Physics - Paper II - For Class XII (Science Group) - Model papers With Solved MCQs 2020 -2021

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Physics - Paper II
For Class XII (Science Group)
Model papers 2020 -2021

The correct answer of number mcqs no. (xix) is waxed paper

SECTION "B" (SHORT ANSWER QUESTION) (28 Marks)

Q.2(i):
Data:
R_g = 50 Ω
I_g = 5 mA = 5 x 10^-3 A
R_x = 0.1 MΩ = 0.1 x 10⁶ Ω
v =?

Formula:
R_x = ^-V / _Ig - R_g
R_x + R_g = ^-V / _Ig
Or
V = I_g(R_x + R_g)

Solution:
V = 5 x 10^-3 (0.1 x 10⁶ + 50)
V = 5 x 10^-3 (100000 + 50)
V = 5 x 10^-3 (100050)
V = 500250 x 10^-3
V = 500.25 V Ans

Q.2(ii):
Data:
d_i = 5 cm
ΔL = 0.01 mm = 0.01 / 10 = 0.001 cm
ΔT = 30 °C
α_i= 1.2 x 10^-5 C^-1
α_b= 1.9 x 10^-5 C^-1

Formula:
ΔL = αL_o ΔT

Solution:
ΔL_i = α_i d_i ΔT
ΔL_b = α_b d_b ΔT
ΔL = ΔL_i - ΔL_b
d_i ≅ d_b

0.001 = α_i d_i ΔT - α_b d_i ΔT
0.001 = d_i (α_i - α_b) ΔT
0.001 = 5 (1.2 x 10^-5 - 1.9 x 10^-5) ΔT
ΔT = 0.001 / 5 x 0.7 x 10^-5
ΔT = 28.5 °C

Final temperature
T = 28.5 + 30
T + 58.5 °C Ans


Q.2(iv):
Data:
R_H = 1.0967800 x 10⁷ m = 1.097 x 10⁷ m
n_f = 2 (Balmer)
a) n_i = 3 (Largest)
b) n_i = ∞ (Shortest)
λ_max = ?
λ_min = ?

Formula:
¹/ _λmax = R_H ( ¹/_nf² - ¹/_ni²)
¹/ _λmin = R_H ( ¹/_nf² - ¹/_∞)

Solution:
For Longest Wavelength

Q.2(v):
Data:
m = 2m_o
a) V =?
b) P = ?

Q.2(vi):
Data:
Number Of Bulbs = b = 5
Power Of Bulb = P_b = 100 W
Number Of Fans = f = 10
Power Of Fan = P_f = 60 W
Number Of Lights = l = 10
Power Of Light = P_l = 40 W
Number Of Iron = i = 1
Power Of Iron = P_i = 1000 W
Time = t = 4 x 30 x 60 x 60 = 432000 sec
Energy = E = ? (Kwh)
R = 90 paisa / unit
Cost = ?

Formula:
E = P x t
P = ^E / _F

Solution:
Total Power:
Bulb = P₁ = b x P_b = 5 x 100 = 500 W
Fan = P₂ = f x P_f = 10 x 60 = 600 W
Light = P₃ = l x P_l = 10 x 40 = 400 W
Iron = P₄ = i x P_i = 1 x 1000 = 1000 W

P = P₁ + P₂ + P₃ + P₄
P = 500 + 600 + 400 + 1000 = 2500 W

E = P x t
E = 2500 x 432000
E = 1.08 x 10⁹ J

1 kwh = 3600 000 J

E = ^{1.08 x 109} / _{36 x 10⁵}
E = 300 kwh Ans.

b) Cost
Cost = Kwh x Rate
Cost = 300 x 0.9
Cost = 270 Rs. Ans.




SOURCE: Board Of Intermediate Education Karachi

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

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Chemistry - Paper II
For Class XII (Science Group)
Model papers 2020 -2021

X11 - CHEMISTRY MODEL PAPER SOLUTION

Section "B“ (Short Answer Questions (Marks: 24)

Note: Attempt any six part questions, Three from organic and Three from inorganic chemistry. All questions carry equal marks.

INORGANIC CHEMISTRY

Q.2: (i) Refer to the list of given compounds.
Ans:
a) Formula:
Formula Of A (Dolomite) = MgCO₃ . CaCO₃
Formula Of B (Whiterite) = BaCO₃

b) Equation when C (Blue Vitriol or Copper sulphate or CuSO₄) heated up to 230 °C.
{Note: Copper sulphate crystalline is blue in color which on heating loses all its water of crystallization slowly changing into colourless from (anhydrous)}

c) Chemical Formula and use of D (Potash Alum or Phitkari)
Chemical Formula of D (Potash Alum or Phitkari) = K₂SO₄.Al₂(SO₄)₃. 24H₂O

Uses Of Potash Alum:
The following are uses of potash alum:
1. Fire Retardant:
The use of potassium alum for textiles, wood and paperless flame resistance is as fire-retardant.
2. Tanning:
For leather tanning, potassium alum is used to extract moisture from hide and avoid rotting. Alum is not covered and can be washed out, as compared to tannic acid.
3. Iron and Steel Dissolving:
This Aluminium solution has the property that steels are dissolved without affecting Aluminium or base metals. For machined castings of steel parts of machinery, alum solutions can be used.

Ans 2: (ii) The IUPAC names of the following:

CHEMICAL FORMULA	IUPAC NAME
K3[Fe(CN)6]	Potassium Hexacyano ferrate (III)
[Zn(OH)4]-2	Tetra hydroxo zincate (II)
[Cr(NH)3Cl3]	Tri amine tri choloro chromium (III)
[Ni(en)2Cl2]	Dichloro bis ethylene di amine nickel (II)

Ans 2 (iii): Hydrogen cannot be placed in Group IA and VII A of the periodic table:

DIFFERENCES / DISSIMILARITIES WITH GROUP IA

1. The physical state of IA group element is solid, where as hydrogen is a gas.
2. Hydrogen is nonmetal but the elements of group IA are all the metals.
3. The alkali metals (Group IA elements) exist in mono atomic state (Li, Na, K etc), while Hydrogen always exist in diatomic state i.e H₂.
4. Hydrogen needs only 1 electron to complete its outer most shell but all alkali metals need 7 electrons to complete their outer most shell.
5. Alkali metals forms ionic bonds but hydrogen can form both ionic and covalent bonds.

DIFFERENCES / DISSIMILARITIES WITH GROUP VIIA:

1. Hydrogen belongs to s-block elements while VII A elements belong to p-block elements.
2. Hydrogen has one electron in their outer most shell but elements of group VII A has seven electrons in their outer most shell.
3. Electron affinity of hydrogen is much less than halogens (Group VIIA elements).
4. Hydride ion (H^-) is unstable whereas halide ions (X^-1 = Cl^-1, Br^-1, I^-1) are stable one.
5. Halogen with carbon form polar covalent while Hydrogen with Carbon form non-polar molecule.
6. Element of group VIIA are oxidizing agent while hydrogen is reducing agent.
(Note: Write down any 4 differences as mentioned in question for each Hydrogen and group IA and Hydrogen and group VIIA)

Ans 2 (iv): Groups of the periodic table that have following ground state electronic configuration in their outer most shell:

Electronic Configuration	Groups / Period / Blocks
3s2, 3p2	Group = IVA / Period = 3rd / Block = p
3s2, 3p6, 4s1	Group = IA / Period = 4th / Block = s
4s2, 3d1	Group = IIIB / Period = 4th / Block = d
4s2, 3d10, 4p5	Group = VIIA / Period = 4th / Block = p

(Note: In question only group is asked.)

Ans 2 (v): Extraction Of Sodium From Rock Salt On Industrial Scale:
Sodium metal is commercially obtained by molten sodium chloride by Down's process.

PRINCIPLE:
The basic principle of Down's Cell is electrolysis of Molten or Fused Sodium Chloride to produce metallic sodium.

CONSTRUCTION OF DOWN'S CELLS:

• Down's Cell is made up of Iron graphite in the center which is used as anode to collect the chlorine gas (Cl₂).
• The cathode is made up of circular bar or bent bar of Cu or Fe, which surround the anode.
• They both are separated by the iron gauze to prevent Na⁺ and Cl^- to come in contact.

CONDITION:

• The melting point of NaCI is high that is 801°C. Calcium Chloride is added to decrease the temperature to 600°C. (Scientific Reason).
• Electrolysis is carried out in the absence of water.

WORKING OF DOWN'S CELL:

• A mixture of NaCl and CaCl₂ is introduced in the cell. As the electric current is passed through electrolyte, molten Sodium Chloride is ionized into Sodium Ion and Chloride Ion.
• Sodium ion being positive in charge moves towards cathode while chloride ion moves towards Anode.
• At Cathode Reduction occurs which result in storage of Sodium. The Na in molten state rises up and stored in inverted through. The 99.9% pure sodium obtained through this procedure.

CHEMICAL REACTION IN CELL:
IONIZATION:

NaC1_(l) ⇌ Na_(l)⁺ + Cl_(l)^-

AT CATHODE: (Reduction)

2Na⁺ +2e^- ⟶ 2Na

AT ANODE: (Oxidation)

2Cl^- ⟶ Cl₂ +2e^-

ADVANTAGES OF DOWN'S CELL:
It has following advantages.

1. An expensive and important byproduct Cl gas is obtained.
2. Liquid sodium can be obtained easily at 600 °C.

EXTRACTION OR METALLURGY OF SODIUM IN DOWN'S CELL

Ans 2 (vi) Equation)
a) Nitric acid react with Phosphorous:
Nitric acid oxidizes non-metals and it is reduced to NO₂.

P + HNO₃ ⟶ H₃PO₄ + 5NO₂ + H₂O

Phosphorous + Nitric acid ⟶ Phosphoric acid + Nitrogen Dioxide + Water

b) Sodium react with oxygen

2Na + O₂ ⟶ Na₂O₂ (Nitrogen per oxide)

c) Carbon monoxide is treated with chlorine
Chlorine give addition reaction with Carbon monoxide

CO + Cl ⟶ COCl₂ (Carbonyl chlorid or Phosgene)

d) Aluminum is treated with Chlorine

2Al + 3Cl₂ ⟶ 2AlCl₃ (Aluminium chloride)

ORGANIC CHEMISTRY

Ans 2 (vii): Definitions:

1. GLYCOSIDIC LINKAGE:
A glycosidic bond or glyeasidIc linkage is a type of covalent bond that joins a carbohydrate (sugar) molecule to another group, which may or may not be another carbohydrate.

2. PLASTICIZER:
A plasticizer is a substance that is added to a material to make it softer and more flexible, to increase its plasticity, to decrease its viscosity or to decrease friction during its handling in manufacture.

3. AROMATICITY:
Aromaticity is a property of cyclic (ring-shaped), planar (flat) structures with pi bonds in resonance (those containing delocalized electrons) that gives increased stability compared to other geometric or connective arrangements with the same set of atoms.

4. HOMOLOGOUS SERIES:
A series of organic compounds that have similar structural features but differ from adjacent members by (-CH₂) group is referred to as homologous series. Each member of homologous series is called homologous.

Ans 2 (viii): Definition Of The Polymerization and Isomerism.
POLYMERIZATION:
DEFINITION: A self addition reaction in which a number of simple molecules (monomers) are joined to form a very large molecule is called polymerization.

OR

A chemical reaction in which monomers are converted into polymer is called as polymerization.

ISOMERISM:
DEFINITION: Compounds having same molecular formula but different structural formula and differ from each other in physical and chemical properties are known as "Isomers" and this phenomenon is called isomerism.' Isomerism is due to the difference in the arrangement of atoms in molecules.

b) Compounds as Isomers and Polymers

COMPOUNDS	Isomers / Polymers
Glucose and Starch	Polymer
CH3-0-CH3 and CH3-CH2-OH	Isomer (Functional Groups)
CH3-CH2-CHO and CH3-CO-CH3	Isomer (Functional Groups)
Vinyl Chloride and PVC	Polymer

Ans 2 (ix): Preparation Of Compounds:
a) Ethylene glycol from ethene:
When ethene is added in dilute and alkaline solution of KMnO₄, ethene is oxidize to ethane 1-2 diol (ethylene glycol) and purple color of KMnO₄ is decolourized.

b) Phenyl hydrazone from formaldehyde:
Formaldehyde (H₂C=O) adds phenyl hydrozine in normal manner to form stable product called phenyl hydrazone (H₂C=N- NH-C₆H₅).

(i) H₂C=O + H₂N-NH-C₆H₅ ⟶ OH-CH₂- NH- NH-C₆H₅ (unstable addition product)

(ii) OH-CH₂- NH- NH-C₆H₅ ⟶ H₂C=N- NH-C₆H₅ (phenyl hydrazone)

c) White solid from Acetylene:
When acetylene is heated with AgNO₃ in Ammonia Silver acetylide (White solid) is formed.

CH=CH + 2CuCl_aq ⟶ AgC=CAg + 2HNO₃

Acetylene (Ethyne) + cuprous chloride ⟶ Silver acetylide (White solid) + Nitric acid

d) Ethane from chloro methane:
When Chloro methane is treated with sodium metal, ethane is formed.

2CH₃ Cl + 2Na ⟶ C₂H₆ + 2NaCl

e) Ethene from ethane:
i) By pyrolysis of ethane:
When ethane is heated to high temperature (500 °C) in the absence of air, a thermal decomposition occur to give ethane and hydrogen.

CH₃-CH₃ ⟶ CH₂=CH₂ + H₂

OR

ii) By halogination of ethane:
First ethane react with halogen (Cl, Br) in the presence of sunlight to produce ethyl halide than ethyl halide react with alcoholic potassium hydroxide (KOH) to give ethene.

Ans 2 (x): IUPAC names of the Componds

Compounds	IUPAC Names
CH3-CH(CH3)-CH(CH3)-CH3	2,3 di methyl butane
CH2=C(CH3)-CH(CH3)=CH	2,3 di methyl 4-yne — 1-pentene
CHI3	Tri iodo methane (Iodoform)
CH3-CH(CH3)-CH(Cl)-CHO	3-methyl 2-chloro butanal
(CH3)3-C.CO-CH2CH3	4.4-dimethyl pent - 3 - one

Ans 2 (xi): Benzene Gives Electrophilic Substitution Reaction:
Benzene molecule consist of three alternative double bond in a cyclic planar structure, having three pairs of delocalized electrons above and below the plane of ring. Hence, it is electron-rich. The resonance energy of benzene is very high 36 Kcal per molecule. As a result, it is highly attractive to electron deficient species i.e. electrophiles. Therefore, it undergoes electrophIlic substitution reactions very easily.

Acylation Of Benzene By Friedel-Craft Reaction:

The introduction of The acyl group (R-C=O), in the benzene ring is generally achieved with the help of Lewis acid catalyst as AlCl₃. This reaction is known as Friedel-Craft Reaction.

Ans 2 (xii): Equations:
a) Acetylene reacts with water in the presence of H₂SO₄ and HgSO₄ at 75 °C.

b) Formaldehyde is polymerized in presence of H₂SO₄.

c) Vapors of acetic acid are passed over MnO₂ at 500 °C.

d) Ethanol in excess, is heated in presence of H₂SO₄.

OR

Short Note On Amino Acid Or Fertilizers

FERTILIZERS:

"Fertilizers are commonly inorganic salts and containing elements such as nitrogen, phosphorus, potassium etc, which are very essential for the growth and development of plants. The yield of agricultural crops can be increased by introducing fertilizers to the soil."

NEED OR USES OF FERTILIZERS:
Fertilizers stimulate the process of metabolism is the plant cells. The need of fertilizers are due to three reasons.

• To make up the deficiency of elements (like N, P, K) and become fertile again.
• To give an additional supply of food.
• To maintain pH of the soil near neutrality or slightly alkalinity i.e from pH 7 to 8. The soil having pH above 10 or below 3 is sterile.

CLASSIFICATION OF FERTILIZERS:
There are two types of fertilizers, depending upon the source and chemical nature:

1. Natural or Organic Fertilizers:
   These are derived from Plant and animals and may be organic or inorganic. They includes manures, peats, hay wastage and rock phosphate and Chillie salt peter etc
   Uses: Natural fertilizers provide more nutrient than synthetic fertilizers.
   
   
2. Synthetic or Mineral Fertilizers:
   Mineral fertilizers are obtained from mineral raw materials. These are preferred on natural fertilizers because they contain exact percentage of elements (e.g. Nitrogen and Phosphorus) as required for an specific crop. They are also called artificial fertilizers.

The Mineral fertilizers are categorized as:

a) Nitrogen Fertilizers:
Fertilizers containing nitrogen as essential elements are called "Nitrogenous Fertilizers".

1. Urea (NH₂ - CO - NH₂)
2. Ammonia NH₃
3. Ammonium Nitrates (NH₄NO₃)
4. Ammonium Sulphate {(NH₄)₂SO₄}

Importance or Uses of Nitrogen:

• It is necessary in the early stage for the rapid growth of plants.
• It is main constituent of proteins.

b) Potassic Fertilizers:
Fertilizers containing potassium as essential element are called "Potassic Fertilizers", such as:

1. Chile saltpeter or Potassium Nitrate, KNO₃.
2. Potassium Sulphate K₂SO₄.

Importance or Uses of Potassium:

• It is required in the formation of starch, sugar and fibrous material.
• It resist plant diseases.
• It makes plant strong by making root healthy.
• It helps in ripening of seeds, fruits and cereals.

c) Phosphatic Fertilizers:
Fertilizers containing phosphorous as essential element are called "Phosphatic Fertilizers."
The important phosphorus fertilizers are given below:

1. Amonium phophate:
   e.g:  (NH₄)₂PO₄
2. Diamonium hydrogen phophate:
   e.g:  (NH₄)₂HPO₄
3. Super Phosphates:
   e.g: Mixture of Ca(H₂.PO₄)₂ and CaSO₄ 
   
4. Triple phosphate:
   e.g: 3Ca(H₂.PO₄)₂ 
   

Importance or Uses of Phosphorus:

• It stimulates the early growth and development of plants.
• It accelerates the seed and fruit formation processes.
• It also resist plant against diseases and frost.

SCOPE OF FERTILIZERS IN PAKISTAN:
Pakistan is an agricultural country It does not only prepare fertilizers of its own demand but also exports in a bulk.

INDUSTRIES IN PAKISTAN:
Some important fertilizer factories of Pakistan are enlisted below:-

1. TSF Plant and urea fertilizer plant, Hazara.
2. Faisalabad Fertilizer Ltd.
3. Pak American Fertilizer Ltd. at Daud Khel.
4. Single Super Phosphate Plant at Jaran Wala.
5. Natural Gas Fertilizer Factory, Multan.
6. Dawood Urea Plant, Lahore.
7. Dhariala Potash Fertilizer Project, Dhariala.
8. Fauji Urea Complex, Sadiqabad.
9. Exxon Fertilizer Co. Dahrki.
10. Urea Plant, Mirpur Mathelo.

(Note: write down only any two industries in notes to save time for others question in exams)

AMINO ACID:

DEFINITION:
Amino acids are bi-functional compounds containing both carboxylic acid group (-COOH) and basic amino group (-NH₂). The general formula of amino acid is R-CH(NH₂)-COOH.
They are building blocks of proteins. They arc linked together by peptide bond

TYPES OF AMINO ACID:
There are two types of amino acids:

1. Essential Amino Acids:
Those which cannot be synthesized by human body but they are essential for:
i) growth of infants
ii) Transmission of impulses in the nervous system
are called "Essential Amino acids".
Example: Leucine, Isoleucine, Methionine etc.
These amino acids must be supplied to our body through diet and their deficiency cause diseases.

2. Non-Essential Amino Acids:
Those which can be synthesized by human body are called "Non-essential Amino Acids". There are 20 amino acids required for synthesis of protein. Out of which 10 are essential and 10 are non essential amino acids.

CLASSIFICATION OF AMINO ACIDS:
Amino acids are classified according to their nature i,e. neutral, acidic or basic.

1. NEUTRAL AMINO ACIDS:
Amino acids containing one amino group and one carboxyl group are called "Neutral Amino Acids". Amino group acts as basic and carboxyl as acidic, both neutralize each other. Hence amino acids exhibit neutral characteristics and amphoteric nature.
For example:
i. Glycine NH₂-CH₂-COOH
ii. Alanine CH₃-CH(NH₂)-COOH
iii. Valine CH₃-CH(CH₃)-CH(NH₂)-COOH

2. ACIDIC AMINO ACIDS:
Amino acids containing one amino group and more than one carboxyl group are called "Acidic Amino Acids". Acidic amino acids exhibit acidic nature.
For example:
i. Aspartic Acid HOOC-CH(NH₂)-CH₂-COOH
ii. Glutamic Acid HOOC-CH(NH₂)-CH₂-CH₂COOH

3. BASIC AMINO ACIDS:
Amino acids containing one hydroxyl group and more than one amino group are called "Basic Amino Acids". Basic amino acids exhibit basic nature.
For example:
i. Arginine HOOC-CH(NH₂)-(CH₂)₃-NH-C(NH)-NH₂
ii. Lysine HOOC-CH(NH₂)-(CH₂)₃-NH₂

ZWITTER (Dipolar Nature of Amino Acids):
A dipolar charged but electrically neutral ion is called Zwitter Ion". In amino acids. carboxyl (acidic) group ionizes to donate proton H⁺ whereas amino (basic) group having lone pair of electron behaves as proton acceptor becoming Lewis Base. Hence amino acids exists as dipolar ion in an un-ionized form called "Zwitter Ion".


Due to Zwitter Ion, amino acids have following properties:
1) Amino acids are soluble in water but insoluble in organic solvents.
2) Amino acids are solids.
3) Amino acids have high melting point and donate or accept proton H⁺.

PEPTIDE BOND:
Amino acids in proteins are linked together through an acid-amide group type of bond known peptide bond. The peptide bond is formed between two amino acid molecules when amino group of one amino acid is linked with the cathoxylic group to other amino acid molecule by the elimination of water molecule.

ROLE OF AMINO ACID IN HUMAN BODY:
When food containing protein is taken, many enzymes act on protein and completely hydrolyse it into amino acids. These amino acids are absorbed in blood which carries them to cells where any one of following action take place:
1) amino acids can be synthesized back to body protein.
2) oxidation may take place to provide energy.
3) body protein may be transformed into carbohydrates or fats in case of low amount in diet or make harmones and other body necessities.




SOURCE: Board Of Intermediate Education Karachi

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

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Physics - Paper I
For Class XI (Science Group)
Model papers 2020 -2021

Max. Marks:43 (25 Marks)

SECTION B (SHORT-ANSWER QUESTION)

NOTE: Attempt any Five part questions from this section. All questions carry equal marks. The use of scientific calculator is allowed. All notations are used in their usual meanings. Draw diagram where necessary.
Q.2 (i): State and prove the law of Conservation of Linear Momentum.
Ans: Law of Conservation of Linear Momentum
Introduction:
This law deals with the momentum of the bodies can collide with each other.

Statement:
"When some bodies in an isolated system acts upon one another, the total momentum of the system remains constant."

OR

"The total momentum of an isolated system of interacting bodies remain constant."

Explanation:
According to this law if two or more bodies in an isolated system collide with one another the total momentum of the system before collision is equal to the total momentum of the system after collision.

Mathematical representation:
Two bodies of masses m₁ and m₂ moving with velocities u₁ and u₂ collide with each other and their velocities becomes v₁ and v₂, then according to the law of momentum.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Proof:
Consider two bodies a and b of masses m₁ and m₂ moving with velocities u₁ and u₂ in same direction as shown in figure.
Let the bodies collide for "t" second and experience force "f". Due to which their velocities become v₁ and v₂.
During collision body a exerts some force f on the body b.The force exerted by body a on body b is equal to the rate of change of momentum of body a i.e.,
Force on body a= rate of change of momentum of body a

m₁v₁ - m₁u₁ = - m₂v₂ + m₂u₂

m₁v₁ + m₂v₂ = m₁u₁ + m₂u₂

(ii) What is difference between static and dynamic equilibrium? State the conditions of equilibrium.
Ans: Difference Between Static And Dynamic Equilibrium
Static Equilibrium:
If the combined effect of all the forces acting on a body is zero and the body is in the state of rest then its equilibrium is termed as static equilibrium.
For example,

• A book lying on a table
• A block hung from a string

Dynamic Equilibrium:
The equilibrium of bodies moving with uniform velocity is called dynamic equilibrium.
For example,

• The jumping of a paratrooper by a parachute is an example of uniform motion, In this case weight is balanced by the reaction of the air on the parachute acts in the vertically upward direction.
• The motion of a small steel ball through a viscous liquid. Initially the ball has acceleration but after covering a certain distance, its velocity becomes uniform because weight of the ball is balanced by upward thrust and viscous force of the liquid. Therefore, ball is in dynamic equilibrium.

Conditions of Equilibrium

There are two conditions of equilibrium:
First Condition of Equilibrium:
According to 1^st condition of equilibrium:-
"An object is said to be in equilibrium if the sum of all the force acting on it in one direction balances the sum of all forces in opposite direction."

Or

"The resultant of the forces acting on the body is zero."

Mathematical Representation:-
If an object is acted upon by forces along x-axis and y-axis then according to 1^st condition of equilibrium:
The sum of all the force acting along x-axis is 0.

∑ F_x = 0
Or  F_1x + F_2x + F_3x .................... F_10x = 0

Similarly, The sum of all the force acting along y-axis is 0 i.e.

∑ F_y = 0

Or  F_1y + F_2y + F_3y .................... F_10y = 0

Since forces along x-axis and y-axis are 0 therefore total force will be zero.

∑ F = 0

Second Condition of Equilibrium:-
Second condition of equilibrium deals with such object where torque can produce.
According to 2_nd condition of equilibrium:-
"If sum of all the torque acting on a body is zero then body is in equilibrium."

Or

"A body is in equilibrium when sum of all the clockwise torque is equal to sum of all the anticlockwise torque."

Mathematical Representation:-

∑ 𝜏 = 0

Clockwise Torque = Anticlockwise Torque

(iii) Drive an expression for the Variation of "g" with depth.
Ans:

(iv) How is the magnifying power of the (i) Astronomical telescope and (ii) compound microscope affected by increasing the focal length of their objectives?
Ans: (i) Astronomical Telescope:
The magnifying power of the Astronomical telescope is given by:

M = f_o / f_e

where
f_o = focal length of objective
f_e = focal length of eyepiece
If the focal length of eyepiece remains the same, then:

M ∝ f_o

Which shows that magnifying power of the Astronomical telescope is directly proportional to the focal length of their objectives. It will increase if the focal length of their objectives increase and the magnifying power of Astronomical telescope will decrease if the focal length of their objectives decrease.

(ii) Compound Microscope:
The magnifying power of the compound microscope is given by:

M = ^L/ _fo (1 + ^d/_f)

where
f_o = focal length of objective
f_e = focal length of eyepiece
d = least distance of distant vision
If the focal length of eyepiece and least distance remains constant, then:

M ∝ ¹ / _fo

Which shows that magnifying power of the compound microscope is inversely proportional to the focal length of their objectives. It will increase if the focal length of their objective decreases and the magnifying power of compound microscope will decreases if the focal length of their objective increases.

(v) Prove that the vectors -------------- can form the sides of a right angled triangle.
Solution:

(vi) Two coherent sources are placed 1.8cm apart. Interference fringes are obtained on screen 80cm away. The fourth bright fringe is at a distance of 1.08 cm from the central fringe. Calculate the wavelength of the light used.
Data:
d = 1.8 cm
L = 80 cm
n = 4
y₄ = 1.08 cm
λ = ?

Solution:
The position of bright fringe with central fringe is given by:
y₄ = ^{n λ L} /_d
1.08 = ^{(4) λ (80)} /_1.8
1.08 x 1.8 = 320 λ
λ = 1.08 x 1.8 / 320
λ = 0. 006075 cm = 0. 006075 / 100
λ = 6.075 x 10^-5 m Ans.

(vii) Find the speed of sound in air at 50 °C and 70 °C (take speed of sound 332 m/s).
Data:
T₁ = 50 °C
T₂ = 70 °C
V_o = 332 m/s
V₅₀ = ?
V₇₀ = ?

Solution:
The speed of sound at 1 °C
V_t = V_o + 0.61t

At 50 °C:
V₅₀ = 332 + (0.61 x 50)
V₅₀ = 362.5 m/s Ans.

At 70 °C:
V₅₀ = 332 + (0.61 x 70)
V₅₀ = 374.7 m/s Ans.

(viii) A truck starts from rest at the top of a slope which is 1 m high and 49 m long. Find its acceleration and speed at the bottom of the slope assuming that friction is negligible. ( Chapter 3: Motion -  example 3.7 - from text book)
Data:

• V_i = 0
• h = 1 m
• l = 49 m
• a = ?
• V_f = ?

Solution:
Since there is no motion perpendicular to the plane for R and Wcos θ must balanced each other as shown in figure:

∴ R - Wcos θ = 0

When friction is negligible, The only unbalanced force acting on the truck is Wsinθ  acting along the X-axis or parallel to the plane. If produces an acceleration (a) and (m) be the mass of the truck, then by Newton's 2nd Law of motion.

W sin θ = ma
But W = mg
∴ mg sin θ = ma
or a = g sin θ
Sin θ = ^h / _l = ¹ / ₄₉
(as sin θ = ^{perpendicular} / _hypotenuse = ^h / _l)
∴ a = 9.8 x ¹ / ₄₉ = 0.2 m/s²

To determine the speed at the bottom of the slope we make use of the equation of motion. We have,

• V_i = 0
• a = 0.2 m/s²
• S = 49 m
• V_f = ?

∴ V_f² - V_i² = 2as
V_f² = V_i² + 2as
V_f² = (0)² + 2 x 0.2 x 49
V_f² = 19.6
Taking square root on both side
∴ V_f = √ 19.6
V_f = 4.4 ms^-1 Ans.

(ix) A diver leaps from a tower with an initial horizontal velocity component of 7 m/s and upward velocity component of 3 m/s. Find the component of her position after 1 second.
Data:
V_ox = 7 m/s
V_oy = 3 m/s
t = 1sec
x-component = ?
y-component = ?

Solution:
For x-component:
S = V x t
x = V_ox x t
x = 7 x 1
x = 7 m Ans.

For y-component:
S = V_it + ¹/₂ at²
y = V_oyt + ¹/₂ (-g)t²
y = 3 x 1 - ¹/₂ x 9.8 (1)²
y = 3 - 4.9
y = - 1.9 m Ans.
The value of y-component is negative, which show downward direction.



SOURCE: Board Of Intermediate Education Karachi

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

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Mathematics - Paper I
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SOURCE: Board Of Intermediate Education Karachi

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

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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