(Released on 16 september 2022
Class 10
MATHS BASIC
MATHS STD
SOCIAl SCIENCE
SCIENCE
HINDI A
HINDI B
SANSKRIT
COMPUTER APPLICATION
Class 12
ENGLISH CORE
ACCOUNTANCY
BUSINESS STuDIES
ECONOMICS
COMPUTER SCIENCE
INFORMATICS PRACTICES
PHYSICAL EDUCATION
HISTORY
POLITICAL SCIENCE
Central board secondary examination 2022
TERM I 2022-23
CHEMISTRY M.Marks:70
General Instructions:
Read the following instructions carefully
a) There are 35 questions in this question paper with internal choice.
b) SECTION A consists of 18 multiple-choice questions carrying 1 mark each.
c) SECTION B consists of 7 very short answer questions carrying 2 marks each.
d) SECTION C consists of 5 short answer questions carrying 3 marks each.
e) SECTION D consists of 2 case- based questions carrying 4 marks each.
f) SECTION E consists of 3 long answer questions carrying 5 marks each.
g) All questions are compulsory. h) Use of log tables and calculators is not allowed
SECTION A
1. The number of significant figures in 2.004 is.
i 3 ii. 4 iii. 5 iv. 1
2. Photoelectric effect is shown by
a. Alkali metals b. Halogens c. Noble gases d. None
3. The number of sigma bonds in ethylene molecule is
a.4 b.6 c.5 d. 2
4. Isoelectronic species have same number of
a. electrons b. Protons c. Neutrons d. None
5. A light radiation of wavelength 400nm has a frequency of
a. 8x1014 b. 7.5x 1014 c. 6 d. none
6. Molarity depends on
a. Temperature
b. Pressure
c. both
d, none
7.An orbital can accommodate a maximum of 2 electrons with opposite spins. This is
a. Hund’s rule
b. Pauli’s exclusion Principle
‘c. Heisenberg’s Uncertainity Principle
d. none
8.The energy of bonding molecular orbital is ..........than antibonding molecular orbital.
a. greater
b. smaller
c. equal
d. none of these
9. In PCl5 molecule the bond angles are ..............
a. 1200 and 900
b. 900 and 1040
c. 1070
d. none
10. The shape of ammonia molecule is
a. Tetrahedral
b. Octahedral
c. Pyramidal
d. None
11. The properties which are independent of the amount of the substance contained in the system but depend upon the nature opf the substance only are called
a. Extensive properties
b. Intensive properties
c. State properties
d. none
12. In Open system there is an exchange of ................with the surroundings.
a.Matter and energy
b. Energy
c. Matter
d. None
13.For a spontaneous process the value of Gibb’s free energy is
a. Negative
b. Positive
c. Zero
d. None
14. The enthalpies of all elements in their standard states are
a. unity
b. zero
c. <0
d. different for each element
15. In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?
a. 305 J
b. 307 J
c. 404J
d. none
16. Sigma bond is .........than pie bond
a. stronger
b. weaker
c. equal in strength
d. none
17.Aufbau’s Principle is based on
a. n+l rule
b.2l+1 rule
c. n-l rule
d.none
18.Which of the following species will have the largest and the smallest size
Mg,Mg+2,Al,Al3+
a. Mg and Al3+
b. Al and Mg2+
c. Al and Mg
d. Mg2+ and Al3+
SECTION B
19State Heisenberg’s Uncertainity Principle . Write its mathematical expression.
20. Define Coordinate bond and give one example.
21. A golf ball has a mass of 40g and a speed of 45 m/s. If the speed can be measured within accuracy of 2%, calculate the uncertainity in the position.
22. Define Photoelectric Effect and write its expression.
23. Half filled and fully filled orbitals are more stable than any other electronic configuration. Why?
24. The ionization enthalpy of Nitrogen is more than that of Oxygen. Why?
25. Draw the Lewis dot structure of CO32- ion
SECTION C
26. What are the differences between sigma and pie bond?
27. Explain the geometry of NH3 molecule on the basis of VSEPR theory.;
28. Derive the relation between Cp and Cv
29. Calculate the energy needed to raise the temperature of 10 g of iron from 250C to 5000C if specific heat capacity of iron is 0.45 J/K/g.
30. Define
a. Ionization enthalpy
b. Electron gain enthalpy
c. Atomic radius
SECTION D
31. Molecular orbitals are formed by the linear combination of wave functions of atoms and Antibonding molecular orbitals are formed by the subtraction of wave functions of atoms. The no of molecular orbitals and antibonding molecular orbitals formed are equal to the number of atomic orbitals participated. The energy of antibonding molecular orbital is more than molecular bonding orbital.
a. What is the formula for bond order according to molecular orbital theory ?
b. How is bond order related to stability?
c. He2 molecule does not exist Why?
d. Hoe is bbopnd order rel;ated to bond length?
32. The first law of Thermodynamics is also called law of conservation of energy . During a physical or chemical change the energy may be converted from one form into another but the total energy remains constant.
a. Write the expression for I law of Thermodynamics.
b. What are the sign conventions for heat energy?
c. What is the sign conventions for work done?
d. What is the value of q for an adiabatic change ?
SECTION –E
33. (a) Arrange the following species in the increasing order of stability on the basis of Molecular Orbital Theory
O2, O2+,O-2
(b) Draw the molecular orbital diagram for O2-
34. What is Hybridisation ? Explain it in Ethylene molecule.
35. What are the frequency and wavelength of a photon emitted during a transition from n1=5 to n2 = 2 state in the hydrogen atom ?
ANSWER KEY
1. ii
2. a
3. c
4. a
5. b
6. a
7. b
8. b
9. a
10. c
11. b
12. a
13. a
14. b
15. b
16. a
17. a
18. a
19. Principle, Mathematical Expression
20. Correct Definition
21. 6.5x10-5m
22. Definition and expression
23. Symmetry , Exchange of energy
24. Due to half filled (stable) electronic configuration of Nitrogen
25. Correct structure
26. Any 3 differences
27. Pyramidal
28. Cp – Cv =R
29. 2.137 KJ
30. Definitions
31. (a) B.O=1/2 (Nb-Na)
(b)Bond order is directly proportional to stability
(c) B.O=0
(d) Bond order is inversely proportional to bond length
32.(a) ΔU=q+W
(b) Δq=+ve (heat gained by the system)
Δq = -ve (heat lost by the system)(
c) W= = +ve ( work done on the system )
W
= - ve ( work done by the system)
(d) q =0
33.(a) O2- <O2
<O2+
(b) Correct
Diagram
34. Definition and Explanation
35. ΔE =4.58x10-19 J, υ =6.91x1014 s-1 ,λ=434nm
Aim- To prepare double salts: ferrous ammonium sulphate (Mohr’s salt)
Theory
When a mixture containing equimolar proportions of ferrous sulphate and ammonium sulphate is crystallised from its solution, a double salt is formed. The formation of double salt may be shown as follows:
(FeSO4+ (NH4)2SO4+ 6H2O →FeSO4. (NH4)2SO4. 6H2O
Ferrous ammonium sulphate (Mohr’s salt)
Fe2+ ions undergo hydrolysis, therefore, while preparing aqueous solutions of ferrous sulphate and ammonium sulphate in water, 2-3 mL dilute
sulphuric acid is added to prevent the hydrolysis of these salts
Requirements
Two beakers (250 ml), china-dish, funnel, funnel-stand, glass-rod, wash-bottle, tripod stand and wire-gauze. Ferrous sulphate crystals, ammonium sulphate crystals, dilute sulphuric acid and ethyl alcohol.
Procedure
1.Take a 250 ml beaker and wash it with water. Transfer 7.0 g ferrous sulphate and 3.5 g ammonium sulphate crystals to it. Add about 2-3 ml of dilute sulphuric acid to prevent the hydrolysis of ferrous sulphate.
2.In another beaker boil about 20 ml of water for about 5 minutes to expel dissolved air.
3. Add the boiling hot water to the contents in the first beaker in small instalments at a time. Stir with a glass rod until the salts have completely dissolved.
4. Filter the solution to remove undissolved impurities and transfer the filtrate to a china-dish.
5. Heat the solution in the china-dish for some time to concentrate it to the crystallisation point.
6. Place the china-dish containing saturated solution over a beaker full of cold water. On cooling crystals of Mohr’s salt separate out.
7. Decant off the mother liquor quickly. Wash the crystals in the china-dish with a small quantity of alcohol to remove any sulphuric acid sticking to the crystals.
8. Dry the crystals by placing them between filter paper pads.
Observations
Weight of crystals obtained =……….. g
Expected yield = ………..g
Colour of the crystals = …………..
Shape of the crystals =…………
The crystals of Mohr’s salt are monoclinic in shape.
Precautions
Collision theory
☝It states that:
☝According to collision theory the molecules collides with great kinetic energy in order to bring about a chemical reaction.
The molecules of the reacting species collide through the space in a rectilinear motion.
Rate of a chemical reaction is proportional to the number of collisions between the molecules of the reacting species.
The molecules must be properly oriented.
☝Rate of successful collisions ∝ Fraction of successful collisions X Overall collision frequency.
☝The number of collisions per second per unit volume of the molecules in a chemical reaction is called collision frequency (Z).
Let A+B --> C + D
Rate = ZABe-Ea/RT
Here ZAB = collision frequency of A and B.
☝In many reactions Rate = P ZABe-Ea/RT
Where p= steric factor which takes into account the proper orientation of the molecules participating in a chemical reaction.
PROBLEM. The activation energy for the reaction 2HI(g) → H2 + I2(g) is 209.5 kJ mol-1 at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
SOLUTION. Ea = 209.5 kJ mol - 1 = 209500 J mol - 1
T = 581 K
R = 8.314 JK - 1 mol - 1
Fraction of molecules of reactants having energy equal to or greater than activation energy is as follows:
x = e-Ea/RT
ln x = -Ea/RT
log x = Ea/2.303RT
log x = 209500 J mol-1/2.303 X 8.314 J K-1 mol-1 X 581 = 18.8323
x = antilog (18.323)
= antilog 19.1977
= 1.471 X 10-19