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TABLE OF CONTENTSCLASS - X (MATHS)

Page NumberALGEBRA : 02

Chapter 1. Quadratic Equations

Chapter 2. Arithmetic Progressions

GEOMETRY : 06

Chapter 3. Circles

Chapter 4. Constructions

MENSURATION : 11

Chapter 5. Area related to Circles

Chapter 6. Surface Areas and Volumes

TRIGONOMETRY : 19

Chapter 7. Some Applications of Trigonometry

COORDINATE GEOMETRY : 21

Chapter 8. Coordinate Geometry

PROBABILITY : 25

Chapter 9. Probability

CLASS - X (SCIENCE)PHYSICS :

Chapter 1. Light - Reflection and Refraction 28CHEMISTRY :

Chapter 2. Carbon and Its Compounds 37Chapter 3. Periodic Classification of Element 46


SOME BASIC FORMULAE

(i) 2 2 22a b a ab b

(ii) 2 2 22a b a ab b

(iii) 2 2a b a b a b

(iv) 2x a x b x x a b ab

(v) 2 2 2 2 2 2 2a b c a b c ab bc ac

(vi) 3 3 2 2 33 3a b a a b ab b

(vii) 3 3 2 2 33 3a b a a b ab b

(viii) 3 3 2 2a b a b a ab b

(ix) 3 3 2 2a b a b a ab b

(x) 3 3 3 3x y z xyz

= 2 2 2x y z x y z xy yz zx

= 2 2 212

x y z x y y z z x

(xi) If 0x y z , then 3 3 3 3x y z xyz

1


CHAPTER 1: QUADRATIC EQUATIONS

1. ax2 + bx + c = 0, represents a quadratic equation where 0a and a, b, c are realnumbers.

2. The values of x satisfying the equation ax2 + bx + c = 0 are called roots (solu-tions) of the equation.

3. The roots of the equation ax2 + bx + c are D

2b

a

and D

2b

a

whereD = b2 – 4ac and is called discriminant.

4. Applications of Quadratic Equations :

(i) Translating the word problem into symbolic language which means identi-fying relationships existing in the problem and then forming the quadraticequation.

(ii) Solving the quadratic equation thus formed.

(iii) Interpreting the solution of the equation, which means translating theresult of mathematical statement into verbal language.

ALGEBRA

2

Some Key Points


The completion of square method of finding the solution of the quadratic equationhelps us in knowing the nature of roots of the quadratic equation.

SOLUTIONS OF A QUADRATIC EQUATION

3

Method 1 Method 2


FINDING THE NATURE OF ROOTS OF A QUADRATIC EQUATION

4


CHAPTER 2 : ARITHMETIC PROGRESSIONS

also,an = Sn – Sn – 1

1. Arithmetic Progression : A sequence in which the difference obtained by sub-tracting from any term to its preceding term is constant throughout, is called anarithmetic sequence or arithmetic progression and is abbreviated as A.P.

2. The sum of 1st ‘n’ positive integers is given by 1S

2n

n n

Some Key Points

5


GEOMETRY

1. Circle : A circle is the locus of a point which moves in a plane insuch a way that its distance from a given fixed point is alwaysconstant.

2. Locus is a set of points which follow some geometric conditions.3. Secant : - If the line PQ and the circle have two common points, then

the line PQ is termed as a secant of the cirlce.4. Tangent : - If the line PQ and the circle have only one common

point, then the line PQ is termed as a tangent to the cirlce.5. There is one and only one tangent to a circle passing through a

point lying on the circle.6. There are exactly two tangents to a circle through a point lying

outside the circle.

CHAPTER 3: CIRCLES

6

Some Key Points


CHAPTER 4 : CONSTRUCTIONS0. CIR1

CONSTRUCTION 1To divide a line segment internally in a given ratio m : n

(e.g., Let m = 3, n = 2)Steps of Construction :(i) Draw any ray AX, mak-

ing an acute angle withAB.

(ii) Locate 5 (= m + n) pointsA1, A2, A3, A4, and A5 onAX such thatAA1 = A1 A2 = A2 A3 = A3A4 = A4 A5.

(iii) Join BA5

(iv) Through the point A3 (m = 3), draw a line parallel to A5B (by making anangle equal to 5AA B ) at AA3 intersecting AB at the point C.

Then, AC : CB = 3 : 2

Division of line segment AB in the ratio 3 : 2

7

Memory Tips If the ratio is given as m : n, then divide AX into (m + n) parts.

If the question says AC is mn of AB, then divide AX into n parts.


To construct a triangle similar to a given triangle as per the given scale factor.Scale Factor: The ratio of the sides of the triangle to be constructed with thecorresponding sides of the given triangle is known as their Scale Factor.

If the scale factor is 34 , then the corresponding sides will be

34 of the given triangle.

CONSTRUCTION 2

Case I Case II

Scale Factor

34

53

Draw a ray BX making an acute angle with BC on the side opposite to the vertex A.

Draw a ray BX making an acute angle with BC on the side opposite to the vertex A.

Locate 4 points B1, B2, B3 and B4 on BX so that BB1=B1B2=B2B3=B3B4.

Locate 5 points B1, B2, B3, B4 and B5 on BX so that BB1=B1B2=B2B3=B3B4=B4B5.

Jo in B4C and draw a line through B3 parallel to B4C to intersect BC at C.

Join B3 to C and draw a line through B5 parallel to B3C, intersecting the extended line segment BC at C.

Draw a line through C parallel to the line CA to intersect BA at A.

Draw a line through C parallel to CA in tersecting the extended line segment BA at A.

ABC is the required triangle. ABC is the required triangle.

8


CONSTRUCTION 3

To construct a tangent to a circle from a point outside it.Given a circle with centre at O and a point P outside it. Now construct two tangentsfrom P to the circle.

Steps of Construction:1. Join PO and bisect it. Let M be the mid-

point of PO.2. Taking M as centre and MO as radius,

draw a circle. Let it intersect the given circleat the points Q and R.

3. Join PQ and PR.4. Thus, PQ and PR are the required tangents.

CONSTRUCTION 4

To draw tangents to a circle from a point outside the circle without using its centre.

Steps of Construction :

(i) Draw any two non-parallel chords AB and CD ofthe circle.

(ii) Construct bisectors OX and OY of chords ABand

CD respectively, intersecting at point O.

(iii) O is the centre of the circle.

Proceed as in steps 1, 2, 3 of the previous example.

9


CONSTRUCTION 5

To construct pair of tangents to a given circle, inclined to each other at an angle of 60o.

Steps of Construction :

Step 1 : Take a point O on the planeof the paper and draw acircle of radius OA = r cm

Step 2 : Extend OA to B such thatOA = AB = r cm

Step 3 : With A as centre draw acircle of radius OA = AB = r cm. Suppose it intersects the circledrawn in step 1 at the point P and Q.

Step 4 : Join BP and BQ. Then BP and BQ are the required tangents whichare inclined to each other at an angle of 60o.

Scale factor could be given any value say mn .

If m < n, then the required triangle is smaller than the given triangle.

If m > n, then the required triangle is larger than the given triangle.

10


CHAPTER 5 : AREAS RELATED TO CIRCLES CIR1

MENSURATION

11


Perimeter and Area of a Circle

1. Circumference and perimeter are

synonymous.

2. Knowing radius, perimeter and area

can be found.

3. The value of π is 227 or 3.14.

1. To calculate the cost of fencing acircular field, where dimensionsof the field are given.

2. To calculate the revolutionstaken by the wheel.

3. To calculate the area or theperimeter of the circle whichcould be a combination of twoconcentric circles whose radii aregiven.

Points to Remember Types of Questions

Sector and Segment of a Circle

12


Calculations of Angles Related to Clock

1. Angle described by the minute hand in 60 minutes = 360°.

Angle described by the minute hand in

1 minute = 36060

=6°.

2. Angle described by the hour hand in 12hours = 360°.

Angle described by the hour hand in

1 hour = 36012

= 30°.

Angle described by the hour hand in 1 minute =3060

=12

Thus, the hour hand rotates through 12

in 1 minute.

1. The sum of the arcs of major andminor sectors of a circle is equalto the circumference of the circle.

2. The sum of the areas of the majorand minor sectors of a circle isequal to the area of the circle.

1. To calculate the area swept bythe minute hand at a given timeif its length is given.

2. To calculate the area grazed bythe horse tied at one corner of afield.


13


Combination of Plane Figures

1. If two circles touch internally, thenthe distance between their centresis equal to the difference of theirradii.

2. If two circles touch externally, thenthe distance between their centresis equal to the sum of their radii.

3. Distance moved by the rotatingwheel in one revolution is equal tothe circumference of the wheel.

The number of revolut ionscompleted by the rotating wheel in1minute =

1. To calculate the area ofcombination of two differentfigures.

2. To calculate the area of theflower bed, which could be asemi-circle, attached to a squarelawn.

3. To calculate the area of theshaded region from a givenfigure.


Distance moved in 1 minuteCircumference

14


CHAPTER 6 : SURFACE AREAS AND VOLUMES

IMPORTANT FORMULAE AT A GLANCE

SOLID

CSA (Open)

in square units

TSA (Closed) in square

units

Volume in cubic

units Illustration

Cube 4a2 6a2 a3

Cuboid 2 h l b

2 lb bh lh lbh

15


Cylinder 2πrh 2πr r h

2πr h

Hollow cylinder 2h(R + r) 2(R + r)

(R + h – r) h (R2 – r2)

Cone πrl

2 2l h r πr r l 213πr h

Sphere 24πr 24πr 34

3πr

Hemi-sphere

22πr 23πr 323πr

Hemi-spherical shell

Inner = 2r2 Outer = 2R2 +

(R2 – r2)

3 343π R r

16


1. Other Important Formulae

Length of largest diagonal of cuboid = 2 2 2l b h units

Diagonal of a cube = 3 unitsa Density = Mass

Volume2. Combination of Solids : Consider the combination of

cylinder and a cone.Total surface area = Curved surface area of the cylinder + Curvedsurface area of the cone.Required volume = volume of Cylinder + volume of cone

3. Conversion of a solid from one shape to another.(i) Volume remains same. (ii) Surface area changes.

4. Metric Conversions

Physical Qantity

10 10 10 10 10 10 10

Length Km hm dam m dm cm mm

Volume Kl hl dal l dl cl ml

Mass Kg hg dag g dg cg mg

10 10 10 10 10 10 10

Small to Big

Units

Big to Small

King Henry Died, Mother Did’t Cried MuchUsing above sentence you can remember this table.

17

Some Key Points


To Convert One Unit to OtherCase I: From big to smallMultiply the quantity to be changed byn times 10, where n is number of timesyou move position/place. Eg. 2 Km = .......... cm.from Km to cm you have to move 5 places towards right side hence2 km × 5 times 10 = 2 × 10 × 10 × 10 × 10 × 10 = 200000 cm

Case II: From small to bigDivide the quantity to be changed by n times 10, where n is number of times youmove position/place. Eg. 5 mm = ................... m.from mm to m you have to more 3 places towards left side hence

5 mm ¸ 3 times 10 = 5 0.005m

10 10 10

1 litre 1000 cm 3 For liquids & gases1 m 3 1000 litres (1 K l) For liquids & gases1 m 3 100,00,00 cm 3 For solids1 cm 3 1000 mm 3 For solids

C onvertions for volum e

Area: is the size of the region enclosed within a plane figure. Area is measuredin square units. The most frequently used units of area are mm2, cm2 and m2.Land areas are often given in hectares (ha)

1 are = 100 m2, 1 ha = 10,000 m2

Embankment of Well

Area of the embankment (shaded region)= (R2 – r2)

Height of the embankment :

= Volume of the earth dug out

Area of the embankment

18

Rr


TRIGONOMETRY

CHAPTER 7 : SOME APPLICATIONS OF TRIGONOMETRY

Definitions

(i) The Line of Sight is the line drawn from the eye of an observerto the object being viewed by the observer.

(ii) The Angle of Elevation of an object viewed is the angleformed by the line of sight with the horizontal when theobject is above the horizontal level, i.e., the case when weraise our head to look at the object.

(iii) The Angle of Depression of an object viewed is the angleformed by the line of sight with the horizontal when theobject is below the horizontal level, i.e., the case when welower our head to look at the object.

To solveheight anddistanceproblems,consider aright-angledtriangle.

Tip

19


A

CC

O

DA

REMARKS

1. Since the alternate angles made by a transversalon two parallel lines are equal, angle of depres-sion of P as seen fromO = angle of elevation of O as seen from P.

AOP = OPB

2. The height or length of an object or the distance between two distant objects canbe determined with the help of trigonometric ratios.

TO FIND THE HEIGHT OF AN OBJECT

To find the height of an object BC, the followinginformation is needed:-

1. The distance AB

2. The angle of elevation, BAC

Now use trigonometric ratios of BAC to calculate BC.

Original height of tree = AC + DC

Where C is the pointfrom where tree breaks

TO FIND HEIGHT OF TREE BROKEN BY WIND

20


CHAPTER 8 : COORDINATE GEOMETRY

CO-ORDINATE GEOMETRY

Note the subscripts of x and y and observe that they follow the followingsequence: 1 2 3, 2 3 1, 3 1 2

21


Position of any point in a plane, is represented by a pair of coordinate axes.

Thus, the Cartesian co-ordinates of a point are given by

(i) Abscissa

(ii) Ordinates

1. Abscissa

The distance of a point from the y-axis is called its x-coordinate or abscissa.

2. Ordinate

The distance of a point from the x-axis is called its y-coordinate or ordinate.

‘x’ comes before ‘y’ and ‘a’ comes before ‘o’ so

x - coordinate = abscissa , y - coordinate = ordinate

The coordinates of a point on the x-axis are of the form (x, 0) and the coordinatesof the point from the y-axis are of the form (0, y).

In cartesian co-ordinate system, the orthogonal axes partition the plane intofour disjoint parts each of which is a quadrant.

Table showing the signs of the coordinates of a point in different quadrants:

Quadrant x-coordinate y-coordinate Example

First +ve +ve (4,7)

Second –ve +ve (–3,2)

Third –ve –ve (–5,–6)

Fourth +ve –ve (6,–3)

22

Some Key Points


4. Distance Formula

(i) Distance between two points P (x1, y1)and Q(x2, y2) is

PQ = 2 22 1 2 1x x y y units

(ii) Distance of a point P (x, y) from the origin

O (0, 0) is given by OP = 2 2x y

(iii) A (x1, y1), B(x2, y2) and C (x3, y3) are three collinear points if AB + BC = AC

5. Section Formula

(i) The coordinates of the point which divides theline segment joining the points, (x1 , y1) and (x2,y2) in the ratio m : n internally are given by

2 1 2 1,mx nx my nym n m n

3. Quadrants and Sign Convention

As shown in the figure, theco- ordinate axes divide aco-ordinate plane into fourequal parts which areknown as quadrants.

n

X

Naming is anti clockwise

+,+–,+

23


(ii) Mid point of a line segment joining the points(x1, y1) and (x2, y2) is given by

1 2 1 2,2 2

x x y y

(iii) Centroid (G) of the trianlge with verticesA(x1, y1), B(x2, y2) and C(x3, y3) is given by

1 2 3 1 2 3,3 3

x x x y y y

6. Area of a ΔABC with vertices A (x1, y1),B (x2, y2), and C (x3, y3),

= 1 2 3 2 3 1 3 1 212

x y y x y y x y y

7. If x y , then , ,x y y x and if x = y, then (x, y) = (y, x)

24


PROBABILITY

CHAPTER 9 : PROBABILITY

1. Probability : It is defined as the ratio of number of favourable outcomes to thenumber of total outcomes.If m is the number of favourable outcomes and n the number of total outcomes,

then mp ,0 p 1n

. Also p can’t be negative.

Probability

Random Experiments

Theoretical ProbabilityExperimental Probability

Events

Sure Event Probability is 1

Impossible EventProbability is 0

Elementary EventEvent having only 1 outcome

Combination of 2 or more elementary events

Compound Event

Complement of a Event Not A or A

Quantitative measure of certainity

It is based on what has actually happened

It is based on the prediction of what will happen on the basis of certain assumptionsAny activity which is associated

with a certain outcome

The collection of all or some of the possible outcomes

Definitions

25


2. Complementary Event: Let E be an event. The complementary event of Edenoted by E or E is the event describing the non-occurrence of E.

Clearly, P(E) + P(not E) = 1 or P(E) + P( E ) = 1

P(E) = 1 – P(not E). or P(not E) = 1 – P(E)

3. Compound Event - Combination of 2 or more elementary events4. The probability of a certain event is 1; and the probability of an impossible

event is 0.5. An event having only one outcome is called an elementary event.

Key Points

1. Probability of occurrence of an event:

Number of outcomes favourable to EP(E) =Total number of possible outcomes

2. Occurrence of an event: Any one of the elementary events associated to theevent A is an outcome.

3. For any event P(A) + P(A) = 1 .

4. Sum of the probabilities of all the elementary events of an experiment is 1.

5. The number of outcomes favourable to the event E is always less than orequal to the number of all possible outcomes.

Or numerator is always less than or equal to the denominator.

Or 0 P(E) 1

6. As the number of trials in the experiment goes on increasing, the experimentaland theoretical probability tends to get same.

7. Probability of an event always lies between 0 and 1, i.e. it cannot be negative.

26


Playing Cards

(i) It consists of 52 cards which are divided into 4 suits of 13 cards each –spades ( ), hearts ( ), diamonds ( ) and clubs ( ).

(ii) Clubs and spades are of black colour.(iii) Hearts and diamonds are of red colour.(iv) Cards in each suit are ace, king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3, and 2.(v) Kings, queen and jack are called face cards.

27

Sample Space


CHAPTER 1 : LIGHT REFLECTION AND REFRACTION

1. Laws of Reflection : The incident ray,the normal at the point of incidence andthe reflected ray, all lie in the same plane.i.e. AO, ON and OB all lie on the sameplane.For all mirrors

i r

i = incident ray, r = reflected ray2. Nature of Images Formed by Plane Mirror :

PHYSICS

Mirror rorriM

Erect and virtual

Same size as the object

Laterally inverted

28

Some Key Points


Concave

mirror:

Convex

mirror

Spherical mirror whosereflecting surface is curved

inwards.Definition

Spherical mirror whosereflecting surface is curved

outward.Definition

Principalaxis

Reflection in a Concave mirror

Reflection in a Convexmirror

Principal axis

29

3. Spherical Mirror

Pole (P) : Centre of the reflecting surface of a spherical mirror.

Centre of Curvature (C) : The centre point of the sphere that the mirror is apart of, lying outside the mirror surface. For a concave mirror C is in front of itand for a convex one C lies behind it.

Length of the Aperture (MN) : Determines the periphery of the reflectingsurface.

Principal Axis (CP) : The imaginary straight line passing through the pole andthe centre of curvature, normal to the mirror at its pole.

The Principal Focus (F) : of a concave mirror is the point on the principal axiswhere a beam of light parallel to the principal axis converges to, after reflection.The principal focus of a convex mirror is the point on the principal axis fromwhere all the individual reflected rays appear to come from.

Focal Length (f) : The distance between the pole and the principal focus,denoted by f. i.e. 2 f = R (where R is radius of curvature)


Object is always placed to the left of

the mirror.

All distances are measured from the

pole of the mirror.

All distances measured to the rightof the origin (along the +x axis) aretaken as positive and those to theleft of the origin (along –x axis) aretaken as negative.

Distances measured perpendicular toand above the principal axis (along +y axis) are taken as positive and thoseperpendicular to and below theprincipal axis (along -y axis) are takenas negative.

5. Image formation by Spherical Mirrors

Note : Sign convention for all kinds of mirrors, whether plane or spherical is the same.

Position of the object

Position of the image

Size & Nature of Image

Ray diagram

At infinity At F Highly diminished

Real and inverted

Beyond C Between F and C

Diminished

Real and inverted

30

4. Sign Convention for Spherical Mirrors


At C At C Same size

Real and inverted

Between C and F

Beyond C Enlarged

Real and inverted

At F At infinity Highly enlarged

Real and inverted

Between P and F

Behind the mirror

Enlarged

Virtual and erect

31

6. Mirror Formula :

1 1 1f v u ,

f = focal length, = Radius of Curvature(R)

2v = image distance and u = object distance

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7. Magnification of a Mirror (m)

m = 'h

h

also m = vu

where 'h = height of image, h = height of object,

8. Refraction : The phenomenon due to which a ray of light travelling from onemedium to another, bends or deviates slightly from its actual path.

Refraction occurs due to the difference in the speed of light in the two media.

While travelling from optically rarer to denser medium, light bends towards the

normal.

From optically denser to rarer medium light bends away from the normal.9. Laws of Refraction

(a) The incident ray, the refracted ray and the normal to the interface of twotransparent media at the point of incidence, all lie in the same plane.

(b) Snell’s law: The ratio of sine angle of incidence to the sine angle ofrefraction is a constant for the light of a given colour and for given pair of

media. ie sin i constantsin r

n . This constant is the refractive index of

the second medium with respect to the first.


Concave lensof Diverging

lens :

Convex lensof

Converginglens

Def inition

Light rays on passingthrough convex lensconverge to a point,

so these are alsocalled converging

lens.

Def inition

Light rays on passingthrough concave lensdiverge from a point,

so these are alsocalled diverging lens.

n21 = Refractive index of medium 2

with respect to medium 1

Medium 1 has the incident ray

medium 2 has the refracted ray

121 21

2 12

v 1n = , nv n

121

2

speedof light in medium 1 vn =speed of light in medium 2 v

212

1

speedof light in medium 2 vn =speed of light in medium 1 v

33

10. Refractive Index of a Medium with respect to another Medium

11. Spherical Lens


Object

pos i tion

Image

pos i tion

Image size &

Image Type

Atinfinity

Atfocus

F1

Highlydiminished,point sizes

Erect andvirtual

Betweeninfinity

andopticalcentre

Betweenfocusand

opticalcentre

Diminished

Erect andvirtual

Object

pos i tion

Image

pos i tion

Image size &

Image Type

BetweenF1 and 2F1

Beyond2F1

Magnified

Inverted andreal

Diagram

Diagram

34

12. Image Formation by Concave Lenses

13. Image Formation by Convex Lenses


Object

Pos iti on

Image

Pos ition

Image size &

Image Type

At focus Atinfinity

Vastlymagnified

Inverted andreal

Betweenfocusand

opticalcentre

On thesame

side of thelens as the

object

Magnified

Erect andvirtual

Atinfinity

Atfocus

Highlydiminished,point sized

Inverted andreal

Beyond2F1

BetweenF2 and

2F2

Diminished

Inverted andreal

At2F1

At 2F2Same size asof the object

Inverted andreal

Diagram

35


14. Lens Formula 1 1 1= –f v u

15. Magnification of a Lens (m) : m = 'h

h also m = vu

16. Power of Lens (P) : P = 1

f in m , for convex lens, power is + ve

where f is in metres, P is in dioptres for concave lens, power is – ve

Combination of power. 1 2 3P = P + P + P .....

(i) SI unit of power : dioptre (D)

(ii) 1 dioptre is the power of a lens whose focal length is 1 m.

(iii) To calculate power, focal length must be in metre

(iv) Power of convex lens is positive and that of concave lens is negative.17. Important Formulae

Concave Mirror

Convex Mirror

Concave lens

Convex lens

Formula 1 1 1+ =v u f

1 1 1+ =v u f

1 1 1– =v u f

1 1 1– =v u f

u Negative Negative Negative Negative

f Negative Positive Negative Positive

m vu v

u – vu – v

u

P - -

1f

1f

P - - Negative Positive

where, u = object distance, v = image distance, f = focal length, R = 2f,

R = radius of curvature; f = focal length , h' = height of image,

h= height of object, P = power, m = magnification

36


CHEMISTRY

CHAPTER 2 : CARBON AND ITS COMPOUNDS

Symbol : C Atomic No: 6 Atomic Mass : 12

Position in Modern Periodic table : 4th Group, 2nd Period Type : Non Metal

1. Unique Properties of Carbon: Remember, these three properties enable

carbon to form a very large number of compounds to the mutual exclusion ofother elements. Such compounds are called organic compounds because suchcompounds are present in all life forms and are vital to their existence.

2. Bonding in Carbon: Carbon atoms form covalent compounds by mutualsharing of its valence electrons with valence electrons of other atoms.

(a) Straight Chain Compounds: In these compounds, carbon atoms arearranged in linear fashion leading to a straight chain of carbon atoms.

37

i t


10 carbon atoms are arranged in a linear fashion in Decane.

(b) Branched Chain Compounds: In thesecompounds, the carbon chain splits off in one ormore directions. In the compound shown belowthe carbon chain has three branches, i.e., theparent chain splits into three segments.

(c) Cyclic Compounds or Rings: In these compounds, thecarbon backbone is linked so as to form a ring or loop andthe terminus of the backbone is not present.

Straight chain, branched chain and cycliccarbon compounds, all may be saturatedand unsaturated. For example, benzene is

an unsaturated cyclic compound.

(d) Structural Isomers: These are compounds having same molecular formulabut different structures.

Pentane can have two different structures which are as follows:

Straight chain structure:

Branched chain structure:

38


3. Important Functional Groups: Functional group is that specific group ofatoms (or a single atom in certain cases) in an organic compound that determinesthe specific properties of that compound.

5. Homologous Series: A homologous series is a series of organic

compounds with a similar general formula, possessing similar chemical propertiesdue to the presence of the same functional group.

Class of Organic Compounds

General Formula of the Homologous Series

Suffix

Alkanes CnH2n + 2 –ane

Alkenes CnH2n –ene

Alkynes CnH2n – 2 –yne 4. Nomenclature of Organic Compunds (IUPAC naming): The general rules for

naming organic compounds containing functional groups are summarised asfollows:

Halogens Alcohol Aldehyde Ketone Carboxylic acid

– Cl, – Br – OH O

C H C = O

O

C OH Prefix: Chloro, Bromo

Suffix: -ol Suffix: -al Suffix: -one Suffix: -oic acid

39


Fu nctiona l g roup P re fix suf fix Exa mple

H aloge n R – X

P re fix-c h loro , b r om o, etc . S uf fix – a ne

C hloropr opa ne

B rom opr opa ne

A lc ohol R – OH Suf fix – ol

P ropa nol

A ldeh yde

S uffix – al

P ro pana l

K e to ne

S uffix - one

P ropan one

Carboxylic acid Suffix - oic

acid Propanoic acid

Single bond(Alkane) Suffix-ane

Methane

Double bond (alkenes) Suffix - ene

Propene

Triple bond (alkynes) Suffix - yne

Propyne

40


5. Summary of Systematic Approach: The approach you should take issummarised in the following flowchart:

6. Chemical Properties of Carbon Componds

(i) Combustion : 2 2C O CO + heat & light

• Carbon, in all its allotropic form, burns in O2 to give CO2 alongwitha large amount of heat and light.

• Saturated hydrocarbons generally give a clean, blue fiame whileunsaturated carbon compounds give a yellow flame with lots of blacksmoke

• Limiting the supply of air results in incomplete combustion, of evensaturated hydrocarbons, giving a sooty flame.

(ii) Oxidation :• Complete combustion (oxidation) of carbon forms carbon dioxide, CO2

• Incomplete combustion of carbon forms carbon monoxide, CO• Alcohols are converted to carboxylic acids on oxidation

4

2 2 73 2 3

Ethanol ethanoicacid acetic acidCr

alkaline KMnO Heator acidified K O HeatCH CH OH CH COOH

41


(iii) Addition Reactions :

Unsaturated hydrocarbons catalysts saturated hydrocarbon

2

Ni / PalladiumH

(iv) Substitution Reactions :

In the presence of sunlight chlorine can replace the hydrogen atoms or asaturated hydrocarbon, one by one.

7. Some Important Carbon Compounds

Compound Structure Common Name

State at room

temperature

Solubility

Ethanol C2H5OH

H

C O HH

H

C

H

H

Ethyl alcohol liquid Soluble in

water

Ethanoic Acid

H

C O HH

H

C

O

acetic acid or vinegar

Liquid , freezes during winter

Soluble in water

3CH COOH

H H

H

HH

H

H

H

H

OH

O

C C

C C

4 2CH Cl 3CH Cl HCl

sunlight

sunlight

Cl2

42

Vegetable oil (liquid) Vegetable ghee (solid)


8. Uses of Carbon Compounds

Ethanol : 1. Is an active ingredient of all the alcoholic drinks.2. In medicineslike tincture iodine, cough syrups & many tonics. As a solvent.

Ethanoic acid : As a preservative in pickles.

9. Chemical Reactions

Compound Chemical Reactions

Reaction with sodium: Ethanol on reaction with sodium evolves

hydrogen and results in the formation of sodium ethoxide. – +

2 5 2 5 2Ethanol Sodium ethoxide

2C H OH + 2Na 2C H O N a H

Reaction with conc. sulphuric acid: Ethanol looses water on

reaction with hot and concentrated H2SO4 (which acts as a

dehydrating agent) and gets converted into alkene. Heating should

be done

at 443 K. 2 4Hot conc. H SO

3 2 2 2 2etheneethanol

CH CH OH CH CH H O

Ethanol

Denatured Alcohol Denatured alcohol is ethyl alcohol which has

been made unfit for drinking purposes by adding small amounts of

poisonous substances like methanol, pyridine, copper sulphate etc.

43


The two reactions are acid-base reactions. The H-atom attached to the O-atom of ethanoic acid is replaced by Na atom.

3 3 2Sodium ethanoate

CH COOH + NaOH CH COONa H O

3 2 3 3 2 2Sodium acetate

2CH COOH + Na CO 2CH COONa CO H O

Esterification: Reaction between a carboxylic acid and an alcohol in

presence of a mineral acid. The final product is known as ester (sweet

smelling compounds). acid

3 3 2 3 2 3CH COOH + CH CH OH CH COOCH CH

Ethanoic acid

Saponification is the reverse of esterification. It is used for

preparing soaps. acid/base

3 2 3 3 3 2CH COOCH CH CH COOH + CH CH OH

10. Soaps and Detergents

Soaps Detergents

They are sodium salts of long chain carboxylic acids.

They are ammonium or sulphonate salts of long chain carboxylic acids.

They are not suitable for washing purposes in hard water because they form scum with Ca and Mg salts in it.

They are suitable for use in hard water since they do not form scum with Ca and Mg salts in it.

44


11. Steps Involved in Cleaning Action of Soaps

1. Soap molecules gather to form micelles.

2. Dirt goes inside the hydrophobic (water

repelling) centre of the micelle.

3. Clothes are made free form dirt by rinsing

away the micelles.Micelle (molecularaggregates fomed

by surfactants in solution)

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CHAPTER 3 : PERIODIC CLASSIFICATION OF ELEMENTS

1. Approach to Periodic Classification of Elements

Dobereiner’s Newland’s Mendeleev’s Modern PeriodicTraids Law of Octaves Periodic Table Table or Long Form

of Periodic Table

1817 1866 1869 1913

2.

3. Newland’s Law of Octaves

Concept / Law Drawbacks or Limitations

If the known elements were arranged in increasing order of their atomic masses, every 8th element displayed the properties of the 1st element, like the 8th note in the octave of music (Sa, Re, Ga, Ma, Pa , Da, Ni, Sa).

1. The elements after calcium did not follow the law of octaves.

2. The properties of the newly discovered elements could not be explained by the law of octaves.

3. At places, more than one element was put in the same slot but these elements had different properties compared to the other elements of the same column.

Dobereiner’s Triads

Concept / Law Success Drawbacks or Limitations

In the triads (group of 3 elements having similar properties), the atomic mass of the middle element was approximately the arithmetic mean of the other two.

Three such triads were identified which are: 1. Li, Na, K 2. Ca, Sr, Ba 3. Cl, Br, I

Other elements were not able to form such triads and thus, th is system of classification of elements was rejected.

46


4. Mendeleev’s Periodic Table

Features

63 elements known at that time were arranged into 8 vertical columns (groups) and 6 horizontal rows (periods) based on their atomic mass and similarity of their properties.

Mendeleev also took in to account the formulae of the hydrides and oxides formed by elements.

Mendeleev’s Periodic Law

The physical and chemical properties of the elements are the periodic function of their atomic masses.

Achievements

1. Mendeleev left few gaps in his periodic table and predicted that in fu ture some elements will be discovered which, according to their properties, will fit exactly in the gaps.

2. Noble gases were not known at the time of Mendeleev. But later on, they were discovered and they could be placed in the Mendeleev’s periodic table.

Limitations 1. Position of hydrogen was not justified. Hydrogen resembled both metals and non-metals in its properties and thus, it couldn’t get a fixed place in the periodic table.

2. Positioning of isotopes were questioned because despite being the atoms of the same element, isotopes were put in different groups.

3. Though Mendeleev arranged the elements in the increasing order of their atomic masses, in some p laces he had put certain elements with higher atomic masses before those with lower atomic masses.

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5. Important Periodic Trends

Non-

Non-metals gain valence e-

48


6. To locate the position of an element in the Modern Periodic Table

• Period = the total number of shells, K, L, M etc. in the electronicconfiguration of the element

• Group = (C) the number of valence electron (in other most shell )

Gradiation in Properties Modern Periodic Table

Property Dow n the groupAcross the per iod

(left to right)

1 Valency Co nstantIncreases con secu tively

from 1 to 4 and then decreases to 1

2 A to mic size Increases D ecrease

3 Metallic pro per ties

Increases

D ecreases ; beco mes metalloids (or semi-

meta ls) as w e reach the middle and th en we

encou nter non-metals.4

Non metallic characte r D ecreases In creases

5 Chemica l reactivity

Increases

6 Natu re of oxide No

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Seize The Day

Seize the day, for time is passing,

Seize the day, your days amassing.

Your time to live, and do, and be,

Is now for you a mystery.

To solve each day as it unfolds,

And watch the treasure it behonds.

Now understand your precious time,

Will one day just be over.

And understand the time to live,

Is now, not under clover!

So seize the moment, seize the time,

Seize that precious day.

And fill your life with happiness,

For this is your today !

------ Marlene Rose

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Notes

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