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Science

TERM IIBy

KhoslaKapoor

Wadhawan

Edited ByJaya Sharma

L.B. MishraSurabhi Jain & Sonika Tyagi

Price of Term I & II : ` 400.00

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MBD BOOKS FOR X (C.B.S.E.) MBD Super Refresher English Communicative MBD Super Refresher Social Science MBD Super Refresher English Language & Literature MBD Super Refresher Science MBD Super Refresher Hindi Course ‘A’ MBD Sanskrit MBD Super Refresher Hindi Course ‘B’ MBD MBD Super Refresher Mathematics

SYLLABUSSCIENCE (CLASS–X)

Second Term Marks: 90

Unit No. Unit Marks

I Chemical Substances -Nature and Behaviour 23

II World of Living 30

III Natural Phenomena 29

V Natural Resources 08

Total 90

Theme: Materials (25 Periods)Unit I: Chemical Substances–Nature and Behaviour

Carbon compounds: Covalent bonding in carbon compounds. Versatile nature of carbon. Homologous series. Nomenclature of carbon compounds containing functional groups (halogens, alcohol, ketones, aldehydes, alkanes and alkynes), difference between saturated hydrocarbons and unsaturated hydrocarbons. Chemical properties of carbon compounds (combustion, oxidation, addition and substitution reaction). Ethanol and Ethanoic acid (only properties and uses), soaps and detergents.

valency, atomic number, metallic and non-metallic properties.

Theme: The World of the Living (30 Periods)Unit II: World of Living

Reproduction: Reproduction in animals and plants (asexual and sexual) reproductive health-need and methods of family planning. Safe sex vs HIV/AIDS. Child bearing and women's health.

Heredity and Evolution:brief introduction; Basic concepts of evolution.

Theme: Natural Phenomena (23 Periods)Unit III: Natural Phenomena

Refraction; Laws of refraction, refractive index.

Refraction of light by spherical lens; Image formed by spherical lenses; Lens formula (Derivation not required);

applications of spherical mirrors and lenses.

Refraction of light through a prism, dispersion of light, scattering of light, applications in daily life.

Theme: Natural Resources (12 Periods)Unit III: Natural Resources

Conservation of natural resources.

Coal and Petroleum conservation. Examples of people's participation for conservation of natural resources.

Regional environment: Big dams: advantages and limitations; alternatives, if any. Water harvesting. Sustainability of natural resources.

Our environment: Eco-system, Environmental problems, Ozone depletion, waste production and their solutions. Biodegradable and non-biodegradable substances.

PRACTICALS—SECOND TERMPracticals should be conducted alongside the concepts taught in theory classes.

LIST OF EXPERIMENTS

1. To study the following properties of acetic acid (ethanoic acid): (i) odour (ii) solubility in water (iii) effect on litmus (iv) reaction with sodium bicarbonate

2.

3. To study the comparative cleaning capacity of a sample of soap in soft and hard water.

4. To determine the focal length of: (i) Concave mirror (ii) Convex lens by

5. To trace the path of a ray of light passing through a rectangular glass slab for different angles of incidence. result.

6.

7. To trace the path of the rays of light through a glass prism.

8. ray diagrams to show the nature of image formed.

9. To study homology and analogy with the help of models/charts of animals and models/charts/specimens of plants.

10. To identify the different parts of an embryo of a dicot seed (Pea, gram or red kidney bean).

QUESTION PAPER DESIGN(CODE NO. 086/090)

Class–X

Time: 3 Hours Max. Marks: 90

S. No. Typology of Questions

Very

Short

Answer

(VSA)

1 Mark

Short

Answer

–I (SAI)

2 Marks

Short

Answer

–II (SAII)

3 Marks

Long

Answer

(LA)

5 Marks

Total

Marks

%

Weight

age

1

Remembering (Knowledge based simple

terms, concepts, principles, or theories, 3 – 1 1 11 15%

2

Understanding (Comprehension - to be familiar with meaning and to understand conceptual ly , interpret , compare, contrast, explain, paraphrase, or interpret information)

– 1 4 1 19 25%

3

Application (Use abstract information in concrete situation, to apply knowledge to new situations, use given content to interpret a situation, provide an example, or solve a problem)

– – 4 1 17 23%

4

High Order Thinking Skills (Analysis & Synthesis - Classify, compare, contrast, or differentiate between different pieces of information, Organize and/or integrate unique pieces of information from a variety of sources)

– 2 – 1 9 12%

5

Inferential and Evaluative (Appraise,

of a decision or outcome, or to predict outcomes based on values)

– – 2+1* 2 19 25%

Total (Theory Based Questions) 3x1=3 3x2=6 12x3= 36 6x5=30 75(24) 100%

Practical Based Questions (PBQs) 9x1=9 3x2=6 – – 15(12)

Total 12x1=12 6x2=12 12x3= 36 6x5=30 90(36)

*One question of 3 marks will be included to assess the values inherent in the texts.

36

There are 118 elements known till today. Among these elements, 98 are naturally occurring.

discovered elements.

their atomic mass.

on the right hand side.

IMPORTANT TERMS AND DEFINITIONS

Dobereiner’s triads:

elements.

Newlands’ Law of Octaves:

CHAPTER IN A NUTSHELL

(NCERT Textbook Chapter 5)

2 CONCEPTS

3.1 Structural Organisation of a Cell

3.2 Cell Organelles

CONCEPTS

2.2 Mendeleev’s Periodic Table

2.3 The Modern Periodic Table

2.4 Periodic Trends

2/18/2016 5:06:31 PM

Super RefresherAll chapters as per NCERTTextbook

Every chapter dividedinto CONCEPTS andeach CONCEPT dealtwith as a completetopic

Chapter in a Nutshell and Important Terms and Definitions provide a complete and comprehensive summary of the chapter

Highlights essential information which must be remembered

Each sub-topic dealt with as a complete unit

Includes NCERT Textbook Activities and Exercises with answers

ach sub-topic dealt witha complete unit

MBD Super Refresher Science-X

38

2.2 Mendeleev’s Periodic Table

NCERT Activity 1 – Page 84

Aim:Procedure:

Observation:Conclusion:

NCERT IN-TEXT QUESTIONS

Page 81

Li

Be

F Cl

K

Ca

3. What were the limitations of Newlands’ law of octaves?BSE_SCI_G10_C02_TII.indd 38

2/1

Science

HOTS questions with answers in every chapter

Self Assessment with answers at the end of the book

Value Based Questions to assess thestudents for social responsibilities

Model Test Papersof 90 marks each

Practice Exercise equaling 20 marks for every chapter

MODEL QUESTION PAPERS FOR PRACTICE

MODEL QUESTION PAPER—1

CLASS—X

SUBJECT—SCIENCE (THEORY)

SECOND TERM (SA—II)

Time Allowed : 3 Hours

Max. Marks : 90

General Instructions :

(i) The question paper comprises two sections, A and B. You are to attempt both sections.

(ii) All questions are compulsory.

(iii) There is no overall choice. However internal choice has been provided in all the three questions of five

marks category. Only one option in each question is to be attempted.

(iv) All questions of Section A and all questions of Section B are to be attempted separately.

(v) Question numbers 1 to 3 in Section A are one mark questions.

(vi) Question numbers 4 to 6 are two marks questions, to be answered in about 30 words each.

(vii) Question numbers 7 to 18 are three marks questions, to be answered in about 50 words each.

(viii) Question numbers 19 to 24 are five marks questions, to be answered in about 70 words each.

(ix) Question numbers 25 to 36 in Section B are practical based questions.

(x) Question numbers 25 to 33 are of one mark and 34 to 36 are of two marks each.

SECTION—A

1. What is far point of the eye ?

2. Give reason as to why the colour of the sky is blue ?

3. Give two examples of biodegradable wastes.

4. 5. We do not clean ponds or lakes but an aquarium needs to be cleaned. Why ?

6. (i) What is term for release of ovum from ovary ?

(ii) 7.

8. Atomic numbers of few elements are given below :

10, 20, 7 and 14.

(a) Identify the elements.

(b) Identify the group numbers of these elements in periodic table.

(c) Identify the periods of these elements in the periodic table.

9. of ecosystem ?

10. What is asexual reproduction ? Discuss spore formation in fungi.M-1

M.indd 1

2/1

MBD Super Refresher Science-X

50

HOTS CORNER

1. The following table shows the position of six elements A, B, C, D, E and F in the periodic table.

Using the above table answer the following questions:

Periods Groups 1 2 3 – 12 13 14 15 16 17 18

2

A

B

C

3

DE

(a) Which element will form only covalent compounds?

(b) Which element is a metal with valency 2?

(c) Which element is a non-metal with valency of 4?

(d) Out of D and E, which one has a bigger atomic radius and why?

(e) Write a common name for the family of elements C and F.

Answer: (a) B (b) D

(c) B

(d) D, because in a period atomic radii decreases from left to right due to increase in

effective nuclear charge. (e) Noble gases

PRACTICE EXERCISE

Maximum Marks–20

Objective Type Questions

1 mark each

A. Fill in the blanks.

1. There are _________ groups and _________ periods in the modern periodic table.

2. The position of __________ in the periodic table is doubtful.

Answers: 1. 18, 7 2. hydrogen 3. atomic number

SELF ASSESSMENT

Objective Type Questions

1 mark each

A. Multiple Choice Questions

1. Which one of the following does not increase while moving down the group of the periodic table?

(a) Atomic radius (b) Metallic character (c) Valency

(d) Number of shells

2. Among the following elements which one will lose an electron most easily?

(a) Li (b) Na

(c) Ca (d) K

VALUE BASED QUESTION

A. Anuj is class-X student. He arranges his books and notebooks subject-wise. How does this

of periodic table makes the study of elements simple and systematic.

2/18/2016 7:41:19 PM

CONTENTS

1 Carbon and its Compounds 1–35

7 Our Environment 171–191

Answers for Self Assessments 232–239

Carbon and its Compounds

1

1.1 Bonding in Carbon

1.2 Versatile Nature of Carbon

1.3 Chemical Properties of Carbon Compounds

1.4 Soaps and Detergents

� Chemical bonds are the attractive forces which hold the atoms or ions together in a chemical

species.

� The major reason of chemical bonding is the tendency of the atoms of the various elements to

acquire stable configuration i.e. to complete their octet and to acquire a state of minimum

energy.

� Noble gases are chemically less reactive gases and have eight electrons in their valence shells

(except helium).

� Octet rule depicts the tendency of an atom of the element to have eight electrons in its valence

shell.

� Chemical bonds are mainly of three types: ionic or electrovalent bond, covalent bond and co-

ordinate bond.

� Covalent bond is the bond formed by equal contribution and mutual sharing of electrons between

two atoms.

� The compounds containing covalent bonds are called covalent compounds.

� Covalency is the number of electrons contributed by an atom of the element for mutual sharing

during the formation of a covalent bond.

� A pair of electrons shared between two atoms is called a bond pair of electrons.

� A pair of electrons present on one atom which does not take part in sharing is termed a lone pair

of electrons.

� Single, double and triple covalent bonds are formed by the mutual sharing of one, two and three

electron pairs respectively between two atoms.

� Organic compounds are made of C, H, O, N, S, P and halogens and have covalent bonds.

� Organic compounds have covalent bonds. They include hydrocarbons and their derivatives.

� Hydrocarbons are the organic compounds made up of carbon and hydrogen.

� Catenation is the property by the virtue of which atoms of the same element get linked together

through covalent bonds so as to form long straight, branched or closed chains or rings.

· � A carbon atom shows tetracovalency or tetravalency because it has four electrons in its valence

shell.

� A large number of compounds are formed by carbon because of its tetravalency and the property

of catenation.

(NCERT Textbook Chapter 4)

MBD Super Refresher Science-X2

� Structural formula of a compound gives us the relative arrangements of bonded atoms in a

molecule of it.

� An atom or a group of atoms which when present in a molecule gives certain special properties

to the compound is called a functional group; e.g. – OH or alcoholic group, – CHO or aldehydic

group.

� A Homologous series of compounds have compounds with the same functional group, and

similar chemical properties. The successive members of a homologous series differ by CH2 unit

and 14 mass units.

� IUPAC System stands for International Union of Pure and Applied Chemistry System.

� Saturated hydrocarbons are the hydrocarbons, which contain only single bonds between carbon

atoms in their molecules.

� Unsaturated hydrocarbons are the hydrocarbons, which contain one or more double or triple

bonds between carbon atoms in their molecules.

� Pyrolysis is the decomposition of higher molecular weight hydrocarbons into lower molecular

weight hydrocarbons on heating at a high temperature in the absence of air.

� Compounds having same molecular formula but different properties are called isomers and this

phenomenon is called isomerism.

� Carbon and its compounds come under the category of major source of fuels.

� Soaps and detergents are cleansing agents.

Allotropy: It is the phenomenon of existence of an element in different forms having different

physical properties but same or slightly different chemical properties.

Soaps: These are sodium or potassium salts of higher fatty acids.

Detergents: These are sodium alkyl benzene sulphonates.

Saponification: It is the process of alkaline hydrolysis of fat to give soap and glycerol.

Ester: Carboxylic acids react with alcohols to form esters and the reaction is known as esterification

reaction.

Absolute alcohol: Pure ethanol or ethanol with less than 1% water is called absolute alcohol.

Denatured alcohol: It is made by adding poisonous substances such as copper sulphate or methanol

to ethyl alcohol or ethanol.

1.1 Bonding in Carbon

Carbon has four electrons in its valence shell and has a tendency to form four covalent bonds resulting

in the formation of covalent compounds like methane (CH4), ethane (C

2H6), acetic acid (CH

3COOH)

etc. These compounds are called organic compounds. Carbon forms covalent bonds with other

C–atoms and some other elements such as H, O, S, N, F, Cl, Br or I. Carbon can form double or triple

bonds (multiple bonds).

NCERT Activity 1 – Page 58

Aim: To sort out the things made up of carbon compounds

Carbon and its Compounds 3

Procedure:

1. Make a list of ten things you have used or consumed since morning.

2. Compile this list with the lists made by your classmates and then sort the items into the table given

below.

Things made of metal Things made of glass/clay Others

Bucket Water

Cooking pan Furniture

Newspaper, books

Soap

Bread

Cup (clay) Tea

Medicine

Observation:

On observing the items listed in the last column of the table and consulting with the teacher, one can

say that most of the items listed in this column are made up of carbon compounds.

Conclusion:

From this activity it can be concluded that most of the things that we use in our daily life are made up

of carbon compounds.

Objective Type Questions 1 mark each

A. Answer the Following Questions

1. Which allotropic form of carbon is smooth as well as slippery?

Answer: 1. Graphite

B. Multiple Choice Questions

1. Which of the following crystalline forms of carbon has the composition C60

to C350

?

(a) Graphite (b) Fullerene

(c) Diamond (d) Coal

2. Which among the following is an allotropic form of carbon?

(a) Diamond (b) Graphite

(c) Fullerene (d) All of these

Answers: 1. (b) Fullerene 2. (d) All of these

Short Answer Type Questions 2 – 3 marks each

A. Describe the structure of Fullerene.

Fullerenes are allotropes of carbon. The fullerene C60

has carbon atoms arranged in the form of a

football and it looks like the geodesic dome designed by the US architect Buckminster Fuller. Thus,

the molecule was named fullerene.

MBD Super Refresher Science-X4

B. Why do covalent molecules have definite shapes?

A covalent bond is formed by the sharing of electrons between two atoms. The bonds are represented

by dashes (—) and are directional in nature. The directional nature of the covalent bonds gives

covalent molecules having two or more such bonds, a definite shape. This is also called definite

geometry of the covalent molecule.

C. Compare the structure of diamond and graphite and explain any one property of each based

upon the structure.

In diamond, each carbon atom is bonded to four other carbon atoms by covalent bonds resulting

in the formation of a three-dimensional network structure. Due to the presence of strong covalent

bonds, diamond is hard and has high melting point. In graphite each C–atom is bonded to

three other carbon atoms to form hexagonal rings, which are held together by weak van

der Waals’ forces of attraction. Therefore, graphite has a two-dimensional sheet-like structure. Due

to the presence of one free electron left with each C–atom, graphite is a good conductor

of electricity.

1.2 Versatile Nature of Carbon

Carbon is a versatile element that forms the basis of all living organisms and many non-living objects.

Carbon forms long linear chains, branched chains or rings with other C–atoms or functional groups. The

ability of carbon to form chains gives rise to homologous series of compounds in which all the members

contain the same functional group and show similar properties. The adjacent members of a homologous

series differ by a CH2 unit. The names of organic compounds are written on the basis of IUPAC system.

NCERT Activity 2 – Page 67

Aim: To calculate the difference in the formulae and molecular masses of

(a) CH3OH and C

2H5OH

(b) C2H5OH and C

3H7OH

(c) C3H7OH and C

4H9OH

Calculations:

(a) C2H5OH – CH

3OH = CH

2

(b) C3H7OH – C

2H5OH = CH

2

(c) C4H9OH – C

3H7OH = CH

2

Difference in molecular masses

(a) [2 × 12 + 5 × 1 + 1 × 16 + 1 × 1] – [1 × 12 + 3 × 1 + 1 × 16 + 1 × 1]

= 46 – 32

= 14 u

Structure of graphite Structure of diamond

Carbon and its Compounds 5

(b) [3 × 12 + 7 × 1 + 1 × 16 + 1 × 1] – [2 × 12 + 5 × 1 + 1 × 16 + 1 × 1]

= 60 – 46

= 14 u

(c) [4 × 12 + 9 × 1 + 1 × 16 + 1× 1] – [3 × 12 + 7 × 1 + 1 × 16 + 1 × 1]

= 74 – 60

= 14 u

Objective Type Questions 1 mark each

Multiple Choice Questions

1. What is the functional group in HCHO?

(a) Ketone (b) Aldehyde

(c) Alcohol (d) Carboxylic acid

2. Butanone is a four-carbon compound with which functional group?

(a) Carboxylic acid (b) Aldehyde

(c) Ketone (d) Alcohol

Answers: 1. (b) Aldehyde 2. (c) Ketone

Short Answer Type Question 2 – 3 marks each

A. What is a homologous series of compounds? List any two characteristics of a homologous series.

A series of compounds having similar structural formulae, same functional group and hence similar

chemical properties are called homologous series of compounds.

Characteristics of homologous series:

1. The members of a homologous series have similar chemical properties.

2. Any two adjacent members of a homologous series differ by a CH2 unit in their molecular formulae.

B. Give the functional group and their suffixes present in alkenes, alkynes and aldehydes.

S. No. Homologous Series Functions Group Suffix

1 Alkenes�����

�����������ene

2. Alkynes���� ����≡��������� �

yne

3. Aldehydes – CHO al

or

���

�

�

���������� ����

Long Answer Type Question 5 marks each

Give general formulae and IUPAC names of first two members of:

1. Monohydric alcohols

2. Haloalkanes

3. Monocarboxylic acids

4. Alkanones

MBD Super Refresher Science-X6

Also give their common names.

S. Homologons General First two IUPAC names Common

No. Series Formula members names

1. Monohydric CnH

2n + 1 – OH CH

3 – OH Methanol Methyl

alcohols where alochol

(Alkanols) n = 1, 2, 3 C2H

5 – OH Ethanol Ethyl

or or alcohol

R – OH CH3 – CH

2 – OH

2. Haloalkanes CnH

2n + 1 – X CH

3 – Cl Chloro- Methyl

or where C2H

5Cl methane chloride

Alkyl halides n = 1, 2, 3 or Chloro- Ethyl

X – F, Cl, Br, l CH3 – CH

2 – Cl ethane chloride

3. Monocarbon- CnH

2n + 1 – COOH H – COOH Methanoic Formic

xdic acid acid acid

n = 0, 1, 2, 3 CH3 – COOH Ethanoic Acedic

acid acid

4. Ketones � � ������� ������

�

CH3 ��������

�

�

Propanone Dimethyl

(Alkanones) Cm

H2m + 1

ketone or

Acetone

n = m simple ketone CH3 – CH

2 ��������

�

�

Butanone Ethyl methyl

n ≠ m mixed ketone Ketone

where

n, m = 1, 2, 3

1.3 Chemical Properties of Carbon Compounds

The important properties of carbon compounds are combustion, addition and substitution reactions.

Carbon and its compounds are some of our major sources of fuels. Ethanol and ethanoic acid are the

carbon compounds which are very important in our daily lives.

NCERT Activity 3 – Page 69

Aim: To observe the nature of flame produced by burning naphthalene, camphor and alcohol

Procedure:

1. Take some carbon compounds (naphthalene, camphor, alcohol) one by one on a spatula; burn

them and observe the flame.

Observation Table:

Compound Flame produced Deposition

Alcohol Non-luminous flame Carbon not deposited

Camphor Smoky flame Carbon deposited

Naphthalene Smoky flame Carbon deposited

MBD Super Refresher Science Class-XTerm-II CBSE /NCERT

Publisher : MBD GroupPublishers

ISBN : 9789385905872Author : Khosla, kapoor,Wadhawan

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