MALHOTRA BOOK DEPOT(Producers of Quality Books)

Based on the latest syllabus and textbook(s) issued by CBSE/NCERT

Science10

TERM IIBy

KhoslaKapoor

Wadhawan

Edited ByJaya Sharma

L.B. MishraSurabhi Jain & Sonika Tyagi

Price of Term I & II : ` 430.00Term I : ` 240.00Term II : ` 220.00

OUR ADDRESSES IN INDIA New Delhi : MBD House, Gulab Bhawan, 6, Bahadur Shah Zafar Marg Ph. 30912330, 30912301, 23318301 Mumbai : A-683, T.T.C. Industrial Area, M.I.D.C. Off. Thane-Belapur Road, Navi Mumbai Ph. 32996410, 27780821, 8691053365 Chennai : No. 26 B/2 SIDCO Estate, North Phase, Pataravakkam Ambattur Industrial Estate, Ambattur Ph. 26359376, 26242350 Chennai : Plot No. 3018, Old Y Block, 3rd Street, 12th Main Road, Anna Nagar West Ph. 23741471 Kolkata : Satyam Building, 46-D, Rafi Ahmed Kidwai Marg Ph. 22296863, 22161670 Jalandhar City : MBD House, Railway Road Ph. 2458388, 2457160, 2455663 Bengaluru : 124/31, 1st Main, Industrial Town (Near Chowdeshwari Kalyan Mantap), West of Chord Road, Rajajinagar Ph. 23103329, 23104667 Hyderabad : 3-4-492, Varun Towers, Barkatpura Ph. 27564788, 9985820001 Ernakulam : Surabhi Building, South Janatha Road, Palarivattom Ph. 2338107, 2347371 Pune : Survey No. 44, Behind Matoshree Garden, Kondhwa - Khadi Machine, Pisoli Road, At. Post-Pisoli Ph. 65271413, 65275071 Nagpur : ‘Chandrakor’, Plot No. G-15, Aath Rasta Chowk, West High Court Road, Laxmi Nagar Ph. 2248104, 2248106, 2248649, 2245648 Ahmedabad : Godown No. 10, Vedant Prabha Estate, Opp. ONGC Pumping Station, Sarkhej Sanand Road, Sarkhej Ph. 26890336 Cuttack : Badambadi, Link Road Ph. 2367277, 2367279, 2313013 Guwahati : Chancellor Commercial, Hem Baruah Road, Paan Bazar Ph. 2131476, 8822857385 Lucknow : 173/15, Dr. B. N. Verma Road, Old 30 Kutchery Road Ph. 4010992, 4010993 Patna : Ist Floor, Annapurna Complex, Naya Tola Ph. 2672732, 2686994, 2662472 Bhopal : Plot No. 137, 138, 139, Sector-I, Special Industrial Area, Govindpura Ph. 2581540, 2601535 Jabalpur : 840, Palash Chamber, Malviya Chowk Ph. 2405854 Goa : H. No. 932, Plot No. 66, Kranti Nagar (Behind Azad Bhawan), Alto Porvorim, Bardez Ph. 2413982, 2414394 Jaipur : C-66A, In front of Malpani Hospital, Road No.1, V.K. Industrial Area, Sikar Road Ph. 4050309, 4020168 Raipur : Behind Aligarh Safe Steel Industries, Vidhan Sabha Road, Avanti Bai Chowk, Lodhi Para Pandri Ph. 2445370, 4052529 Karnal : Plot No. 203, Sector-3, HSIDC, Near Namaste Chowk, Opp. New World Ph. 2220006, 2220009 Shimla (H.P.) : C-89, Sector-I, New Shimla-9 Ph. 2670221,2670618 Jammu (J&K) : Guru Nanak College of Education, Jallo Chak, Bari Brahmana Ph. 2467376, 9419104035 Ranchi (Jharkhand) : Shivani Complex, 2nd Floor, Jyoti Sangam Lane, Upper Bazar Ph. 9431257111 Sahibabad (U.P.) : B-9 & 10, Site IV, Industrial Area Ph. 3100045, 2896939 Dehradun (Uttarakhand) : Plot No. 37, Bhagirathipuram, Niranjanpur, GMS Road Ph. 2520360, 2107214

DELHI LOCAL OFFICES:Delhi (Shakarpur) : MB 161, Street No. 4 Ph. 22546557, 22518122Delhi (Daryaganj) : MBD House, 4587/15, Opp. Times of India Ph. 23245676Delhi (Patparganj) : Plot No. 225, Industrial Area Ph. 22149691, 22147073

We are committed to serve students with best of our knowledge and resources. We have taken utmost care and attention while editing and printing this book but we would beg to state that Authors and Publishers should not be held responsible for unintentional mistakes that might have crept in. However, errors brought to our notice, shall be gratefully acknowledged and attended to.

© All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise without the prior written permission of the publisher. Any breach will entail legal action and prosecution without further notice.

Published by: MALHOTRA BOOK DEPOT MBD House, Railway Road, Jalandhar City.Printed at: M. GULAB SINGH & SONS (P) LTD. B-5/14, Site IV, Industrial Area, Shahibabad (U.P.)

ContributorsProject Manager: Manish Sharma Composition & Layout: Dhirender Singh Negi and Team

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 Punjabi 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.

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 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. 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

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 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 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

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 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

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

Q Elements are classified on the basis of similarities in their properties.

Q Dobereiner grouped the elements into triads.

Q Newlands gave the law of octaves.

Q Mendeleev arranged the elements in the increasing order of their atomic masses and according to

their chemical properties and formulated a table called Mendeleev’s periodic table.

Q Mendeleev left some gaps in his periodic table and predicted the existence of some yet to be

discovered elements.

Q There were some anomalies in Mendeleev’s periodic table.

Q Henry Moseley showed that the atomic number of elements is a more fundamental property than

their atomic mass.

Q In the modern periodic table, the elements are arranged in the order of increasing atomic numbers

and it is based upon modern periodic law.

Q The modern periodic table has 7 horizontal rows called periods and 18 vertical columns called groups.

Q Elements belonging to same group in the modern periodic table have similar properties.

Q Metals are present in the left hand side of the modern periodic table while non-metals are present

on the right hand side.

Q There is a zig-zag line of elements separating metals and non-metals in the modern periodic table.

These elements have intermediate properties of metals and non-metals and are called metalloids.

Q Elements arranged in the modern periodic table show periodicity in the properties such as atomic

size, valency, metallic and non-metallic character.

IMPORTANT TERMS AND DEFINITIONS

Dobereiner’s triads: It is a group of three elements having similar chemical properties in which

the atomic weight of the middle element is the average of the atomic weight of the other two

elements.

Newlands’ Law of Octaves: It states that when elements are arranged in the ascending order of

their atomic weights, every eighth element has properties similar to the first element like the notes of

an octave of music.

CHAPTER IN A NUTSHELL

(NCERT Textbook Chapter 5)

Periodic Classification of Elements

2 CONCEPTS

3.1 Structural Organisation of a Cell

3.2 Cell Organelles

CONCEPTS

2.1 EarlyAttempt

sofClassification

ofElements

2.2 Mendeleev’s Periodic Table

2.3 The Modern Periodic Table

2.4 Periodic Trends

CBSE_SCI_G10_C02_TII.indd 36

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

36

Each sub-topic dealt with as a complete unit

MBD Super Refresher Science-X

38

2.2 Mendeleev’s Periodic Table

Dmitri Ivanovich Mendeleev classified elements on the basis of their atomic masses. Mendeleev classified

63 elements known at that time in a table called Mendeleev’s periodic law. Mendeleev left some gaps for

the elements undiscovered at that time in his periodic table and thus achieved a much better format for

the classification of elements. However, Mendeleev’s classification had many drawbacks. These are:

(i) Mendeleev could not assign a correct position for hydrogen.

(ii) Isotopes of elements were not given separate positions.

(iii) Position of isotopes of many known elements is also not justified.

(iv) Many similar elements were placed together while many dissimilar elements were placed together.

NCERT Activity 1 – Page 84

Aim: To assign a correct position of hydrogen in Mendeleev’s periodic table

Procedure:1. Try to assign hydrogen a correct position in Mendeleev’s Periodic table based on its resemblance

with alkali metals and halogens.

2. Identify the group and period to which hydrogen should be assigned.

Observation: It is observed that due to its lower mass (1.008 u), hydrogen is placed in group-1along with

alkali metals. However it’s resemblance to halogens cannot be justified by this position.

Conclusion: It is concluded that a correct position cannot be assigned to hydrogen in Mendeleev’s period

table.

NCERT IN-TEXT QUESTIONS

Page 81 1.DidDobereiner’striadsalsoexistinthecolumnsofNewlands’octaves?Compareandfindout.

Yes, Dobereiner’s triads also existed in columns of Newlands’ octaves. These are:

H Li

Be

F Na

Mg

Cl

K

Ca

2.WhatwerethelimitationsofDobereiner’sclassification?

(1) Dobereiner could find only three triads from the elements known at that time.

(2) It was applicable to a few elements.

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

(1) It was applicable only up to calcium.

(2) At many places, two elements were put in the same slot in order to adjust elements in the

Newland’s table.

(3) After the discovery of noble gas, law of octave was not found to be valid.

CBSE_SCI_G10_C02_TII.indd 38

2/18/2016 7:36:54 PM

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. Define periodicity and give its cause.

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) Define implantation.

7. Define metallic character of an element. How does it vary along a period and down a group ?

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. What is biological magnification ? Will the levels of this magnification be different at different levels

of ecosystem ?

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

BM.indd 1

2/18/2016 5:36:29 PM

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.

3. Thebasisofm

odernclassification

ofelementsis___

_______.

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

arrangement help Anuj? Explain the importance of arrangement and classification of elements.

Sucharrangement

ofbooksandno

tebooksmakesit

easyforAnujtofin

dthemeasilyand

thatmakes

hislifesimpleands

ystematic.Similarly

,arrangementand

classificationofkn

ownelementsinth

eform

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

CBSE_SCI_G10_C02_TII.indd 50

2/18/2016 7:41:19 PM

CONTENTS

1 Carbon and its Compounds 1–37

2 Periodic Classification of Elements 38–63

3 How Do Organisms Reproduce? 64–91

4 Heredity and Evolution 92–114

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 M-1–M-10

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 chemicalspecies.

� The major reason of chemical bonding is the tendency of the atoms of the various elements toacquire stable configuration i.e. to complete their octet and to acquire a state of minimumenergy.

� 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 valenceshell.

� 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 betweentwo 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 sharingduring 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 pairof electrons.

� Single, double and triple covalent bonds are formed by the mutual sharing of one, two and threeelectron 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 togetherthrough 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 valenceshell.

� A large number of compounds are formed by carbon because of its tetravalency and the propertyof catenation.

(NCERT Textbook Chapter 4)

MBD Super Refresher Science-X2

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

� An atom or a group of atoms which when present in a molecule gives certain special propertiesto the compound is called a functional group; e.g. – OH or alcoholic group, – CHO or aldehydicgroup.

� A Homologous series of compounds have compounds with the same functional group, andsimilar chemical properties. The successive members of a homologous series differ by CH

2 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 carbonatoms in their molecules.

� Unsaturated hydrocarbons are the hydrocarbons, which contain one or more double or triplebonds between carbon atoms in their molecules.

� Pyrolysis is the decomposition of higher molecular weight hydrocarbons into lower molecularweight hydrocarbons on heating at a high temperature in the absence of air.

� Compounds having same molecular formula but different properties are called isomers and thisphenomenon 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 differentphysical 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 esterificationreaction.

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 methanolto 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 resultingin the formation of covalent compounds like methane (CH

4), ethane (C

2H

6), acetic acid (CH

3COOH)

etc. These compounds are called organic compounds. Carbon forms covalent bonds with otherC–atoms and some other elements such as H, O, S, N, F, Cl, Br or I. Carbon can form double or triplebonds (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 givenbelow.

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 cansay 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 upof 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 afootball 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 representedby dashes (—) and are directional in nature. The directional nature of the covalent bonds givescovalent molecules having two or more such bonds, a definite shape. This is also called definitegeometry 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 resultingin the formation of a three-dimensional network structure. Due to the presence of strong covalentbonds, diamond is hard and has high melting point. In graphite each C–atom is bonded tothree other carbon atoms to form hexagonal rings, which are held together by weak vander Waals’ forces of attraction. Therefore, graphite has a two-dimensional sheet-like structure. Dueto the presence of one free electron left with each C–atom, graphite is a good conductorof 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. Theability of carbon to form chains gives rise to homologous series of compounds in which all the memberscontain the same functional group and show similar properties. The adjacent members of a homologousseries differ by a CH

2 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

2H

5OH

(b) C2H

5OH and C

3H

7OH

(c) C3H

7OH and C

4H

9OH

Calculations:

(a) C2H

5OH – CH

3OH = CH

2

(b) C3H

7OH – C

2H

5OH = CH

2

(c) C4H

9OH – C

3H

7OH = 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 similarchemical 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 C = C(Double bond)

ene

2. Alkynes – C C –≡(Triple bond) yne

3. Aldehydes – CHO al

or

– CO

H(Aldehydic group)

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 C H – C –n 2n + 1

O

CH3 – C – CH

3

O

Propanone Dimethyl

(Alkanones) Cm

H2m + 1

ketone or

Acetone

n = m simple ketone CH3 – CH

2 – C – CH

3

O

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 thecarbon 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; burnthem 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

Conclusion:

Alcohol produces non-luminous flame while camphor and naphthalene produce smoky flame.

MBD CBSE Super Refresher ScienceClass 10 Term-2

Publisher : MBD GroupPublishers

Author : Jaya Sharma

Type the URL : http://www.kopykitab.com/product/10016

Get this eBook

40%OFF