Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2013 Delhi to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2013 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.
\(\dagger\) Deleted from Syllabus.
Set I
\(\dagger\) Question 1.
How many atoms constitute one unit cell of a face-centred cubic crystal? [1]
Answer:
4 atoms constitute one unit cell of a fee crystal.
Question 2.
Name the method used for refining of Nickel metal. [1]
Answer:
Mond process is used for refining of nickel metal.
\(\dagger\)Question 3.
What is the covalency of nitrogen in N
2
O
5
? [1]
Answer:
The covalance of nitrogen in N
2
O
5
is 4 because each nitrogen atom has four shared pairs of electrons
Question 4.
Write the IUPAC name of: [1]
Answer:
IUPAC name : 4-chloropent-1-ene
Question 5.
What happens when CH
3
-Br is treated with KCN? [1]
Answer:
Question 6.
Write the structure of 3-methyl butanal. [1]
Answer:
Question 7.
Arrange the following in increasing order of their basic strength in aqueous solution:
CH
3
NH
2
, (CH
3
)
3
N, (CH
3
)
2
NH. [1]
Answer:
Question 8.
What are the types of an RNA molecule which perform different functions? [1]
Answer:
m-RNA, t-RNA, r-RNA
Question 9.
18 g of glucose, C
6
H
12
O
6
(Molar Mass = 180g/ mol) is dissolved in 1 kg of water in a saucepan. At what temperature will this solution boil? [2] (K
b
for water = 0.52 K kg mol
-1
, boiling point of pure water = 373.15 K)
Answer:
We know that:
Elevation of boiling point ∆T
b
= \(\frac{-W_B}{M_B} \times \frac{1000 \times K_b}{\text { wt. of solvent }}\)
Given: W
B
= 18 g M,
M
B
= Formula of glucose is C
6
H
12
O
6
= 6 × 12 + 12 + 6 × 16 = 180
Wt. of solvent = 1 kg or 1000 g, K
b
= 0.52 K kg mol
-1
Hence, ∆T
b
= \(\frac{18 \mathrm{~g}}{180} \times \frac{1000 \times 0.52}{1000 \mathrm{~g}}\) = 0.052 K
∴ B.P of the solution = 373.15 + 0.052 = 373.202 K
Question 10.
The conductivity of 0.20 M solution of KCl at 298 K is 0.025 S cm
-1
. Calculate its molar conductivity. [2]
Answer:
Molar conductivity Λ
m
= \(\frac{1000 \times K}{M}\)
Given: \(K\) = 0.025 S cm
-1
, M = 0.20 M
Hence, Λ
m
= \(\frac{0.025 \times 1000}{0.20}\) ∴ Λ
m
= 125 S cm
2
mol
-1
Question 11.
Write the dispersed phase and dispersion medium of the following colloidal system: [2]
(i) Smoke
(ii) Milk.
OR
What are lyophilic and lyophobic colloids? Which of these sols can be easily coagulated on the addition of small amounts of electrolytes?
Answer:
(i) Smoke: Dispersed phase → Solid; Dispersion medium → Gas
(ii) Milk: Dispersed phase → Fat (Liquid); Dispersion medium → Liquid
OR
Lyophobic sols : Substances like metals, their sulphides, 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 12.
Write the differences between physisorption and chemisorption with respect to the following: [2]
(i) Specificity
(ii) Temperature dependence
(iii) Reversibility and
(iv) Enthalpy change Answer:
Answer:
| Basis | Physisorption | Chemisorption |
| (i) Specificity | It is not specific in nature i.e. all gases are adsorbed on all solids to some extent. | It is highly specific in nature and occurs only when there is some possibility of compound formation between the gas being adsorbed and the solid being adsorbent. |
| (ii) Temperature dependence | It is independent of temperature as it occurs at low temperature and decreases with increase in temperature. | It is temperature dependent and increases with increase in temperature. |
| (iii) Reversibility | It is reversible i.e. desorption of gas takes place by increasing the temperature or decreasing the pressure. | It is irreversible in nature as it involves formation of compound instead of release of gas. |
| (iv) Enthalpy change | It has low enthalpy of adsorption i.e., 20-40 kj mol 4 . | It has high enthalpy of adsorption i.e., 40-240 kj/mol. |
Question 13.
(a) Which solution is used for the leaching of Silver metal in the presence of air in the metallurgy of silver?
(b) Out of ‘C’ and ‘CO’, which is a better reducing agent at the lower temperature range in the blast furnace to extract iron from the oxide ore? [2]
Answer:
(a) The 0.5% solution of sodium or potassium cyanide is used for the leaching of silver metal.
(b) CO will be a better reducing agent at the lower temperature range in the blast furnace to extract iron from the oxide ore because in Ellingham diagwn the CO
2
line lies below
Fe, FeO line. Hence CO reduces Fe
2
O, Fe
3
O, FeO etc.
Question 14.
What happens when [2]
(i) PCl
5
is heated?
(ii) H
3
PO
3
is heated?
Write the reaction involved.
Answer:
(i) On heating PCl
5
decomposes into PCl
3
+ Cl
2
(ii) Ortho phosphorous acid on heating disproportionates to give orthophosphoric acid and phosphine.
Question 15.
(a) Which metal in the first transition series (3d series) exhibits +1 oxidation state most frequently and why?
(b) Which of the following cations are coloured in aqueous solutions and why? [2]
Sc
3+
, V
3+
, Ti
4+
, Mn
2+
(Atomic Nos. Sc = 21, V = 23, Ti = 22, Mn = 25)
Answer:
(a) Copper exhibits +1 oxidation state more frequently i.e., Cu
+1
because of its electronic
configuration 3d
10
4s
1
. It can easily lose 4s1 electron to give stable 3d1° configuration.
(b) SC
3+
= 4s° 3d° = no unpaired electron
V
3+
= 3d
2
4s° = 2 unpaired electrons
Ti
4+
= 3d° 4s° = no unpaired electron
Mn
2+
= 3d
5
4s° = 5 unpaired electrons
Thus V
3+
and Mn
2+
are coloured in their aqueous solution due to presence of unpaired electron.
Question 16.
Chlorobenzene is extremely less reactive towards a nucleophilic substitution reaction. Give two reasons for the same. [2]
Answer:
The reasons are:
(i) Due to resonance/diagrammatic représentation, C—Cl bond acquires a partial double bond character which is difficult to cleave.
(ii) Due to repulsion between nucleophile and electron rich arenes.
Question 17.
Explain the mechanism of the following reaction: [2]
Answer:
Question 18.
How will you convert:
(i) Propene to Propan-2-ol?
(ii) Phenol to 2, 4, 6 – trinitrophenol?
Answer:
(i) Propene to propan-2-ol :
(ii) Phenol to 2,4,6-trinitrophenol
Question 19.
(a) What type of semiconductor is obtained when silicon is doped with boron? [3]
(b) What type of magnetism is shown in the following alignment of magnetic moments?
(c) What type of point defect is produced when AgCl is doped with CdCl
2
?
Answer:
(a) p-type semi-conductor is obtained when silicon is doped with boron.
(b) Ferromagnetism is shown when the alignment of magnetic moments will be
(c) Impurity defect of ionic solids is produced when AgCl is doped with CdCl
2
. Due to this defect vacancies are created that result in higher electrical conductivity of the solid.
Question 20.
Determine the osmotic pressure of a solution prepared by dissolving 2.5 × 10
-2
g of K
2
SO
4
in 2 L of water at 25°C, assuming that it is completely dissociated. [3]
(R = 0.0821 L atm K
-1
mol
-1
, Molar mass of K
2
SO
4
= 174 g mol
-1
)
Answer:
Question 21.
Calculate the emf of the following cell at 298 K: [3]
Fe(s) | Fe
2+
(0.001 M) || H
+
(1M) | H
2
(g) (1 bar), Pt(s)
(Given E
0
cell
= +0.44 V)
Answer:
As Fe + 2H
+
→ Fe
2+
+ H
2
(n = 2)
According to Nernst equation
E
cell
= E°
cell
– \(\frac{0.0591}{2}\)log\(\frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{H}^{+}\right]^2}\)
E
cell
= 0.44 – \(\frac{0.0591}{2}\)log\(\frac{10^{-3}}{1^2}\)
∴ E
cell
= 0.44 – \(\frac{0.0591}{2}\) × (-3) = 0.44 + 0.0887 = 0.529 V
Question 22.
How would you account for the following? [3]
(i) Transition metals exhibit variable oxidation states.
(ii) Zr (Z = 40) and Hf (Z = 72) have almost identical radii.
(iii) Transition metals and their compounds act as a catalyst.
OR
Complete the following chemical equations:
(i) Cr
2
O
2-
7
+ 6Fe
2+
+ 14H
+
→
(ii) 2CrO
–
4
+ 2H
+
→
(iii) 2MnO
–
4
+ 5C
2
O
-2
4
+ 16H
+
→
Answer:
(i) The variability of oxidation state of transition elements is due to incompletely filled d-orbitals and presence of unpaired electrons.
(ii) Zr and Hf have almost identical radii due to lanthanoid contraction which is due to weak shielding of d-electrons
(iii) The catalytic properties of the transition elements are due to the presence of unpaired electrons in their incomplete d-orbitals and variable oxidation states.
OR
(i) Cr
2
\(\mathrm{O}_7^{2-}\) (aq) + 6Fe
2+
+ 14H
+
→ 2Cr
3+
+ 6Fe
3+
+ 7H
2
O
(ii) 2CrO\(\mathrm{O}_4^{2-}\) + 2H
+
→ Cr
2
\(\mathrm{O}_7^{2-}\) + H
2
O
(iii) 2Mn\(\mathrm{O}_4^{-}\) + 5C
2
\(\mathrm{O}_4^{2-}\) + 16H
+
→ 2Mn
2+
+ 8H
2
O + 10CO
2
Question 23.
Write the IUPAC names of the following coordination compounds: [3]
(i) [Cr(NH
3
)
3
Cl
3
]
(ii) K
3
[Fe(CN)
6
]
(iii) [CoBr
2
(en)
2
]
+
, (en = ehylenediamine)
Answer:
(i) IUPAC name : [Cr(NH
3
) Cl
3
] : Triammine trichlorido chromium (III)
(ii) IUPAC name : K
3
[Fe(CN)
6
] : Potassium hexacyanoferrate (III)
(iii) IUPAC name : [CoBr
2
(en
2
)] : Dibromidobis (ethane-1, 2-diamine) cobalt (III)
Question 24.
Give the structures of A, B and C in the following reactions: [3]
Answer:
Question 25.
Write the names and structures of the monomers of the following polymers: [3]
(i) Buna-S
(ii) Neoprene
(iii) Nylon-6, 6
Answer:
(i) Buna-S
(ii) Neoprene
(iii) Nylon-6, 6 : Monomers of Nylon-6, 6 are hexamethylene diamine and adipic acid.
Question 26.
After watching a programme on TV about the adverse effects of junk food and soft drinks on the health of school children, Sonali, a student of Class XII, discussed the issue with the principal. The principal immediately instructed the canteen contractor to replace the fast food with the fibre and vitamins rich food like sprouts, salad, fruits etc. This decision was welcomed by the parents and the students. [3]
After reading the above passage, answer the following questions:
(a) What values are expressed by Sonali and the Principal of the school?
(b) Give two examples of water-soluble vitamins.
Answer:
The two water soluble vitamins are Vitamin B
2
(Riboflavin) and Vitamin C (Ascorbic acid).
Question 27.
(a) Which one of the following is a food preservative? [3]
Equanil, Morphine, Sodium benzoate
(b) Why is bithional added to soap?
(c) Which class of drugs is used in sleeping pills?
Answer:
(a) Sodium Benzoate : It is a food preservative.
(b) 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.
(c) Tranquilizers like barbiturates are used in sleeping pills.
Question 28.
(a) A reaction is second order in A and first order in B. [5]
(i) Write the differential rate equation.
(ii) How is the rate affected increasing the concentration of A three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?
(b) A first-order reaction takes 40 minutes for 30% decomposition. Calculate t
1/2
for this reaction.
(Given: log 1.428 = 0.1548)
Answer:
(a) (i) Differential rate equation : \(\frac{d x}{d t}\) = K [A]
2
[B]
(ii) When concentration of A is increased to three times, the rate of reaction becomes 9 times
r = K[3A]
2
B ∴ r = 9KA
2
B i.e. = 9 times
(iii) r = K[2A]
2
[2B] ∴ r = 8KA
2
B i.e. 8 times
(b)
Or
(a) For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
(b) Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the equation:
log k = log A – \(\frac{E_a}{2.303 R}\left(\frac{1}{T}\right)\)
where E
a
is the activation energy. When a graph is plotted for log k vs. \(\frac{1}{\mathrm{~T}}\), a straight line with a slope of -4250 K is obtained. Calculate ‘E
a
’ for the reaction.. (R = 8.314 JK
-1
mol
-1
) [5]
Answer:
(a)
(b) Slope of the line = \(\frac{-E_a}{2.303 \mathrm{R}}\) = -4250 K
∴ E
a
= 2.303 × 8.314 (JK
-1
mol
-1
) × 4250 K = 81.375 J mol
-1
or 81.375 kJ mol
-1
Question 29.
(a) Give reasons for the following: [5]
(i) Bond enthalpy of F
2
is lower than that of Cl
2
.
(ii) PH
3
has a lower boiling point than NH
3
.
(b) Draw the structures of the following molecules:
(i) BrF
3
(ii) (HPO
3
)
3
(iii) XeF
4
OR
(a) Account for the following:
(i) Helium is used in diving apparatus.
(ii) Fluorine does not exhibit a positive oxidation state.
(iii) Oxygen shows catenation behaviour less than sulphur.
(b) Draw the structures of the following molecules.
(i) XeF
2
(ii) H
2
S
2
O
8
Answer:
(a) (i) Due to smaller size of F than CI as a result of which electron-electron repulsions between the lone pairs of electrons are very large than that of Cl, hence bond dissociation energy of F
2
is less than that of Cl
2
.
(ii) PH
3
has lower boiling point than NH
3
because PH
3
cannot form hydrogen bonds like NH
3
.
(b) (i) BrF
3
:
(ii)(HPO
3
)
3
: Cyclotrimetaphosphoriç. acid:
(iii)
OR
(a) (i) Helium is used in diving apparatus because of its very low solubility in blood and therefore an oxygen-helium mixture is used for artificial respiration.
(ii) Because it is most electronegative element and does not have d-orbitals for octet expansion, therefore it shows only a negative oxidation state of -1.
(iii) The greater catenatin 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.
(b) (i)
(ii) H
2
S
2
O
8
:
Question 30.
(a) Although phenoxide ion has a number of resonating structures than carboxylate ion, the carboxylic acid is a stronger acid than phenol. Give two reasons. [5]
(b) How will you bring about the following conversions?
(i) Propanone to propane
(ii) Benzoyl chloride to benzaldehyde
(iii) Ethanal to but-2-enal.
OR
(a) Complete the following reactions:
(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Ethanal and Propanal
(ii) Benzoic acid and Phenol.
Answer:
(a) Carboxylic acid is a stronger acid than phenol because:
(i) In the resonating structure of phenol and carboxylic acid, the negative charge on the carboxylate ion is delocalised over two oxygen atoms while they are localized on oxygen atom.
(b)
(i) Propane to propane
(ii) Benzoyl chloride to benzaldehyde (Rosenmund reduction)
(iii) Ethanal to but-2-enal:
Or
(a)
(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) Benzoic acid and Phenol :
Set II
Note: Except for the following questions, all the remaining questions have been asked in the previous set.
Question 1.
What type of stoichiometric defect is shown by AgCl? [1]
Answer:
Frenkel defect is shown by AgCl.
Question 2.
Write the IUPAC name of [1]
Answer:
IUPAC name : 4-bromo-4-methylpent-2-ene
Question 4.
What type of bonding helps in stabilizing the a-helix structure of proteins? [1]
Answer:
α-helix formation → Intramolecular hydrogen bonding.
Question 6.
What inspired N.Bartlett for carrying out a reaction between Xe and PtF
6
? [1]
Answer:
Since PtF
6
oxidises O
2
to \(\mathrm{O}_2{ }^{+}\), Bartlett thought that PtF
6
should also oxidise Xe to Xe
+
because the ionization enthalpies of O
2
(1175 kJ mol
-1
) and Xe (1170 kJ mol
-1
) are quite close.
Question 7.
What happens when ethyl chloride is treated with aqueous KOH? [1]
Answer:
Question 8.
Write the structure of 4-chloropentan-2-one. [1]
Answer:
Question 9.
How will you convert the following?
(i) Propan-2-ol to propanone.
(ii) Phenol to 2,4,6 – tribromophenol?
Answer:
(i) Propane-2-ol to propanone
(ii) Phenol to 2, 4, 6-tribromophenol : Phenol to 2, 4, 6-tribromophenol : Phenol reacts bromine in presence of polar solvent H
2
O to form 2, 4, 6—tribromophenol (white ppt.)
Question 11.
What is the difference between oil/water (O/W) type and water/oil (W/O) type emulsions? Give an example of each type. [2]
Answer:
Difference between two types of emulsions are:
Oil/Water type emulsion: Emulsions of oil in water in which oil is the dispersed phase and water is thé dispersion medium.
e.g. Milk is an emulsion of liquid fat dispersed in water.
Water/Oil type emulsion : Emulsions of water in oil in which water is the dispersed phase and oil is the dispersion medium.
e.g. Cold liver oil is an emulsion of oil i.e. water is the dispersed phase and oil is the dispersion medium.
Two applications of emulsion are:
- The digestion of fats in the intestines takes place by the process of emulsification.
- Several oily drug prepapred in the form of emulsion.
Question 17.
(a) Which of the following ores can be concentrated by froth floatation method and why?
Fe
2
O
3
, ZnS, Al
2
O
3
(b) What is the role of silica in the metallurgy of Copper? [2]
Answer:
(a) ZnS ore can be concentrated by froth floatation method because it is a sulphide ore.
(b) 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 18.
(a) Why does p-dichlorobenzene have a higher m.p. than its o- and m- isomers? [2]
(b) Why is (-) – Butan-2-ol optically inactive?
Answer:
(a) p-isomers are comparatively more symmetrical and fit closely in the crystal lattice, thus require more heat to break these strong forces of attraction. Therefore higher melting point than o- and rn-isomers.
(b) (±)-Butan-2-ol is optically inactive because in racemic mix one type of rotation is cancelled by other.
Question 23.
Write the names and structures of the monomers of the following polymers: [3]
(i) Polystyrene
(ii) Dacron
(iii) Teflon
Answer:
(i) Polystyrene: Naine: Monomer is styrene
Structure: C
6
H
5
CH = CH
2
(ii) Dacron : Name : Monomers are ethylene, glycol and terephthalic acid
Structure:
(iii) Teflon : Name : Monomer is tetrafluoroethylene
Question 27.
Write the types of isomerism exhibited by the following complexes: [3]
(i) [CO(NH
3
)
5
Cl]SO
4
(ii) [Co(en)
3
]
3+
(iii) [CO(NH
3
)
6
] [Cr(CN)
6
]
Answer:
(i) [CO(NH
3
)
5
Cl]SO
4
— Ionisation isomerism
(ii) [Co(en)
3
]
3+
— Optical isomerism
(iii) [CO(NH
3
)
6
] [Cr(CN)
6
] — Coordination isomerism
Set III
Note: Except for the following questions, all the remaining questions have been asked in previous sets.
Question 1.
What type of substances would make better Permanent Magnets, Ferromagnetic or Ferrimagnetic? [1]
Answer:
Ferromagnetic substances would make better permanent magnets Example : Fe, Co, Ni etc.
Question 3.
What is the composition of ‘Copper matte’? [1]
Answer:
Copper matte chiefly consists of Cuprous Sulphide (Cu
2
S) and some uncharged Ferrous Sulphide (FeS).
Question 5.
What is a glycosidic linkage? [1]
Answer:
The two monosaccharide units are 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.
Question 6.
Write the IUPAC name of (CH
3
)
2
CH.CH(Cl)CH
3
[1]
Answer:
IUPAC name : 2-chloro-3-methylbutane
Question 7.
Which compound in the following pair undergoes faster S
N
1 reaction? [1]
Answer:
reacts faster by S
N
1 mechanism as it is a teritiary halide and it produces a stable tertiary carbonation.
Question 8.
Write the structure of a p-Methylbenzaldehyde molecule. [1]
Answer:
Question 9.
What is the difference between multi-molecular and macromolecular colloids? Give one example of each. [2]
Answer:
Multi-molecular colloid is aggregation of large number of atoms or smaller molecules of a substance having size in the colloidal range. Whereas macromolecular colloid is the solution containing macromolecules in the colloidal range.
Example: Multimolecular colloids: Gold sol, Sulphur sol
Macromolecular colloids: Proteins, cellulose (any one)
Question 14.
(a) Give an example of zone refining of metals
(b) What is the role of cryolite in the metallurgy of aluminium? [2]
Answer:
(a) Zone refining is based on thiprinciple that the impurities are more soluble in the melt than in the solid state of the metal.
(b) The role of cryolitë (Na
3
AIF
6
) is to lower the melting point of the mixture and brings conductivity.
Question 17.
Account for the following: [2]
(i) The C—Cl bond length in chlorobenzene is shorter than that in CH
3
-Cl.
(ii) Chloroform is stored in closed dark brown bottles.
Answer:
(i) In haloalkanes, the halogen atom is attached
to sp
3
-hyhridized carbon while in haloarenes it is attached to sp’-hvhridized carbon whose size is smaller than sp
3
orbital carbon. Therefore, C—Cl bond in chlrobenzene is shorter than alkyl chloride.
(ii) CHCl
3
is stored in dark coloured bottles to cut off light because CHCl
3
is slowly oxidised by air in presence of light to form an extremely poisonous gas, carbonyl chloride, popularly known as phosgene.
Question 18.
How will you convert: [2]
(i) Propene to Propan-1-ol?
(ii) Ethanal to Propan-2-ol?
Answer:
(i) Propene to Propane-1-ol
(ii) Ethanal to Propan-2-ol
Question 23.
Give the structures of products A, B and C in the following reactions: [3]
Answer:
Question 27.
Write the names and structures of the monomers of the following polymers: [3]
(i) Bakelite
(ii) Nylon-6
(iii) Polythene
Answer:
(i) Bakelite : Name: Monomers are
(a) Phenol; Structure : (Ç
6
H
5
OH) and
(b) Formaldehyde; Structure : (HCHÇ)).
(ii) Nylon-6: Name : Monomer unit, which is derived from Caprolactuin, is
(iii)
Polyethene : Name — Monomer is Ethene
Structure : CH
2
=CH
2
