CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Chemistry with Solutions Set 4 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Chemistry Set 4 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 types of hybrid orbitals of nitrogen in NO2+, N03 and NH4+ respectively are expected to be
(A) sp, sp3 and sp2
(B) sp, sp2, and sp3
(C) sp2, sp and sp3
(D) sp2, sp3 and sp
OR
Hydrogen bonds are formed in many compounds, e.g., H2O, HF, NH3. The boiling point of such compounds depends to a large extent on the strength of hydrogen bond and the number of hydrogen bonds. The correct decreasing order of the boiling points of above compounds is
(A) HF > H2O > NH3
(B) H2O > HF > NH3
(C) NH3 > HF > H2O
(D) NH3 > H2O > HF
Answer:
Option (B) is correct.

Explanation:
Number of orbitals involved in hybridization (H) = 1/2 [V+M-C+A] where
V = valence electrons of central atom
M = no. of monovalent atoms linked with central atom
C = charge of cation
A = charge of anion
N02+ = 1/2 [5+0-1+0] = 2or sp
N03 = 1/2 [5+0-0+1] = 3 or sp2
N04+ = 1/2 [5+4-1+0] = 4 or sp3
OR
Option (B) is correct.

Explanation:
Strength of hydrogen bonding:
HF > H2O > NH3
Also, each H2O molecule is linked with four H2O molecule through H-bonds while each HF molecule is linked only to two HF molecules. Therefore, the decreasing order of boiling points will be H2O > HF >NH3.

CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions

Question 2.
One mole of any substance contains 6.022 × 1023 atoms/molecules. Number of molecules of H2S04 present in 100 mL of 0.02 M H2S04 solution is: [1]
(A) 12.044 × 1020 molecules
(B) 6.022 × 1023 molecules
(C) 1 × 1023 molecules
(D) 12.044 × 1023 molecules
Answer:
Option (A) is correct.

Explanation:
Number of moles of H2S04
= molarity × volume in mL = 0.02 × 100 = 2 millimoles = 2 × 102 mol

Number of molecules of H2S04
= Number of moles × NA = 2 × 103 × 6.022 × 1023 = 12.044 × 1020 molecules

Question 3.
Isostructural species are those which have the same shape and hybridisation. Among the given species identify the isostructural pairs.
(A) [NF3 and BF3]
(B) [BF4 and NH4+]
(C) [BCl3 and BrCl3]
(D) [NH2 and NO3]
OR
Which of the following species has tetrahedral geometry?
(A) BH4
(B) NH2
(C) C032-
(D) H3O+
Answer:
Option (B) is correct.

Explanation:
(i) NF3 is pyramidal whereas BF3 is planar triangular.
(ii) BF4 and NH4+ are tetrahedral.
(iii) BCl3 is triangular planar whereas BrCl3 is pyramidal.
(iv) NH3 is pyramidal whereas NO3 is triangular planar.

Commonly Made Error
Students get confuse with the shape of the given species and do mistakes.

Answering Tip
Make a list of species having different shapes in this chapter and learn them and also understand why these species have specific shapes.
OR

Option (A) is correct.

Question 4.
Thermodynamics is not concerned about ____________. [1]
(A) energy changes involved in a chemical reaction.
(B) the extent to which a chemical reaction proceeds.
(C) the rate at which a reaction proceeds.
(D) the feasibility of a chemical reaction.
Answer:
Option (C) is correct.

Explanation:
Thermodynamics is not concerned about how and at what rate these energy transfor¬mations are carried out, but is based on initial and final states of a system undergoing the change.

Question 5.
The volume of gas is reduced to half from its original volume. The specific heat will be ____________. [1]
(A) reduce to half
(B) be doubled
(C) remain constant
(D) increase four times
Answer:
Option (C) is correct.

Explanation:
Specific heat is an intensive property and is independent of the volume of the substance.

CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions

Question 6.
Which of the following has a minimum wavelength? [1]
(A) Gamma rays
(B) Blue lights
(C) Infrared rays
(D) Micro wave
Answer:
Option (A) is correct.

Read the following text and answer the following questions on the basis of the same:
Originally, the term oxidation was used to describe the addition of oxygen to an element or a compound. Because of the presence of dioxygen in the atmosphere (—20%), many elements combine with it and this is the principal reason why they commonly occur on the earth in the form of their oxides, A careful examination of some reactions in which hydrogen has been replaced by oxygen prompted chemists to reinterpret oxidation in terms of removal of hydrogen from it and, therefore, the scope of term oxidation was broadened to include the removal of hydrogen from a substance.
Answer the questions (7) to (10) given below:

Question 7.
Oxidation is explained as: [1]
(A) addition of hydrogen
(B) addition of oxygen
(C) removal of hydrogen
(D) both (B) and (C)
Answer:
Option (D) is correct.

Explanation:
It involves addition of oxygen and removal of hydrogen.

Question 8.
Choose the reduction reaction/s from the following: [1]
(A) Mg + F2 → MgF2
(B) 2FeCl3 + H2 → 2FeCl2 + 2 HCl
(C) Both (A) and (B)
(D) None of these
Answer:
Option (B) is correct.

Explanation :
The reaction involves removal of electronegative element chlorine form ferric chloride.

Question 9.
H2S + Cl2 → 2HCl + S
Choose the statement which is correct for the above reaction, [1]
(A) H2S is reduced.
(B) Cl is oxidised.
(C) Cl an electronegative element is added to hydrogen.
(D) Cl an electropositive element is added to hydrogen
Answer:
Option (C) is correct.

Explanation :
In the given reaction
H2S + Cl2 → 2HCl + S
Chlorine is electronegative element which gets added to hydrogen.

CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions

Question 10.
Which of the following is not an example of redox reaction? [1]
(A) CuO + H2 → CU + H2O
(B) Fe2O3 + 3CO → 2Fe + 3CO2
(C) 2K + F2 → 2KF
(D) BaCl2 + H2S04 → BaSO4 + 2HCl
Answer:
Option (D) is correct.

Explanation:
It is not a redox reaction as there is no change in oxidation number of any of the reactants. It is an example of double displacement reaction.

Question 11.
Why arenes are called aromatic hydrocarbons? [1]
OR
n-Pentane has higher boiling point than neo-pentane. Why ?
Answer:
Since most of the arenes possess pleasant odour
(Greek; aroma meaning pleasant smelling), the class of compounds was named as ‘aromatic compounds’
OR
(i) CH3—CH2—CH2—CH2—CH3
n-pentane
CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions 1
With the increase in number of branched chain, the molecules attain the shape of sphere, thereby decreasing surface area and weak van Der Waals forces. Therefore, pentane has higher boiling point than neo-pentane.

Commonly Made Error
Students confuse the reason for higher boiling point and write wrong answer.

Answering Tip
Understand the concept, the reason behind the variation of boiling point among alkanes. Understand how increase in the carbon atom and branching in the alkanes affect the boiling point. Knowing the general concept, helps students to answer easily with out confusion.

Question 12.
Assertion (A): Energy of resonance hybrid is equal to the average of energies of all canonical forms.
Reason (R): Resonance hybrid cannot be presented by a single structure. [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 (D) is correct.

Explanation:
Canonical structures always have more energy than resonance hybrid. Resonance hybrids are always more stable than any of the canonical structures. The delocalisation of electrons lowers the orbitals energy and gives stability.

Question 13.
Assertion (A): All isotopes of a given element show the same type of chemical behaviour.
Reason (R): The chemical properties of an atom are controlled by the number of electrons in the atom. [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.

Explanation:
Chemical properties of atoms are controlled by the number of electrons, which are determined by the number of protons in the nucleus. Number of neutrons present in the nucleus have very little effect on the chemical properties of an element.
Therefore, all the isotopes of a given element show same chemical behaviour.

CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions

Question 14.
Write the Ksp expression for Ag2CrO4. [1]
Answer:
CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions 2
Ksp = [Ag+]2 [CrO42-]
= (2S)2 (S)
= 4S3

Question 15.
If the length of the crest of a wave is 4 pm. Write the wavelength of this wave. [1]
Answer:
Wavelength = 2 × length of crest
= 2 × 4
= 8 pm

Question 16.
Name the indicator used in redox titrations involving K2Cr2O7 as an oxidising agent. [1]
Answer:
Potassium ferricyanide K3(Fe(CN)6 is used as external indicator in redox titrations involving K2Cr2O7 as an oxidising agent.

Question 17.
What is benzenoid and non-benzenoid? [1]
OR
How many isomers are possible for monosubstituted and disubstituted benzene?
Answer:
Aromatic compounds containing benzene ring are known as benzenoids and those not containing a benzene ring are known as non-benzenoids.
OR
There is one mono substituted benzene and three di-substituted benzene, i.e., ortho, para and met a.

Question 18.
Explain on the basis of molecular orbital diagram why O2 should be paramagnetic ? [1]
Answer:
On the basis of molecular orbital diagram, O2 molecule contains two unpaired electrons in the n*2px and n*2py orbitals. Due to the presence of unpaired electrons oxygen show paramagnetic behaviour.

Question 19.
The solubility of A2X3 is y mol dm-3. Calculate its solubility product. [1]
Answer:
CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions 3

Question 20.
Is it safe to stir 1 M AgN03 solution with copper spoon ? [1]
Given, E°Ag+/Ag = + 0.80 V, E°Cu2+/Cu = + 0.34V
OR
The standard reduction potentials of three metals cations X, Y and Z are + 0.52, – 3.03 and – 1.18V respectively. Arrange X, Y and Z in order of increasing reducing power.
Answer:
As we know that a metal having low value of E° can displace the metal having high value of E°. The standard reduction potential of Cu is less than Ag. So, it is not safe to stir 1 M AgNO3 solution with copper spoon as copper can displace silver from 1 M AgNO3 solution.

Commonly Made Error:
Students confuse the reactivity order and give wrong answer.

Answering Tip:
Remember the order of reactivity through electrochemical series.
OR
Greater the value of more negative standard reduction potential of the metal, greater will be its reducing power.

Cation Standard Reduction Potential (V)
X +0.52
Y -3.03
Z -1.18

∴ Increasing order of reducing power :
X < Z < Y

Commonly Made Error
Students confuse with the variation of reduction potential values and give wrong order.

Answering Tip
Remember that reducing power increases with increase in the value of negative standard reduction potential.

Section – B

Question 21.
What are the developments leading to the Bohr’s model of atom? [2]
OR
What is the Hund’s rule of maximum multiplicity ? Explain with suitable example.
Answer:
In order to improve Rutherford’s atomic model, two new concepts played a major role, these are :
(i) Dual nature of electromagnetic spectrum.
(ii) Experimental results regarding atomic spectra.
OR
According to this rule, “pairing of electrons in the orbitals belonging to the same sub-shell (p, d or f) does not take place until each orbital belonging to that sub shell has got on electron each, i.e., it is singly occupied.
Example: Electronic configuration of 7N
= 2, 5 = 1s2, 2s22p3
CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions 4

CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions

Question 22.
To which orbit the electron in H-atom will jump on absorbing 12.1 eV energy? [2]
Answer:
According to Bohr’s theory, Energy of electron in nth orbital of H-atom with atomic number
(Z) = -13.6\(\frac{z^2}{n^2}\)eV n
The energy of ground state electron in H-atom = -13.6 eV
Now, it absorbs 12.1 eV energy.
So, its energy increases to -13.6 +12.1= – 1.5 eV
∴ E = -13.6 \(\frac{z^2}{n^2}\) eV
-1.5 = -13.6 × \(\frac{(1)^2}{n^2}\) (for hydrogen Z = 1)
n = \(\frac{13.6}{1.5}\) = 9.06 ≈ 9
∴ n = 3
Hence, the electron in H-atom jumps to 3rd orbit on absorbing 12.1 eV energy.

Question 23.
Draw the cis- and trans- structures of but-2-ene. Which isomer will have a dipole moment? [2]
OR
(i) How will you convert: ethyne to but-2-yne
(ii) Draw the geometrical isomers of 2, 3-dichlorobut-2-ene
Answer:
The cis-and-fmns- structures of but-2-ene are as follows-
CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions 5
Although, both ds-but-2-ene and frans-but-2-ene has non- polar compounds but ds-but-2-ene has a small dipole moment. In frans-but-2-ene, the vectors of any small bond dipoles must cancel each other because of its shape. So, the dipole moment of frans-but-2-ene is zero. (1)
OR
CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions 6

Question 24.
Write IUPAC names of the products obtained by addition reactions of HBr to hex-l-ene. [2]
(i) in the absence of peroxide and
(ii) in the presence of peroxide.
Answer:
CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions 7
Commonly Made Errors
Some students make mistake while writing IUPAC name of hex-l-ene.
They also do mistake during addition of hydrogen bromide across the double bond.

Answering Tips
Understand the concept of addition of hydrogen halides.
The position of bond must be understood clearly.
They should understand which products are formed for the presence and absence of peroxides without getting confused.

Question 25.
Give one example each of a molecule in which empirical formula and molecular formula are: [2]
(i) same (ii) different
Answer:
(i) same molecular formula and empirical formula: carbon dioxide both is CO2
(ii) molecular formula and empirical formula are different: Hydrogen peroxide H2O2 and empirical formula id HO

Question 26.
Write IUPAC names of the following: [2]
CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions 8
Answer:
(i) 1- chloroethene
(ii) 1-methylcyclobutane

CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions

Question 27.
Split the reaction 2K(s) + Cl2(g) → 2KCl(s) into oxidation and reduction half reactions. [2]
Answer:
CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions 9
Oxidation half reaction
K → K+ + e (Loss of electron)
Reduction half reaction
Cl2 + 2e → 2Cl (Gain of electron)

Commonly Made Error
Students do mistakes while writing half reactions.

Answering Tip
Understand concepts of oxidation and reduction. Understanding the concept, it will be easy to write half reactions and balance these equations.

Section – C

Question 28.
Calculate:
(i) Mass in gram of 5.8 mol N2O
(ii) Number of moles in 8.0 g of O2
(iii) Molar mass if 11.2 L at STP weigh 8.5 g. [3]
Answer:
(i) Molecular mass of N2O = 2(14) + 16 = 44
Mass = Molecular mass × No. of moles
= 44 × 5.8
= 255.2 g

(ii) Number of moles of
O2 = \(\frac{\text { Mass of } \mathrm{O}_2}{\text { Molecular mass of } \mathrm{O}_2}\)
= \(\frac{8.0}{32}\)
= 0.25 mol

(iii) 11.2 L at STP weigh = 8.5 g
22.4 L at STP will weigh = \(\frac{8.5}{11.2}\) × 22.4
= 17 g mol-1

Commonly Made Error:
Students often forget to mention units in the final answer, thus losing marks.

Answering Tips:
Do the calculations carefully.
Do not forget to write units with the final answer.

Question 29.
In what respect the Mendeleev’s classification is superior to the other classification given earlier? [3]
OR
Comment on the need to classify elements. Explain the earliest attempts to classify elements by Dobereiner and Newlands.
Answer:
Mendeleev’s classification is superior to the earlier classifications:
(i) Mendeleev’s developed a periodic table of elements wherein the elements were arranged on the basis of their atomic mass and also on the similarity in chemical properties.
(ii) Among chemical properties, he took the formulae of the hydrides and oxides formed by an element as one of the basic properties of an element for classification.
OR
There was need to classify elements for systematic study of properties of different elements and their compounds. The need gave rise to the classification of properties of element and led to the formulation of periodic table.
Dobereiner classified certain elements into group of three, called triad. If elements of a triad were arranged in increasing order of their atomic weights, the atomic weight of the middle element was approximately the arithmetic mean of other two elements. It is referred to as law of triads. (1) Newlands suggested that when elements are arranged in increasing order of their atomic weights, every 8th element resembles its properties, with the first one just like the 8th note of a musical scale. This is known as Newlands law of Octaves.

Question 30.
For oxidation of iron,
4Fe(s) + 3O2(g) → 2Fe2O3(s)
Entropy change is – 549.4 J K-1 mol-1 at 298 K. Inspite of negative entropy change of this reaction, why is the reaction spontaneous?
fH° for this reaction is – 1648 × 103 J mol-1) [3]
Answer:
One decides the spontaneity of a reaction by considering:
ΔStotal = ΔSsys + ΔSsurr
For calculating ΔSsurr, we have to consider the heat absorbed by the surroundings which is equal to -ΔrH°. At temperature T, entropy change of the surroundings is:
DSsurr = \(-\frac{\Delta_r \mathrm{H}^{\Theta}}{\mathrm{T}}\) (at constant pressure)
= \(\frac{-\left(-1648 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}\right)}{298 \mathrm{~K}}\)
= 5530JK-1mol-1
So, total entropy change for this reaction
DrStotal =5530 JK-1mol-1 + (-549.4 JK-1 mol-1)
= 4980.6 J K-1 mol-1
This show that the above reaction is spontaneous.

CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions

Question 31.
(a) Which out of NH3 and NF3 has higher dipole moment and why ?
(b) Give reason
He2 molecule is not formed (on the basis of MO theory). [3]
Answer:
(a) The dipole moment of NH3 is higher than NF3, although both the molecules are pyramidal in shape.
According to structure of both molecules, in NH3, the orbital dipole due to lone pair is in the same direction
CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions 10
as the resultant dipole moment of the three N – H bonds, while in NF3 the orbital dipole due to lone pair is in the opposite direction to the resultant dipole moment of three N – F bonds.
Hence, the resultant dipole of NH3 is higher than NF3.

Commonly Made Error
Students forget to write the dipole moment arrow for lone pair of NF3 in opposite direction while drawing the structure.

Answering Tip
Understand the question and reason behind the higher dipole moment of NH3 than NF3. Accordingly draw the structures.

(b) Helium atom has electronic configuration of 1s2. According to molecular orbital theory, the configuration for He2 molecule is σ1s2, σ1s2
Here, both bonding and antibonding orbitals have 2 electrons each.
Bond order = \(\frac{2-2}{2}\) = 0
∴ Helium molecule does not exist.

Question 32.
Calculate the pH of a solution obtained by mixing 100 mL of 0.10 M CH3COOH and 100 mL of 0.05 M NaOH solution. (Ka for CH3COOH is 1.8 × 10-5). [3]
OR
If 35.0 cm3 of 0.050 M Ba(NO3)2 are mixed with 25.0 cm3 of 0.020 M NaF2 will any BaF2 precipitate ? (Ksp of BaF2 is 1.7 × 10-6 at 298 K).
Answer:
NaOH + CH3COOH → CH3COONa + H2O
100 ml of 0.10 M CH3COOH contains = 100 x 0.1
= 10 m mol CH3COOH 100 ml of 0.05 M
NaOH contains = 100 × 0.05
= 5 m mol of NaOH
5 m mol NaOH react with 5 m mol CH3COOH to form 5 mol of CH3COONa.
5 m mol of CH3COOH remain unreacted. After mixing,
CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions 11

OR

Total volume of solution after mixing
= 35 + 25 = 60 cm3

Cone, of Ba(NO3)2 after mixing
= 0.05 × \(\frac{35}{60}\) = 2.9 × 10-2M

Cone, of NaF after mixing
= 0.02 × \(\frac{25}{60}\) = 8.3 × 10-3 M

Then, ionic product = 2.9 × 10-2 × 8.3 × 10-3
= 2.4 × 10-4

Ksp for BaF2 = 1.7 × 10-6
Since ionic product is greater than Ksp, precipitation of BaF2 will occur.

Question 33.
What is photoelectric effect? Explain it on the basis of quantum theory. [3]
Answer:
Photoelectric effect: It is defined as the emission of electrons from metal surface when radiation strikes it. Electrons emitted in this manner are called photoelectrons.

Photoelectric effect on the basis of quantum theory:
(i) Electrons can only absorb or emit energy in discrete amounts called quanta (packets)
(ii) Energy of each quanta (photon) is directly proportional to frequency of radiation.
E = hv
where E = Energy of photon
h = Planck’s constant
v = Frequency of radiation

Question 34.
Explain briefly the various factors on which ionisation enthalpy depends. [3]
Answer:
(a) Atomic size : With the increase in atomic size, number of electron shell increases. IE decreases with increase in atomic size.
(b) Nuclear charge: As the magnitude of positive charge on the nucleus of an atom increases, the attraction with the electron also increases Therefore, IE increases with increase in magnitude of nuclear charge.
Screening effect : Greater the magnitude of Screening effect, less will be the value of IE.

CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions

Section – D

Question 35.
What is resonance? What are the conditions for writing the resonating structures ? [5]
Answer:
Resonance: When a molecule cannot be represented by a single structure but its characteristic properties can be described by two or more than two canonical structures, then actual structure is said to be a resonance hybrid of these structures. This phenomenon is known as resonance.

Conditions for Writing Resonating Structures:
The following are essential conditions for writing resonating structures:

  • The contributing structures should have same atomic positions. They should differ only in electronic arrangements.
  • These structures should have same number of unpaired electrons.
  • These structures should have nearly same energy.
  • These structures should be written such that negative charge is present on an electronegative atom and positive charge is present on an electropositive atom.
  • In contributing structures, like charges should not reside on adjacent atoms.

Question 36.
(a) What is standard enthalpy of formation ?
(b) Given
N2(g) + 3H2(g) → 2NH3(g); ΔrH° = – 92.4 kj/mol. What is the standard enthalpy of formation of ammonia gas?
(c) Differentiate between intensive property and extensive property.
(d) Enthalpies of formation of CO(g), C02(g), N2O(y), and N2O4(g) are -110, 393, 81, and 9.7 kj/mol respectively.
Find the value of ΔrH for the reaction N2O4(g) + 3CO(g) → N2O(y) + 3CO2(g). [5]
OR
(a) Define Gibb’s energy. Give its mathematical expression. What is Gibb’s energy criteria of spontaneity.
(b) Predict the sign of AS for the following changes:
(i) Freezing of water
(b) C(graphite) → C(diamond)
Answer:
(a) Standard enthalpy of formation (ΔfH°) is enthalpy change accompanying the formation of 1 mole of the substance from its constituent elements in their standard state. ΔfH° can be > 0 or < 0.

(b) Given: N2(g) + 3H2(g) → 2NH3(g); ΔfH° = -92.4 kj mol-1
This is heat evolved for 2 moles for NH3(g)
∴ Heat evolved for 1 mole of NH3(g)
Hence, ΔfH° of NH3 gas = – 46.2 kj mol-1

(c) The properties that depend on the quantity of matter contained in the system are called extensive properties. For example, mass, volume, etc. On the other hand, the properties which depend on the nature of the substance and not on the amount of substance are called intensive properties. For example, viscosity, etc.

(d) N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)
ΔfH = ?
ΔfH = ΣΔfH(product) – ΣΔfH(reactants)
= [ΔfH(N2O) + 3 × ΔfHCO2] – [ΔfH(N2O4) + 3 × ΔfCo]
= [81 + 3 × (-393)] – [9.7 + 3(-110)
=-777.7 kj/mol

OR

(a) Gibbs Energy: It is defined as the maximum amount of energy available to a system, during a process that can be converted into useful work.
Mathematical Expression of Gibbs Energy: Gibbs energy is mathematically expressed as:
G = H – TS
where G = Gibbs energy
H = enthalpy of system
S = entropy of system
T = absolute temperature

Change in free energy (DG) for the system :
DG = DH – TDS
where DH = change in enthalpy
DS = change in entropy

Gibbs Energy Criteria of Spontaneity : For a spontaneous reaction, DG is negative, i.e., DG < 0

(b) (i) Freezing of Water:
Here, entropy decreases (i.e., DS = -ve) because when liquid (water) freezes to form solid (ice), the molecules attain an ordered state.
(ii) C(graphite) → C(diamond)
Here, entropy decreases (i.e., DS = -ve) because when C(graphite) is converted in to C(diamond), it attains more ordered state due to presence of covalent bonding in diamond.

Commonly Made Error:
Students often confuse in expressions/formulas and the sign convention while writing the answer.

Answering Tip :
Understand that the mathematical expression is different than change in free energy.

CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions

Question 37.
(i) Using curved-arrow notation, show the formation of reactive intermediates when the following covalent bonds undergo heterolytic cleavage.
(a) CH3-SCH3,
(b) CH3-CN,
(c) CH3-CU
(ii) Giving justification, categorise the following molecules/ions as nucleophile or electrophile :
HS, BF3, C2H5O, (CH3)3N:, CI+, CH3-C+ = O, H2N, N+2O. [5]
OR
(i) Write resonance structures of CH2=CH-CHO. Indicate relative stability of the contributing structures.
(ii) In which C-C bond of CH3CH2CH2Br, the inductive effect is expected to be the least?
Answer:
CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions 12
(ii) Nucleophiles: HS, C2H5O, (CH3)3N:, H2N:-
These species have unshared pair of electrons, which can be donated and shared with an electrophile. Electrophiles: BF3, Cl+,
CH3-C+ = O, NO2
Reactive sites have only six valence electrons; can accept electron pair from a nucleophile.

Commonly Made Error
While writing nucleophiles, some students forget to show negative ion.

Answering Tip
Students must remember which is giving electron pair and which is accepting electron pair.

OR

(i) Structure I: Most stable, more number of covalent bonds, each carbon and oxygen atom has an octet and no separation of opposite charge.
Structure II, negative charge on more electronegative atom and positive charge on more electropositive atom;
Structure III does not contribute as oxygen has positive charge and carbon has negative charge, hence least stable.
CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions 13

(ii) Magnitude of inductive effect diminishes as the number of intervening bonds increases. Hence, the effect is least in the bond between carbon-3 and hydrogen.
CBSE Sample Papers for Class 11 Chemistry Set 4 with Solutions 14