Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2020 (Series HMJ/5) to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2020 (Series HMJ/5) with Solutions
Time allowed: 3 hours
Maximum Marks: 70
General Instructions:
- All questions are compulsory.
- Section A : Q. no. 1 fo 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. 56/5/1
Section – A
Read the given passage and answer the questions 1 to 5 that follow: [20 × 1 = 20]
The halogens have the smallest atomic radii in their respective periods. The atomic radius of fluorine is extremely small. All halogens exhibit -1 oxidation state. They are strong oxidising agents and have maximum negative electron gain enthalpy. Among halogens, fluorine shows anomalous behaviour in many properties. For example, electro negativity and ionization enthalpy are higher for fluorine than expected whereas bond dissociation enthalpy, m.p and b.p. and electron gain enthalpy are quite lower than expected. Halogens react with hydrogen to give hydrogen halides (HX) and combine amongst themselves to form a number of compounds of the type XX’, XX’ 3 , XX’ 5 and XX’7 called inter-halogens.
Question 1.
Why halogens have maximum negative electron gain enthalpy?
Answer:
Because of their high electronegativity they have strong tendency to accept electrons which is due to their small size and high effective nuclear charge in their corresponding periods.
Question 2.
Why fluorine shows anomalous behaviour as compared to other halogens?
Answer:
Fluorine shows anomalous behaviour because of its small size, interelectronic repulsions and absence of d-orbitals in valency shell.
Question 3.
Arrange the hydrogen halides (HF to HI) in the decreasing order of their reducing character.
Answer:
HI > HBr > HCl > HF
Question 4.
Why fluorine is a stronger oxidizing agent than chlorine?
Answer:
Because of its low bond dissociation enthalpy and highest eletronegativity it can accept electron pair easily and undergoes reduction.
Question 5.
What are the sizes of X and X’ in the interhalogen compounds?
Answer:
X is bigger in size than that of X’ in the interhalogen compounds.
Questions 6 to 10 are one ivord answers:
Question 6.
Name the cell used in hearing aids and watches.
Answer:
Mercury cell
Question 7.
How much charge in terms of Faraday is required to reduce one mol of Mn\(\mathrm{O}_4^{-}\) to Mn
2+
?
Answer:
Mn\(\mathrm{O}_4^{-}\) → Mn
2+
Mn\(\mathrm{O}_4^{-}\) + 8H+ 5e
–
→ Mn
2+
+ 4H
2
O
1 mol of e = IF = 96,500 C
So 5 moles of e = 5 × 96,500 C = 4,82,500 C
∴ Required charge = 4,82,500 C
Question 8.
Write the slope value obtained in the plot of log [R
0
] / [R] vs. time for a first order reaction.
Answer:
Question 9.
Name the sweetening agent used in the cooking of sweets for a diabetic patient.
Answer:
Saccharin is used for a diabetic patient for preparation of sweets.
Question 10.
Name the polymer which is used for making electrical switches and combs.
Answer:
Bakelite, as it is a poor conductor of heat and electricity.
Questions 11 to 15 are Multiple Choice Questions:
Question 11.
In the Mond’s process the gas used for the refining of a metal is
(a) H
2
(b) CO
2
(c) CO
(d) N
2
Answer:
(c) CO
Question 12.
The conversion of an alkyl halide into an alcohol by aqueous NaOH is classified as
(a) a dehydrohalogenation reaction
(b) a substitution reaction
(c) an addition reaction
(d) a dehydration reaction
Answer:
(b) a substitution reaction
Question 13.
CH
3
CONH
2
on reaction with NaOH and Br
2
in alcoholic medium gives
(a) CH
3
CH
2
NH
2
(b) CH
3
CH
2
Br
(c) CH
3
NH
2
(d) CH
3
COONa
Answer:
(c) CH
3
NH
2
Question 14.
The oxidation state of Ni in [Ni(CO)
4
] is
(a) 0
(b) 2
(c) 3
(d) 4
Answer:
(a) 0
Question 15.
Amino acids are
(a) acidic
(b) basic
(c) amphoteric
(d) neutral
Answer:
(c) amphoteric
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) : Conductivity of an electrolyte increases with decrease in concentration.
Reason (R) : Number of ions per unit volume decreases on dilution.
Answer:
(D) Assertion (A) is wrong, but Reason (R) is correct statement.
Mn\(\mathrm{O}_4{ }^{-}\) + 8H
+
+ 5e
–
→ Mn
2+
+ 4H
2
O
1 mol of e = IF = 96,500 C
So 5 moles of e = 5 × 96,500 C = 4,82,500 C
∴ Required charge = 4,82,500 C
Question 17.
Assertion (A) : The C-O-C bond angle in ethers is slightly less than tetrahedral angle.
Reason (R) : Due to the repulsive interaction between the two alkyl groups in ethers.
Answer:
(D) Assertion (A) is wrong, but Reason (R) is correct statement.
Question 18.
Assertion (A) : Low spin tetrahedral complexes are rarely observed.
Reason (R) : Crystal field splitting energy is less than pairing energy for tetrahedral complexes.
Answer:
(A) Both Assertion (A) and Reason (R) are correct statements, and Reason (R) is the correct explanation of the Assertion (A).
Question 19.
Assertion (A) : Elevation in boilirig point is a colligative property.
Reason (R) : Elevation in boiling point is directly proportional to molarity.
Answer:
(B) Both Assertion (A) and Reason (R) are correct statements, but Reason (R) is not the correct explanation of the Assertion (A).
Question 20.
Assertion (A) : Oxidation of ketones is easier than aldehydes.
Reason (R) : C-C bond of ketones is stronger than C-H bond of aldehydes.
Answer:
(D) Assertion (A) is wrong, but Reason (R) is correct statement.
Section – ‘B’
Question 21.
State Raoult’s law for a solution containing volatile components. What is the similarity between Raoult’s law and Henry’s law? [2]
Answer:
Raoult’s law. “In a solution, the vapour pressure of a component at a given temperature is equal to the mole fraction of that component in the solution multiplied by the vapour pressure of that component in pure state.”
Similarity between Raoult’s law and Henry’s law is that the partial pressure or vapour pressure of the volatile component (gas) is directly proportional to the mole fraction of that component in the solution.
Question 22.
Write the role of
(a) Dilute NaCN in the extraction of Gold.
(b) CO in the extraction of Iron. [1 + 1 = 2]
Answer:
(a) NaCN is used in leaching of gold from its ore by formation of a soluble complex cyanide leaving behind impurities.
4Au + 8KCN + 2H
2
O + O
2
→ 4K[Au(CN)
2
] + 4KOH
(b) Coke, i.e., CO acts as a reducing agent in the extraction of iron from its oxides in the zone of reduction in blast furnance at 1123 K.
Or
How is leaching carried out in the case of low grade copper ores? Name the method used for refining of copper metal. [2]
Answer:
In case of low grade copper ores, leaching is carried out using acid or bacteria in the presence of air. In this process Cu goes into the solution as Cu
2+
ions and the resulting solution is treated with scrap iron or hydrogen to get metallic copper.
Cu + 2H
+
+ \(\frac{1}{2}\)O
2
→ Cu
2+
+ 2H
2
O
Cu
2+
+ H
2
→ Cu + 2H
+
Electrolytic refining is used for refining of copper metal.
Question 23.
Define adsorption with an example. What is the role of adsorption in heterogeneous catalysis? [2]
Answer:
Adsorption is defined as the deposition of molecular species onto the surface of adsorbent.
Example: Adsorption of methylene blue dye on animal charcoal.
In heterogeneous catalysis, gaseous or liquid reactants are adsorbed on the surface of solid catalyst by physisorption or chemisorption, due to which, the concentration of the reactant molecules increases on the surface and also the rate of reaction.
Or
Define Brownian movement. What is the cause of Brownian movement in colloidal particles? How is it responsible for the stability of Colloidal Sol?
Answer:
Brownian movement is the zig-zag or random motion of the colloidal particles. It is due to the collision of these particles by the molecules of dispersion medium with different forces from different directions.
During Brownian motion, a stirring effect is produced which does not allow the sol particles to settle down. Hence, it affects the stability of the colloidal sol.
Question 24.
(a) Write the IUPAC name and hybridisation of the complex [Fe(CN)
6
]
3-
.
(Given: Atomic number of Fe=26)
(b) What is the difference between an ambidentate ligand and a chelating ligand? [1 + 1 = 2]
Answer:
(a) IUPAC name: Hexacyanoferrate (III) ion
Hybridisation: d
2
sp
3
(b) An ambidentate ligand possesses two donor sites but can ligate using one donor site at a time. For example, CN can ligate through either ‘C’ or ‘N’ atom of it.
A chelating ligand has multiple donor sites and can ligate using atleast two donor sites at a time and can form cyclic complexes. For example, ethylene diamine, EDTA, etc.
Question 25.
How do antiseptics differ from disinfectants? Name a substance which can be used as a disinfectant as well as an antiseptic. [2]
Answer:
Difference between antiseptics and disinfectants
| Antiseptics | Disinfectants |
| They are chemical substances which prevent the growth of micro-organisms and may even kill them. | They are chemical substances which kill micro-organisms. |
| They are safe to be applied to the living tissues. | They are not safe to be applied to the living tissues. |
| They are generally applied on wounds, cuts, ulcers and diseased skin surfaces. Example: Furacin, soframycin, dettol and savlon. | They are used to kill micro-organisms present in the drains, toilets, floors etc. Example: Phenol (> 1% solution) and chlorine (0.2 to 0.4 ppm). |
Phenol can be used as an antiseptic as well as a disinfectant. Its 0.2% solution is used as an antiseptic while 1 % solution as a disinfectant.
Question 26.
Identify the monomers in the following polymers:
Answer:
(i) The monomers in the given polymer are: Ethylene glycol (HOCH
2
CH
2
OH) and Phthalic acid
(ii) The monomer in the given polymer is: Acrylonitrile
Question 27.
Draw the structures of the following:
(i) H
2
S
2
O
8
(ii) XeF
6
[1 + 1 = 2]
Answer:
(i) Structure of H
2
S
2
O
8
:
(ii) Structure of XeF
6
:
Section – ‘C’
Question 28.
A 0.01 m aqueous solution of AlCl
3
freezes at -0.068° C. Calculate the percentage of dissociation. [Given: K
f
for Water = 1.86 K Kg mol
-1
] [3]
Answer:
Using (i) and (ii),
3.65 = 1 + 3α
α = \(\frac{2.65}{3}\)
∴ α = 0.883
∴ Percentage of disassociation = 0.883 × 100 = 88.3%
Question 29.
When a steady current of 2 A was passed through two electrolytic cells A and B containing electrolytes ZnSO
4
and CuSO
4
connected in series, 2 g of Cu were deposited at the cathode of cell B. How long did the current flow? What mass of Zn was deposited
at cathode of cell A? [3]
[Atomic mass: Cu = 63.5 g mol
-1
, Zn = 65 g mol
-1
; 1F = 96500 C mol
-1
]
Answer:
Given: I = 2 A
Weight of Cu deposited at cathode of cell B = 2 g
n = 2
So, Weight of Cu deposited, W \(=\frac{\text { Atomic mass } \times \mathrm{I} \times t}{n \mathrm{~F}}\)
2 = \(\frac{63.5 \times 2 \times t}{2 \times 96,500}\) ∴ t = \(\frac{2 \times 96,500 \times 2}{63.5 \times 2}\)
∴ t = \(\frac{3,86,000}{127}\) = 3039 seconds
Now, weight of zinc deposited, W = \(\frac{\mathrm{ZI} t}{n \mathrm{~F}}\)
= \(\frac{65.0 \times 2 \times 3,039}{2 \times 96,500}\) = 2.05 g
Question 30.
Differentiate between following:
(i) Amylose and Amylopectin
(ii) Globular protein and Fibrous protein
(iii) Nucleotide and Nucleoside [1 + 1 + 1 = 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-glycosidic linkages between C
1
of one and C
4
of the other which is further linked to C
6
of other glucose.
(ii)
| Globular Proteins | Fibrous Proteins |
| 1. Globular proteins have almost spheroidal shape due to folding of the polypeptide chain. | 1. Polypeptide chains of fibrous proteins consist of thread like molecules which tend to lie side by side to form fibres. |
| 2. Globular proteins are soluble in water. | 2. Fibrous proteins are insoluble in water. |
| 3. Globular proteins are sensitive to small changes of temperature and pH. Therefore they undergo denaturation on heating or on treatment with acids/bases. | 3. Fibrous proteins are stable to moderate changes of temperature and pH. |
|
4. They possess biological activity that’s why they act as enzymes.
Example: Maltase, invertase etc., hormones (insulin) antibodies, transport agents (haemoglobin), etc. |
4. 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. |
(iii)
Nucleotides : A nucleotide contains all the three basic components of nucleic acids, i.e. a phosphoric acid group, a pentose sugar and a nitrogenous base. These are formed by esterification of C
5
’ — OH of the sugar of the nucleoside with phosphoric acid.
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 β-linkage.
Question 31.
Identify A, B, C, D, E and F in the following: [6 × 1/2 = 3]
Answer:
‘A’ = 2-Methylpropene
‘B’ = 2-Bromo-2-methylpropane
‘C’ = 2,2,3,3-tetramethylbutane
‘D’ = Isobutylmagnesium bromide
‘E’ = 2-Methylpropane
‘F’ = 3-Ethoxy-2-Methylpropane
Question 32.
Give the structures of final products expected from the following reactions:
(i) Hydroboration of propene followed by oxidation with H
2
O
2
in alkaline medium.
(ii) Dehydration of (CH
3
)
3
C — OH by heating it with 20% H
3
PO
4
at 358 K. [3 × 1 = 3]
(iii) Heating of
Answer:
(i) Hydroboration of propene followed by oxidation with H
2
O
2
in alkaline medium.
(ii) Dehydration of (CH
3
)
3
C—OH by heating it with 20% H
3
PO
4
at 358 K
(iii)
Or
How can you convert the following?
(i) Phenol to o-hydroxybenzaldehyde.
(ii) Methanal to ethanol.
(iii) Phenol to phenyl ethanoate. [1 + 1 + 1 = 3]
Answer:
(i) Phejiol to o-hydroxybenzaldehyde (Reimer-Tiemann reaction):
(ii) Methanal to ethanol:
(iii) Phenol to phenyl ethanoate (Acetylation):
Question 33.
Give reasons:
(i) Aniline does not undergo Friedal-Crafts reaction.
(ii) Aromatic primary amines cannot be prepared by Gabriel’s phthalimide synthesis.
(iii) Aliphatic amines are stronger bases than ammonia. [3 × 1 = 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) Aromatic primary amines cannot be prepared by Gabriel’s phthalimide synthesis as aryl halides do not undergo nucleophilic substitution reaction with the potassium salt formed by reaction between phthalimide and ethanolic potassium hydroxide.
(iii) Aliphatic amines are stronger bases than ammonia due to +I (electron releasing effect) effect of the alkyl groups leading to high electron density on the nitrogen atom. Hence, lone pair of electrons can be easily donated.
Question 34.
Write three differences between lyophobic sol and lyophilic sol. [3]
Answer:
Difference between lyophobic sol and lyophilic sol
| Lyophobic sol | Lyophilic sol |
| They are thermodynamically unstable. | They are thermodynamically stable. |
| Precipitation in the lyophobic sol is an irreversible process. | Precipitation in the lyophilic sol is a reversible process. |
| There is less or no attraction between colloids and the liquid. | There is a strong attraction between colloids and the liquid. |
Or
Define the following terms: [1 + 1 + 1 = 3]
(i) Protective colloid
(ii) Zeta potential
(iii) Emulsifying agent
Answer:
(i) Protective colloid. They are lyophillic colloids such as gums, soaps, gelatine etc., that when added to lyophobic colloids, protect them to get precipitated by coagulating action of electrolytes.
(ii) Zeta potential. When one type of the ions of the electrolyte are adsorbed on the surface of colloidal particles it forms a fixed layer which attracts another layer of opposite ions thus forming a Helmholtz electrical double layer whose potential difference between the two layers is termed as Zeta potential.
(iii) Emulsifying agent. The substances which are added in added quantities to stabilize the emulsions hence prevent them from separating are called emulsifying agents, for example: soaps, proteins, agar etc.
Section – D
Question 35.
Give reasons:
(a)
(i) Transition metals and their compounds show catalytic activities.
(ii) Separation of a mixture of Lanthanoid elements is difficult.
(iii) Zn, Cd and Hg are soft and have low melting point.
(b) Write the preparation of the following:
(i) Na
2
Cr
2
O
7
from Na
2
CrO
4
(ii) K
2
MnO
4
from MnO
2
[3 + 2 = 5]
Answer:
(a) (i) The catalytic properties of the transition elements are due to the presence of unpaired electrons in their incomplete d-orbitals and variable oxidation states.
(ii) Separation of a mixture of Lanthanoid elements is difficult because of similarities in their physical and chemical properties due to Lanthanoid contraction.
(iii) Zn, Cd and Hg are soft and have low melting point because of absence of any unpaired electrons in their d-orbitals and very weak metallic bonding.
(b)
(i) Na
2
Cr
2
O
7
from Na
2
CrO
4
(ii) Finely powdered pyrolusite is fused with KOH in the presence of air to give green coloured potassium manganate.
Or
(a) Account for the following:
(i) Ti
3+
is coloured whereas Sc
3+
is colourless in aqueous solution.
(ii) Cr
2+
is a strong reducing agent.
(b) Write tivo similarities between chemistry of lanthanoids and actinoids.
(c) Complete the following ionic equation:
3Mn\(\mathrm{O}_4^{2-}\) + 4H
+
→ [2 + 2 + 1 = 5]
Answer:
(a) (i) Sc
3+
(3d°4s°) is colourless in aqueous solution whereas Ti
3+
(3d
1
4s°) is coloured because of d-d transistion by one unpaired electron in Ti
3+
while in Sc
3+
no unpaired electron is present.
(ii) Cr
2+
exists in the d
4
system and is easily oxidized to Cr
3+
by losing one electron which has the stable d
3
/\(\left[t_{2 g}^3\right]\) orbital configuration. So, Cr
2+
is a strong reducing agent.
(b) (i) Both lanthanoids and actinoids show contraction in size and irregularity in
their electronic configuration.
(ii) Both show +3 as their common oxidation state.
(c) 3Mn\(\mathrm{O}_4^{2-}\) + 4H
+
→ 2Mn\(\mathrm{O}_4^{-}\) + MnO
2
+ 2H
2
O
Question 36.
(a) Write the products formed when benzaldehyde reacts with the following reagents: (i) CH3CHO in presence of dilute NaOH
(ii)
(iii) Conc. NaOH
(b) Distinguish between the following:
(i) CH
3
-CH = CH-CO-CH
3
and CH
3
-CH
2
-CO-CH = CH
2
(ii) Benzaldehyde and Benzoic acid [3 + (1 + 1) = 5]
Answer:
(a) (i) Benzaldehyde reacts with CH
3
CHO in presence of dilute NaOH (Cross-Aldol
condensation):
(ii)
(iii) Benzaldehyde reacts with Conc. NaOH (Cannizzaro reaction):
(b)
(i) Distinction between CH
3
—CH = CH—CO—CH
3
and CH
3
—CH
2
—CO—CH = CH
2
Pent-3-ene-2-one being methyl ketone gives ellow ppt of iodoform while pent-1-ene-3-one will not.
(ii) Distinction between Benzaldehyde and Benzoic acid
Or
(a) Write the final products in the following:
(i)
(b) Arrange the following in the increasing order of their reactivity towards nucleophilic addition reaction:
(c) Draw the structure of 2, 4 DNP derivative of acetaldehyde.
Answer:
(a)
Question 37.
(a) A first order reaction is 25% complete in 40 minutes. Calculate the value of rate constant. In what time will the reaction be 80% completed?
(b) Define order of reaction. Write the condition under which a bimolecular reaction follows first order kinetics. [3 + 2 = 5]
Answer:
(a) For the first order reaction, K = \(\frac{2.303}{t}\) log \(\frac{a}{a-0.25 a}\)
When reaction is 25% complete:
Given. X = \(\frac{25 a}{100}\) = 0.25a; t = 40 min.
K = \(\frac{2.303}{40}\) log \(\frac{a}{a-0.25 a}\) ⇒ K = \(\frac{2.303}{40}\) log \(\frac{a}{0.75 a}\)
K = 0.057575 (log 100 – log 75)
K = 0.057575 (2 – 1.87506) = 0.057575 × 0.12494 = 0.007193 min
-1
∴ Value of rate constant (K) = 0.007193 min
-1
When reaction is 80% complete
∴ x = 0.8a, t = ?
K = 0.007193 min
-1
∴ t = \(\frac{2.303}{\mathrm{~K}}\) log \(\frac{a}{a-x}\) = \(\frac{2.303}{0.007193}\) log \(\frac{a}{a-0.8 a}\)
= 320.1723 log \(\frac{a}{0.2 a}\) = 320.1723 (log 10 – log 2)
= 320.1723(1 – 0.3010) = 320.1723 × 0.699 = 223.80 min.
∴ Time required for 80°Á completion is 223.80 minutes.
(b) The sum of powers to which the molar concentration in the rate law equation are raised to express the observed rate of the reaction is called the order of reaction.
A bimolecular reaction may become kinetically of first order if one of the reactants is present in large excess.
Or
(a) A first order reaction is 50% complete in 30 minutes at 300 K and in 10 minutes at 320 K. Calculate activation energy (Ea) for the reaction.
(R = 8.314J K
-1
mol
-1
)
[Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021, log 5 = 0.6991]
(b) Write the two conditions for collisions to be effective collisions.
(c) How order of reaction and molecularity differ towards a complex reaction? [3 + 1 + 1 = 5]
Answer:
(a)
(b) Two conditions for collisions to be.effective are:
(i) The colliding molecules must have energy greater than the threshold energy.
(ii) The colliding molecules must have proper orientation at the Ulme of collision.
(c) Order of reaction. Order of reaction is applicable to elementary as well as complex reactions.
Molecularity of reaction. Molecularity is applicable only to elementary reactions and for the overall complex reaction molecularity has no significance.
SET II Code No. 56/5/2
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.
Out of
which will undergo S
N
1 reaction faster with OH
–
?
Answer:
Question 7.
Write the IUPAC name of
Answer:
N, N-Dimethvlaniline
Question 8.
What type of linkage is present in polysaccharides?
Answer:
Glycosidic linkage is present in polysaccharides.
Question 9.
Name an artificial sweetener whose use is limited to cold drinks.
Answer:
Aspartame
Question 10.
Name the polymer which is used for making non-stick utensils.
Answer:
Teflon
Questions 11 to 15 are Multiple Choice Questions:
Question 11.
Kohlrausch has given the following relation for strong electrolytes:
\(\wedge\) = \(\Lambda_0\) – A\(\sqrt{C}\)
Which of the following equality holds?
(a) \(\wedge\) = \(\Lambda_0\), as C → \(\sqrt{\mathbf{A}}\)
(b) \(\wedge\) = \(\Lambda_0\) as C → 0
(c) \(\wedge\) = \(\Lambda_0\) as C → 0
(d) \(\wedge\) = \(\Lambda_0\) as C → 1
Answer:
(c) \(\wedge\) = \(\Lambda_0\) as C → 0
Question 12.
In an electrochemical process, a salt bridge is used
(a) as a reducing agent.
(b) as an oxidizing agent.
(c) to complete the circuit so that current can flow.
(d) None of these.
Answer:
(c) to complete the circuit so that current can flow.
Question 13.
In a chemical reaction X → Y, it is found that the rate of reaction doubles when the concentration of X is increased four times. The order of the reaction with respect to X is
(a) 1
(b) 0
(c) 2
(d) 1/2
Answer:
(d) 1/2
Question 14.
Which of the following will give a white precipitate upon reacting with AgNO
3
?
(a) K
2
[Pt(en)
2
Cl
2
]
(b) [Co(NH
3
)
3
Cl
3
]
(c) [Cr(H
2
O)
6
]Cl
3
(d) [Fe(H
2
O)
3
Cl
3
]
Answer:
(c) [Cr(H
2
O)
6
]Cl
3
Question 15.
Copper matte contains
(a) Cu
2
S, Cu
2
O and silica
(b) Cu
2
S, CuO and silica
(c) Cu
2
S, FeO and silica
(d) Cu
2
S, FeS and silica
Answer:
(d) Cu
2
S, FeS and silica
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) : 0.1 M solution of KCl has greater osmotic pressure than 0.1 M solution of glucose at same temperature.
Reason (R) : In solution, KCl dissociates to produce more number of particles.
Answer:
(A) Both Assertion (A) and Reason (R) are correct statements, and Reason (R) is the correct
explanation of the Assertion (A).
Question 17.
Assertion (A) : Conductivity of an electrolyte increases with decrease in concentration.
Reason (R) : Number of ions per unit volume decreases on dilution.
Answer:
Assertion (A) is wrong, but Reason (R) is correct statement.
Question 18.
Assertion (A) : Ortho and para-nitrophenols can be separated by steam distillation.
Reason (R) : Ortho isomer associates through intermolecular hydrogen bonding while Para isomer associates through intramolecular hydrogen bonding.
Answer:
(C) Assertion (A) is correct, but Reason (R) is wrong statement.
Question 19.
Assertion (A) : Oxidation of ketones is easier than aldehydes.
Reason (R) : C-C bond of ketones is stronger than C-H bond of aldehydes.
Answer:
(D) Assertion (A) is wrong, but Reason (R) is correct statement.
Question 20.
Assertion (A) : Low spin tetrahedral complexes are rarely observed.
Reason (R) : Crystal field splitting energy is less than pairing energy for tetrahedral complexes.
Answer:
(A) Both Assertion (A) and Reason (R) are correct statements, and Reason (R) is the correct explanation of the Assertion (A).
Question 23.
Draw the structures of the following:
(i) H
2
S
2
O
7
(ii) BrF
5
[2]
Answer:
(i) Structure of H
2
S
2
O
7
:
(ii) Structure of BrF
5
:
Question 25.
Identify the monomers in the following polymers: [2]
Answer:
(i) The monomer is — o-Hydroxymethyl (Phenol)
(ii) The monomers are — Hexamethylene diamine (H
2
N — (CH
2
)
6
— NH
2
) and Adipic acid (HOOC—(CH
2
)
4
—COOH)
Question 26.
Discuss the nature of bonding in metal carbonyls. [2]
Answer:
The metal-carbon bonds in metal carbonvis have both a and it characters. A weak σ bond is formed by donation of e
–
pair from irhon to metal (M ← C ≡ 0) and a stronger π bond is formed by back donation of electrons from filled d-orbital of metal into empty antibonding π* orbital of carbon (πM → C bond). The total bonding is thus M = C = O and bond order of C—O bond is reduced from triple bond to double bond. This is synergic effect.
Question 30.
Define the following terms with a suitable example in each: [1 + 1 + 1 = 3]
(i) Polysacchandes
(ii) Denatured protein
(iii) Fibrous protein
Answer:
(i) Polysaccha rides. They give a large number of molecules long chains of monosaccharides linked by glycosidic linkage on hydrolysis. For example, starch, cellulose, etc.
(ii) Denatured protein. The protein which has lost its biological activity due to cleavage of hydrogen bonds and unfolding of protein molecular through change in temperature, pH and coagulation is called denatured protein. For example, coagulation of white of an egg on boiling.
(iii) Fibrous protein. These are linear thread-like molecules which tend to lie side by side to form fibres and their polypeptide chains are held together by hydrogen and disuiphide bonds. For example, Keratin in skin, hair, etc.
SET III Code No. 56/5/3
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.
A hydrocarbon C
5
H
12
gives only one monochioride on photochemical chlorination. Identify the compound.
Answer:
Neopentene gives only one monochioride.
Question 7.
Out of (CH
3
)
3
N and (CH
3
)
2
NH, which one is more basic in aqueous solution?
Answer:
2° amine (CH
3
)
2
NH is more basic than 3° amine (CH
3
)
3
N in aqueous solution.
Question 8.
Out of Cis — [Pt(en)
2
Cl
2
]
2+
and Trans – [Pt(en
2
)Cl
2
]
2+
which one is optically active?
Answer:
Cis — [Pt(en)
2
Cl
2
]
2+
is optically active.
Question 9.
Name the method of refining used to obtain semiconductor of very high purity.
Answer:
Zone-refining method is used to obtain semiconductor of very high purity.
Question 10.
Is
a homopolymer or copolymer?
Answer:
Questions 11 to 15 are multiple choice questions:
Question 11.
The amount of electricity required to produce one mole of Zn from ZnSo
4 solution will be:
(a) 3F
(b) 2F
(c) 1F
(d) 4F
Answer:
(b) 2F
Question 12.
Zinc is coated over iron to prevent rusting of iron because
Answer:
(b)
\(\mathbf{E}^{\mathrm{o}} \mathrm{Zn}^{2+} / \mathrm{Zn}\) < \(\mathrm{E}^{\mathrm{o}} \mathrm{Fe}^{2+} / \mathrm{Fe}\)
Question 13.
The unit of rate constant depends upon the
(a) molecularity of the reaction.
(b) activation energy of the reaction.
(c) order of the reaction.
(d) temperature of the reaction.
Answer:
(c) order of the reaction.
Question 14.
The formula of the complex traimminetri(nitrito-O) Cobalt (III) is
(a) [Co(ONO)
3
(NH
3
)
3
]
(b) [Co(NO
2
)
3
(NH
3
)
3
]
(c) [Co(ONO
2
)
3
(NH
3
)
3
]
(d) [Co(NO
2
)(NH
3
)
3
]
Answer:
(a) [Co(ONO)
3
(NH
3
)
3
]
Question 15.
Which of the following is a disaccharide?
(a) Glucose
(b) Starch
(c) Cellulose
(d) Lactose
Answer:
(d) Lactose
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) : An ideal solution obeys Henry’s law.
Reason (R) : In an ideal solution, solute-solute as well as solvent-solvent interactions are similar to solute-solvent interaction.
Answer:
(D) Assertion (A) is wrong, but Reason (R) is correct statement.
Question 17.
Assertion (A) : Conductivity of an electrolyte increases with decrease in concentration.
Reason (R) : Number of ions per unit volume decreases on dilution.
Answer:
(D) Assertion (A) is wrong, but Reason (R) is correct statement.
Question 18.
Assertion (A) : Benzaldehyde is less reactive than ethanol towards nucleophilic addition reactions.
Reason (R) : Ethanal is more sterically hindered.
Answer:
(C) Assertion (A) is correct, but Reason (R) is wrong statement.
Question 19.
Assertion (A) : Low spin tetrahedral complexes are rarely observed.
Reason (R) : Crystal field splitting energy is less than pairing energy for tetrahedral complexes.
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) : The C-O-C bond angle in ethers is slightly less than tetrahedral angle.
Reason (R) : Due to the repulsive interaction between the two alkyl groups in ethers.
Answer:
(D) Assertion (A) is wrong, but Reason (R) is correct statement.
Question 22.
Draw the structures of the following:
(i) HClO
4
(ii) XeOF
4
[2]
Answer:
(i) Structure of HClO
4
: (Perchloric acid)
(ii) Structure of XeOF
4
:
Question 24.
Identify the monomers in the following polymers: [1 + 1 = 2]
Answer:
(i) The monomers of given structure (Dacron) are Ethylene glycol (HOCH
2
CH
2
OH and Terephthalic acid
(ii) The monomers of the given structure (Mellamine-formaldehyde) are Melamine and Formaldehyde (HCHO)
Question 27.
Define the followings terms with a suitable example in each: [1 + 1 = 2]
(i) Bacteriocidal antibiotics
(ii) Food preservatives
Answer:
(i) Bacteriocidal antibiotics. Those drugs which kill bacteria directly are known as bacteriocidal antiboitics. For example. Penicillin, ofloxacin etc.
(ii) Food preservatives. Those chemical substances which are used to protect food against
microbial growth such as bacteria, yeasts and moulds are called food preservatives.
For example. Sodium benzoate, Sodium metabisuiphite, table salts, vegetable oils etc.
Question 31.
(i) What are the hydrolysis products of DNA?
(ii) What happens when D-glucose is treated with Bromine water?
(iii) What is the effect of denaturation on the structure of proteins? 1+1+1=3
Answer:
(i) Hydrolysis products of DNA are a pentose sugar, i.e., 2-deoxy-D-ribose, phosphoric acid and nitrogenous bases like Adenine, Guanine, Cytosine and Thymine.
(ii) When D-glocose is treated with bromine water D-gluconic acid is formed.
(iii) Due to denaturation, the secondary and tertiary structures of protein are destroyed as their globules get unfolded and helixes get uncoiled and thus enzyme loses its activity. However, primary structure of protein remains unaffected.