Class 11 Question Paper 2024 Chemistry (with Answers)

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Question Paper
2024

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Annual Exam Question Paper 2024

SESSION ENDING EXAMINATION (2023-24)
CLASS-XI
SUBJECT- CHEMISTRY (043)
MM: 70 Time: 3 Hours
General Instructions:
Read the following instructions carefully.
a) There are 33 questions in this question paper with internal choice.
b) Section A consists of 16 multiple choice questions carrying 1 mark each.
c) Section B consists of 5 very short answer questions carrying 2 mark each.
d) Section C consists of 7 short answer questions carrying 3 marks each.
e) Section D consists of 2 case study 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 calculators is not permitted.
SECTION A
The following questions are multiple choice questions with one correct answer. Each
questions carry one mark. There is no internal choice in this section
1. What are the number of oxygen molecules in 0.5 mol of CaCO3 ? 1
(A) 1.0 x 6.022 X 1023 (B) 1.5 x 6.022 X1023
(C) 0.5 x 6.022 X 1023 (D) 3.0 x 6.022 X1023
2. Which of the following group of species are isoelectronic? 1
(A) O2- , F - , Na , Mg 2+ (B) O - , F - , Na +, Mg +
(C) O2- , F - , Na + , Mg 2+ (D) O2- , F - , Na , Mg 2+

3. Energy of an electron is given by En = 1
(A) - 2.18 x 10 -18 J [Z 2 / n 2] (B) - 2.18 x 10 -14 J [Z 2 / n 2]
(C) 2.18 x 10 -18 J [Z / n 2] (D) - 2.18 x 10 -18 J [Z 2 / n ]

4 The statement that is not correct for periodic classification of element is : 1
(A) The properties of elements are periodic function of their atomic numbers.
(B) Nonmetallic elements are less in number than metallic elements.
(C) For transition elements, the 3d- orbitals are filled with electrons after 3p- orbitals and
before 4s- orbitals.
(D) The first ionization enthalpies of elements generally increase with increase in atomic
number as we go along a period.
5 Which of the following statements are correct? 1
(A) Helium has the highest first ionization enthalpy in the periodic tables.
(B) Chlorine has less negative electron gain enthalpy than fluorine.

(C) Mercury and bromine are gases at room temperature.
(D) In any period , atomic radius of alkali metal is the highest.
6 The strongest conjugate base is: 1
- - - -
(A) Cl (B) Br (C) I (D) F
7 Which one of the following pairs of solutions not an acidic buffer ? 1
(A) H2CO3 + Na2CO3 (B) H3PO4 + Na3PO4

(C) HClO4 + NaClO4 (D) CH3COOH + CH3COONa
8 Identify the correct statement in relation to the following reaction : 1
Zn + 2 HCl ZnCl2 + H2
(A) Zinc is acting as an oxidant (B) Chlorine is acting as a reductant
(C) Hydrogen is acting as an reductant (D) Zinc is acting as a reductant
9 Acetone (CH3COCH3) and Propanal (CH3CH2CHO) are 1
(A)Position isomers (B) Functional group isomers
(C) Geometrical isomers (D) Optical isomers
10 A mixture of o-nitrophenol and p-nitrophenol can be separated by 1
(A) Sublimation (B) Steam distillation
(C)Fractional crystallization (D) Simple distillation
11 Number of bonds in benzene is 1
(A) 6 -sigma and 3- pi bonds (B) 12- sigma and 3-pi bonds
(C) 3-sigma and 12- pi bonds (D) 6- sigma and 6- pi bonds
12 Heating a mixture of sodium acetate and soda-lime gives 1
(A) Benzene (B) Methane (C) Ethane (D) Ethyne
From Q.N. 13 to16, a statement of Assertion (A) is given followed by a
corresponding statement of reason (R) just below it. Of the statements,
mark the correct answer as :
(A) Both A and R are true and R is correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
13 Assertion : All isotopes of a given element show the same type of chemical behavior. 1
Reason : The chemical properties of an element are controlled by the number of electrons
in the atom.

14 Assertion : The number of radial nodes in 3s and 4p orbitals is not equal. 1
Reason : The number of radial nodes in any orbital depends upon the value of ‘n’ and ‘l’.

15 Assertion : Electron gain enthalpy of elements become less negative as we go down a 1
group.
Reason : Size of the atom increases on going down the group and the added electron
would be farther from the nucleus

16 Assertion : Lassaigne’s extract is boiled with dil. HNO3 before testing for halogens by 1
AgNO3.
Reason : CN – and S 2 – ions present in the extract interfere with the test of halide ions by
AgNO3.

SECTION - B
This section contains 5 questions with internal choice in one question. The following questions are very
short answer type and carry two marks each.
17. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with 2
aqueous hydrochloric acid according to the reaction:
4HCl + MnO2 (S) 2H2O (l) +MnCl2 (aq) +Cl2 (g)

How many grams of HCl react with 5.0 g of manganese dioxide ?
Atomic mass of Mn = 55 u.
18. Yellow light emitted from a sodium lamp has a wavelength(ʎ) of 580 nm. Calculate the 2
frequency(v) and the wave number (⊽) of the yellow light.
19. Describe hybridisation in case of PCl5. Why are the axial bond longer as compared to the 2
equatorial bond ?
OR
What is meant by the term bond order? Calculate the bond order of O2+.
20. For the reaction , 2 A(g) + B (g) 2 D(g) , ∆U0 = -10.5KJ and ∆S0 2
-1 0
= - 44.10JK . Calculate ∆G for the reaction and predict whether the reaction may occur
spontaneously.
21. What are electrophiles and nucleophiles ? Explain with examples. 2

SECTION C
This section contains 7 questions with internal choice in one question. The following questions are short
answer type questions and carry 3 marks each.
22 (a) State Pauli’s exclusion principle. 1 3
(b) Write all the four quantum numbers of the last electron of chlorine. 1
(c) Write electronic configuration of Cr (24). 1
23 Explain why ? 3
(a) Electron gain enthalpy of fluorine is less negative than that of chlorine.
(b) First ionization enthalpy of nitrogen is greater than that of oxygen.
(c) Write IUPAC name and symbol of element having atomic number 118.
Calculate the standard enthalpy of formation of CH3OH(l) from the following data : 3
0 -1
(i) CH3OH(l) + 3/2 O2(g) CO2(g) + 2H2O (l); ∆r H = -726KJmol
24 (ii) C(S) + O2(g) CO2(g) : ∆C H0 = - 393 KJmol-1
(iii) H2(g) + 1/2 O2(g) H2O (l) : ∆f H0 = - 286 KJmol-1
OR
State Hess’s Law of constant heat summation. 1
Using Hess’s Law, Calculate the enthalpy of formation of carbon monoxide(CO) from the
following data: 2
0 -1
(i) C(s) + O2(g) CO2(g) ; ∆rH = - 393.5kJ mol
(ii) CO(g) +1/2O2(g) CO2(g) ; ∆rH0 = -283.0 kJ mol-
25 Permanganate (VII) ion, in basic solution oxidises iodide ion I – to produce molecular
iodine (I2) and manganese (IV) oxide MnO2 . Write a balanced ionic equation to represent
this redox reaction.
OR
Balance the following equation by ion electron method

P4 + OH- (aq.) ---🡪 PH3 + HPO2- [In basic medium]
26 (a) Define Inductive effect. 3
(b) Why tertiary carbocation is more stable than primary and secondary carbocation?
(c) Write the name and structures of geometrical isomers of 2-Butene.
27 (a)Write IUPAC of the following compound: 1 3
(i) CH3 CH(CH3)CH2CH2 CHO (ii) CH3 CH(CH3)CH(CH3)CH2OH

b) On complete combustion , 0.246 g of an organic compound gave 0.198 g of 2
carbon dioxide and 0.1014 g of water . Determine the percentage composition of
carbon and hydrogen in the compound
28 (a) An alkene “A” on ozonolysis gives a mixture of ethanal and pentan-3-one 1 3
.Write the structure of ‘A’.
b) Complete the following reaction: 2
(i) CH3CH2-Br + Na Dry Ether
(ii) CH3-CH2Cl alc. KOH, heat

Section D
The following questions are case-based questions. Each question has an internal choice
and carries 4 (1+1+2) marks each. Read the passage carefully and answer the questions
that follow.

29 Read the text carefully and answer the questions : 1+2+
1
A system in thermodynamics refers to that part of universe in which observations are
made and remaining universe constitutes the surroundings. The wall that separates the
system from the surroundings is called boundary.
The State of the System must be described in order to make any useful calculations by
specifying quantitatively each of the properties such as its pressure (p), volume (V), and
temperature (T) as well as the composition of the system. In thermodynamics, the state of
a system is introduced.The state of a thermodynamic system is described by its
measurable or macroscopic (bulk) properties. We can describe the state of a gas by
quoting its pressure (p), volume (V), temperature (T), amount (n) etc. Variables like p, V,
T are called state variables or state functions because their values depend only on the state
of the system and not on how it is reached.
By conventions of IUPAC in chemical thermodynamics. The q is positive, when heat is
transferred from the surroundings to the system and the internal energy of the system
increases and q is negative when heat is transferred from system to the surroundings
resulting in decrease of the internal energy of the system.
Let us consider the general case in which a change of state is brought about both by doing
work and by transfer of heat. We write change in internal energy for this case as: ∆U = q
+ w The equation i.e., ∆U = q + w is mathematical statement of the first law of
thermodynamics, which states that the energy of an isolated system is constant. It is
commonly stated as the law of conservation of energy .
Q.1. State first law of thermodynamics and write its mathematical form. 1
Q.2. What are extensive and intensive properties ? Explain with examples. 2
Q.3.What are state functions ? Give examples. 1

30 Read the text carefully and answer the questions : 4

After having some idea about the terms atoms and molecules, One atomic mass unit is
defined as a mass exactly equal to one-twelfth of the mass of one carbon – 12 atom.
Molecular mass is the sum of atomic masses of the elements present in a molecule. The
mole, symbol mol, is the SI unit of amount of substance. One mole contains exactly 6.022
× 1023 elementary entities. This number is the fixed numerical value of the Avogadro
constant, NA , when expressed in the unit mol–1 and is called the Avogadro number. The
mass of a carbon–12 atom was determined by a mass spectrometer and found to be equal
to 1.992648 × 10–23 g. Knowing that one mole of carbon weighs 12 g, the number of atoms

in it is equal to 12g/moiC-12 / 1.992648×1023 g / C- 12 atom. = 6.0221367 × 1023
atoms/mol.
The mass of one mole of a substance in grams is called its molar mass. The molar mass in
grams is numerically equal to atomic molecular/formula mass in u.
An empirical formula represents the simplest whole number ratio of various atoms present
in a compound, whereas, the molecular formula shows the exact number of different types
of atoms present in a molecule of a compound. Many a time, reactions are carried out
with the amounts of reactants that are different than the amounts as required by a balanced
chemical reaction.
1) One atomic mass unit (amu) is defined as a mass exactly equal to one-twelfth of the
mass of one ……atom.
(a) Hydrogen -1 (b) Carbon -12
(c) Oxygen -12 (d) Chlorine -35
2) The mass of one mole of a substance in grams is called its.
(a) Atomic mass (b) Molecular Weight
(c) Molecular mass (d) molar mass.
3)The empirical formula of ethanoic acid is
(a) CHO (b) C2HO2
(c) CH2O (d) CH2O2
4) Two mole contains exactly …elementary entities.
(a) 6. 02214076 × 1021 (b) 6.02214076 × 1022
(c) 12.044 × 1023 (d) 0.2214076 × 1024

SECTION E
The following questions are long answer questions. All questions have an internal choice
and carry 5 marks each.
31 (a) State Le-Chatelier’s principle. 1+2+
(b) Describe the effect of : 2
(i) Addition of H2 (ii) Addition of CH3OH
(iii) Removal of CO (iv) Removal of CH3OH
On the equilibrium of the reaction : 2H2(g) + CO (g) CH3OH (g)
(c) Classify the following as Lewis acids and bases- (a) OH- (b) F- (c) BF3 (d) NH4+
OR

(a) What is meant by common ion effect ?
(b) Calculate the pH of a 1.0 x 10-8 M solution of HCl. (given:- log1.10 = 0.0414). 1+2+2
(c) Arrange the following in increasing order of acidic
strength-
(i) HF, HCl. HBr, HI
(ii) CH4, NH3, H2O, HF

32 (a) State Huckel’s rule of aromaticity. What are necessary and sufficient conditions for a 2
molecule to be aromatic?
(b) Give the following reactions-
1
(i) Friedal Craft acylation of benzene with acetyl chloride
(ii) Reduction of phenol with zinc dust. 1
(iii)Sodium acetate react with soda lime. 1
OR
(a) State Markovnikov’s rule. 1
(b) Write the mechanism of addition of HBr to propene. 2

(c) Perform the following conversion:
2
(i) Ethyne to Ethanal
(ii) Benzene to Nitrobenzene

33 Answer any five question: 5×1
Accounts for the following:
(i) Sigma bonds, is stronger than pi bond.
(ii) NF3 is pyramidal while BF3 is triangular planar.
(iii) O2 is paramagnetic .
(iv) H2O is liquid but H2S is gas .
(v) P-Nitrophenol has higher boiling point than o- Nitrophenol.
(vi) Dipole moment of BeH2 is zero.
(vii) Shape of H2O is angular.

SESSION ENDING EXAMINATION (2023-24)
CLASS-XI
SUBJECT: CHEMISTRY
MARKING SCHEME

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
B C A C A D C D B B B B A D A A

17 Molar mass of MnO2 =55 +32 = 87 g. 1
Molar mass of 4 moles of HCl = 4 x 36.5 = 146g.
So, mass of HCl will react with 5.0g of MnO2 = 146 /87 x 5 = 8.40g.1 1

18 ʎ =580 nm = 580 x 10 -9 m. 1
Frequency, v = c / ʎ = 3.0 x10 8 m/s / 580 x 10 -9 m = 5.17 x 10 14 s -1.
Wave number, (⊽) = 1 / ʎ = 1 / 580 x 10 -9 m = 1.72 x 10 6 m-1 . 1
19 Hybridisation is sp3d. Shape – Trigonal bipyramidal 1
Because, Axial bonds experience more repulsion than equatorial bonds. 1
Or
No. of bonds formed between two atoms is called bond order. 1
Bond order = ½ (B.e - A.B.e)
Bond order of O2+ = 1/2(B.e - A.B.e) = 1/2 (8-3) =2.5 1

20 For given reaction, ∆ng = 2- (3) = -1 1
∆H0 = ∆U0 + ∆ng RT = -10.5Kj + (-1) (8.314 x 10-3Kj) (298) = -10.5 – 2.48 = -
12.98Kj
∆G0 = ∆H0 - T∆S0 = -12.98Kj – 298 (-44.1 x 10-3kJ) = -12.98kJ + 13.14kJ = 0.16kJ. 1
As ∆G0 comes out to be +ve, the reaction will not occur spontaneously.
21 Electrophiles are electron deficient species and can accept an electron pair from 1
electron rich species. Example:- H+, Br+,NO2+, BF3,etc.
A nucliophiles are electron rich species and can donate an electron pair to electron 1
deficient species. Eg:- OH-, OR-, CN-, NH3, etc.
22 (a) No two electrons of an atom can have all the four quantum numbers 1
identical.
(b) n= 3, l=1, m=0, s= -1/2 1
(c) 1s2 2s2 2p6 3s2 3p6 4s1 3d5 . 1
23 (a) Due to small size of fluorine than chlorine and more inter electronic 1
repulsion in fluorine than chlorine.
(b) Due to half filled 2p orbital of nitrogen. 1
(c) Ununoctium, Uuo 1
24 Aim: C(s) + 2H2(g) + 1/2O2(g) CH3OH (l), ∆fH0 =? 1
Eqn. (ii) + 2 Eqn. (iii) – Eqn. (i) gives the required eqn. with 1
∆H = -393 + 2 (-286) – (-726) kJ mol-1 = -239 kJ mol-1. 1
OR
Correct statements
We aim at: C(s) + ½O2 (g) CO (g); ∆fH0 =?
Subtracting eqn. (ii) from eqn. (i), we get 1
C(s) + 1/2 O2(g) – CO (g) 0 ; ∆fH0 = -393.5 – (-283.0) = -110.5 kJ mol-
1 2
.
Or C(s) + 1/2 O2(g) CO (g); ∆fH0 = -110.5 kJ.

So, heat of formation of CO is ; ∆fH = -110.5 kJ mol-1.
0

25 Indicating oxidation number and oxidation & reduction 1
Balancing charge and hydrogen atom 1
Balancing number of electron and writing final balanced equation as
1

2MnO-4 (aq) + 6I- (aq) + 4H2O(l) 2MnO2 (s) + 3I2 (s) + 8OH- (aq)

26 (a) The polarization of a sigma bond due to electron withdrawing or electron
donating effect of adjacent groups pr atoms in a carbon chain is called
inductive effect.
(b) Due to hyperconjugation effect.
(c)

27 (a) (i) 4-methyl pentanal (ii) 2,3-dimethyl butan-1-ol ½+1/2
(b) Percentage of carbon =12/44 x 0.198/0.246 x100 = 21.95 % 1
Percentage of hydrogen = 2/18 x 0.1014/0.246 x 100 = 4.48% 1
28 1

(a)
3 – Ethyl pent -2-ene 1
(b) (i)CH3 CH2 CH2 CH3 1
(ii) CH2=CH2
29 1. Energy can neither be created nor destroyed. Any other statement. ∆U = q 1
+ w
2. Correct definition with example
3. a state function is a property whose value is independent of the path taken to 1
reach that value.
Examples of state functions include: Internal energy, Enthalpy, Entropy. 1
30 1—B, 2---D, 3-----C, 4 ---C
31 a. Correct statement 1
b. (i) forward direction (ii) Backward direction (iii) Backward direction (iv)
forward direction 2
c. Lewis acids – BF3, NH4+. Lewis bases -OH-,F- 2
OR
a)Common ion effect is a phenomenon in which degree of dissociation of any weak
electrolyte is supressed by addition of small amount of strong electrolyte 1
b) Water has 10−7 M H+ and 10−7 M OH−.
When 10−8 M HCl is added the concentration of proton in water is not neglected. 1
So, Total [H+] = 10−7 + 10−8 = 10-8[ 1 + 10 ] = 11 x 10−8 M
pH = − log (H+) = − log( 11 X 10−8 ) = 8 − log11 = 6.598
(C) (i) HF<HCl<HBr<HI 1
(ii) CH4<NH3<H2O<HF

1
1
32 (a) If a planar ring molecule would have aromatic properties. His rule states that 2
if a cyclic, planar molecule has 4n+2 π electrons, it is considered aromatic.
Condition:- (i) Organic compound should have a single cyclic cloud of delocalized π
electrons above and below the plane of the molecule.
(ii) It should be planer.
(iii) It should contain (4n+2 π) electron where n = 1,2,3…etc.

(b) (i)

3×1

(ii)

(iii) Correct reaction
OR
(a) In addition to alkenes, the negative part addendum goes to that carbon atom
which has less number of hydrogen atom.
(b)
1

2

(c) (i)

2×1

(ii)

33 (i) Because sigms bond is formed by axial overlapping of two orbitals 5×1
where as π bond is formed by side wise overlapping of two orbitals.
(ii) In NF3 nitrogen is sp3 hybridized where as in BF3 boron is sp2
hybridized.
(iii) O2 is paramagnetic due to presence of two unpaired electrons in anti-
bonding molecular orbitals.
(iv) B/c water molecules are held together with intense inter molecular
hydrogen bonding but in H2S molecule there is no hydrogen bonding.
(v) P-nitrophenol molecules are associated with inter molecular hydrogen
bonding where as in O-nitrophenol intra molecular hydrogen bonding is
present.
(vi) Due to linear shape of BeH2 dipole moment of each B-H bond are in
opposite direction, they cancel each other and net diploe moment is 0.
(vii) In water molecule oxygen is sp3 hybridized. Due to presence of two
lone pair of electrons shape of water molecular is angular.
THE END

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