Chemistry - Paper I - For Class XI (Science Group) - Model papers With Solved MCQs 2020 -2021

Go To Index
Chemistry - Paper I
For Class XI (Science Group)
Model papers 2020 -2021

Section 'B' (Short Answer Questions)

Note: Attempt any five part questions. (Marks = 25)
Q.2 (i): 1.367g of an organic compound containing C, H and 0 was combusted in a stream of air yield 3.002g CO₂ and 1.64g H₂O. what is the empirical formula.
Solution:
The masses of CO₂ and H₂O are used to find out the masses of carbon and hydrogen:

Thus the empirical formula of the compound = C₃H₈O

(ii) Define the following:

1. Significant figure
2. System
3. Viscosity
4. Gay-Lussac Law 

Ans: DEFINITIONS:
1. SIGNIFICANT FIGURE:
Those reliable digits in a number which are known with certainty in which last digit remain uncertain with + 1.
All Non-zero numbers are significant.
Example:

• 324 have 3 significant.
• Zero may be or may not be significant.
• 137 has 3 significant
• 0.0034 has 2 significant
• 200 has 1 significant
• 2.000 has 4 significant
• 32016 has 5 significant

2. SYSTEM:
Any real or imaginary portion of the universe which is under consideration is called system. There are three types of systems:

• (a) Open System:
  The system in which, mass as well as energy can either enter into or leave the system is known as an open system.
• (b) Closed System:
  The system in which, exchange of energy in between system and surroundings is possible but mass can neither enter into nor leave the system, is called a closed system.
• (c) Isolated System:
  The system in which no exchange of either mass or energy in between the system and surrounding is possible is called an isolated system.

3. VISCOSITY:
"Viscosity of a liquid is the measure of its resistance to flow. It comes into being due to internal frictional force among molecules." It is represented by η (eta).
The instrument used to measure viscosity is called OSTWALD VISCOMETER.
Factors:
It depends on:

• Size of molecules
• Shape of molecules
• Intermolecular attraction and
• Temperature

Units Of Viscosity:

• Poise (1 poise = 1 g/cm)
• Centipoise
• Millippoise
• N.sec/m²

4. GAY- LUSSAC'S LAW OF COMBINING VOLUME:
According to this law
"Gases react in the ratio of small whole number by volume, measured under same condition of temperature and pressure".
For Example in:

CH_4(g) + 2O_2(g) ⟶ CO_2(g)+ 2HO_2(g)

One volume of CH₄ reacts with two volumes of O₂ to form one volume of CO₂ and two volumes of H₂O_(g).
As at S.T.P. one mole of a gas has a volume of 22.4 dm³, we can say that, "22.4 dm³ of CH₄ reacts with 44.8 dm³ O₂ to from 22.4 dm³ CO₂ and 44.8 dm³ of H₂O.
Thus we can find out the volume of any gas.


(iii) State Boyle's Law, Charles's law" and prove them in term of Kinetic Molecular Theory.

B0YLE'S LAW:
Introduction:
Gases have the property of compression and expansion. This behaviour is summarised in Boyle's law. Robert Boyle in 1662 published the results of his experiments on the variation of volume with pressure for various gases.
Statement:

"The volume of a given mass of a gas is inversely proportional to its pressure provided temperature is kept constant".

Mathematical Form:

V ∝ ¹/_P (at constant 'T')

Or V = K . ¹/_P

Or PV = K

Or PV = constant ....... (i)

"At constant temperature, the product of "P" and "V" of a certain mass of a gas is always constant".

For initial state of gas P₁V₁ = K

For final state of gas P₂V₂ = K

∴ P₁V₁ = P₂V₂ .............. (ii)

Explanation:
At constant temperature if the pressure of gas is doubled, its volume decreases to one-half and if the pressure is increased to four times, the volume is reduced to one-fourth of original volume. Similarly if the pressure on the gas is reduced to half its volume becomes double.

Proof of Boyle's Law On The Basis Of Kinetic Molecular Theory:
According to kinetic molecular theory of gases, there is large distance between the molecules of gases, Hence at constant temperature when pressure of gas is increased, the molecules come closer, and as a result volume decreases. Conversely we can say, when volume is decreased, the molecules now collide more frequently with the walls of container. Hence pressure is increased.

CHARLES'S LAW:
Introduction:
Gases expand on heating and contract on cooling under constant pressure. Charles', a French Physicist in 1787 observed the relationship between volume of a given mass of gas and temperature at constant pressure.
Statement:

"At constant pressure the volume of given mass of a gas is directly proportional to the absolute temperature OR Kelvin temperature of the gas at constant pressure."

Explanation:
Charles found that all gases expand or contract by same amount when heated or cooled by same amount of temperature or in Kelvin. He calculated that for every degree rise or fall of temperature, the increase or decrease of volume is ¹/₂₇₃ is called FRACTIONAL CHANGE IN VOLUME.
Mathematical form:

V ∝ T (at constant 'P')

OR V = KT

OR ^V/_T = K

The volume at constant `T' depends upon the quantity of gas and its pressure.

For initial state of gas ^V₁/_T1 = K

For final state of gas. ^V₂/_T2 = K

∴ ^V₁/_T1 = ^V₂/_T2

V₁ x T₂ = V₂ x T₁

Proof Charles' Law On The Basis Of Kinetic Molecular Theory:
According to Charles' law the volume of a given mass of a gas is directly proportional to its absolute temperature at constant pressure. According to kinetic molecular theory the average kinetic energy of gas molecules is directly proportional to its absolute temperature, so if the temperature of the gas is increased the average kinetic energy of the gas molecules also increases, due to which the sample of the gas expands to keep the pressure constant as a result the volume of gas increases.

(iv) Write down the electronic configuration for ground states of each of the following.

• Cl (Z = 17)
• Ca⁺² (Z = 20)
• Fe (Z = 26)
• N^-3 (Z = 7)

Ans: Electronic Configuration For Ground States Of Elements:
(i) Cl (Z = 17)
The electronic configuration of the neutral chlorine atom is: 1s² 2s² 2p⁶ 3s² 3p⁵

(ii) Ca⁺² (Z = 20)
A calcium 2+ ion has lost its two valence electrons, and now has 18 electrons.
Thus, the electronic configuration of a Ca⁺² ion is : 1s² 2s² 2p⁶ 3s² 3p⁶

(iii) N^-3 (Z = 7)
A Nitrogen 3- ion has gained three valence electrons, and now has 10 electrons.
Thus, the electronic configuration of a N^-3 (Z = 7) ion is: 1s² 2s² 2p⁶

(iv) Fe (Z = 26)
The electronic configuration of neutral Fe atom is: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s²

(v) Differentiate between the following (write only two differences for each).

• Sigma and Pi bond
• Hydration and Hydrolysis

Ans: Difference Between Sigma and Pi Bond

S.NO.	Sigma Bond	Pi Bond
1.	It is formed by the head to head (end to end) overlapping of half- filled atomic orbitals (i.e. s-s, s-p and p-p) along their nuclear axis	It is formed by the side wise (parallel) overlapping of half- filled p-orbitals.
2.	Sigma bond is considered to be strong i.e. more stable and less reactive because of maximum overlapping of orbitals.	Pi - bond is considered to be weak i.e. less stable and more reactive because of lesser overlapping or orbitals.
3.	Bond energy is high.	Bond energy is low.
4.	Sigma bond can form between s-orbitals.	S-orbitals do not take part in making pi -bond.
5.	It does not restrict the rotation of atom along their bond axis.	They restrict the rotation of atom along their bond axis.
6.	Only one σ bond is formed between the two atoms.	More than l π bonds are formed between the two atoms.
7.	Only one sigma bond exist between two atoms.	One on two pi - bond can be formed between two σ-bonded atoms.
8.	Sigma electrons in a sigma bond are localized.	Pi electrons in a pi-bond may be localized or delocalized.
9.	All single bonds are sigma bonds.	A double bond consists of one sigma and one pi -bonds, a triple bond consists of one sigma and two pi -bonds.
10.	It has symmetrical electron cloud density around the bond axis.	It has unsymmetrical electron cloud density around the bond axis.

(ii) Difference Between Hydration and Hydrolysis

S.NO.	HYDROLYSIS	HYDRATION
1.	Hydrolysis is a chemical process	Hydration is a physical process.
2.	When salt is dissolved in water and pH of water changes is called Hydrolysis. NH4Cl + H2O ⟶ NH4OH + HCl	When salt is dissolved in water and pH of water does not change is called Hydration. eg. CuSO4. 5H2O FeSO4.7H2O
3.	It is a process in which H-OH bond breaks and new is formed.	It is a process in which water molecules are added to a substance without the breaking of H-OH bond.
4.	Acid and base are formed	No new substance is formed.
5.	It does not depend upon charge density.	Depends upon charge density.
6.	Water dissociate into H+ and OH- ions.	Water does not dissociate into ions.
7.	Products are formed	Mixture is formed.

(Note: write only two differences for each, as mentioned in question.)

(vi) The ratio of rates of diffusion of two gasses A and B is 1.5:1. If relative molecular mass of gas A is 16, find out the relative molecular mass of gas B.
Solution:
Data:
Mass Of Gas A = 16
Mass of gas B = M = ?
Rate of Diffusion Of gas A = 1.5
Rate of Diffusion Of gas B: 1

Formula:

(vii) State First Law of Thermodynamic. In a certain process, 500 J of work is done on a system which gives off 200 J of heat. What is the value of change in Internal energy for the process.
Ans: First Law of Thermodynamic:
This law was given by Helmholtz in 1847. According to this law:
Statement:

" Energy can neither be created nor destroyed, but it can be changed from one form to another."

In other words,

"The total energy of a system and its surrounding must remain constant and sum of all energies is zero."

OR

"The amount of heat absorbed by a system is equal to the sum of change in internal energy and work done."

Mathematically:

• Suppose a system whose internal energy in the initial state is "E₁"
• Let the quantity of heat absorbed by the system from the surrounding is "q"
• Heat does some work on the surrounding is "w"
• Due to heat and work done, the internal energy change to E₂
• The change in internal energy is ΔE

According to law:

q = E₂ - E₁ + w

Or q = ΔE + w

Or ΔE = q - w

ΔE depends only on the initial and final state of the system.

Numerical:
Solution:
Data:
Change in internal energy ΔE =?
Work done = w = - 500 J
Heat (give off by the system) q = - 200 J
(Hint: The reaction is exothermic because heat is give off so the value of w and q are negative)

Formula:
ΔE = q - w
ΔE = (-200) - (- 500)
ΔE = -200 + 500
ΔE = 300 J Ans

(viii) Explain the effects of surface area and concentration of reactant on the rate of reaction.
Ans: FACTORS AFFECTING RATE OF REACTION:-
(i) Effect Of Concentration Of Reactants On The Rate Of Reaction:-
Rate of reaction varies with the concentration of reactant because according to the "Law of Mass Action" the rate of a chemical reaction is directly proportional to the multiplication of concentrations of reactants.

A + B ⟶ C + D

Rate ∝ [A][B]

Rate = K [A][B]

It means, by increasing the concentrations of reactants, the collision frequency increases, so the number of effective collisions per second increases, as a result more product is formed. The greater the concentration of reactant the higher will be the product.

(ii) Effect Of Surface Area of Reactants On The Rate Of Reaction :-
The physical state of reacting substances is very important in determining their reactivity. The surface area of the reactant plays an important role with respect to rate of reaction. Greater the surface area, the higher is the rate of reaction.
Example:-
(i) Reaction between a piece of marble (CaCO₃) and an acid is slow.

CaCO₃ + HCl ⟶ CaCl₂ + H₂O + CO₂

But finally didvided marble reacts vigorously because the powdered marble offer greater surface area for HCl to act upon.

(ii) Powdered zinc reacts more rapidly with water than the pieces of zinc because powdered zinc provided more surface area for the reaction, as a result more the collision frequency increases, so the number of effective collisions increases and more product will be formed.

(iii) Amorphous boron is much more reactive than crystalline boron, because the amorphous boron offer greater surface area or the reaction.

(iv) In case of liquid, the greater surface area, increases their rate of evaporation. For this reason spilled milk or a liquid evaporated faster than present in a glass or cup.

(Give any one or two examples with definition)

Hence we can say that if the surface area of the reactant increases then the rate of reaction also increases.

(ix) Define Dipole moment. Why dipole moment of CO₂ and CCI₄ is zero.
Ans: DIPOLE MOMENT:
The tendency of a polar molecule to be oriented under the influence of electric field is called Dipole moment.

OR

It measures the net molecular polarity.

OR

The polarity of a molecule i.e. magnitude of positive or negative poles of a molecule expressed quantitatively by the term known as Dipole Moment. A diatomic polar molecule has two oppositely charged ends or poles. The extent or tendency of a polar molecule to turn or orient in an electric field is called Dipole Moment.

OR

The product of magnitude of charge at each pole and the inter-nuclear distance between two opposite charges is called dipole moment.

MATHEMATICAL REPRESENTATION:
The product of the electric charge (q) and the distance between the positive and negative centers (r) is called Dipole moment. Dipole moment is represented by (μ) Mathematically.

Dipole moment = Charge x distance between two opposite charge

Or μ = Q x d

Or μ = e x d

Where,
μ = Dipole moment
e = Charge
d = Distance between two opposite charge

The dipole moment is a vector quantity. The direction of the polarity of a covalent bond is indicated by using a crossed tail arrow (|⟶) pointing towards negative end of the dipole.

UNIT:
The S.I. unit for dipole moment is Coulomb.meter (C.m)
It is commonly expressed in Debye (D).
Debye is related to S.I. unit C - m as follow:
1D = 3.34 x 10^{- 30} C.m

Dipole Moment In CO₂:
Structure Of CO₂ (Triatomic Linear molecule)
The molecular structure of CO₂ is linear structure in which angle between carbon and oxygen is 180°.

O = C = O

The dipole moment for each C = O bond is 2.3 debye, but due to the linear structure both the dipole moments of C = O cancel each other and net dipole moment of the molecule is zero.

Dipole Moment In CCl₄:
Structure Of CCl₄ (Polyatomic Symmetrical molecule)
The polyatomic symmetrical molecules have μ = 0.
CCl₄ contains four C — Cl polar bonds which are oriented in opposite direction, thereby cancelling out each other. Thus CCl₄ has zero dipole moment due to its symmetrical structure.

(x) Predict the effect of increase in temperature and pressure on the following system at equilibrium state (only predict the direction)

• N₂ + 3H₂ ⇌ 2NH₃ + Heat
• N₂ + O₂ + Heat ⇌ 2NO

Ans: Effect Of Increase In Temperature And Pressure:

(i) N₂ + 3H₂ ⇌ 2NH₃ + Heat

Formation of Ammonia is an exothermic reaction.
(a) Effect Of Increase In Temperature:
Direction: Equilibrium Shifts towards left i.e more N₂ + 3H₂ are present in equilibrium mixture.
Reason:
Synthesis of ammonia is exothermic reaction, that is heat is liberated in the forward reaction. The low temperature favours the formation of ammonia. If at equilibrium state, temperature is increased, according to Le-Chatelier's principle, the equilibrium shifts to left (in opposite direction). Thus high temperature, at equilibrium favour the reverse direction and reduce the formation of ammonia.

(b) Effect Of Increase In Pressure
Direction: Equilibrium Shifts towards right to give high yield of products (NH₃).
Reason:
The formation of ammonia proceed with the reduction in volume (4 volume on the reactant side and 2 volume on the product side). The reaction, is therefore, carried out under high pressure, the equilibrium is forced to right as the formation of ammonia lower the volume according to Le-Chatelier's principle, relieves the pressure.

(ii) N₂ + O₂ + Heat ⇌ 2NO

Formation of NO is an endothermic reaction.
(a) Effect Of Increase In Temperature:
Direction: Equilibrium moves towards right to give high yield of products (NO).

(b) Effect Of Increase In Pressure
Direction: No effect
Reason:
Volume (no. of moles) of reactant is equal to volume of product.

SECTION "C"
(DETAILED-ANSWER QUESTIONS) (Max Marks: 17)

Note: Attempt any one questions from this section.
Q.3 (a): Write the postulates of Bohr's atomic. Derive the formula for the radius of nth orbit of hydrogen atom by using Bohr's atomic model. (9)
Ans: BOHR'S THEORY:
I Rutherford's model of atom fails to explain the stability of atom and appearance of the line spectra. A Danish physicist Neils Bohr in 1913 was the first to present a simple theory of the atom based on the "Quantum Theory", which explained the appearance of line spectra.

Basic Postulates Of Bohr's Theory:
The postulates of Bohr's a theory are given below:
i) An atom has a number of stable orbits or stationary states in which an electron can reside without emission or absorption of energy.
ii) An electron may pass from one of these non-radiating states to another of lower energy with the emission of radiations whose energy equals the energy difference between the initial and final states

 ΔE = E₂ - E₁.

iii) In any of these states the electrons move in a circular path around the nucleus.
iv) The motion of the electrons in these states is governed by the ordinary laws of mechanics and electrostatic, which provides its angular momentum is an integral multiple of h/π.
It can be written as:

mvr = n ^h / _2π

Here mvr is the angular momentum of the electron. Thus Bohr's first condition defining the stationary states could be stated as:
"Only those orbits were possible in which the angular momentum of the electrons would be an integral multiple of  h /2π." These stationary states correspond to energy levels in the atom.
(v) When an electron revolves around the nucleus in an orbit, a centrifugal force is produced due to the circular motion, which is balanced by electrostatic force of attraction between electron and nucleus.

CALCULATION OF RADIUS OF n^TH BOHR'S ORBITS:
Simple Atom:
To derive an expression for radius, consider an atom of hydrogen consisting of a single electron revolving around a nucleus.
Suppositions:
Consider an electron revolving around its nucleus.
e = charge on the proton
Ze = Total charge on the nucleus
Z = Atomic number of atom
m = Mass of the electron
r = The radius of the orbit and
v = the tangential velocity of the revolving electron.
The two types of electrostatics forces acting on the revolving electron simultaneously.
Electrostatic Or Centripetal Force
According to coulomb's law, the electrostatic force of attraction "Fe" between the nucleus of charge "+Ze" and the electron of charge "-e" separated by a distance "r" is given by the expression:

(b) Write the postulates of electron pair repulsion theory. Explain the shape of H₂O and NH₃ on the basis of electron pair repulsion theory.
Ans: ELECTRON PAIR REPULSION THEORY:
In 1940 SIGWICK and POWELL pointed out that the shapes of molecules can be determined by the repulsion between the electron pairs present in valency shell of central atom.
The basic concept of VSEPRT (Valance Shell Electron Pair Repulsion theory) is that the valance electron pair (lone pairs) and the bond pairs) are arranged around the central atom to remain at a maximum distance apart to keep repulsion at minimum.

POSTULATES OF ELECTRON PAIR REPULSION THEORY:
The main postulates of VSEPRT are:
1. There may be two types of electron pairs surrounding the central atom
(a) Bond Pairs: These are the result of the sharing of unpaired electrons of central atom with unpaired electrons of surrounding atoms. These are also called ACTIVE SET OF ELECTRONS.
(b) Lone Pairs : These are the paired electrons, which have not taken part in sharing. They are also called NON-BONDING PAIRS. They are also considered to be ACTIVE SET OF ELECTRONS.

2. Being similarly charged (i.e. negative) the bond pairs as well as the lone pairs repel each other.
3. Due to repulsion, the electron pairs of central atom try to be as far apart as possible, hence they orient themselves in space in such a manner that force of repulsion between them is minimized.
4. The force of repulsion between lone pairs and bond pairs is not the same. The order of repulsion is as follows:
Lone pair — Lone pair repulsion > Lone pair — Bond pair repulsion > Bond pair — Bond pair repulsion. these repilsion are called Vander Waals Repulsion.
5. In case of molecules with double and triple bonds, the π electron pairs are not considered to be an active set of electrons, hence not included in the count of total electron pairs.
6. The shape of molecule depends upon total number of electron pairs (bonding and lone pairs),It is summarized as follows:

The Shape of NH₃ On The Basis Of Electron Pair Repulsion Theory:
The Lewis Structure of NH₃ shows that the central atom Nitrogen, has five electrons out of which three are unpaired an'd two are paired. The unpaired electrons shared with three Hydrogen atoms, hence, Nitrogen is surrounded by four electron pairs, three are paired and one is lone pair, this lone pair repels the bond pair, as a result the bond angle reduces from 1.09.5° to 107°.

The Shape of H₂O On The Basis Of Electron Pair Repulsion Theory:
The Lewis structure of water shows that the central atom Oxygen, is surrounded by four active sets of electrons, out of which two are bond paired and two are lone pairs. According to Electron Pair. Repulsion model the tetrahedral arrangement would give maximum separation, however two lone pair repels the bond pair hence the bond angle reduces from. 109.5° to 104.5°.

SOURCE: Board Of Intermediate Education Karachi