Based on the Term I syllabus, the model question paper is prepared. The chapter that covers syllabus are

1.The Solid State

2. Solution

3. Electrochemistry

4. Chemical Kinetics

Download CBSE Class 12th Standard CBSE Chemistry question papers, sample papers, important questions, and previous year solved papers in PDF format. Get free study materials, NCERT solutions, and exam preparation resources for Class 12th Standard CBSE Chemistry

Questions:

Generic 2

11 x 2 = 22

Give the main difference between an electrochemical cell and an electrolytic cell.

Define conductivity and give its units. Why is alternating current not used in place of direct current for measuring the electrolytic.

If half-life period of a reaction is inversely proiportional to initial concentration of the reactant, what is the order of reaction ?

For a first order reaction, time taken for half of the reaction to complete is t₁whereas that for 3/4 th of the reaction to complete id t₂. how are t₁ and t₂ related to each other?

A reaction is first order in A and second order in B
(i) Write differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times ?
(iii) How is the rate affected when concentration of both A and B is doubled ?

(a) Define Azeotropes and explain broefly minimum boiling azeotrope by taking suitable example.
(b) The vapour pressures of pure liquids A and B are 450 mm 700 mm of Hg respectively at 350 K. Calculate the composition of the liquid mixture if total vapour pressure is 600 mm of Hg. Also find the composition of the mixture in the vapour phase.

Classify the following solids in different categories based on the nature of intermolecular forces operating in them. Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide.

How does amorphous silica differ from quartz?

Explain the following with suitable example:An n-type semiconductor.

10.

What is physical significance of energy of activation? Explain with diagram.

11.

With the help of diagram explain the role of activated complex in a reaction.

Generic 3

10 x 3 = 30

12.

Predict the structure of MgO crystal and coordination number of its cation in which cation and anion radii are equal to 65 pm and 140 pm respectively.

13.

Br- ions form a close-packed structure. If the radius of Br^- ion is 195 pm, calculate the radius of the cation that just fits into the tetrahedral hole. Can a cation A⁺ having a radius of 82pm be slipped into the octahedral hole of the crystal A⁺ Br^- ?

14.

The two ions A⁺ and B^- have radii 88 and 200pm respectively. In the close packed crystal of compound AB, predict the coordination number of A⁺

15.

In a solid oxide ions are arranged in ccp. One-sixth of the tetrahedral voids are occupied by the cations A while one-third of the octahedral voids are occupied by the cations B. What is the formula of the compound?

16.

In corrundum, oxide ions are arranged in hexagonal close packing and aluminum ions ocupy two-third of the octahedral voids. What is the formula of corrundum?

17.

A cubic solid is made up of two elements A and Y are present at the corners of the cube and atoms A at the body centre. What is the formula of the compound? What are the coordination numbers of X and Y?

18.

A compound formed by elements A and B has a cubic structure in which A atoms are at the corners of the cube and B atoms are at the face centres. Derive the formula of the compound.

19.

(a) Assuming complete ionization, calculate the expected freezing point of solution prepared by dissolving 6.00 g of Glauber's salt, Na₂SO₄. 10H₂O in 0.1 kg of H₂O. (K _f for H₂O = 1.86 K kg mol^-1) [At. mass of Na = 23, S = 32, O= 16, H = 1 u].
(b) Two liquids X and Y boil at 110 ^oC and 130 ^oC respectively. Which of them has higher vapor pressure at 50^oC ?

20.

Assign reasons for the following:
(i) Phosphorus doped silicon is a semiconductor.
(ii) Schottky defect lowers the density of a solid.
(iii) Some of the very old glass objects appear slightly milky instead of being transparent.

21.

(a) What type of semiconductor is obtained when silicon is doped with boron?
(b) What type of magnetism is shown in the following alignment of magnetic moments?

(c) What type of point defects are produced when AgCl is doped with CdCl₂?

Multiple Choice Question

15 x 1 = 15

22.

Standard reduction potentials of the half reactions are given below :
\({ F }_{ 2 }(g)+{ 2e }^{ - }\longrightarrow 2{ F }^{ - }(aq);{ E }^{ 0 }=+2.85V\)
\({ Cl }_{ 2 }(g)+{ 2e }^{ - }\longrightarrow 2{ Cl }^{ - }(aq);{ E }^{ 0 }=+1.36V\)
\({ Br }_{ 2 }(s)+{ 2e }^{ - }\longrightarrow 2{ Br }^{ - }(aq);{ E }^{ 0 }=+1.06V\)
\({ I }_{ 2 }(s)+{ 2e }^{ - }\longrightarrow 2{ I }^{ - }(aq);{ E }^{ 0 }=+0.53V\)
The strongest oxidizing and reducing agents respectively are :

(a)

Cl₂ and Br^-

(b)

Cl₂ and I₂

(c)

F₂ and I^-

(d)

Br₂ and Cl^-

23.

Given
\({ E }_{ { Cr }^{ 3+ }/{ Cr } }^{ 0 }=-0.74V,\\ { E }_{ { MnO }_{ 4 }^{ - }/{ Mn }^{ 2+ } }^{ 0 }=1.51V\)
\({ E }_{ { CrO }_{ 7 }^{ - }/{ Cr }^{ 3+ } }^{ 0 }=1.33V,{ E }_{ Cl/{ Cl }^{ - } }^{ 0 }=1.36V\)
Based on the data given above, strongest oxidizing agent will be

(a)

MnO₄^-

(b)

Cl^-

(c)

Cr³⁺

(d)

Mn²⁺

24.

When measured against a standard calomel electrode, an electrode is found to have a standard reduction potential of 0.100 V. If standard reduction potential of calomel electrode is + 0.244 V and it acts as anode, the standard electrode potential of the same electrode against standard hydrogen electrode will be

(a)

- 0.144 V

(b)

+ 0.100 V

(c)

- 0.344 V

(d)

- 0.100 V

25.

The reaction taking place in the cell
Pt | H₂ (g) | HCl (1.0 M) | AgCl | Ag is 1 atm

(a)

\(AgCl+\left( 1/2 \right) { H }_{ 2 }\longrightarrow Ag+{ H }^{ + }+{ Cl }^{ - }\)

(b)

\(Ag+{ H }^{ + }+{ Cl }^{ - }\longrightarrow AgCl+\left( 1/2 \right) { H }_{ 2 }\)

(c)

\(2{ Ag }^{ + }+{ H }_{ 2 }\longrightarrow 2Ag+2{ H }^{ + }\)

(d)

\(2Ag+2{ H }^{ + }\longrightarrow 2{ Ag }^{ + }+{ H }_{ 2 }\)

26.

Equivalent conductivity at infinite dilution for sodium potassium oxalate, (COO^-)₂ Na⁺ K⁺ , will be (given, molar conductivities of oxalate, K⁺ and Na⁺ ions at infinite dilution are 148.2, 50.1, 73.5 S cm² mol^-1 respectively)

(a)

271.8 S cm² eq^-1

(b)

67.95 S cm² eq^-1

(c)

543.6 S cm² eq^-1

(d)

135.9 S cm² eq^-1

27.

Schottky defect is observed in crystal when ..................................... .

(a)

some cations move from their lattice sites to interstitial sites.

(b)

equal number of cations and anions aremissing from the lattice.

(c)

some lattice sites are occupied by electrons.

(d)

some impurity is present in the lattice.

28.

Cations are present in the interstital site in............................... .

(a)

Frenkel defect

(b)

Schottky defect

(c)

Vacancy defect

(d)

Metal deficiency defect

29.

Which of the following is a network solid?

(a)

SO₂ (Solid)

(b)

I₂

(c)

Diamond

(d)

H₂O (Ice)

30.

For which crystal anion-anion contact is valid?

(a)

NaF

(b)

NaI

(c)

CsBr

(d)

KCl

31.

The energy gap(E₈)between valence band and conduction band for diamond,silicon and germanium are in the order

(a)

E₈(diamond)>E₈(silicon)>E₈(germanium)

(b)

E₈(diamond)>E₈(silicon)<E₈(germanium)

(c)

E₈(diamond)=E₈(silicon)=E₈(germanium)

(d)

E₈(diamond)>E₈(silicon)>E₈(germanium)

32.

A metal crystallize with a face-centred cubic lattice.The edge of the unit cell is 408pm.The diameter of the metal atom is

(a)

288pm

(b)

408pm

(c)

144pm

(d)

204pm

33.

A solid compound XY has NaCl structure.If the radius of the cation is 100pm,the radius of the anion(Y^-) will be

(a)

275.1

(b)

322.5pm

(c)

241pm

(d)

165.7pm

34.

In an ionic compound A⁺X^-,the radii of A⁺ and X^- ions are 1.0pm and 2.0pm respectively.The volume of the unit cell of the crystal AX will be

(a)

27pm³

(b)

64pm³

(c)

125pm³

(d)

216pm³

35.

The arrangement of X^- ions around A⁺ions in solid AX is given in the fig(not drawn to scale).If the radius of X^- is 250pm,the radius of A⁺ is

(a)

104pm

(b)

125pm

(c)

183pm

(d)

57pm

36.

Wax is an example of

(a)

ionic crystal

(b)

covalent crystal

(c)

molecular crystal

(d)

amorphous crystal

Generic 5

3 x 5 = 15

37.

Calculate the standard electrode potential of Cu⁺/Cu half cell. Given that the standard reduction potentials of Cu²⁺/Cu and Cu²⁺/Cu⁺ are 0.337 V and 0.153 V respectively.

38.

When inversion of surcose is studied at pH = 5, the half-life period is always found to be 500 minutes irrespective of any initial concentration but when it is studied at pH = 6, the half-life period is found to be 50 minutes. Derive the rate law expression for the inversion of surcose.

39.

The half time of first order decomposition of nitramide is 2.1 hour at 15^oC. NH₂NO₂(aq) \(\longrightarrow\) N₂O(g) + H₂O (I)
If 6.2 g of MH₂NO₂ is allowed to decompose, calculate
(i) time taken for NH₂NO₂ to decompose 99% and
(ii) volume of dry N₂O produced at this point, measured at STP.

Visit QB365-SLMS - https://www.qb365.in/slms/

Answers Key:

Generic 2

11 x 2 = 22

Differences between electrochemical cell and electrolytic cell.

Electrochemical cell	Electrolytic cell
1. It is a device which converts chemical energy into electrical energy	1. It is a device which converts electrical energy into chemical energy
2. In this case, the redox reaction is spontaneous.	2. In this case, the redox reaction is non-spontaneous and electrical energy
3. In electrochemical cell, anode is negative and cathode is positive.	3. In electrolytic cell, anode is positive and cathode is negative electrode.

Conductivity is the conductance of a solution of 1 cm length and having 1 sq.cm as the area of cross-section i.e., of one-centimeter cube.
Direct current results in the electrolysis of the electrolytic solution. As a result, the concentration of the electrolyte near the electrodes changes and these result change in the resistance of the solution.

Two. This is because for reaction of nth order, \({ t }_{ { 1 }/{ 2 } }\propto \frac { 1 }{ { \left[ { A }_{ 0 } \right] }^{ n-1 } } .Hence\quad { t }_{ { 1 }/{ 2 } }\propto \frac { 1 }{ { \left[ { A }_{ 0 } \right] } } when \ n=2\)

\({ t }_{ 2 }=2{ t }_{ 1 }\) This is because, for 3/4th of the reaction to complete, time required = two half-lives.

\((i) \ \frac { dx }{ dt } =k{ \left[ A \right] }{ \left[ B \right] }^{ 2 }.\)
\((ii) \ Rate=k{ \ ab }^{ 2 }\)
If [B] is tripled, Rate = \(ka{ \left( 3b \right) }^{ 2 }=9k \ { ab }^{ 2 }=9 \ times \)
(iii) If both [A]and[B] are doubled, Rate = \(k(2a){ \left( 2b \right) }^{ 2 }=8k \ { ab }^{ 2 }=8 \ times\)

(b) If x_A is mol fraction of A in liquid phase, then x_A (450)+(1 - x_A)(700) = 600 or x_A = 0.4, x_B = 0.6.
Hence, \({ p }_{ A }=0.4\times 450=180\quad mm,{ p }_{ B }=0.6\times 700=420\quad mm,{ y }_{ A }=\frac { 180 }{ 600 } =0.3,{ y }_{ B }=\frac { 420 }{ 600 } =0.7\)

Ionic solids: Potassium sulphate, Zinc sulphide.
Metallic solids: Tin, Rubidium.
Molecular solids: Benzene, Urea, Ammonia, Water, Argon.
Covalent solids: Graphite, Silicon carbide.

Amorphous silica is obtained when molten silica is cooled rapidly because the solid silica so obtained lacks the internal order of repeating SiO₄ tetrahedra. Since the degree of internal order known as crystallinity is a variable quantity, it is possible to increase the crystallinity of amorphous solids by employing special techniques. Quartz has greater degree of crystallinity in its structure.

When group 14 element is doped with group 15 elements, we get n-type semiconductor, e.g. silicon doped with arsenic.

10.

Energy of activation is the extra energy which must be supplied to the reactants so that these can change into products. It is difference between energy of reactants and energy of activated complex as shown in diagram below.

The reactants molecules which possess activation energy and collide in proper orientation lead to formation of product molecules.

11.

Activated complex is intermediate compound between reactants and products as shown above. It is highly unstable as it has highest energy. It readily charges into product. Those molecules which can form activated complex can lead to formation of products, e.g

Generic 3

10 x 3 = 30

12.

6,octahedral

13.

80.73 pm

14.

15.

ABO₃

16.

Al₂O₃

17.

Coordination number of each of X and Y = 8

18.

AB₃

19.

(i) Dispersion forces
(ii) Dispersion forces
(iii) Ion-dipole attraction
(iv) H-bound and dipole-dipole attraction
(v) Dipole -dipole attraction.

20.

(i) It is because its conductance is intermediate between conductor and insulator.
(ii) In Schottky defect, both cations and anions are missing which lead to lowering in the density of a solid.
(iii) The glass is supercooled liquid, therefore, very old glass objects become slightly milky because of heating during the day and cooling at nights, i.e. annealing over a number of years, glass acquires some crystalline structure.

21.

(a) p-type semiconductor
(b) Ferromagnetism
(c) Impurity defects lead to vacancy defects.

Multiple Choice Question

15 x 1 = 15

22.

23.

(a)
MnO₄^-

24.

(b)
+ 0.100 V

25.

(a)
\(AgCl+\left( 1/2 \right) { H }_{ 2 }\longrightarrow Ag+{ H }^{ + }+{ Cl }^{ - }\)

26.

(d)
135.9 S cm² eq^-1

27.

(b)
equal number of cations and anions aremissing from the lattice.

28.

(a)
Frenkel defect

29.

30.

Iodide ion is largest in size and Na⁺ion is small in size.Hence,I^--I^- contact is possible in Na⁺I^-(Na⁺ion present in the void)

31.

(A)is the correct option because diamond is an insulator and non-metallic character decreases down the group

32.

(a)
288pm

33.

34.

(d)
216pm³

35.

(a)
104pm

36.

Generic 5

3 x 5 = 15

37.

Cu⁺ + e^-⇾ Cu; ∆G^o₃=?
(i)------(ii) gives the required result, i.e.,ΔG₃^° = ∆G⁰₁ - ΔG₂^° = [-0.674 - (- 0·153)] F = - 0·521 F
\(-nFE^0_{Cu^+/CU}=-0.5621F\ or\ E^0_{CU^+/Cu}=0.521V\)

38.

At pH = 5, as half-life period is found to be independent of initial concentration of sucrose, this means with respect to sucrose, it is a reaction of first order, i.e., Rate = k [Sucrose].
If n is the order with respect to H⁺ion, t_1/2 ∝ [H⁺]^I-n,
i.e., 500 ∝ (10^-5)^I-n [PH = 5 means [H⁺] = 10^-5 M] .......(I)
and 50 ∝ (l0^-6)I-n [pH = 6 means [H⁺] = 10^-6 M] .......(ii)
Dividing (i) by (ii), 10 = (lo)l-n i.e. 1 - n = 1 or n = 0, i.e., order with respect to H+ion = O.Hence, overall rate law is Rate = k [Sucrose] [H⁺]^o.

39.

\(k={0.693\over t_{1/2}}={0.693\over 2.1hr}=0.33hr^{-1}\)
x = 99% of a = 0·99 a
\(t={2.303\over k}log{a\over a-x}={2.303\over 0.33hr^{-1}}log{a\over a-0.99a}={2.303\over 0.33}log 10^2=13.69hours\)
(ii) Amount decomposed = 99% of 6.2 g =\({99\over 100}\times 6.2g=6.138g\)
1 mol NH₂NO₂ (63g) produce N₂O at STP = 22·4 L
6.138 g will produce N₂O at STP =\({22.4\over 63}\times6.138\ L=2.2176L\)