CBSE Sample Papers for Class 11 Chemistry Set 3 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Chemistry with Solutions Set 3 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Chemistry Set 3 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 I type questions carrying 3 marks each.
• Questions 35-37 in Section D are long-answer II 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.
If the concentration of glucose (C₆H₁₂O₆) in blood is 0.9 g L^-1 what will be the molarity of glucose in blood? [1]
(A) 5 M
(B) 50 M
(C) 0.005 M
(D) 0.5 M
Answer:
(C) 0.005 M

Explanation:
Molecular weight of glucose = 180 g/ mol
So number of moles of glucose
= \(\frac{0.9}{180}\) mol
Now molarity = \(\frac{\text { no. of moles }}{\text { volume of solution in (L) }}\)
= \(\frac{0.005}{1}\) = 0.005 M

Question 2.
Electromagnetic radiations differ from each other in _____________ and wavelength. [1]
(A) frequency
(B) time period
(C) wave velocity
(D) none of these
Answer:
(A) frequency

Explanation:
Electromagnetic radiations differ from each other in frequency and wavelength.

Question 3.
We know that the relationship between K_c and K_p is [1]
K_p = K_c(RT)^∆n
What would be the value of ∆n for the reaction
NH₄Cl(S) → NH₃ (g) + HCl(g)
(A) 1
(B) 0.5
(C) 1.5
(D) 2
Answer:
(B) 0.5

Explanation:
Δn.= (Number of moles of gaseous products) – (Number of moles of gaseous reactants) = 2 – 0 = 2 (1)

Commonly Made Error:
Students sometimes forget that only gaseous components are to be taken into account for calculation.

Answering Tip:
Remember that Δn is calculated only for gaseous products.

Question 4.
For the reaction H₂(g) + I₂ (g) → 2HI(g), the standard free energy is Δ\(\mathrm{G}^{\Theta}\) > 0.
The equilibrium constant (K) would be [1]
(A) K = 0
(B) K > 1
(C) K = 1
(D) K < 1
OR
PCl₅/ PCl₃ and Cl₂ are at equilibrium at 500 K in a closed container and their concentrations are 0.8 × 10^-3 mol L^-1, 1.2 × 10^-3 mol L^-1 and 1.2 × 10^-3 mol L^-1 respectively. The value of K_c for the reaction PCl₅ (g) → PCl₃ (g) + Cl₂(y) will be
(A) 1.8 × 10³ mol L^-1
(B) 1.8 × 10^-3
(C) 1.8 × 10^-3 L mol^-1
(D) 0.55 × 10⁴
Answer:
(B) K > 1

Explanation:
ΔG° = -RT In K. ΔG° > 0 means ΔG° is positive. This is possible only if InK is negative, i.e., K < 1.

OR

(B) 1.8 × 10^-3

Explanation:
K_c = \(\frac{\left[\mathrm{PCl}_3\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}\) = \(\frac{1.2 \times 10^{-3} \times 1.2 \times 10^{-3}}{0.8 \times 10^{-3}}\)
= 1.8 × 10^-3

Question 5.
A hydrocarbon with carbon-carbon single bond is called [1]
(A) alkene
(B) alkane
(C) alkyne
(D) benzene
Answer:
(B) alkane

Explanation:
A hydrocarbon with carbon-carbon single bond is called alkane. It is a saturated hydrocarbon.

Question 6.
The sign of Δ S in the reaction, N₂(g) + O₂(g) → 2NO(g) is [1]
(A) +ve
(B) -ve
(C) Zero
(D) None of these
OR
Which of the following has S = – ve?
(A) Adsorption
(B) Dissolution of NH₄Cl in water
(C) H₂ → 2H
(D) 2NaHCO₃ (s) → Na₂CO₃ + CO₂ + H₂O
Answer:
(C) Zero

Explanation:
It will be spontaneous only if ΔS = +ve.

OR

(A) Adsorption

Explanation:
In case of adsorption, entropy decreases due to attraction between adsorbent and adsorbate.

Read the following text and answer the following questions on the basis of the same:
We can classify the elements into four blocks viz., s-block, p-block, d-block and /-block depending on the type of atomic orbitals that are being filled with electrons. Two exceptions are observed to this categorisation. Strictly, helium belongs to the s-block but its positioning in the p-block along with other group 18 elements is justified because it has a completely filled valence shell (1s²) and as a result, exhibits properties characteristic of other noble gases.

Answer the questions (7) to (10) given below:

Question 7.
Which is the other exception which does not justify its position in periodic table? [1]
(A) Neon
(B) Sodium
(C) Hydrogen
(D) Carbon
Answer:
(C) Hydrogen

Explanation:
Hydrogen is an exception as it is a nonmetal and is placed above group I in the periodic table because it has ns1 electron configuration like the alkali metals.

Question 8.
The electronic configuration of scandium (Z = 21) is [1]
(A) 3d¹4s²
(B) 3d²4s¹
(C) 3s¹4d²
(D) 3s²4d¹
Answer:
(A) 3d¹4s²

Explanation:
The electronic configuration of Scandium (Z-21) is [Ar]3d¹ 4s².

Question 9.
Number of elements in 6th period are [1]
(A) 16
(B) 8
(C) 32
(D) 40
Answer:
(C) 32

Explanation:
6^th period contains 32 elements.

Question 10.
An element belongs to 3rd period and group-13 of the periodic table. Which of the following properties will be shown by the element? [1]
(A) Poor conductor of electricity
(B) Liquid, metallic
(C) Solid, metallic
(D) Solid, non metallic
Answer:
(C) Solid, metallic

Explanation:
This element of 3^rd period and group 13 is Aluminium which is good conductor of electricity and is a solid metal.

Question 11.
Assertion (A): The empirical mass of Ethene is half of its molecular mass.
Reason (R): The empirical formula represents the simplest whole number ratio of various atoms present in a compound. [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) A is true but R is false.
(D) Both A and R are false.
Answer:
(A) Both A and R are correct and R is the correct explanation of A.

Explanation:
As molecular formula gives exact number of atoms present in the compound, so molecular formula is more informative.

Question 12.
Assertion (A): Among the two O-H bonds in H₂O molecule, the energy required to break the first O-H bond and the other O-H bond is the same.
Reason (R): This is because the electronic environment around oxygen is the same even after breakage of one O-H bond. [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) A is true but R is false.
(D) Both A and R are false.
Answer:
(D) Both A and R are false.

Explanation:
In case of H₂O molecule, the enthalpy needed to break the two O-H bonds is not the same. The difference in the ∆_aH° value shows that the second O-H bond undergoes some change because of changed chemical environment.

Question 13.
What is Pauli’s exclusion principle? [1]
OR
What are the four quantum numbers of 19^th electron of copper?
Answer:
According to this principle, “No two electrons in an atom can have the same set of four quantum numbers.

OR

29^Cu = 2, 8, 18, 1
= 1s², 2s²2p⁶, 3s² 3p⁶ 3d¹⁰, 4s¹
19^th electron of copper lies in 4s orbital.
So, n = 4, l = 0, m = 0, s = +½

Question 14.
Second ionisation energy is always more than first ionization energy, why ? [1]
Answer:
Second IE is always more than first IE because positive ion (cation) has more effective nuclear charge which holds the remaining electron more firmly and therefore, a more of energy is required to remove than before the electron from it as compared to a neutral atom.

Question 15.
For an isolated system ΔU = 0, what will be ΔS? [1]
Answer:
Since ΔU = 0
∴ T = 0
As we know that Δs = \(\frac{q_{\text {rev }}}{T}=\frac{q_{\text {rev }}}{0}\)
∴ ΔS is more than zero, (ΔS > 0) i.e., positive.

Question 16.
The boiling point of H₂O is higher than that of H₂S. Why ? [1]
Answer:
Due to intermolecular H-bonding in H₂O, an extra energy in the form of heat is required to break these H-bonds due to which H₂O boils at a higher temperature than H₂S, which does not have H-bonding.

Question 17.
Which quantum number is not obtained from solution of Schrodinger wave equation? [1]
Answer:
Spin quantum number is not obtained from solution of Schrodinger wave equation.

Question 18.
How will you convert: Acetylene to chlorobenzene. [1]
OR
Write IUPAC names of the products obtained by the ozonolysis of 3,4-dimethyl hept-3-ene.
Answer:
Acetylene on heating at very high temperature gets converted to benzene, which on further reaction with chlorine gives chlorobenzene.

OR

The products obtained by the ozonolysis of 3,4-dimethyl hept-3-ene are butanone and pentan- 2-one.

Commonly Made Error:
Students usually do mistake in the ozone addition at the double bond and write the product of the reaction wrongly.

Answering Tips:

• Understand the reaction and how the ozone is added at the double bond.
• Please take care, generally these reactions give ketones, but few alkenes also give aldehydes.

Question 19.
Why -NO₂ is a meta directing group ? [1]
Answer:
-NO₂ group is meta directing because it is an electron withdrawing group. It decreases electron density at ortho- and para-positions thereby directs the incoming group towards meta-position.

Question 20.
Why is the solution of an alkali metal in liquid ammonia appears blue ? [1]
OR
Amongst all alkali metals, why is lithium regarded as most powerful reducing agent in aqueous solution form?
Answer:
Alkali metal dissolve in liquid ammonia to give blue colour solution which conduct electricity.
The ammoniated electrons absorb energy corresponding to a red region of visible light an that is why emits blue color

OR

Lithium is the best reducing agent because it has the lowest reducing potential so has great tendency to lose electrons.
Li^– + e^– → Li(s), E^–_Red = -3.07V.

Section – B

Question 21.
Why are electron gain enthalpies of Be and Mg are positive ? [2]
OR
Write down the electronic configuration of Fe³⁺ and Ni²⁺. How many unpaired electrons are present in their outermost orbit? (Given, Atomic number, Fe = 26, Ni = 28)
Answer:
(a) They have fully filled s-orbital and hence no tendency to accept an additional electron.
(b) That’s why energy is needed if an extra electron is added.
Therefore, electron gain enthalpies of be and Mg are positive

OR

Electronic configuration of Fe³⁺
= 2, 8, 13
= 1s², 2s²2p⁶, 3S²3p⁶ 3d⁵
Number of unpaired electrons in Fe³⁺ = 5
Electronic configuration of Ni²⁺
= 2, 8, 16
= 1s², 2s²2p⁶, 3s²3p⁶3d⁵
Number of unpaired electrons in Ni²⁺ = 2

Commonly Made Error:
Students do mistakes while writing electronic configurations.

Answering Tip:
Take care that for anions, we need to add the electrons to the atomic number and for cations, we need to subtract the electrons from the atomic number of that element.

Question 22.
Define black body and black body radiations. [2]
Answer:
Black body: The ideal body, which emits and absorbs radiations of all frequencies, is called a black body.

Black body radiation: The radiation emitted by such a body is called black body radiation.

Question 23.
(i) Draw Lewis structure of : [2]
(a) CO₃^2- ions
(b) NH₄⁺ ions.
(ii) Why H₂SO₄ has an exceptional Lewis structure ?
Answer:
(i) Lewis dot structure for:
(a) CO₃^2-

(b) NH₄⁺

(ii) H₂SO₄

Because it has more than 8 valence electrons around the central atom, i.e., 12. Thus, exceeds the octet rule. It is an exception of Lewis structure.

Commonly Made Error:
Generally, students do mistake while writing Lewis structure.

Answering Tips:

• Understand the concept behind writing Lewis structure.
• Make note of valence electrons and analyze how sharing of electrons has taken place in the given compound and then write structures.
• Do not miss out dots.

Question 24.
Density of water at room temperature is 1.0 g/cc. How many molecules are there in one drop of water if its volume is 0.1 cc ? [2]
Answer:
Mass of 0.1 cc water = 1 × 0.1 = 0.1 gram
= \(\frac{0.1}{18}\) mol.
Thus, number of molecules in one drop of water
= \(\frac{0.1}{18}\) × 6.023 × 10²³s
= 0.033 × 10²³ molecules

Question 25.
Indicate the σ and π bonds in the following molecules: [2]
(i) C₆H₆,
(ii) C(,H12
OR
What are hybridisation states of each carbon atom in the following compounds?
(i) (CH₃)₂CO,
(ii) CH₂=CHCN
Answer:
(i)

There are six C – C sigma bonds, six C-H sigma bonds, and three C=C pi resonating bonds in the given compound.

(ii)

There are 12 C-H and 6 C-C bonds in the given compound.

Commonly Made Error:
Students get confuse with sigma and pi bonds and do mistakes.

Answering Tip:
Generally single bonds are sigma bonds, double bonds contain both sigma and pi bonds.

OR

(i)

(ii)

Question 26.
Draw cis – and trans – isomers of the following compounds. Also write their IUPAC names 2
(i) CHCl = CHCl
(ii) C₂H₅C(CH₃) = C(CH₃)C₂H₅
Answer:
(i)

(ii)

Commonly Made Errors:

• Some students get confused in cis and trans forms.
• Some students do not write IUPAC name correctly.
• Some students make mistake in writing the position of double bond.

Answering Tips:

• Be sure that placing groups to same side of the double bond is cis form and placing groups to opposite sides of the double bond is trans form.
• Number the compound properly by keeping double bond in mind and name the compound correctly.

Question 27.
Identify the oxidant and reductant in the reaction:
I₂(aq) + 2S₂O₃^2-(aq) → 2I^–(aq) + S₄O₆^2-(aq) [2]
Answer:
I₂ (aq) + 2S₂O₃^2-(aq) → 2I^– (aq) + S₄O₆^2-(aq)
For the identification of oxidant and reductant, the oxidation numbers of I and S in each species are calculated as :
Oxidation number of I in I₂ = 0
Oxidation number of S in S₂O₃^2-
= 2(x) + 3(-2)
= – 2
2x – 6 = -2
2x = – 2 + 6 = 4
x = \(\frac{4}{2}\) = +2
Oxidation number of I in I^– = -1
Oxidation number of S in S₄O₆^2-
= 4(x) + 6(-2) = -2
4x – 12 = – 2
4x = – 2 + 12 = 10
x = \(\frac{10}{4}\) = + 2.5
Now,

Since, in this reaction, the oxidation number of I decreases from I₂ to I^–. So, it acts as oxidant which oxidises S₂O₃^2-. Whereas, the oxidation number of S increases from S₂O₃^2- to S₄O₆^2-. So, it acts as reductant which reduces I₂.
Oxidant = I₂
Reductant = S₂O₃2^–

Commonly Made Error:
Students often write definition in their own word missing out on one part or the other.

Answering Tip:
Definition is not complete with just one statement. Second statement is equally important.

Section – C

Question 28.
If a man submits to a diet of 9500 kj per day and expands energy in all forms to a total of 12000 kj per day. What is the change in internal energy per day ? If the energy lost was stored as sucrose (1632 kj per 100 g), how many days should it take to lose 1 kg ? (Ignore water loss for this problem). [3]
Answer:
AE = 12000 – 9500 = 2500 kj
100 g of sucrose has energy = 1632 kJ
∴ 1000 g of sucrose has energy
=1632 × 10 = 16320 kj
Now 2500 kj of energy is lost in one day.
∴ 16320 kj of energy is lost in
= \(\frac{1}{2500}\) × 16320
= 6.5 days.

Question 29.
A photon of wave length 4 × 10^-7m strikes on metal surface, work function of the metal being 2.13 eV.
Calculate :
(i) The energy of the photon (eV).
(ii) The kinetic energy of the emission (1 eV = 1.6020 × 10^-19J). [3]
Answer:
(i)

Energy of the photon in eV = 3.10

(ii) Kinetic energy of the emission of photoelectrons (K. E)
= hv – ω₀ (where ω₀ = work function)
= 3.10 eV – 2.13 eV = 0.97 eV

Question 30.
In sulphur estimation, 0.157 g, an organic compound gave 0.4813 g of barium sulphate. What is the percentage of sulphur in the compound ?
[At. mass of (A) Barium = 137 u (B) Sulphur = 32 u (C) Oxygen = 16 u] 3
OR
Two oxides of a metal contain 27.6% and 30.0% of oxygen respectively. If the formula of the first oxide is M₃O₄, find that of the second.
Answer:
Mass of organic compound = 0.157 g
Mass of BaSO₄ = 0.4813 g
Molecular mass of BaSO₄ = 137 + 32 + 64 = 233 u
As one molecule BaS04 contains 1 sulphur that is 32 g sulphur
Percentage of Sulphur
= \(\frac{32}{233} \times \frac{\text { Mass of } \mathrm{BaSO}_4}{\text { Mass of organic compound }} \times 100\)
= \(\frac{32}{233} \times \frac{0.4813}{0.157} \times 100\)
= 42.10%

OR

In the first oxide,
% of oxygen = 27.6
% of metal = 100 – 27.6 = 72.4 parts by mass
Formula of first oxide = M₃O₄
So, 72.4 parts by mass of metal = 3 atoms of metal and
4 atoms of oxygen = 27.6 parts by mass of oxygen
In the second oxide,
Oxygen = 30.0 parts by mass
Metal = 100 – 30 = 70 parts by mass
But 72.4 parts by mass of metal = 3 atoms of metal
∴ 70 parts by mass of metal = \(\frac{3}{72.4}\) × 70
= 2.90 atoms of metal
276 parts by mass oxygen = 4 atoms oxygen
∴ 30 parts by mass of oxygen = \(\frac{4}{27.6}\) × 30
= 4.35 atoms of oxygen (1)
Ratio of M : O in the second oxide
= 2.90 : 4.35 = 1 :1.5 or 2 : 3
∴ Formula of second oxide = M₂O₃

Commonly Made Error:
Students do mistake in finding part by mass value and get confused while writing the values for the M:0 ratio.

Answering Tip:
Understand the problem correctly and note down the given values. Keeping in mind the formula of first oxide carefully find the part by mass value and M:0 ratio.

Question 31.
Give reason :
(i) He₂ molecule is not formed (on the basis of MO theory).
(ii) Water is a liquid and hydrogen sulphide is a gas though O and S belong to the same group.
(iii) HF is polar though it has covalent bond. [3]
Answer:
(i) Helium atom has electronic configuration of 1s². According to molecular orbital theory, the configuration for He₂ molecule is σ1s², σ*1S²
Here, both bonding and antibonding orbitals have 2 electrons each.
∴ Bond order = \(\frac{2-2}{2}\) = 0
∴ Helium molecule does not exist.

(ii) Oxygen (O) and sulphur (S) belong to the same group and forms hydrides H₂S and H₂O. Oxygen being more electronegative forms hydrogen bonding with hydrogen atom of another water molecule. While sulphur can’t form hydrogen bonds. Hence, boiling point of water increases and it is liquid at room temperature.

(iii) In hydrogen fluoride, HF, hydrogen atom forms a weak bond, i.e., hydrogen bond with fluorine atom of the other molecule while hydrogen is covalently bonded to its fluorine atom, this may be shown as :
H^δ+ ……….. F^δ- ………….. H^δ+ -F^δ- ………….. H^δ+ – F^δ- ……………..
In other words, hydrogen atom acts as a bridge between two atoms, holding one atom by a covalent bond and the other atom by a hydrogen bond. Due to hydrogen bonding, H-atom acquires partial positive charge (^δ+) and F-atom acquires partial negative charge (^δ-), so it is polar in nature.

Question 32.
(i) Lil is covalent in nature. Why ?
(ii) Li is kept wrapped in paraffin wax and not stored in kerosene. Why ?
(iii) Why BeCl₂ in aqueous solution exists as [Be(H₂O)₄]²⁺ ? [3]
OR
(i) Why alkaline earth metals, cannot be obtained by reduction of their oxides ?
(ii) Why the elements of group 2 are known as alkaline earth metals?
Answer:
(i) Because of the small size of cation and higher polarising power, Lil is covalent in nature lithium halides except LiF are predominantly covalent.
(ii) Lithium cannot be stored in kerosene as it floats on the surface of kerosene because of its low density. Therefore, lithium is generally kept wrapped in paraffin wax.
(iii) Due to small size and high ionisation enthalpy, Be forms coordination compound.
BeCl₂ + H₂O → [Be(H₂O)₄]²⁺ + 2Cl^– s (i)

OR

(i) The alkaline earth metals cannot be obtained by reduction of their oxides because the enthalpies of formation of these oxides are quite high and consequently they are very stable to heat.
(ii) The elements of group 2 are known as alkaline earth metals because their oxides and hydroxides are alkaline in nature and these metal oxides are found in the earth’s crust.

Question 33.
Explain why alkyl groups act as electron donors when attached to a n system. [3]
Answer:
When an alkyl group is attached to a n system, it acts as an electron-donor group by the process of hyperconjugation. To understand this concept better, let us take the example of propene. In hyperconjugation, the sigma electrons of the C-H bond of an alkyl group are delocalised which is attached to an unsaturated system. This results in delocalisation due to partial overlap of a sp3-s sigma bond orbital with an empty p-orbital of the pi bond of adjacent carbon atom. This kind of overlap leads to delocalisation of the pi electrons, making it more stable.

Commonly Made Error:
Students explain this answer with out example and do mistakes while writing hyper conjugation structures.

Answering Tip:
Always explain reason with examples. Understand the concept of hyperconjugation and avoid mistakes while writing structures give ‘+’ and ‘-‘ signs properly.

Question 34.
If water vapour is assumed to be a perfect gas, molar enthalpy change for vaporisation of 1 mol of water at 1 bar and 100°C is 41 kj mol^-1. Calculate the internal energy change, when:
(i) 1 mol of water is vaporized at 1 bar pressure and 100°C.
(ii) 1 mol of water is converted into ice. [1]
OR
50 mL of a liquid is taken in an insulated container at a pressure of 1 bar. The pressure is steeply increased to 100 bar. The volume of the liquid is lowered by 1 mL at this constant pressure. Determine AH and AU.
Answer:
(i) H₂O(1) → H₂O(g)
Δn_(g) = 1 – 0 = 1
∵ ΔH = ΔU + Δn_gRT
or ΔU = ΔH – Δn_gRT
Given, ΔH = 41 kj mol^-1,
T = 100 + 273 = 373 K
∴ ΔU = 41 kj mol^-1 – 1 × 8.3 × 10^-3 mol^-1 K^-1 × 373 K
= 37.904 kj mol^-1

(ii) H₂O(l) → H₂O(S)
There is negligible change in volume.
So, pΔV = Δn_g RT = 0
∴ ΔH = ΔU
or ΔU = 41 kj mol^-1

OR

Given, V₁ = 50 mL
P₁ = 1 bar
p₂ = 100 bar
V₂ = 49 mL
For adiabatic process q = 0
According to first law of thermodynamics,
ΔU = q + W
ΔU = 0 + W
∴ ΔU = W
ΔU = W = – pΔV = – 100(49 – 50)
= -100(- 1)
= 100 bar mL (1)

∴ ΔH = ΔU + Δ_pV
= ΔU + p₂V₂ – P₁V₁
= 100 + (100 × 49) – (1 × 50)
= 100 + 4900 – 50
= 5000 – 50 = 4950 bar mL

Section – D

Question 35.
At 450 K, K_P = 2.0 × 10¹⁰/bar for the given reaction at equilibrium.
2SO₂(g) + O₂(g) → 2SO₃(g) + 189.4 kj
(a) What is K_c at this temperature?
(b) What is the value of K_c for the reverse reaction at the same temperature?
(c) What would be the effect on equilibrium if :
(i) more SO₂ is added ?
(H) pressure is increased ?
(iii) the temperature is increased ? [5]
OR
The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each of them.
Answer:
(a) 2SO₂ (g) + O₂ (g) → 2SO₃ (g)
For the given reaction
∆n_g = n_p – n_r = 2 – 3 = -1
K_p = K_c (RT)^∆n
K_p = 2 × 10¹⁰ bar^-1
K_c = \(\frac{\mathrm{K}_p}{(\mathrm{RT})^{\Delta n}}\)
= 2 × 10¹⁰ bar^-1 (0.0831 L bar K^-1 mol¹) × (450 K)
= 7.48 × 10¹¹ moL^-1

(b) For reverse reaction,
Kc = \(\frac{1}{\mathrm{~K}_c(\text { forward reaction) }}\)
= \(\frac{1}{7.48 \times 10^{11}}\)
= 1.34 × 10^-12 L mol^-1

(c) (i) If more SO₂ is added, rate of forward reaction increases and more SO₃ will be formed.
(ii) If more pressure is increased, the reaction will shift in forward direction, i.e., towards lesser number of moles.
(iii) Increase in temperature will favour backward reaction.

OR

For milk –
pH = 6.8
∵ pH = – log[H⁺]
6.8 = – log [H⁺]
or log [H⁺] = – 6.8 = 7̄.20
[H⁺] = antilog 7̄.20
= 1.585 × 10^-7 M
For black coffee –
pH = 5.0
∵ PH = – log [H⁺]
5.0 = -log [H⁺]
or log [H⁺] = – 5.0
[H⁺] = 10^-5 M
For tomato juice –
pH = 4.2
∵ pH = – log [H⁺]
4.2 = – log [H⁺]
or log [H⁺] = – 4.2 = 5̄.80
[H⁺] = antilog 5̄.80
= 6.310 × 10^-5 M (1)
For lemon juice
pH = 2.2
∵ PH = – log [H⁺]
2.2= – log [H⁺]
or log [H⁺] = – 2.2 = 3̄.80
[H⁺]= antilog 3̄.80
= 6.310 × 10^-3 M
For egg white –
pH = 7.8
∵ PH = – log [H⁺]
7.8 = – log [H⁺]
or log [H⁺] = – 7.8 = 8̄.20
[H⁺] = antilog 8̄.20
= 1.585 × 10^-8 M

Question 36.
What is meant by hybridisation? Compound CH₂ = C = CH₂ contains sp or sp² hybridized carbon atoms. Will it be a planar molecule? [5]
Answer:
The atomic orbitals combine to form new set of equivalent orbitals known as hybrid orbitals. Unlike pure orbitals, the hybrid orbitals are used in bond formation. The phenomenon is known as hybridization which can be defined as the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of new set of orbitals of equivalent energies and shape. No. It is not a planar molecule.

Central carbon atom is sp hybridised and its two unhybridised p-orbitals are perpendicular to each other. The p-orbitals in one plane overlap with one of the p-orbital of left terminal carbon atom and the p-orbital in other plane overlaps with p-orbital of right side terminal carbon atom. This fixed the position of two terminal carbon atoms and the hydrogen atoms attached to them in planes perpendicular to each other. Due to this the pair of hydrogen atoms attached to terminal carbon atoms are present in different planes.

Commonly Made Error:
Sometimes students get confused in carbon . atom hybridisation.

Answering Tip:
Always give a brief explanation of how hybrid orbitals are formed along with the definition.

Question 37.
Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yiekis 1-bromopropane. Explain giving proper reasons and give proper step by step mechanism for both the reactions. [5]
OR
(i) Arrange the following set of compounds in the order of their decreasing relative reactivity with an electrophile. Give reason.

(ii) Give one important structural difference between cyclohexanol and phenol.
Answer:
Addition of HBr to propene (unsymmetrical alkene) follows Markovnikov’s rule according to which the negative part of the addendum gets attached to that C atom which possesses lesser number of hydrogen atoms.

Mechanism:
Hydrogen bromide provides an electrophile, H⁺, which attacks the double bond to form carbocation as:

Secondary carbocations are more stable than primary carbocations. Therefore, the former predominates as it will form at a faster rate. Thus, in the next step, Br” attacks the carbocation to form 2-bromopropane as the major product.

Addition of HBr to unsymmetrical alkenes like propene in the presence of light or peroxide takes place contrary to the Markovnikov’s rule. This so happens only with HBr but not with HC1 and HI. This addition of HBr to propene in the presence of benzoyl peroxide follows anti-Markovnikov’s rule or peroxide effect or Kharasch effect.

Secondary free radicals are more stable than primary radicals. Therefore, the former predominates since it forms at a faster rate. Thus, 1-bromopropane is obtained as the major product.

Commonly Made Errors:

• Some students write the reaction, but forget to explain the mechanism.
• Some students forget to explain the attack of Br- on carbocation.

Answering Tip
Students must explain addition using Markovnikov’s rule as well as anti – I Markovriikov’s rule.

OR

(i) The methoxy group (-OCH₃) is electron releasing group. It increases the electron density in benzene nucleus due to resonance effect (+R-effect). Hence, it makes anisole more reactive than benzene towards the electrophile.
In case of alkyl halides, the electron density para positions due to +R

However, the halogen atom also withdraws electrons from the ring because of its -I effect. Since the -I effect is stronger than the +R effect, the halogens are moderately deactivating. Thus, overall electron density on benzene ring decreases, which makes further substitution difficult. -N02 group is electron withdrawing group. It decreases the electron density in benzene nucleus due to its strong -R-effect and strong -I-effect. Hence, it makes nitrobenzene less reactive.
Therefore, overall reactivity of these three compounds towards electrophiles decreases in the following order:
Anisole > Chlorobenzene > Nitrobenzene

Commonly Made Error:
Some students got confused with methoxy and methyl group and make mistake while writing names.

Answering Tips:
The basic nature of substituted group must be discussed.
Write your answer based on inductive and resonance effect.
Show diagrammatically resonance structures to justify your answer.

(ii) In phenol, benzene ring has alternate single and double bonds while cyclohexanol is alicyclic compound.