Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2020 (Series HMJ/4) to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2020 (Series HMJ/4) with Solutions
Time allowed: 3 hours
Maximum Marks: 70
General Instructions
- Question paper comprises four Sections-A, B, C and D.
- There are 37 questions in the question paper. All questions are compulsory.
- Section A : Q. no. 1 to 20 are very short answer type questions carrying one mark each. Answer these questions in one word or one sentence.
- Section B : Q. no. 21 to 27 are short answer type questions carrying two marks each.
- Section C : Q. no. 28 to 34 are long answer type-I questions carrying three marks each.
- Section D : Q. no. 35 to 37 are long answer type-II questions carrying five marks each.
- There is NO overall choice in the question paper. However, an internal choice has been provided in 2 questions of two marks, 2 questions of three marks and all the 3 questions of five marks. You have to attempt only one of the choices in such questions.
- However, separate instructions are given with each section and question, wherever necessary.
- Use of calculators and log tables is NOT permitted.
Set I Code No. 56/4/1
Section – ‘A’
Read the given passage and answer the questions 1 to 5 that follow:
Colloidal particles always carry an electric charge which may be either positive or negative. For example, when AgNO 3 solution is added to KI solution, a negatively charged colloidal sol is obtained. The presence of equal and similar charges on colloidal particles provides stability to the colloidal sol and if, somehow, charge is removed, coagulation of sol occurs. Lyophobic sols are readily coagulated as compared to lyophilic sols.
Question 1.
What is the reason for the charge on sol particles?
Answer:
The charge on sol particles is due to the preferential adsorption of either positive or negative ions on the surface.
Question 2.
Why the presence of equal and similar charges on colloidal particles provides stability?
Answer:
The presence of equal and similar charges on colloidal particles provides stability to it because the repulsive forces between similarly charged particles prevent them from aggregating when they come closer to each other.
Question 3.
Why a negatively charged sol is obtained on adding AgNO
3
solution to KI solution?
Answer:
A negatively charged sol is obtained on adding AgNO
3
solution to KI solution because the precipitated Agl adsorbs I” ions from the dispersion medium.
Question 4.
Name one method by which coagulation of lyophobic sol can be carried out.
Answer:
Electrophoresis
Question 5.
Out of KI or K
2
SO
4
which electrolyte is better in the coagulation of positive sol?
Answer:
K
2
SO
4
is better in the coagulation of positive sol.
Questions 6 to 10 are one word answers:
Question 6.
Name the method applied for the concentration of Bauxite ore in the extraction of Aluminium.
Answer:
Leaching
Question 7.
Out of
,
which one is more reactive towards S
N
1 reaction?
Answer:
is more reactive towards S
N
1 reaction.
Question 8.
Write an isomer of C
3
H
9
N which gives foul smell of isocyanide when treated with chloroform and ethanolic NaOH.
Answer:
Propanamine (C
3
H
9
N) gives carbylamine reaction.
Question 9.
Which one of the following is an antidepressant drug?
Chloramphenicol, Luminal, Bithional
Answer:
Luminal is an antidepressant drug.
Question 10.
Write the name of component of starch which is water soluble.
Answer:
Amylose
Questions 11 to 15 are Multiple Choice Questions:
Question 11.
How many ions are produced from the complex [Co(NH
3
)
5
Cl] Cl
2
in solution?
(a) 4
(b) 2
(c) 3
(d) 5
Answer:
(c) 3
Question 12.
In a lead storage battery
(a) PbO
2
is reduced to PbSO
4
at the cathode.
(b) Pb is oxidised to PbSO
4
at the anode.
(c) Both electrodes are immersed in the same aqueous solution of H
2
SO
4
.
(d) All the above are true.
Answer:
(d) All the above are true.
Question 13.
The slope in the plot of In [R] vs. time gives
(a) +k
(b) \(\frac{+k}{2.303}\)
(c) -k
(d) \(\frac{-k}{2.303}\)
(Where [R] is the final concentration of reactant.)
Answer:
(c) -k
Question 14.
The pair [Co(NH
3
)
4
Cl
2
] Br
2
and [Co(NH
3
)
4
Br
2
] Cl
2
will show
(a) Linkage isomerism
(b) Hydrate isomerism
(c) Ionization isomerism
(d) Coordinate isomerism
Answer:
(c) Ionization isomerism
Question 15.
An α-helix is a structural feature of
(a) Sucrose
(b) Polypeptides
(c) Nucleotides
(d) Starch
Answer:
(b) Polypeptides
Questions 16 to 20:
(A) Both Assertion (A) and Reason (R) are correct statements, and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are correct statements, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is correct, but Reason (R) is wrong statement.
(D) Assertion (A) is wrong, but Reason (R) is correct statement.
Question 16.
Assertion (A) : F
2
is a strong oxidising agent.
Reason (R) : Electron gain enthalpy of fluorine is less negative.
Answer:
(B) Both Assertion (A) and Reason (R) are correct statements, but Reason (R) is not the correct explanation of the Assertion (A).
Question 17.
Assertion (A) : (CH
3
)
3
C — O — CH
3
gives (CH
3
)
3
C —I and CH
3
OH on treatment with HI.
Reason (R) : The reaction occurs by S
N
1 mechanism.
Answer:
(A) Both Assertion (A) and Reason (R) are correct statements, and Reason (R) is the correct explanation of the Assertion (A).
Question 18.
Assertion (A) : Transition metals have low melting points.
Reason (R) : The involvement of greater number of (n – 1)d and ns electrons in the interatomic metallic bonding.
Answer:
(D) Assertion (A) is wrong, but Reason (R) is correct statement.
Question 19.
Assertion (A) : Hydrolysis of an ester follows first order kinetics.
Reason (R) : Concentration of water remains nearly constant during the course of the reaction.
Answer:
(A) Both Assertion (A) and Reason (R) are correct statements, and Reason (R) is the correct explanation of the Assertion (A).
Question 20.
Assertion (A) : Benzoic acid does not undergo Friedal-Crafts reaction.
Reason (R) : The carboxyl group is activating and undergo electrophilic substitution reaction.
Answer:
(C) Assertion (A) is correct, but Reason (R) is wrong statement.
Section B
Question 21.
What happens when
(i) a pressure greater than osmotic pressure is applied on the solution side separated from solvent by a semipermeable membrane?
(ii) acetone is added to pure ethanol? [1 + 1 = 2]
Answer:
(i) When the external pressure applied becomes more than the osmotic pressure of the solution, then the solvent will flow from the solution into the pure solvent through the semi-permeable membrane. The process is called Reverse Osmosis (RO).
(ii) When acetone molecules are added to pure ethanol, they get in between the molecules of ethanol and break some of the hydrogen bond due to which inter-molelcular attractive forces weaken and vapour pressure of solution increases because of this mixture of ethanol and acetone. It shows positive deviation in vapour pressure from Raoult’s law.
Question 22.
Write the principle of the following refining methods:
(a) vapour phase refining
(b) chromatography [1 + 1 = 2]
Answer:
(a) Vapour phase refining.
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.
(b) Chromatography. This is the method used for the separation and purification of elements. It can also be used for testing the purity of a compound. The principle behind the chromatography is that different components of a mixture are differently adsorbed on an adsorbent.
Or
Write chemical equations involved to obtain: [1 + 1 = 2]
(i) Cu from Cu
2
S
(ii) Ag from [Ag(CN)
2
]
–
complex
Answer:
Question 23.
Write the balanced chemical equations involved in the preparation of KMnO
4
from pyrolusite ore (MnO
2
). [2]
Answer:
Potassium Permangante (KMnO
4
) is prepared from pyrolusite ore (MnO
2
). The ore (MnO
2
) is fused with an alkali metal hydroxide like KOH in the presence of air or an oxidising agent like KNO
3
to give dark green potassium manganate (K
2
MnO
4
). K
2
MnO
4
disproportionates in a neutral or acidic solution to give potassium permanganate.
Or
Write the balanced ionic equations showing the oxidising action of acidified dichromate \(\left(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\right)\) solution with (i) Iron (II) ion and (ii) tin (II) ion. 1+1=2
Answer:
(i) Oxidising action of acidified dichromate with Iron (II) ion:
(ii) Oxidising action of acidified dichromate with tin (II) ion:
Question 24.
Write the IUPAC name and hybridisation of the following complexes:
(i) [Ni(CN)
4
]
2-
(ii) [Fe(H
2
OH)
6
]
2+
(Given: Atomic number of Ni = 28, Fe = 26) 1+1=2
Answer:
(i) [Ni(CN)
4
]
2-
: IUPAC name — Tetracyanonickelate (II) ion; Hybridisation —dsp
2
(ii) [Fe(H
2
O)
6
]
2+
: IUPAC name — Hexaaquairon (II) ion; Hybridisation — sp
3
d
2
Question 25.
Define the following terms with a suitable example in each:
(i) Antibiotics
(ii) Antiseptics [1 + 1 = 2]
Answer:
(i) Antibiotics. The chemical substances which are produced completely or partially by chemical synthesis in low concentration and either’kill or inhibit the growth of microorganisms by intervening in their metabolic processes, are known as antibiotics. Example. Tetracycline, Vancomycin, Pencillin ofloxacin etc.
(ii) Antiseptics. These are chemical substances that are applied on living tissues. Antiseptics prevent the growth of micro-organisms and may even kill them. Example. Furacin, Soframycin, Dettol etc.
Question 26.
Write the reactions showing the presence of following in the open structure of glucose:
(i) a carbonyl group
(ii) Straight chain with six carbon atoms. [1 + 1 = 2]
Answer:
(i) Reaction showing presence of a carbonyl group in glucose:
(ii) Straight chain with six carbon atoms. On prolonged heating with HI, it forms n-hexane, shows that all the six carbon atoms are linked in a straight chain.
Question 27.
State Henry’s law. Calculate the solubility of CO
2
in water at 298 K under 760 mm Hg.
(K
H
for CO
2
in water at 298 K is 1.25 × 10
6
mm Hg) [2]
Answer:
Henry’s law states, “the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution”.
Given: T = 298 K; K
H
= 1.25 × 10
6
mm Hg; p = 760 mm Hg
According to Henry’s law, p = K
H
× x ……where[x = Mole fraction of the gas Substituting the values, 760 = 1.25 × 10
6
× x
x = \(\frac{760}{1.25 \times 10^6}\) = 608 × 10
-6
We know, x \(=\frac{760}{1.25 \times 10^6}\)
608 × 10
-6
\(=\frac{\text { Moles of gas }}{55.5}\) [∵ Moles of gas <<< Moles of H
2
O; Moles of H
2
O in 1 L = 55.50
Moles of gas = 608 × 10
-6
× 55.5 = 33,744 × 10 = 33.75 × 10 mol
-1
∴ Solubility of CO
2
in water is 33.75 × 10
-3
moles.
Section – ‘C’
Question 28.
The freezing point of a solution containing 5g of benzoic acid (M = 122 g mol
-1
) in 35g of benzene is depressed by 2.94 K, What is the percentage association of benzoic acid if it forms a dimer in solution? (K
f
for benzene 4.9 K kg mol
-1
) [3]
Answer:
Given: ω
2
(Mass of solute, i.e., benzoic acid) = 5 g
ω
1
(Mass of benzene, i.e., solvent) = 35 g
∆T
f
= 2.94 K; K
f
= 4.9 Kg mol
-1
; M
2
= ?, i = ?, α = ?
Using, M
2
= \(\frac{1000 \times \mathrm{K}_f \times w_2}{\Delta \mathrm{T}_f \times w_1}\) = \(\frac{1000 \times 4.9 \times 5}{2.94 \times 35}\)
= \(\frac{24,500}{102.9}\) = 238.09 g mol
-1
Calculated molar mass of benzoic acid 122 g mol
-1
Van’t Hoff factor, i \(=\frac{\text { Calculated molar mass }}{\text { Observed molar mass }}\) = \(\frac{122}{238.09}\)
∴ i = 0.512
If α is degree of association, then
Question 29.
The rate constant for the first order decomposition of N
2
O
5
is given by the following
equation:
k = (2.5 × 10
14
s
-1
) e
(-25000K)/T
Calculate Ea for this reaction and rate constant if its half-life period be 300 minutes. [3]
Answer:
Given k = (2.5 × 10
14
s
-1
) e
(-25000K)/T
According to Arrhenius equation
k = Ae
-Ea/RT
∴ \(\frac{\mathrm{Ea}}{\mathrm{RT}}\) = \(\frac{-25,000 k}{T}\)
Ea = 25,000 k × R = 25,000 k × 8.314 J mol
-1
∴ Ea = 207850 J mol
-1
or 207.85 KJ mol
-1
When t
1/2
= 300 min
k = \(\frac{0.693}{300 \times 60 \mathrm{~s}}[latex] = [latex]\frac{0.693}{18,000}[latex] = 0.0000385 ∴ k = 3.85 × 10
-5
s
-1
Question 30.
Write the name and structures of monomer(s) in the following polymers: [1 + 1 + 1 = 3]
(i) Nylon-6
(ii) PVC
(iii) Neoprene
Answer:
Question 31.
Following ions are given:
Cr
2+
, Cu
2+
, Cu
+
, Fe
2+
, Fe
3+
, Mn
3+
Identify the ion which is
(i) a strong reducing agent.
(ii) unstable in aqueous solution.
(iii) a strong oxidising agent.
Give suitable reason in each. [1 + 1 + 1 = 3]
Answer:
(i) Cr
2+
is a strong reducing agent because of its half-filled stable t
2g
configuration when its configuration changes from 3d
4
to 3d
3
.
(ii) Cu
+
is unstable in aqueous solution because it disproportionates in its solution due
to high second ionisation enthalpy of Cu
2+
and more negative hydration enthalpy
than that of Cu
+
.
(iii) Mn
3+
is a strong oxidising agent because it acquires half-filled stable d
5
configuration on gain of one electron.
Question 32.
(i) Write the structure of major alkene formed by β-elimination of 2, 2, 3-trimethyl-3-bromopentane with sodium ethoxide in ethanol.
(ii) Which one of the compounds in the following pairs is chiral?
(iii) Identify (A) and (B) in the following:
Answer:
Or
How can you convert the following?
(i) But-1-ene to 1-iodobutane
(ii) Benzene to acetophenone
(iii) Ethanol to propanenitrile [1 + 1 + 1 = 3]
Answer:
Question 33.
Arrange the following compounds as directed:
(i) In increasing order of solubility in water:
(CH
3
)
2
NH, CH3NH2, C6H5NH2
(ii) In decreasing order of basic strength in aqueous solution:
(CH
3
)
3
N, (CH
3
)
2
NH
2
, CH
3
NH
2
(iii) In increasing order of boiling point:
(C
2
H
5
)
2
NH, (C
2
H
5
)
3
N, C
2
H
5
NH
2
[1 + 1 + 1 = 3]
Answer:
(i) Solubility order in water:
C
6
H
5
NH
2
< (CH
3
)
2
NH <CH
3
NH
2
(ii) Basic strength order in aqueous solution:
(CH
3
)
2
NH > CH
3
NH
2
> (CH
3
)
3
N
(iii) Boiling point order
(C
2
H
5
)
3
N < (C
2
H
5
)
2
NH < C
2
H
5
NH
2
Question 34.
Write the product(s) of the following reactions:
Answer:
(a) Write the mechanism of the following S
N
1 reaction:
(b) Write the equation for the preparation of 2-methyl-2-methoxypropane by Williamson synthesis. [2 + 1 = 3]
Answer:
(a) Mechanism of S
N
1 reaction for
(b) Preparation of 2-methyl-2-methoxypropane by Williamson synthesis:
Question 36.
(a) An orgnic compound (A) having molecular formula C
4
H
8
O gives an orange-red precipitate with 2, 4-DNP reagent. It does not reduce Tollens’ reagent but gives yellow precipitate of iodoform on heating with NaOH and I
2
. Compound (A) on reduction with NaBH
4
gives compound (B) which undergoes dehydration reaction on heating with conc. H
2
SO
4
to form Compound (C). Compound (C) on Ozonolysis gives two molecules of ethanal.
Identify (A), (B) and (C) and write their structures. Write the reactions of Compound (A) with
(i) NaOH/I
2
and
(ii) NaBH
4
.
(b) Give reasons:
(i) Oxidation of propanal is easier than propanone.
(ii) α-hydrogen of aldehydes and ketones is acidic in nature. [3 + 2 = 5]
Answer:
(a) Since compound (A) with molecular formula C
4
H
8
O forms a 2, 4-DM’ it must be either aldehyde or ketone.
Since it does not reduce Tollens’ reagent, then it must be a ketone.
Since it gives yellow ppt. with NaOH and I
2
, it must have a methyl ketone.
Thus the probable structure of Ketone will be —
Thus, A = Butan-2-one
B = Butan-2-ol
C = But-2-ene
(b) (i) Oxidation of propanal is easier than propanone due to presence of a hydrogen atom on the carbonyl group which can be converted into OH group without involving the cleavage of any other bond while in propanone it is not possible without cleavage of C—C bonds.
(ii) α-hydrogen of aldehydes and ketones are acidic in nature due to -I effect of the carbonyl group which weakens the CaH bond and partly due to the resonance stabilization of the resulting enolate anion.
Or
(a) Draw structures of the following derivatives:
(i) Cyanohydrin of cyclobutanone
(ii) Hemiacetal of ethanal.
(b) Write the major product(s) in the following:
(c) How can you distinguish between propanal and propanone? [2 + 2 + 1 = 5]
Answer:
(c) Propanal and pro panone. Propanal being an aldehyde gives positive test with Fehling solution in which a red brown ppt. of cuprous oxide is obtained while propanone being a ketone does not respond to this test.
Question 37.
(a) Account for the following:
(i) Tendency to show -2 oxidation state decreases from oxygen to tellurium.
(ii) Acidic character increases from HF to HI.
(iii) Moist SO
2
gas acts as a reducing agent.
(b) Draw the structure of an oxoacid of sulphur containing S—O—S linkage.
(c) Complete the following equation:
XeF
2
+ H
2
O → [3 + 1 + 1 = 5]
Answer:
(a) (i) Tendency to show —2 oxidation state decreases from oxygen to tellurium because of decrease in electronegativity in the group.
(ii) Acidic character increases from HF to HI because bond dissociation enthalpy decreases from HF to HI.
(iii) Moist SO
2
gas is a reducing agent because sulphur can expand its covalency upto +6 from -14 due to presence of empty d-orbital, but as we move down the group, the stability of +6 oxidation state (O.S.) decreases ar.d of +4 O.S. increases due to inert pair effect.
(b) Oxoacid of sulphur containing S—O—S linkage is pyrosuiphuric acid (oleum) H
2
S
2
O
7
.
(c)
Or
(a) Among the hydrides of group 16, write the hydride
(i) Which is a strong reducing agent.
(ii) Which has maximum bond angle.
(iii) Which is most thermally stable.
Give suitable reason in each.
(b) Complete the following equations:
[3 + 1 + 1 = 5]
Answer:
(a) (i) H
2
Te is a strong reducing agent among its group members due to decrease in their thermal stability due to increased size of terrarium (central atom) which makes breaking of H—Te easier on heating.
(ii) H
2
O has maximum bond angle due to stronger Ione pair — Ione pair repulsion
than bond pair—bond pair repulsion.
(iii) H
2
O is most thermally stable among its members because of its high bond
dissociation energy due to its small size.
(b)
Set II Code No. 56/4/2
Except for the following questions, all the remaining questions have been asked in Set I.
Questions 6 to 10 are one word answers:
Question 6.
Name the depressant which is used to seperate PbS and ZnS containing ore in froth floatation process.
Answer:
Sodium cyanide, NaCN
Question 7.
Out of
which will react faster in S
N
1 reaction with OH
–
?
Answer:
Question 8.
Out of CH
3
NH
2
and CH
3
OH, which has higher boiling point?
Answer:
CH
3
OH (Methanol) has higher boiling point than CH
3
NH
2
.
Question 9.
Which one of the following is a narcotic analgesic?
Penicillin, Codeine, Ranitidine
Answer:
Codeine is a narcotic analgesic.
Question 10.
Write the name of linkage joining two monosaccharides.
Answer:
Glycosidic linkage joins two monosaccharides.
Questions 11 to 15 are Multiple Choice Questions:
Question 11.
The coordination number of ‘Co’ in the complex [Co(en)
3
]
3+
is
(a) 3
(b) 6
(c) 4
(d) 5
Answer:
(b) 6
Question 12.
An electrochemical cell behaves like an electrolytic cell when
(a) E
cell
= E
external
(b) E
cell
= 0
(c) E
external
> E
cell
(d) E
external
E
cell
Answer:
(c) E
external
> E
cell
Question 13.
The half-life period for a zero order reaction is equal to
(a) [latex]\frac{0.693}{k}\)
(b) \(\frac{2 k}{[R]_0}\)
(c) \(\frac{2.303}{k}\)
(d) \(\frac{[\mathrm{R}]_0}{2 k}\)
(where [R]
o
is initial concentration of reactant and k is rate constant.)
Answer:
(d) \(\frac{[R]_0}{2 k}\)
Question 14.
The crystal field splitting energy for octahedral (∆
0
) and tetrahedral (∆
t
) complexes is related as
(a) ∆
t
= \(\frac{2}{9}\)∆
0
(b) ∆
t
= \(\frac{5}{9}\)∆
0
(c) ∆
t
= \(\frac{4}{9}\)∆
0
(d) ∆
t
= 2∆
0
Answer:
(c) ∆
t
= \(\frac{4}{9}\)∆
0
Question 15.
α-D(+) glucose and β-D(+) glucose are
(a) Geometrical isomers
(b) Enantiomers
(c) Anomers
(d) Optical isomers
Answer:
(c) Anomers
Questions 16 to 20:
(A) Both Assertion (A) and Reason (R) are correct statements, and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are correct statements, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is correct, but Reason (R) is wrong statement.
(D) Assertion (A) is wrong, but Reason (R) is correct statement.
Question 16.
Assertion (A) : F
2
has lower bond dissociation enthalpy than Cl
2
.
Reason (R) : Fluorine is more electronegative than chlorine.
Answer:
(B) Both Assertion (A) and Reason (R) are correct statements, but Reason (R) is not the correct explanation of the Assertion (A).
Question 17.
Assertion (A) : Hydrolysis of an ester follows first order kinetics.
Reason (R) : Concentration of water remains nearly constant during the course of the reaction.
Ans.
(A) Both Assertion (A) and Reason (R) are correct statements, and Reason (R) is the correct explanation of the Assertion (A).
Question 18.
Assertion (A): Transition metals have high melting point.
Reason (R) : Transition metals have completely filled d-orbitals.
Answer:
(C) Assertion (A) is correct, but Reason (R) is wrong statement.
Question 19.
Assertion (A) : (CH
3
)
3
C—O—CH
3
gives (CH
3
)
3
C—I and CH
3
OH on treatment with HI.
Reason (R) : The reaction occurs by S
N
1 mechanism.
Answer:
(A) Both Assertion (A) and Reason (R) are correct statements, and Reason (R) is the correct
explanation of the Assertion (A).
Question 20.
Assertion (A) : Benzoic acid does not undergo Friedal-Crafts reaction.
Reason (R) : The carboxyl group is activating and undergo electrophilic substitution reaction.
Answer:
(C) Assertion (A) is correct, but Reason (R) is wrong statement.
Question 23.
Write TUPAC name and hybridization of the following complexes:
(i) [Ni(CO)
4
]
(ii) [C0F
6
]
3-
(Atomic number Ni = 28, Co = 27) [1 + 1 = 2]
Answer:
(i) [Ni(CO)
4
]: IUPAC name—Tetracarbonylnickel (0); Hybridization—sp
3
(ii) [CoF
6
]
3-
: IUPA C name — Hexafluoridocobaltate (III) ion; Hybridization —sp
3
d
2
Question 25.
Define the following terms with a suitable example in each: [1 + 1 = 2]
(i) Tranquilizers
(ii) Anionic detergent
Answer:
(i) Tranquilizers. Tranquilizers are a class of chemical compounds used for the treatment of stress and mild or even severe mental disease. For example, Equanil, meprobamate, veronal.
(ii) Anionic detergent:
Anionic deterjents. Those detergeits in which large part of their molecules are anions and used in cleansing action, are called anionic detergents.
Example: Sodium alkyl sulphates
These are obtained from long straight chain alcohols containing 12-18 carbon atoms by treatment with conc. H
2
SO
4
followed by neutralization with NaOH.
Example: Sodium lauryl sulphate.
Question 26.
Write the reactions showing the presence of following in the open structure of glucose: [1 + 1 = 2]
(i) an aldehyde group
(ii) a primary alcohol
Answer:
(i) Reaction showing presence of an aldehyde group in open structure of glucose and its bromination
(ii) Reaction showing presence of primary alcohol is its reaction with nitric acid
Question 31.
Write the name and structures of monomers ‘a the fòllowing polymers: [1 + 1 + 1 = 3]
(i) Nylon 6, 6
(ii) Terylene.
(iii) PHBV
Answer:
(i) Nylon 6,6:
It has two repeating monomers
(ii) Terylene:
SET III Code No. 56/4/3
Except for the following questions, all the remaining questions have been asked in Set I and Set II.
Questions 6 to 10 are one word answers:
Question 6.
Name the method used for the refining of Zinc.
Answer:
Electrolytic refining
Question 7.
Out of CH
3
CH
2
CH
2
Cl and CH
2
= CH—CH
2
—Cl, which one is more reactive towards S
N
1 reaction?
Answer:
CH
2
= CH — CH
2
— Cl (allyl chloride) is more reactive than propyl chloride (CH
3
CH
2
CH
2
Cl) towards S
N
1 reaction.
Question 8.
Write an isomer of C
3
H
9
N which does not react with Hinsberg reagent.
Answer:
Isomer of C
3
H
9
N which does not react with Hinsberg reagent is Tertiary amine, i.e., (CH
3
)
3
N (Trimethylamine).
Question 9.
What type of protein is present in Keratin?
Answer:
Fibrous protein is present in Keratin.
Question 10.
Name the compound which is added to soap to provide antiseptic properties.
Answer:
Bithional, as it acts as deodorant in soaps.
Question 11 to 15 are Multiple Choice Questions:
Question 11.
Which of the followIng is the most stable complex?
(a) [Fe(CO)
5
]
(b) [Fe(H
2
O)
6
]
3+
(c) [Fe(C
2
O
4
)
3
]
3-
(d) (Fe(CN)
6
]
3-
Answer:
(c) [Fe(C
2
O
4
)
3
]
3-
Question 12.
Which of the following is correct for spontaneity of a cell?
(a) ∆G = -ve E° = +ve
(b) G = +ve E° = 0
(c) ∆G = -ve E° = +ve
(d) ∆G = +ve E° = -ve
Answer:
(a) ∆G = -ve E° = +ve
Question 13.
For a zero order reaction, the slope in the plot of [R] vs. time is
(a) \(\frac{-k}{2.303}\)
(b) -k
(c) \(\frac{+k}{2.303}\)
(d) +k
(where [R] is the final concentration of reactant)
Answer:
(b) -k
Question 14.
What type of isomerism is shown by the pair [Cr(H
2
O)
6
]Cl
3
and [Cr(H
2
O)
5
Cl] Cl
2
H
2
O?
(a) Ionization isomerism
(b) Coordination isomerism
(c) Solvate isomerism
(d) Linkage isomerism
Answer:
(c) Solvate isomerism
Question 15.
Which one is the complementary base of cytosine in one strand to that in other strand of DNA?
(a) Adenine
(b) Guanine
(c) Thymine
(d) Uracil
Answer:
(b) Guanine
Questions 16 to 20:
(A) Both Assertion (A) and Reason (R) are correct statements, and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are correct statements, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is correct, but Reason (R) is wrong statement.
(D) Assertion (A) is wrong, but Reason (R) is correct statement.
Question 16.
Assertion (A) : F
2
has low reactivity.
Reason (R) : F—F bond has low ∆
bond
H°.
Answer:
(D) Assertion (A) is wrong, but Reason (R) is correct statement.
Question 17.
(CH
3
)
3
C—O—CH
3
gives (CH
3
)
3
C—I and CH
3
OH on treatment with HI.
Reason (R) : The reaction occurs by S
N
1 mechanism.
Answer:
(A) Both Assertion (A) and Reason (R) are correct statements, and Reason (R) is the correct explanation of the Assertion (A).
Question 18.
Assertion (A) : Transition metals have low melting points.
Reason (R) : The involvement of greater number of (n – 1)d and ns electrons in the interatomic metallic bonding.
Answer:
(D) Assertion (A) is wrong, but Reason (R) is correct statement.
Question 19.
Assertion (A) : (CH
3
)
3
C—O— CH
3
gives (CH
3
)
3
C — I and CH
3
OH on treatment with HI.
Reason (R) : The reaction occurs by S
N
I mechanism.
Answer:
(A) Both Assertion (A) and Reason (R) are correct statements, and Reason (R) is the correct explanation of the Assertion (A).
Question 20.
Assertion (A): For complex reactions molecularity and order are not same.
Reason (R) : Order of reaction may be zero.
Answer:
(B) Both Assertion (A) and Reason (R) are correct statements, but Reason (R) is not the correct explanation of the Assertion (A).
Question 23.
Write the IUPAC name and hybridisation of the following complexes:
(i) [Co(NH
3
)
6
]
3+
(ii) [NiCl
4
]
2-
(Given: Atomic number: Ni = 28, Co = 27) [1 + 1 = 2]
Answer:
(i) [Co(NH
3
)
6
]
3+
: IUPAC name — Hexaamminecobalt (III) ion; Hybridization — d
2
sp
2
(ii) [NiCl
4
]
2-
: IUPAC name—Tetrachloridonickelate (II) ion; Hybridization—sp
3
Question 25.
Write the reactions showing the presence of following in the open structure of glucose:
(i) five — OH groups
(ii) a carbonyl group 1+12
Answer:
(i) Reaction showing presence of five -OH groups:
(ii) Reaction showing presence of carbonyl group
Question 27.
Define the following terms with a suitable example in each: [1 + 1 = 2]
(i) Antacids
(ii) Artificial Sweetener
Answer:
(i) Antacids. The substances which neutralize the excess acid and raise the pH to an appropriate level in stomach are called antacids.
Example: Sodium bicarbonate, Mg(OH)
2
, etc.
(ii) Artficial Sweetener. The 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.
Question 31.
Write the names and structures of monomers in the tollowing polymers: [1 + 1 + 1 = 3]
(i) Buna-S
(ii) Glyptal
(iii) Bakelite
Answer:
(i)
(ii)
(iii) Bakelite. Phenol, Structure: [C
6
H
5
OH] and
Formaldehyde Structure: [HCHO]