CBSE Sample Papers for Class 11 Chemistry Set 2 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Chemistry with Solutions Set 2 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Chemistry Set 2 with Solutions

Time Allowed: 3 hours
Maximum Marks: 70

General Instructions:

• All questions are compulsory.
• This question paper contains 37 questions.
• Questions 1-20 in Section A are objective type-very short answer type questions carrying 1 mark each.
• Questions 21 – 27 in Section B are short answer type questions carrying 2 marks each.
• Questions 28 – 34 in Section C are long-answer 1 type questions carrying 3 marks each.
• Questions 35 – 37 in Section D are long-answer 11 type questions carrying 5 marks each.
• There is no overall choice. However, an internal choice has been provided in six questions of one mark, two questions of two marks, two questions of three marks and two questions of five marks. You must attempt only one of the choices in such questions.
• Use log tables, if necessary. Use of calculator is not allowed.

Section – A

Question 1.
The number of atoms present in one mole of an element is equal to Avogadro number. Which of the following element contains the greatest number of atoms ? [1]
(A) 4 g He
(B) 46 g Na
(C) 0.40 gCa
(D) 12 g He
Answer:
Option (D) is correct.

Explanation :
Greater the number of moles, greater will be the number of atoms in an element.
Number of moles = \(\frac{\text { Given Mass }}{\text { Molecular Mass }}\)
Number of moles of 4 g He = \(\frac{4}{4}\) = 1 mol
Number of moles of46gNa = \(\frac{46}{23}\) = 2 mol
Number of moles of 0.40 g Ca = \(\frac{0.40}{40}\) = 0.01 mol
Number of moles of 12 g He = \(\frac{12}{4}\) = 3 mol
As, number of moles is highest in 12 g He. So it contains greatest number of atoms which is equal to no. of moles × avogadros number.

Commonly Made Error
Students usually do wrong calculations and : do mistake in writing the atomic mass of the elements in the calculation.

Answering Tips
Learn the atomic mass of elements and do not get confused with atomic number values of elements. Take care while doing calculations.

Question 2.
Number of n bonds and a bonds in the following structure is [1]

(A) 6,19
(B) 4,20
(C) 5,19
(D) 5,20
Answer:
Option (C) is correct.

Explanation :
The given compound is naphthalene. Each double bond in a molecule consists of a sigma bond and a pl bond, therefore in naphthalene there are total 19 sigma bonds (6 c-c single bonds, 8 C-H single bonds and 5 from C=C bonds) and 5 pi bonds.

Commonly Mad. Error
Students make mistake in counting sigma bonds. They miss the sigma bond included in the pi bonds, which involve in the double bonds.

Answering Tip
Remember each double bond in a molecule consists of a sigma bond and a pi bond and all single bonds are sigma bonds.

Question 3.
The electronic configuration of gadolinium (Atomic number 64) is [1]
(A) [Xe] 4/3 5d⁵ 6s²
(B) [Xe] 4/7 5d² 6s¹
(C) [Xe] 4/7 5d¹ 6s²
(D) [Xe] 4/8 5d⁶ 6s²
Answer:
Option (C) is correct.

Explanation :
Gadolinium (Gd, Z = 64), belongs to 6th period, / block. The expected electronic configuration of Gadolinium [Xe]4/86s, but the observed electronic configuration is [Xe]4f 5dl6s2. This is because half filled orbital is more stable than the partially filled orbital. As 5d and 6/ orbitals are close with less energy difference, 1 electron from 4/ orbital (expected to have 8e) jumps to 5d orbital making it 5d1 making 4/ orbital half filled with high stability (as per Hund’s rule exactly of maximum multiplicity).

Commonly Made Error
Students usually select wrong choice having in mind Gadolinium’s expected configuration.

Answering Tips
Some elements deviate from their expected electronic configuration to attain stable configuration as per Hund’s rule of maximum multiplicity. Please make a list of such elements and study them to avoid such mistakes.

Question 4.
Which of the following statements about the electron is incorrect ? [1]
(A) It is a negatively charged particle.
(B) The mass of electron is equal to the mass of neutron.
(C) It is a basic constituent of all atoms.
(D) It is a constituent of cathode rays.
OR
Which of the following is responsible to rule out the existence of definite paths or trajectories of electrons?
(A) Pauli’s exclusion Principle.
(B) Heisenberg’s uncertainty principle.
(C) Hund’s rule of maximum multiplicity.
(D) Aufbau principle.
Answer:
Option (B) is correct.

Explanation :
The mass of electron is very small in comparison to the mass of neutron.

OR

Option (B) is correct.

Explanation:
According to Heisenberg’s uncertainty principle, the position and velocity of an electron cannot be determined simultaneously with accuracy which rules out the existence of fixed paths.

Question 5.
The oxidation state of elements Cl₂, O₃, P₄, S₈, Na, Mg, Al is [1]
(A) +2
(B) +3
(C) 0
(D) -1
Answer:
Option (C) is correct.

Explanation:
Elements in free or uncombine state has oxidation state “0”.

Question 6.
In the reaction, when H₂ combines with O₂ to form water, there is an electron transfer from: [1]
(A) Oz to H₂O
(B) H to O
(C) O to H
(D) No electron transfer
Answer:
Option (B) is correct.

Explanation:
The H atom goes from a neutral (zero) state in H2 to a positive state in H₂O.

Read the passage given below and answer the questions given below it: (1 × 4 = 4)
The branch of science dealing with the relations between energy, heat, work and accompanying changes in the nature and behaviour of various substances around us is called thermodynamics. The main aim of the study of chemical thermodynamics is to learn (i) transformation of energy from one form into another form, (ii) utilization of various forms of energy and (iii) changes in the properties of systems produced by chemical or physical effects. Laws of thermodynamics apply only when a system is in equilibrium or moves from one equilibrium state to another equilibrium state. Macroscopic properties like pressure and temperature do not change with time for a system in equilibrium state.
Answer the questions from (7) to (10):

Question 7.
A system which can neither exchange matter nor energy with the surroundings is called : [1]
(A) Open system
(B) Isolated system
(C) Closed system
(D) Ideal system
Answer:
Option (B) is correct.

Explanation:
In an isolated system, there is no exchange of energy or matter between the system and the surroundings.

Question 8.
Which of the following is intensive property? [1]
(A) Molarity
(B) Temperature
(C) Density
(D) All of these
Answer:
Option (D) is correct.

Explanation :
An intensive property, is a physical property of a system that does not depend on the system size or the amount of material in the system.

Question 9.
Temperature and heat are: [1]
(A) Extensive properties
(B) Intensive properties
(C) Intensive and extensive properties respectively
(D) Extensive and intensive properties respectively
Answer:
Option (C) is correct.

Explanation:
An extensive property is a property that depends on the amount of matter in a sample, e.g., Mass, volume and heat are examples of extensive properties. An intensive property is a property of matter that depends only on the type of matter in a sample and not on the amount, e.g., Pressure and temperature.

Question 10.
In adiabatic process: [1]
(A) q > 0
(B) q = 1
(C) q = 0
(D) q < 0
Answer:
Option (C) is correct.

Explanation:
Adiabatic process is the change occurring within a system as a result of transfer of energy to or from the system in the form of work only, i.e., no heat is transferred.

Question 11.
Assertion (A): A solution containing a mixture of acetic acid and sodium acetate maintains a constant value of pH on addition of small amounts of acid or alkali.
Reason (R): A solution containing a mixture of acetic acid and sodium acetate acts as a buffer solution around pH 4.75. [1]
(A) Both A and R are correct and R is the correct explanation of A.
(B) Both A and R are correct but R is not the correct explanation of A.
(C) Both A and R are not correct.
(D) A is not correct but R is correct.
Answer:
Option (A) is correct.

Question 12.
Assertion (A): Electron gain enthalpy becomes less negative as we go down a group.
Reason (R): Size of the atom increases on going down the group and the added electron would be farther from the nucleus. [1]
(A) Both A and R are true and R is the 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.
Answer:
Option (D) is correct.

Explanation:
Electron gain enthalpy becomes less negative as we move down a group except that of O and F. This is because when an electron is added to O or F₁ the added electron goes to the smaller n = 2 quantum level and suffers significant repulsion from the other electrons present in this level.

Question 13.
Arrange the following in increasing order of stability; [1]
(CH₃)₃C+, CH₃CH₂CH₂C⁺H , CH₃CH₂C⁺HCH₃, CH₃C⁺H₂, CH₃CH₂C⁺H₂
Answer:
CH₃C⁺H₂ < CH₃CH₂C⁺H₂ < CH₃CH₂H₂C⁺H₂ < CH₃CH₂C⁺HCH₃ < (CH₃)₃C⁺

Question 14.
Electronegativity of F > Cl > Br > I, why? [1]
Answer:
F, Cl, Br and I exist in a group. As we know that electronegativity decreases from top to bottom in a group because the force of attraction between the outermost electron and the nucleus decreases. So, relative order of electronegativity is Cl > F > Br > I.

Commonly Made Error
Students generally confuse with the order of electronegativity and do mistakes.

Answering Tip
Learn and understand thoroughly the periodic properties and trends to avoid mistakes.

Question 15.
Arrange the following in increasing order of -1 effect.
—NO₂, —COOH, —F, —CN, —I.
Answer:
—I < —F < —COOH < —CN < —NO₂.

Question 16.
Give different isomers of C₄H₁₀ with their IUPAC names. [1]
Answer:
The isomers of C4H10 with their IUPAC names are written below.

Question 17.
Write IUPAC names of the products obtained the ozonolysis of 3,4-dimethyl hept-3-ene. [1]
OR
Complete the following equation:
CaC₂ + 2H₂O → ________ + ________
Answer:
The products obtained by the ozonolysis of 3,4-dimethyl hept-3-ene are butanone and pentan- 2-one.

OR
CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂

Question 18.
SO₃^2- is Bronsted base or acid and why ?
Answer:
SO₃^2- is a Bronsted base because it can accept proton (H⁺).

Question 19.
Why is cyclopropane more reactive than propane ? [1]
Answer:
In cyclopropane, the three carbon ring forms an equilateral triangle, with bond angle 60° which is much less compared to the normal tetrahedral bond angle of 109.5° for sp3 hybridised carbon. Therefore, the cyclopropane molecule is very much strained and compressed and hence more reactive.

Question 20.
Write balanced equations for : 1
(i) BF₃ + LiH →
OR
Write balanced equations for :
(ii) B₂H₆ + H₂O →
Answer:
(i) 2BF₃ + 6LiH → B₂H₆ + 6LiF
OR
(ii) B₂H₆ + 6H₂O → 2B(OH)₃(aq) + 6H₂(g)

Commonly Made Error
Usually students forget to balance the equation or make errors while balancing.

Answering Tip
Make a list of important reactions, learn and practice balancing these reactions. Remember to always balance a chemical reaction.

Section – B

Question 21.
How did Bohr’s model of an atom explains:
(i) the stability of the atom
(ii) origin of the spectral lines in H-atom? [2]
Answer:
(i) The electron in the atom can move around the nucleus in a circular path of fixed radius and energy. These paths are called orbits, stationary states or allowed energy states. These orbits are arranged concentrically around the nucleus. Radiation can occur only when the electron jump from one orbit to another. The atom will be completely stable in the state with the smallest orbit, since there is no orbit of lower energy into which the electron can jump.
(ii) According to Bohr’s model, radiation (energy) is absorbed if the electron moves from the orbit of smaller Principal quantum number to the orbit of higher Principal quantum number, whereas the radiation (energy) is emitted if the electron moves from higher orbit to lower orbit. It explains the origin of spectral lines in H-atom.

Question 22.
Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour. [2]
Answer:
Acidic character of a species is defined on the basis of ease with which it can lose its H- atoms. The hybridisation state of carbon in the given compound is :

As the s-character increases, the electronegativity of carbon increases and the electrons of C-H bond pair lie closer to the carbon atom. As a result, partial positive charge of H- atom increases and H+ ions ^are set free. The s-character increases in the order :
sp³ < sp² < sp Hence, the decreasing order of acidic behaviour is Ethyne > Benzene > n-Hexane.

Commonly Made Error
Students confuses themselves and write the decreasing order of acidic behaviour wrongly.

Answering Tip
Understand the concept of s-character in the compound with the help of hybridization carbon atom, which contains acidic hydrogen and then correlate with acidic behaviour.

Question 23.
Give the IUPAC name of :
(a) (CH₃)₂(C₂H₅)C-(CH₂)₃-C(CH₃)₃
(b) ortho-chloro anisole
OR
Write structural formulae for compounds named as: [2]
(a) 1-Bromoheptane
(b) 5-Bromoheptanoic acid
Answer:
(a) 2,2,6, 6 – tetramethyl octane (C₂H₅)C—(CH₂)₃—C(CH₃)₃
Or

Question 24.
Write any two chemical similarities between lithium and magnesium.
OR
What is diagonal relationship? What are its main causes? [2]
Answer:
(i) Both lithium and magnesium react slowly with water. Their oxides and hydroxides are less soluble in water.
(ii) Both are harder and lighter than other elements in the group.
OR
In addition to horizontal and vertical trends in a periodic table; there is a diagonal relationship between certain sets of elements (Li and Mg, Be and Al, B and Si) that have upper left/lower right relative location in the periodic table. These pairs of elements share similar size and electronegativity, resulting in similar properties.
Diagonal relationship is due to the increase in the charge in the ion and decrease in the size of ion as we move along the period.

Question 25.
How many significant figures are present in the answer of the following calculations: [2]
(i) 0.0125 + 0.8250 + 0.025
(ii) \(\frac{0.025 \times 298.15 \times 0.1155}{0.5785}\)
Answer:
(i) 0.0125 + 0.8250 + 0.025 = 0.8625 = 0.863
∵ 0.025 has three digits including zero after the decimal point so, the result should be reported only upto three digits after the decimal point. Therefore, there are three significant figures in the answer.

(ii) \(\frac{0.025 \times 298.15 \times 0.1155}{0.5785}\)
= \(\frac{0.860908125}{0.5785}\)
= 1.48817
= 1.49
∵298.15 has two digits after the decimal point so, the result should be reported only upto two digits after the decimal point.
So, there are three significant figures.

Question 26.
The species H₂O, HCO₃, HSO₄ and NH₃ can act both as Bronsted acid and bases. For each case give the corresponding conjugate acid and base. [2]
Answer:

Commonly Made Error
Students confuse and make mistake in writing the corresponding conjugate base and conjugate acid for the species.

Answering Tip
Be clear that addition of H⁺ ion from the species gives conjugate acids and subtraction of H⁺ ion from the species gives conjugate bases.

Question 27.
Illustrate by taking examples of transition elements and non-transition elements that oxidation states of elements are largely based on electronic configuration. [2]
Answer:
Oxidation state of an element depends upon the number of electrons present in the outermost shell of (8 – no. of valence shell electrons).

Examples:
Alkali metals (Group-1):
General electronic configuration = ns¹
Oxidation state = +1

Alkaline Earth metals (Group-2):
General electronic configuration = ns²
Oxidation state = + 2

Group 13 elements:
General electronic configuration = ns²np³
Oxidation state = + 3 and + 1

Group 14 elements :
General electronic configuration = ns²np²
Oxidation state = + 4, + 2

Group 15 elements:
General electronic configuration = ns²np³
Oxidation states = -3, + 3 and + 5
Nitrogen (N) also shows +1, +2 and +4 oxidation states.

Group 16 elements:
General electronic configuration = ns²np⁴
Oxidation states = -2, +2, + 4 and + 6

Group 17 Elements:
General electronic configuration = ns2np5 Oxidation state = -1
Cl, Br and I also show +1, +3, +5 and + 7 oxidation states.

Group 18 elements:
General electronic configuration = ns2np6 Oxidation state = 0
Transition elements (d-block elements)
General electronic configuration = (n- 1 )d^1-10 ns^1-2
The elements show variable oxidation states.
Most common oxidation states are +2 and +3.

Commonly Made Error
Students get confuse and do mistake in writing the oxidation states.

Answering Tip
Learn thoroughly the electronic configuration of each group and from which it is easy to find the oxidation states. Make a list of some elements which show exceptional oxidation states.

Section – C

Question 28.
Give one example which has the following geometry on the basis of VSEPR theory :
(a) Linear
(b) Trigonal planar
(c) Tetrahedral
(d) Pyramidal
(e) T-shape
(f) Square planar
OR
(a) Discuss the shapes of PH, and AsFs using VSEPR theory.
(b) Define antibonding molecular orbital.
Answer:

Commonly Made Error:
Students give wrong examples for the shape and make mistakes.

Answering Tip
Make a list of types of molecules with their hybridization, shape and their examples in this chapter and learn them.

OR

(a) PH₃ : sp³ hybridisation,
Pyramidal shape.

AsF₅: sp³d Hybridisation
Trigonal bipyramidal shape.

(b) The molecular orbital formed by the subtractive effect of the electron waves of the combining atomic orbitals is called antibonding molecular orbital.

Question 29.
How were cathode rays discovered ? With the help of suitable experiments show that:
(i) Cathode rays travel in straight lines,
(ii) Cathode rays consists of material particles,
(iii) Cathode rays consists of negatively charged particles.
Answer:
Faraday began to study electrical discharge in partially evacuated tubes, known as cathode ray discharge tubes. A cathode ray tube is made of glass containing two thin pieces of metal, called electrodes, sealed in it. The electrical discharge through the gases could be observed only at very low pressures and at very high voltages. When sufficiently high voltage is applied across the electrodes, current starts flowing through a stream of particles moving in the tube from the negative electrode (cathode) to the positive electrode (anode). These were called cathode rays or cathode ray particles.

The results of these experiments are summarised below:
(i) In the absence of electrical or magnetic field, cathode rays travel in straight lines.
(ii) The characteristics of cathode rays (electrons) do not depend upon the material of electrodes and the nature of the gas present in the cathode ray tube.
(iii) In the presence of electrical or magnetic field, the behaviour of cathode rays are similar to that expected from negatively charged particles, suggesting that the cathode rays consist of negatively charged particles, called electrons.

Question 30.
The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, C0₂(g) and H₂O(1) are produced and 3267.0 kj of heat is liberated. Calculate the standard enthalpy of formation, Δ_fH° of benzene. Standard enthalpies of formation of C0₂(g) and H₂O(1) are – 393.5 kJ mol^-1 and – 285.83 kj mol^-1 respectively.
(a) Name the hybridisation involved in –
(b) Why dipole moment of CO₂ is zero while H₂0 is a polar though both have similar formula?
(c) Calculate the bond order for N₂ molecule.
Answer:
Reaction for the formation of benzene,
6C(graphites) + 3H₂(g) > C₆H₆(1), Δ_fH° = ? …………(i)

The enthalpy of combustion of 1 mol of benzene is :
C₆H₆(l) + \(\frac{15}{2}\)0₂(g) → 6C0₂(g) + 3H₂O(l); Δ_CH° = -3267 kj mol^-1 … (ii)

The enthalpy of formation of 1 mol of C0₂(g)
C(graphite,s) + 0₂(g) → CO₂(g);
Δ_fH° = – 393.5 kj mol^-1 …(iii)

The enthalpy of formation of 1 mol of H20(1) is :
H₂(g) + \(\frac{1}{2}\)O₂(g) → H₂O(l);
Δ_fH° = – 285.83 kj mol^-1 …(iv)

Multiplying eq. (iii) by 6 and eq. (iv) by 3, we get
6C(graphite,s) + 6O₂(g) → 6CO₂(g), ΔfH° = -2361 kj mol^-1
3H₂(g) + \(\frac{3}{2}\)O₂(g) → 3H₂O(l); Δ_fH° = – 857.49 kj mol^-1

On adding above equations,
6C(graphite,s) + 3H₂(g) + \(\frac{15}{2}\)O₂(g) → 6CO₂(g) + 3H₂O(l); Δ_fH° = – 3218.49 kj mol^-1 ……(v)

On reversing eq. (ii)
6C0₂(g) + 3H₂O(l) → C₆H₆(1) + \(\frac{15}{2}\) O₂(g); Δ_fH° = 3267kj mol^-1 …(vi)

On adding equation (v) and (vi), we get
6C(graphite,s) + 3H2(g) → C6H6(1); Δ_fH° = 48.51 kj mol^-1

Standard enthalpy of formation of benzene
(Δ_fH°) = 48.51 kj mol^-1

Commonly Made Error
Students make mistake in writing the equations for formation of benzene, enthalpy of combustion of benzene, carbon dioxide and water.

Answering Tip
Understand the problem and practice writing the necessary equations required for the problem with enthalpy values. It is very important. Then practice further steps for the calculation.

Question 31.
(a) Name the hybdisation involved in –
(i) C₂H₂
(ii) SF₆
(b) Why dipole moment of CO₂ is zero while H₂O is a polar though both have similar formula?
(c) Calculate the bond order for N molecule.
Answer:
(a) (i) C₂H₂ – sp hybridisation
(ii) SF₆ – sp³d² hybridisation

(b) Due to sp hybridisation, geometry of CO₂ molecule is linear. So, the resultant dipole moment becomes zero.

But H₂O has sp³ hybridisation and possesses bent structure due to which the dipole moment adds up. Hence, H₂O is polar molecule.

(c) Bond order on N₂⁺:
NN₂⁺: No. of electrons = 14 – 1 = 13
Molecular orbital configuration:

Question 32.
If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in whole number ratio.
(a) Is this statement true?
(b) If yes, according to which law?
(c) Given one example related to this law.
Answer:
(a) Yes, this statement is true.
(b) This statement is according to the law of multiple proportions.
(c) Hydrogen combines with oxygen to form two compounds, namely, water and hydrogen peroxide.

Hydrogen + Oxygen → Water

Hydrogen + Oxygen → Hydrogen peroxide

Here, the masses of oxygen (i.e., 16 g and 32 g) which combine with a fixed mass of hydrogen (2g) bear a simple ratio, i.e. 16 : 32 or 1: 2.

Commonly Made Error:
Students many times do mistake in giving correct examples.

Answering Tip
Understand the basic concept of law of multiple proportions with examples. This will help them to write better.

Question 33.
(i) What is the main difference between electromagnetic wave theory and Planck’s quantum theory ?
(ii) Which mle is violated in the following orbital diagram :

OR
(a) Which has more energy of electron:
(i) Last electron of Cl^– or last electron of O^2-.
(ii) n = 4, l = 3 or n = 5, l = 2.
(b) Write the electronic configuration for:
(i) Mn²⁺
(ii) S^2-
Answer:
(i) Electromagnetic wave theory : Energy is radiated or absorbed continuously.
Planck’s quantum theory: Energy is radiated or absorbed not continuously but discontinuously in the form of small packets called quanta or photons.

(ii) Hund’s rule of maximum multiplicity is being violated.

OR

(a) (i) Electronic Configuration :
Cl^– = 2,8, 8 = 1s²,2s²2p⁶, 3s²3p⁶
O^2- = 2,8 = 1s², 2s²sp⁶
The last electron of Cl^– occurs in 3p sub-shell and last electron of O^2- occurs in 2p sub-shell. As we know that, according to Aufbau principle, 3p sub-shell has more energy than 2p sub-shell. Therefore, last electron of Cl^– has more energy of electron.

(ii) n = 4, l = 3 represent 4f
n = 5, l = 2 represent 5 p
According to Aufbau principle, the energy of 4/ is more than 5p. So, n = 4, l = 3 has more energy

(b) (i) Mn²⁺ = 1s²,2s²2p⁶,3s²3p⁶3d⁵
(ii) S^2- = 1s², 2s²2p⁶, 3s²3I⁶

Question 34.
Calculate the pH of a solution which is 0.1 M in HA and 0.5 M in NaA. Kfl for HA is 1.8 × 10^-6 ?
Answer:

Question 35.
Define the following:
(a) Standard enthalpy of formation (Δ_fH°).
(b) Standard enthalpy of combustion (Δ_cH°).
The reaction 2C + O₂ → 2CO is carried out by taking 24.0 g of carbon and 96.0 of 02.
Find out
(a) Which reactant is left in excess?
(b) How much of it is left?
(c) How many grams of the other reactant should be taken so that nothing is left at the end of the reaction?
OR
Define the following:
(a) Hess’s law of constant heat summation.
(b) First law of thermodynamics.
Find out whether it is possible to reduce MgO using carbon at 298 K. If not, at what temperature it becomes spontaneous. For reaction
MgO(s) + C(s) > Mg(s) + CO(g)
Δ_rH° = 91-18 kj/mol,
Δ_rS° = 197.67 JK^-1 mol^-1.
Answer:
(i) (a) Enthalpy change accompanying the formation of 1 mole of a compound from its constituent elements, all substances being in their standard states (1 bar pressure and 298 K).
(b) It is the enthalpy change accompanying the complete combustion of one mole of a substance in excess of oxygen or air.

(ii) (a)

24 g of carbon will react with 32 g of 0₂ to from 56 g of CO.
So, oxygen is left in excess. (1)
(b) Amount of oxygen left = 96 – 32 = 64 g
(c) Mass of other reactant = 96 – 24 = 72 g
OR
(i) (a) Hess’s law of constant heat summation :
According to Hess’s law, If a reaction takes place in several steps then its standard reaction enthalpy is the . sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.
(b) According to first law of thermodynamics, energy can neither be created nor be destroyed although it may be changed from one form to another.

(ii) ΔG = ΔH – TΔS
= 91.18 × 10³ – 298 × 197.67
= 91180 – 58905.66
= 32274.34 J/mol
For reaction to be spontaneous, ΔG should be negative. Temperature should be greater than 298 K for reaction to be spontaneous.

Question 36.
(i) Why 3° carbocation are more stable than 1° carbocation ?
(ii) Compare inductive and electromeric effects.
(iii) Why CCl₃COOH is a stronger acid than (CH₃)₃CCOOH ?
(iv) Define the term nucleophile.
(v) Draw the resonance structure of aniline, using curved arrow for electronic movements. [5]
Answer:
(i) Tertiary carbocation has three electron repelling alkyl groups. This increases +1 effect on carbon and reduces the positive charge making it more stable.

(ii) Inductive effect describes the transmission of electrical charges between atoms in a molecule whereas electromeric effect describes the transmission of electron pairs between atoms in a molecule. Inductive effect is affected by electronegativity values of atoms whereas electromeric effect is caused by the number of double bonds and their arrangement.

(iii) CCl₃COOH is a stronger acid than (CH₃)₃CCOOH due to presence of three -I effect causing group (-Cl) which increases the acidic strength. Whereas (CH₃)₃COOH contains three +1 effect causing group (-CH₃) which decreases the acidic strength.

(iv) Nucleophile: A reagent that brings an electron pair to the reactive site is called a nucleophile (Nu:), i.e., nucleus seeking.
Negatively charged nucleophile- hydroxide (OH”), cyanide (CN”), etc. Neutral nucleophile-

Section – D

Question 37.
(a) Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene ?
(b) Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why ? [5]
OR
(a) Write down the products of ozonolysis of 1, 2-dimethylbenzene (o-xylene). How does the result support Kekule structure for benzene ?
(b) Which of the following compounds will show cis-trans isomerism ?
(i) (CH₃)₂C = CH-C₂H₅
(ii) H₂C = CBr₂
(iii) C₆H₅CH = CHCH₃
(iv) H₃C – CH = (CClCH₃)
Answer:
(a) As per the given information, propanal and pentan-3-one are the ozonolysis products of an alkene. Let the given alkene be ‘A’. Writing the reverse of the ozonolysis reaction, we get:

The products are obtained on the cleavage of ozonide ‘X’. Hence, ‘X’ contains both products in the cyclic form. The possible structure of ozonide can be represented as:

Now, ‘X’ is an addition product of alkene ‘A’ with ozone. Therefore, the possible structure of alkene ‘A’ is:

Commonly Made Error
Some students do not understand the concept and write wrong structure for ozonide.

Answering Tip
Students must understand the concept of Ozonolysis and reverse of Ozonolysis.

(b) Geometrical isomers of hex-2-ene are :
im-23
The dipole moment of ds-compound is a sum of the dipole moments of C-CH₃ and C- CH₂ CH₂CH₂ bonds acting in the same direction. The dipole moment of frans-compound is the resultant of the dipole moments of C-CH₃ and C-CH₂CH₂CH₃ bonds acting in opposite directions. Hence, cis- isomer is more polar than trans-isomer. The higher the polarity, the greater is the intermolecular dipole-dipole interaction and the higher will be the boiling point. Hence, ds-isomer will have a higher boiling point than fraws-isomer.

Commonly Made Error
Some students draw incorrect isomer structures.

Answering Tip
Understand how the dipole moment is affected when the substituents are on the same side and on the different sides.

OR
(a) o-xylene has two resonance structures :
im-24
All three products, i.e., methyl glyoxal, 1, 2-dimethylglyoxal and glyoxal are obtained from two Kekule structures. Since, all three products cannot be obtained from any one of the two structures, this proves that o-xylene is a resonance hybrid of two Kekule structures.

(b) The compounds which show cis-trans isomerism are (iii) and (iv).
im-25
Alkenes (i) and (ii) have identical atoms or groups on one of the carbon atoms of the double bond and hence do not show geometrical isomerism. In contrast, alkenes (iii) and (iv) have different atoms or groups on each carbon atom of the double bond and hence, exhibit geometrical isomerism.

Commonly Made Error
Students get confused and select incorrect choice.

Answering Tip
In cis -trans isomerism different groups are ! attached on carbon atom.