Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2018 Comptt (Delhi & Outside Delhi) to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2018 Comptt (Delhi & Outside Delhi) with Solutions
Question 1.
For the reaction A -» B, the rate of reaction becomes three times when the concentration of A is increased by nine times. What is the order of reaction? [1]
Answer:
r = k[A]
n
……(i) 3r = k[9A]
n
……..(ii)
\(\frac{3 r}{r}\) = \(\frac{k[9 \mathrm{~A}]^n}{k[\mathrm{~A}]^n}\) (Divide (ii) by (i))
3 = 9
n
⇒ 3 = 3
2n
2n = 1 ⇒ n = 1/2 ∴ The order of reaction will be half.
Question 2.
Why is adsorption always exothermic? [1]
Answer:
Adsorption leads to a decrease in the residual forces on the surface of the adsorbent. This causes a decrease in the surface energy of the adsorbent. Therefore, adsorption is always exothermic.
Question 3.
Write the coordination isomer of [Cu(NH
3
)
4
][PtCl
4
]. [1]
Answer:
Coordination isomer is [Pt (NH
3
)
4
][CUCl
4
].
Question 4.
An aromatic organic compound ‘A’ with molecular formula C
8
H
8
O gives positive DNP and iodoform tests. It neither reduces Tollens’ reagent nor does it decolourise bromine water. Write the structure of ‘A’. [1]
Answer:
Since compound ‘A’ does not reduce Tollens’ reagent and gives positive DNP and iodoform test so it may be Ketone.
So, the ketone with molecular formula C
8
H
8
O is
Question 5.
Predict the major product formed when sodium ethoxide reacts with tert. Butyl chloride. [1]
Answer:
Question 6.
Why a mixture of Carbon disulphide and acetone shows positive deviation from Raoult’s law? What type of azeotrope is formed by this mixture? [2]
Answer:
When Carbon disulphide (CS
2
) is added to acetone the solute-solvent dipolar interactions are weaker than solute-solute and solvent-solvent interactions. Hence it shows positive deviation. In positive deviations, we get minimum boiling azeotropes at a specific composition.
Question 7.
Which one of the following compounds is more reactive towards S
N
2 reaction and why?
CH
3
CH(Cl)CH
2
CH
3
or CH
3
CH
2
CH
2
Cl [2]
Answer:
CH
3
CH
2
CH
2
Cl (1-chloropropane) is more reactive towards S
N
2 reaction because 1° halide has less steric hindrance.
Question 8.
A current of 1.50 A was passed through an electrolytic cell containing AgNO
3
solution with inert electrodes. The weight of silver deposited was 1.50 g. How long did the current flow? (Molar mass of Ag = 108 g mol
-1
, IF = 96500 C mol
-1
). [2]
Answer:
Ag
+
+ e
–
→ Ag
1 mole of Ag(108 g) are deposited by 1F = 96500 C
∴ 1.5 g of Ag will be deposited by electricity (Q) = \(\frac{96500}{108}\) × 1.50 C = 1340.3 C
Q = I × t ⇒ t = \(\frac{\mathrm{Q}}{\mathrm{I}}\)
t = \(\frac{1340.3}{1.50}\) ∴ t = 893.5 s
The conductivity of a 0.01 M solution of acetic acid at 298 K is 1.65 × 10
-4
S cm
-1
. Calculate molar conductivity (Λ
m
) of the solution.
Answer:
Given: k = 1.65 × 10
-4
S cm
-1
, M = 0.01 M, Λ
m
= ?
Using formula, Λ
m
= \(\frac{1000 \times k}{\mathrm{M}}\)
or Λ
m
= \(\frac{1000 \times 1.65 \times 10^{-4}}{0.01}\) ∴ Λ
m
= 16.5 S mol
-1
Question 9.
Draw the structures of the following: [2]
(i) XeF
2
(ii) BrF
5
Answer:
(i) Structure of XeF
2
:
(ii) Structure of BrF
5
:
Question 10.
Identify the following: [2]
(i) Transition metal of 3d series that exhibits the maximum number of oxidation states.
(ii) An alloy consisting of approximately 95% lanthanoid metal used to produce bullet, shell and lighter flint.
Answer:
(i) Manganese (Mn) exhibits the maximum number of oxidation states i.e., upto +7.
(ii) Misch metal.
Question 11.
Write the product(s) formed when [3]
(i) 2-Bromopropane undergoes dehydrohalogenation reaction.
(ii) Chlorobenzene undergoes nitration reaction.
(iii) Methylbromide is treated with KCN.
Answer:
Question 12.
Calculate the freezing point of an aqueous solution containing 10.5 g of Magnesium 4. bromide in 200 g of water, assuming complete dissociation of Magnesium bromide.
(Molar mass of Magnesium bromide = 184 g mol
-1
, for water = 1.86 K kg mol
-1
). [3]
Answer:
Since MgBr
2
is an ionic compound, so undergoes complete dissociation
Question 13.
(i) Complete the following reaction and suggest a suitable mechanism for the reaction:
(ii) Why ortho-Nitrophenol is steam volatile while para-Nitrophenol is less volatile? [3]
Answer:
(i)
Mechanism for the reaction: It involves the following three steps:
Step 1 : Formation of protonated alcohol
Step 2: Formation of Carbocation: It is the slowest and rate determining step:
Step 3: Formation of ethene by elimination of a proton
ii) o-Nitrophenol forms intramolecular H-bond whereas molecules of p-Nitrophenol get associated through intermolecular H-bonding and have relatively higher boiling points and are less volatile.
Question 14.
A reaction is first order in A and second order in B [3]
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) How is the rate affected when the concentration of both A and B are doubled?
Answer:
(i) r = k[A][B]
2
⇒ r = kAB
2
(ii) When concentration of B increases to 3 times the rate of reaction becomes 9 times.
r = kA(3B)
2
∴ r = 9kAB
2
= 9 times
(iii) When concentration of both A and B are doubled, the rate of reaction becomes 8 times.
r = k(2A)(2B)
2
∴ r = 8kAB
2
= 8 times
Question 15.
Consider the following reaction: [3]
Cu (s) + 2Ag
+
(aq) → 2Ag (s) + Cu
2+
(aq)
(i) Depict the galvanic cell in which the given reaction takes place.
(ii) Give the direction of flow of current.
(iii) Write the half-cell reactions taking place at cathode and anode.
Answer:
(i) Cu(s)/ Cu
2+
(aq) → Ag
+
(aq) / Ag(s)
(ii) Electrons flow from copper electrode (cathode) to silver electrode (anode) and current flows from silver (anode) to copper electrode (cathode) in the external circuit.
(iii) At anode: Cu(s) → Cu
2+
(aq) + 2e
–
(Oxidation half reaction)
At cathode: 2Ag
+
(aq) + 2e
–
→ 2Ag(s) (Reduction half reaction)
Question 16.
Give reason for the following observations: [3]
(i) When Silver nitrate solution is added to Potassium iodide solution, a negatively charged colloidal solution is formed.
(ii) Finely divided substance is more effective as an adsorbent.
(iii) Lyophilic colloids are also called reversible sols.
Answer:
(i) When silver nitrate solution is added to potassium iodide solution, a negatively charged colloidal solution is formed due to adsorption of iodide ions from KI on to the AgI precipitate.
(ii) Finely divided substance is a more effective adsorbent because its surface area increases and physisorption is directly proportional to the surface area of the adsorbent.
(iii) Lyophillic colloids are also called reversible sols because when the dispersion phase is separated from the dispersion medium of colloid, the sol can be formed again by just mixing the dispersion phase and medium again.
Question 17.
Do as directed: [3]
(i) Arrange the following compounds in the increasing order of their basic strength in aqueous solution:
CH
3
NH
2
, (CH
3
)
3
N, (CH
3
)
2
NH.
(ii) Identify ‘A’ and ‘B’: C
6
H
5
NH
2
(iii) Write equation of carbylamine reaction.
Answer:
Question 18.
Explain the following: [3]
(i) Amino acids behave like salts rather than simple amines or carboxylic acids.
(ii) The two strands of DNA are complementary to each other.
(iii) Reaction of glucose that indicates that the carbonyl group is present as an aldehydic group in the open structure of glicose.
Answer:
(i) Because in aqueous solution, the —COOH group of an amino acid loses a proton and — NH
2
group accepts a proton to form zwitter ion (salt).
(ii) The two strands of DNA have bases which are paired with one another by hydrogen bonding which is very specific. Adenine pairs with only thymine via 2 hydrogen bonds and guanine pairs with cytosine through 3 hydrogen bonds. Thus, the sequence of bases in one strand automatically determines that of the other.
(iii) Glucose reacts with HCN to give a cyanohydrin, or with a Schiffs base to give an oxime. It shows the presence of a carhonyl group.
Question 19.
Give the formula of monomers involved in the formation of the following polymers: [3]
(i) Buna-N
(ii) Nylon-6
(iii) Dacron
Answer:
(i) Monomers of Buna-N are:
1, 3-Butadiene (CH
2
= CH — CH = CH
2
)
Acrylonitrile (CH
2
= CH – CN)
(ii) Nylon-6: The monomeric repeating unit of Nylon-6 is
(iii) Monomers of Dacron are:
Ethylene glycol (HO — CH
2
— CH
2
— OH)
Terephthalic acid (HOOC — (C
6
H
4
) — COOH)
Question 20.
Write the role of: [3]
(i) NaCN in the extraction of gold from its ore.
(ii) Cryolite in the extraction of aluminium from pure alumina.
(iii) CO in the purification of Nickel.
Answer:
(i) 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
(ii) Cryolite lowers the high melting point of alumina (Al
2
O
3
) and makes it a good conductor of electricity.
(iii) In the Mond’s process CO forms a volatile complex with Ni metal which on further decomposition gives pure Nickel metal.
Question 21.
Write IUPAC name for each of the following complexes: [3]
(i) [Ni(NH
3
)
6
]Cl
2
(ii) K
3
[Fe(CN)
6
]
(iii) [Co(en)
3
]
3+
Answer:
(i) IUPAC Naine: Hexaamminenickel(lI) chloride
(ii) IUPA C Name: Potassium hexacyanoferrate(III)
(iii) IUPA C Naine: Tris (ethane-1, 2-drnine) Cobalt(III) ibn
Question 22.
(i) Complete the following equations: [3]
(a) 2Mn\(\mathrm{O}_4^{-}\) + 5S\(\mathrm{O}_3^{2-}\) + 6H
+
→
(b) Cr
2
\(\mathrm{O}_7^{2-}\) + 6Fe
2+
+ 14H
+
→
(ii) Based on the data, arrange Fe
2+
, Mn
2+
and Cr
2+
in the increasing order of stability of +2 oxidation state.
Answer:
Or
Write the preparation of following:
(i) KMnO
4
from K
2
MnO
4
(ii) Na
2
CrO
4
from FeCr
2
O
4
(iii) Cr
2
\(\mathrm{O}_7^{2-}\) from Cr\(\mathrm{O}_4^{2-}\)
Answer:
Question 23.
Mathew works in a multinational company where the working conditions are tough. He started taking sleeping pills without consulting a doctor. When his friend Amit came to know about it he was disturbed and advised Mathew not to do so. He suggested that Mathew should instead practice yoga to be stress free. Mathew is now relaxed and happy after practicing yoga. [4]
(a) Name the class of chemical compounds used in sleeping pills.
(b) Why is it advisable not to take the dose of sleeping pill without consulting a doctor?
(c) Pick out the odd chemical compound on the basis of its different medicinal property: Luminal, Seconal, Phenacetin and Equanil.
*(d) List at least two qualities of Amit that helped Mathew to be happy.
Answer:
(a) Tranquilizers — Barbiturates
(b) If the drugs are taken in higher doses than recommended by a doctor, they may
cause harmful effects and act as poison and cause death.
(c) Phenacetin is an antipyretic while rest are tranquilizers.
*(d) As per Latest CBSE curriculum, Value Based Questions will not be asked in the examination.
Question 24.
(i) What happens when
(a) chlorine gas reacts with cold and dilute solution of NaOH?
(b) XeF
2
undergoes hydrolysis?
(ii) Assign suitable reasons for the following:
(a) SF
6
is inert towards hydrolysis.
(b) H
3
PO
3
is diprotic.
(c) Out of noble gases only Xenon is known to form established chemical compounds.
Answer:
(i)
(ii) (a) In SF
6
, S is protected by 6F atoms and doés not allow H
2
O molecules to attach on it. So SF
6
inerts towards hydrolysis.
(b) H
3
PO
3
is diprotic because it contains only two ionizable H-atoms which are present as —OH groups.
(c) Because Xenon is large in size with higher atomic mass and radius, so the force of attraction between the outer electron and the protons in the nucleus becomes weaker and therefore Xenon is easily available to form compounds. It has the least ionization energy among the noble gases.
Or
(i) Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidizing power of F
2
and Cl
2
.
(ii) Complete the following reactions:
(a) Cu + HNO
3
(dilute) →
(b) Fe
3+
+ SO
2
+ H
2
O →
(c) XeF
4
+ O
2
F
2
→
Answer:
(i) Since the standard reduction potential value of fluorine is more than that of chlorine, so fluorine is a stronger oxidising agent.
- Bond dissociation enthalpy of F 2 is less as compared to that of chlorine.
- The negative electron gain enthalpy of fluorine is slightly less than that of chlorine.
- The hydration enthalpy of flouride ion is much higher than that of Cl – ion due to smaller size.
(ii) (a) 3Cu + 8HN0
3
(dil) → 3Cu(NO
3
)
2
+ 2NO + 4H
2
O
(b) 2Fe
3+
+ SO
2
+ 2H
2
O → 2Fe
2+
+ S\(\mathrm{O}_4^{2-}\) + 4H
+
(c) XeF
4
+ O
2
F
2
→ XeF
6
+ O
2
Question 25.
(i) Give reasons:
(a) HCHO is more reactive than CH
3
— CHO towards addition of HCN.
(b) pKa of O
2
N—CH
2
—COOH is lower than that of CH
3
—COOH.
(c) Alpha hydrogen of aldehydes & ketones is acidic in nature.
(ii) Give simple chemical tests to distinguish between the following pairs of compounds:
(a) Ethanal and Propanal
(b) Pentan-2-one and Pentan-3-one
Answer:
(i) (a) HCHO is more reactive than CH
3
—CHO towards addition of HCN because of absence of +I group on carbonyl group while in CH
3
CHO, one +I group i.e., — CH
3
group is present. As a result, nucleophillic addition reactions occur more readily in HCHO than CH
3
-CHO.
(b) Nitroacetic acid is a stronger acid than acetic acid because — NO
2
group attached to acetic acid due to its high electronegativity weakens the O— H bond and stabilizes the anion making H
+
release easily but in CH
3
COOH it is not possible to donate H
+
.
(c) Because after removal of α-hydrogen from aldeh des and ketones, the conjugated base is obtained which can be stabilized by resonance.
(ii) (a) Ethanal and Propanal: Ethanal and propanal can be distinguished by iodoform test. Warm each compound of iodine and sodium hydroxide solution in water.
Ethanal gives yellow crystal of iodoform while propanal does not respond to iodoform test.
(b) (i) Pentan-2-one and Pentan-3-one
Or
(i) Write structure of the product(s) formed:
(ii) How will you bring the following conversions in not more than two steps:
(a) Propanone to propene
(b) Benzyl chloride to phenyl ethanoic acid
Answer:
\(\dagger\)Question 26.
(i) (a) Following is the schematic alignment of magnetic moments: [5]
What type of magnetism is shown by this substance?
(b) What type of stoichiometric defect is shown by
(i) KCl
(ii) AgCl?
(ii) An element with density 11.2 g cm
-3
forms a fcc lattice with edge length of 4 × 10
-8
cm. Calculate the atomic mass of the element. (N
A
= 6.02 × 10
23
mol
-1
)
Answer:
(i) (a) Antiferroinagnetism is shown by this substance.
(b) (i) KCl shows Schottky defect.
(ii) AgCl shows Frenkel defect.
(ii) Given: a = 4 × 10
-8
cm; d = 11.2 g cm
-3
; z = 4(for fcc); M = ?
Or
\(\dagger\)Silver metal crystallises with a face centred cubic lattice. The length of the unit cell is found to be 3.0 × 10
-8
cm. Calculate atomic radius and density of silver.
(Molar mass of Ag = 108 g mol
-1
, N
A
= 6.02 × 1023 mol
-1
)
Answer:
Given: a = 3.0 × 10
-8
cm; z = 4 (for fcc)
r = ? d = ?
Using formula, r = \(\frac{a}{2 \sqrt{2}}\) ⇒ r = \(\frac{3 \times 10^{-8} \mathrm{~cm}}{2 \times 1.414}\) = \(\frac{3 \times 10^{-8}}{2.828}\) ∴ r = 1.0608 × 10
-8
cm
d = \(\frac{z \times \mathrm{M}}{a^3 \times \mathrm{N}_{\mathrm{A}}}\) = \(\frac{4 \times 108}{\left(3 \times 10^{-8}\right)^3 \times 6.022 \times 10^{23}}\) = \(\frac{432}{27 \times 6.022 \times 10^{-1}}\) = \(\frac{4320}{162.594}\)
d = 26.57 g/cm
3
Please Note: The questions asked in Set II and Set III are identical to Set I. Only the serial numbers of questions were changed.