaldehyde and Acetophenone
(iii) Pentan-2-one and Pentan-3-one [2,3]
Answer:
(a) (i) Two molecules of aldehydes or ketones condense in presence of a dil.alkali to form a ß-hydroxy aldehyde or ß-hydroxyketone respectively CH3OH
(ii) Cannizzaro reaction : Aldehydes, which do not have an a-hydrogen atom undergo self oxidation and reduction on treatment with cone, alkali and produce alcohol and carboxylic acid salt.
(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.
(ii) Acetophenone and Benzophenone : They can be distinguished by iodoform test which is given by only acetophenone with the formation of yellow ppt. while benzophenone does not respond to iodoform test.
(iii) Pentan-2-one and Pentan-3-one
By Iodoform test
Or
(a) Draw the structures of the following compounds:
(i) 4-chloropentan-2-one (ii) But-2-en-l-al
(b) Write the product(s) in the following:
Answer:
SET II
Note : Except for the following questions, all the remaining questions have been asked in Set-I.
Question 1.
What type of stoichiometric defect is shown by NaCl? [1]
Answer:
Schottky defect is shown by NaCl.
Question 2.
Define ‘Emulsions’. [1]
Answer:
Emulsions : An emulsion is a colloidal dispersion in which both the dispersed phase and dispersion medium are liquids.
Question 3.
Draw the structure of XeF
4
. [1]
Solution:
XeF
4
:
SP
3
d
2
Shape → Square Planar
Question 4.
What role is played by CO
2
in getting pure alumina (Al
2
O
3
) in the extraction of aluminium? [1]
Answer:
Role of CO
2
in extraction of Al
2
O
3
: During leaching the precipitated aluminium hydroxide is neutralised by CO
2
to produce hydrated alumina which on heating gives pure alumina
Question 9.
Differentiate between molarity and molality of a solution. How can we change molality value of a solution into molarity value? [2]
Answer:
Difference between molarity and molaljty.
Molarity : It is the number of moles of solute dissolved in 1 litre of solution. It is temperature dependent.
M = \(\frac{\omega \times 1000}{\text { mol.mass } \times V}\)
Molarity: It is the number of moles of solute dissolved in 1 kg of the solvent.
m = \(\frac{\omega \times 1000}{M_2 \times W}\)
The relationship between molarity and molality is M
m = \(\frac{\mathrm{M}}{d-\frac{\mathrm{MM}_2}{1000}}\)
When molality = molarity, we get 1
1 = \(\frac{1}{d-\frac{\mathrm{MM}_2}{1000}}\) or d – \(\frac{\mathrm{MM}_2}{1000}\) = 1
∴ d = 1 + \(\frac{\mathrm{MM}_2}{1000}\)
Molarity is temperature dependent while molality is not.
For very dilute solution, the factor MM2/1000 can be neglected in comparison to 1.
Hence molality will be same to molarity when density d = 1.
Molarity (M) and Molality (m) relationship :
∵ Molarity is M moles of solute present in 1000 mL solution
If density of solution is d g mL
-1
, then
Mass of solution = 1000 d g
Mass of solute = MM
2
,
∴ Mass of solvent = 1000 d – MM
2
g
Question 11.
Describe underlying principles of the following processes :
(i)Froth floatation process of concentration of ores.
Answer:
(i) Froth floataion method : 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.
Question 12.
Draw the structures of the following compounds :
(i) H
2
SO
4
(ii) N
2
O
5
Solution:
Question 14.
Assign reason for each of the following :
(i) Transition elements exhibit paramagnetic behaviour.
(ii) Co
2+
is easily oxidised in the presence of a strong ligand. [2]
Answer:
(0 Because of presence of unpaired electrons in their d-subshell in atomic and ionic state.
(ii) Co
2+
ion is easily oxidised to Co
3+
ion in presence of a strong ligand because of its higher crystal field energy which causes pairing of electrons to give inner orbital complexes (d
2
sp
3
).
Question 22.
Define the following terms :
(i) Sorption
(ii) Tyndall effect
(iii) Electrophoresis [3]
Answer:
(i) Sorption : It is a physical and chemical process by which one substance becomes attached to another.
or, In order to distinguish between adsorption and absorption a common term sorption is used.
(ii) Tyndall effect : When a beam of light is passed through a colloidal solution and viewed perpendicular to the path of the incident light, the path of light becomes visible as a bright streak. The illuminated path is called Tyndall cone and the phenomenon is called Tyndall effect.
(iii) Electrophoresis : When electric current is passed through a colloidal solution, the positively charged particles move towards cathode while negatively charged particles move towards anode where they lose their charge and get coagulated. The phenomenon is known as Electrophoresis.
Question 26.
What are the following? Give one example of each.
(i) Sweetening agents
(ii) Food preservatives
(iii) Antibiotics [3]
Answer:
(i) Sweetening agent : 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.
(ii) Food preservatives : They are used to prevent spoilage of food due to microbial growth.
Example : Table salt, vegetable oils, sodium benzoate etc.
(iii) Antibiotics : Antibiotics 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 Antibiotics
Example: Aspirin
(b) Narcotic Antibiotics
Example: Morphine
Question 27.
Give names of the monomers of the following polymers :
(i) Neoprene
(ii) Polystyrene
(iii) Polypropene [3]
Answer:
(i) Neoprene : Monomer is Chhrirprenc or 2-chiqro-l, 3-butadiene.
(ii) Polystyrene : Monomer is Styrene.
C
6
H
5
CH = CH
2
(iii) Polypropene : Monomer is Propene.
CH
3
CH = CH
2
SET III
Note : Except for the following questions, all the remaining questions have been asked in Set-I and Set-II.
Question 1.
Write a distinguishing feature between a metallic solid and an ionic solid. [1]
Answer:
The electrical conductivity in metallic solid is due to free electrons while in ionic solid the conductivity is due to presence of ions.
Question 2.
What is a ‘shape-selective catalyst’? [1]
Answer:
The catalyst reaction in which small sized molecules are absorbed in the pores and cavities of selective adsorbents like zeolites is known as shape-selective catalysis.
Question 3.
What is the role of NaOH in the metallurgy of aluminium? [1]
Answer:
The powdered bauxite ore is heated with conc. NaOH to remove impurities of Fe
2
O
3
, SiO
2
and TiO
2
.
Question 4.
Draw the structure of XeOF
4
.
Answer:
XeOF
2
:
Question 10.
Define the following terms :
(i) Rate constant (k)
(it) Activation energy (E
a
) [2]
Answer:
(i) Rate constant (k) : It is a proportionality constant and is equal to the rate of reaction when the molar concentration of each of the reactants is unity.
(ii) Activation energy (E
a
): The minimum extra amount of energy absorbed by the reactant molecules to form the activated complex is called activation energy.
Question 11.
Describe the principle involved in each of the following processes :
(i) Zone refining of metals
(ii) Vapour phase refining of metals [2]
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.
(ii) Vapour phase refining of metals: 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.
Question 12.
Complete the following chemical equations :
Answer:
Question 14.
Describe the general trends in the following properties of the first series (3d) of the transition elements :
(i) Number of oxidation states exhibited
(ii) Formation of oxometal ions [2]
Answer:
(i) The number of oxidation states increases upto middle of series i.e. upto +7 and then decreases.
(ii) Oxometal ions are poly atomic ions with oxygen.
Example :\(\mathrm{VO}_2{ }^{+}\),VO
+2
, TiO
2+
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]
(Atomic mass of Fe = 55.84 g mol
-1
)
Answer:
Given
Question 26.
What are biodegradable and non-biodegradable detergents? Give one example of each ?
Answer:
Biodegradable detergents : Detergents, having straight hydrocarbon chains are easily degraded or decomposed by microorganisms, are known as biodegradable detergents.
Example : Sodium lauryl sulphate.
Non-biodegradable detergents : Detergents containing branched hydrocarbon chains^ and are not easily decomposed by the micro-organisms, are known as non-biodegradable detergents.
Example : Sodium 4 – (1, 3, 5, 7-tetramethyloctyl) benzene sulphonate.
Question 27.
Write the names of monomers used fdr getting the following polymers :
(i) Teflon
(ii) Bakelite
(iii) Neoprene [3]
Answer:
(i) Teflon (Polytetrafluoroethene)
(ii) Bakelite → Phenol and formaldehyde → Condensation
(iii) Neoprene :
