Science
10
TERM II
By
Khosla
Kapoor
Wadhawan
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SYLLABUS
SCIENCE (CLASS–X)
Second Term
Unit No.
Unit
Marks: 90
Marks
I
Chemical Substances -Nature and Behaviour
23
II
World of Living
30
III
Natural Phenomena
29
V
Natural Resources
08
Total
Theme: Materials 90
(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.
Periodic classification of elements: Need for classification, Modern periodic table, gradation in properties,
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: Heredity; Mendel's contribution- Laws for inheritance of traits: Sex determination:
brief introduction; Basic concepts of evolution.
Theme: Natural Phenomena (23 Periods)
Unit III: Natural Phenomena
Reflection of light by curved surfaces; Images formed by spherical mirrors, centre of curvature, principal axis,
principal focus, focal length, mirror formula (Derivation not required), magnification.
Refraction; Laws of refraction, refractive index.
Refraction of light by spherical lens; Image formed by spherical lenses; Lens formula (Derivation not required);
Magnification. Power of a lens; Functioning of a lens in human eye, defects of vision and their corrections,
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 : Natural Resources
Conservation of natural resources.
Management of natural resources. Conservation and judicious use of natural resources. Forest and wild life;
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 TERM
Practicals 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.To study saponification reaction for preparation of soap.
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 obtaining the image of a distant object.
5.To trace the path of a ray of light passing through a rectangular glass slab for different angles of incidence.
Measure the angle of incidence, angle of refraction, angle of emergence and interpret the result.
6.To study (a) binary fission in Amoeba, and (b) budding in yeast with the help of prepared slides.
7.To trace the path of the rays of light through a glass prism.
8. To find the image distance for varying object distances in case of a convex lens and draw corresponding 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
Very
Short
S. No.
Typology of Questions
Answer
(VSA)
1 Mark
Short
Short
Long
Answer
Answer
Answer
Total
–I (SAI)
–II (SAII)
(LA)
Marks
2 Marks
3 Marks
5 Marks
%
Weight
age
1
Remembering (Knowledge based simple
recall questions, to know specific facts,
terms, concepts, principles, or theories,
Identify, define or recite, information)
3
–
1
1
11
15%
2
Understanding (Comprehension - to be
familiar with meaning and to understand
conceptually, 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
Higher 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,
judge, and/or justify the value or worth
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
Practical Based Questions (PBQs)
9x1=9
3x2=6
–
Total
12x1=12 6x2=12
12x3= 36
*One question of 3 marks will be included to assess the values inherent in the texts.
6x5=30 75(24)
–
15(12)
6x5=30 90(36)
100%
Super Refresher
All chapters as per NCERT
Textbook
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HELL
UTS
IN A N
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Chapter in a Nutshell
and Important Terms and
Definitions provide a complete
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the chapter
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2/18/2016
5:06:31
PM
38
I.indd
C02_TI
I_G10_
CBSE_SC
2.2
36
Highlights essential
information which
must be remembered
Includes NCERT Textbook
Activities and Exercises with
answers
CB
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Answer
Self Assessment with answers at the
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5:3
6:2
9P
M
CONTENTS
1
Carbon and its Compounds 2
Periodic Classification of Elements 38–63
3
How Do Organisms Reproduce? 64–91
4
Heredity and Evolution 5
Light: Reflection and Refraction 115–160
6
Human Eye and Colourful World 161–185
7
Our Environment 186–206
8
Management of Natural Resources 207–226
Objective Type Questions (Based Upon Practical Skills) 227–246
Answers for Self Assessments 247–254
Model Question Papers for Practice 1–37
92–114
M-1–M-10
Carbon and its Compounds
(NCERT Textbook Chapter 4)
1.1
1.2
1.3
1.4
·
Bonding in Carbon
Versatile Nature of Carbon
Chemical Properties of Carbon Compounds
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 coordinate 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.
1
MBD Super Refresher Science-X
2
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 (C2H6), acetic acid (CH3COOH)
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-X
4
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.
Structure of graphite
Structure of diamond
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 C2H5OH
(b) C2H5OH and C3H7OH
(c) C3H7OH and C4H9OH
Calculations:
(a) C2H5OH – CH3OH = CH2
(b) C3H7OH – C2H5OH = CH2
(c) C4H9OH – C3H7OH = CH2
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
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
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
1 mark each
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
C=C
ene
–C≡C–
1
Alkenes
2.
Alkynes
(Triple bond)
yne
3.
Aldehydes
– CHO
or
al
(Double bond)
–C
O
H
(Aldehydic group)
Long Answer Type Question
Give general formulae and IUPAC names of first two members of:
1. Monohydric alcohols
2. Haloalkanes
3. Monocarboxylic acids
4. Alkanones
5 marks each
MBD Super Refresher Science-X
6
Also give their common names.
S.
No.
1.
2.
3.
Homologons
Series
General
Formula
IUPAC names
Methanol
Common
names
CH3 – OH
alcohols
CnH2n + 1 – OH
where
(Alkanols)
n = 1, 2, 3
C2H5 – OH
or
or
R – OH
Chloro-
Methyl
Monohydric
Methyl
alochol
Ethanol
Ethyl
alcohol
Haloalkanes
CnH2n + 1 – X
CH3 – CH2 – OH
CH3 – Cl
or
where
C2H5Cl
methane
chloride
Alkyl halides
n = 1, 2, 3
or
Chloro-
Ethyl
X – F, Cl, Br, l
CnH2n + 1 – COOH
CH3 – CH2 – Cl
H – COOH
n = 0, 1, 2, 3
CH3 – COOH
CnH2n + 1 – C –
CH3 – C – CH3
ethane
Methanoic
acid
Ethanoic
acid
Propanone
chloride
Formic
acid
Acedic
acid
Dimethyl
ketone or
Acetone
Ethyl methyl
Ketone
Monocarbonxdic acid
O
4.
First two
members
Ketones
(Alkanones)
O
CmH2m + 1
n = m simple ketone
n ≠ m mixed ketone
where
n, m = 1, 2, 3
O
CH3 – CH2 – C – CH3
Butanone
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
Alcohol
Camphor
Naphthalene
Flame produced
Non-luminous flame
Smoky flame
Smoky flame
Deposition
Carbon not deposited
Carbon deposited
Carbon deposited
Conclusion:
Alcohol produces non-luminous flame while camphor and naphthalene produce smoky flame.
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Class 10 Term-2
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