Std.XI : Foundation Course

Std

. X

I -

Fo

un

da

tio

n

Foundation

Physics

Chemistry

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INDEX

Ch. No.

Topics Pg. No.

PHYSICS

1 Use of Log Tables 1

2 Elementary Calculus 5

3 Measurements 14

CHEMISTRY

1 Structure of Atom (Electronic Configuration)

29

2 Redox Reaction (Determination of Oxidation Number)

33

3 IUPAC - Nomenclature 49

XI.23/Phy/FC/Ch.1/Pg.1

MODULE 1

If n = bx, then x is called the logarithm of n to the base b. It is written as bx log n .

In common logarithm the base b = 10.

Formulae used in logarithmic calculations :

1. log mn = log m + log n

2. m

logn

= log m log n

3. logmn = n log m

Characteristic and Mantissa : The integral part of the logarithm of a number is called as the characteristic of the logarithm and the fractional part is called the mantissa of the logarithm.

Rules for finding the characteristics :

1. a. If the number is greater than or equal to 1 and has m digits in its integral part then the characteristic

is (m 1)

b. If the number is less than 1 and if there are p zeros immediately after the decimal point the

characteristic is (p + 1).

e.g. If n = 352.5, characteristic = 2; If n = 0.1251, characteristic is 1.

2. Alternatively write the number in scientific notation i.e., write it as a number between 1 and 10, multiplied by an appropriate power of 10. Then the power of 10 gives the characteristic.

e.g. i. if the number n = 352.5 It can be written as 3.525 102. Hence the characteristic is 2.

ii. if the number n = 0.1251 It can be written as 1.251 101. Hence the characteristic is 1 (written

as 1 )

The mantissa of the logarithm of a number is determined using log tables. For 3525 the mantissa is 0.5471 and for 1251 the mantissa is 0.0972.

Hence, log 352.5 = 2.5471 and log 0.1251 = 1 .0972

To find antilog: i. Separate the characteristic p and mantissa q of the logarithm. ii. The antilog of mantissa q is found out from the antilog table. iii. Let p + 1 = m a. If m is positive, then place the decimal point after m digits. b. If m is negative or zero then put |m| zeroes between the decimal point and the antilog of the

mantissa. e.g. 1. Let the number be 2.2015 Antilog of mantissa from the table = 1591

Antilog of the number = 159.1

2. Let the number be 3.2015

Antilog of the mantissa = 1591 Antilog of the number = 0.001591

Alternatively, we can put a decimal point after the first digit of the antilog of the mantissa and multiply it by 10 raised to the power of the characteristic.

e.g. 1. If the number is 2.2015 Antilog of the mantissa = 1591

Antilog of the number = 1.591 102

2. If the number is 3.2015

Antilog of mantissa = 1591

Antilog of the number = 1.591 103

Use of Log Tables

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(2) Vidyalankar : Foundation - Physics


Solved Examples

1. Find log of (i) 245.3 (ii) 24.53 (iii) 2.453 (iv) 0.2453 (v) 0.02453

Solution :

(i) Since, there are three digits in the integral part, the characteristics is (3 1) = 2. To find the mantissa we see the log tables.

In the first column we see 24 and in that row we see in the column under 5. We find 3892. Then we see under mean difference 3 which gives 5. This is added to 3892 which gives 3897. Hence log245.3 = 2.3897

(ii) There are two digits in the integral part. Hence the characteristics is (2 1) = 1 The mantissa part remains same as in (i). Hence, log 24.53 = 1.3897

(iii) There is one digit in the integral part. Hence the characteristic is (1 1) = 0 The mantissa part remains same as in (i). Hence log 2.453 = 0.3897

(iv) The number is less than 1. There is no zero after the decimal point. Hence the characteristic is

1 or 1 .

log 0.2453 = 1.3897

(v) The number is less than 1. There is one zero after the decimal point. Hence the characteristic is 2

or 2 .

log 0.02453 = 2.3897

2. Find antilog of (i) 2.3897 (ii) 1.3897 (iii) 0.3897 (iv) 1.3897 (v) 2.3897

Solution : (i) Here the mantissa is 0.3897. In the antilog table we see 0.38 in the first column and in that row

under 9 we find 2449. Then we see mean difference under 7 which is 4. This is added to 2449 which gives 2453. Since characteristic is 2. The decimal point is put after three digits.

Hence antilog 2.3897 = 245.3

(ii) Here since mantissa part is same as (i) but the characteristic is 1. The decimal point is but after 2 digits.


(iii) Hence again the mantissa part is same but the characteristics is 0. The decimal point is put after 1 digit.


(iv) Here the characteristic is 1 or 1, there will be 0 to the left of the decimal point and no zero to the right of decimal point.

Hence antilog 1.3897 0.2453

(v) Here the characteristic is 2 or 2. Hence there will be one zero to the right of the decimal point.

Hence antilog 2.3897 0.02453

3. Find the value of 8492 3.75

47.8 52.24

Solution :

Let y = 8492 3.75

47.8 52.24

log y log8492 log3.75 log47.8 log52.24

= 3.9290 + 0.5740 1.6794 1.7180

= 4.5030 3.3974

= 1.1056

y = antilog1.1056 = 12.76

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Notes on Use of Log Table (3)


4. Find 3 0.01562

Solution :

Let y = 3 0.01562

log y = 1

log0.015623

= 1

2.19373

= 1

(3 1.1937)3

[Adding and subtracting 1]

= 1 0.3979

= 1.3979

y = antilog 1.3979 0.2500

PROBLEMS

Classwork Problems

1. Write logarithm of i. 6024 ii. 602.4 iii. 6.024 iv. 0.6024 v. 0.06024 vi. 0.0006024

2. Write antilogarithm of

i. 2.4354 ii. 0.4354 iii. 1.4354 iv. 3.4354 v. 7.4354 vi. 9.4354

3. Evaluate the following :

i. 45.83 0.5432

0.02739

ii.

2(2.41) 2.61

1.374

iii.

35.87 0.0514

0.0578

iv. 2

3

(0.3125)

0.4629 v.

16.23

426.8

4. Find i. sin 15 ii. sin 20.5 iii. Sin 30 15 iv. if sin = 0.4245

5. Find i. cos 35 ii. cos 40.1 iii. Cos 45 34 iv. if cos = 0.8700

6. Find i. tan 25 ii. tan 45.6 iii. tan 50 25 iv. if tan = 1.1530

7. Calculate sini

sinr if i = 70 and r = 45.

8.

mAsin

2

Asin

2

if A = 60, = 1.62, find m.

9. 2 2R P Q 2PQcos and Qsin

tanP Qcos

Find the value of R and for following cases :

i. P = 15, Q = 20, = 60

ii. P = 25, Q = 50, = 120

10. 3r

T 2GM

Calculate T if r = 6.4 107 m, G = 6.67 1011 N.m2/kg2, M = 6 1024kg

11. 1 1 1

f v u Determine the value of f for the following cases (Use reciprocal table)

i. u = 11.5, v = 9.50 ii. u = 21.5, v = 30.4 iii. u = 25.45, v = 12.25

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Homework Problems

1. Find 5.38 0.47

2. 2 2R P Q 2PQcos and 1 Qsintan

P Qcos

find R and if P = 3, Q = 4, = 30

3. sini

sinr . Find r if = 1.5 and i = 60

4. V 2gR Calculate V if g = 9.8 m/s2, R = 6.4 106 m

5. 2

3

2

T GMr

4

, find the value of r if T = 24 60 60 S

G = 6.67 1011 N.m2/kg2, M = 6 1024 kg, = 3.142

6. 1 1 1

.f v u Determine the value of f if u = 30.5 and v = 52.43

Classwork Problems

1. i. 3.7799 ii. 2.7799 iii. 0.7799 iv. 1.7799 v. 2.7799 vi. 4.7799

2. i. 272.6 ii. 2.726 iii. 0.2726 iv. 0.002726 v. 2.726 107 vi. 2.726 109

3. i. 908.903 ii. 11.0329 iii. 5.648 iv. 0.1262 v. 0.1950

4. i. 0.2588 ii. 0.3502 iii. 0.5038 iv. 257

5. i. 0.8192 ii. 0.7649 iii. 0.7001 iv. 2933

6. i. 0.4663 ii. 1.0212 iii. 1.2095 iv. 494

7. 1.33

8. 48 12

9. i. R = 30.41, = 34 42 ii. R = 27.84, = 69

10. 1.608 105s

11. i. 55.44 ii. 73.42 iii. 23.61

Homework Problems

1. 1.591 2. R = 6.766, = 1712 3. r = 35 16

4. 1.12 104 m/s 5. 4.229 107m 6. 19.33

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Elementary Calculus

MODULE 1

Differential calculus:

Suppose we have a quantity y whose value depends on a single variable x and expressed by an equation, for example, y = x2

We say that y is a function of x and write y = f(x)

We can write y = f(x) = x2

If x changes by an amount x, then there is a change in the value of y by amount y.

If y = x2, then y + y = (x + x)2

y = (x + x)2 x2

= x2 + 2 x x + (x)2 x2

= 2 x x + (x)2

y

x =

22 x x ( x)

x = 2x + x

As x approaches (or tends) to zero,

y

x approaches (or tends to) 2x.

Hence we write

x 0

ylim

x = 2x.

The limit of the ratio

y

x as x approaches zero is called the derivative of y with respect to x and is written as

dy

dx. It represents the rate of change of y with respect to x. Hence if y = x2 then

dy

dx = 2x.

The process of finding derivative is called differentiation. Derivatives of some common functions are listed below :

1. d

dx(c) = 0 if c is a constant 2.

d

dx (cu) = c

du

dx

3. If y = xn, dy

dx = nxn1 4. If y = sin x,

dy

dx = cos x

5. If y = cos x, dy

dx = sin x 6. If y = tan x,

dy

dx = sec2 x

7. If y = sec x, dy

dx = tan x sec x 8. If y = cosec x,

dy

dx = cot x cosec x

9. If y = nx then dy

dx =

1

x

Some formulae related to derivatives:

1. d

dx (au) = a

du

dx, where a is a constant and u is a function of x.

2. d

dx (u + ) =

du

dx +

d

dx

3. d

dx (u ) = u

d

dx +

du

dx, where u and are functions of x

4.

d u

dx =

2

du du

dx dx

5. dy

dx =

dy

du

du

dx

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In physics, instantaneous velocity and acceleration are defined as

= t 0

x dxlim

t dt

a =

t 0

dlim

t dt

For a body moving with constant acceleration a and initial velocity u, the displacement is given by

s = ut + 1

2 at2

The velocity can be obtained by differentiating S.

ds

dt = = u + at

The acceleration is obtained by differentiating .

d

dt = a

Graphical interpretation of derivative:

Consider a graph of y = f(x) against x.

Consider a point P on the curve whose co-ordinates are (x, y) and another point Q whose coordinates are (x

+ x, y + y) as shown in fig(a). The slope of the line joining P and Q is given by tan =

y

x.

Suppose the point Q moves along the curve towards P. In this process x and y decrease and approach

zero. The line PQ then approaches the tangent to the curve at point P and y

x

= tan approaches the slope

of the tangent at P. (fig. (b))

slope (m) =

x 0

y dylim

x dx = tan

Hence, the derivative at the point P of the curve is the slope of the tangent to the curve at the point P.

In physics, the slope of the displacement time graph at any point gives the velocity

ds

dt and the slope of

velocity time graph at any point gives the acceleration d

dt

.

Solved Problems : (Differential Calculus)

1. If y 2 x 3 find dy

dx

Solution :

1/2dy d d2x 3

dx dx dx

1

12

12 x 0

2

= 1

21

xx

x

y

Q (x + x, y + y)

P(x, y)

x Fig (a)

y

Fig (b)

P(x, y)

x

y

Vidyala

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Notes on Elementary Calculus (7)


2. If y = sin3x, find dy

dx

Solution :

Let u = 3x then y = sin x

dy dy du

dx du dx

dy du

cosu, 3du dx

dy

3cos3xdx

3. Find dy

dx if y = x. log x

Solution :

dy d

dx dx(x.log x) =

dx.

dx(log x) + log x.

d(x)

dx

= x. 1

xlog x. 1 = 1 + log x. [Ans.]

4. Find dy

dx if y =

3x 1

sinx

Solution :

3dy d x 1

dx dx sinx

=

3 3

2

d dsinx. (x 1) (x 1). (sinx)

dx dx

sin x =

2 3

2

sinx.(3x ) (x 1).cos x

sin x

2 3

2

dy (3x )sinx (x 1)cos x

dx sin x. [Ans.]

5. Find dy

dx if y =

1

4x 5.

Solution :

Method I

dy d 1

dx dx 4x 5

=

2

d d(4x 5) (1) 1. (4x 5)

dx dx

(4x 5) =

2

4

(4x 5) [Ans.]

Method II

y = (4x + 5)1

1dy d(4x 5)

dx dx = 4(1) (4x + 5)2 =

2

4

(4x 5). [Ans.]

6. Find dy

dx if y = x sin x.

Solution :

dy d

dx dx (x sin x) = x

d

dx (sin x) + sin x

d(x)

dx

= x cos x + sin x. [Ans.]

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7. If s = 3 9t t

53 9

, the find v as a function of time.

Solution :

As we know v = ds

dt

v =

3 9d t t5

dt 3 9 =

3 9d t d t d(5)

dt 3 dt 9 dt

= t2 + t8 + 0

v = t2 + t8. [Ans.]

8. If s = A sin t then give relation between acceleration and displacement.

Solution :

a = dv

dt but v =

ds

dt

v = d

dt (A sin t)

v = A cos t

a = d

dt (A cos t) = A (sin t).

= A2sin t = 2 (A sin t)

a = 2 s. [Ans.]

Do you know :

The above result is the equation of Simple Harmonic Motion.

9. a = t3 + 3 Find rate of change of acceleration (a) with respect to time.

Solution : Rate of change of acceleration

w.r.t. time = 3da d(t 3)

dt dt

= 3t2. [Ans.]

PROBLEMS

Exercise 1

Classwork Problems

1. Differentiate the following with respect to x:

i. x3 ii. x iii. 1

x iv. 3x2 v. x + 2

2. Differentiate the following with respect to x:

i. 2x3/2 + 5 ii. x3 + 3

1

x + 9 iii. 34x 5

3. Differentiate the following w.r.t. x:

i. x 1 2x ii.

3x

4x 5

4. i. sin 2x ii. sin x3 iii. cos 5x iv. cos x2

Integral Calculus :

Consider the function g(x) = 3x

3. If we differentiate g(x), then we get

d

dxg(x) = x2. Then

3x

3 is the integral of

x2. Let x2 = f(x).

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Hence if d

dx g(x) = f(x) then g(x) in called the integral of f(x). We write g(x) = f(x) dx .

It represents the limit of the sum of an infinite number of terms.

An integral with lower and upper limits is known as definite integral. It is a number. Indefinite integral has no limits. It is a function.

For definite integral we can write. b

b

aa

f(x) dx g(x) = g(b) g(a)

Here, a and b are the lower and the upper limits.

For example

23

22

11

x 8 1 7x dx

3 3 3 3

Integrals of some common functions are given below:

1. i. nx dx = n 1x

n 1

, provided n 1

ii. ncx dx = cn 1x

n 1

, provided n 1, c is a constant.

2. 1

x dx =

elog x

3. sin x = cos x

4. cos x = sin x

In physics, we have, ds

dt = ds = dt

Total change in displacement = ds = dt

Similarly, d

dt

= a d = a dt

Total change in velocity = d a dt

Geometrical interpretation of an integral:

Consider a body moving with constant acceleration a. Let its velocity at

any time t be given by =u + at where u is the initial velocity. A graph of against t is shown in fig. 2.

If we take a small time interval t, then the displacement in the time

interval will be t. This is (nearly) equal to the area of rectangle of

length and width t. If we divide the entire time interval from zero to t into infinite number of infinitely small time intervals and add up all such areas we will get total displacement from t = 0 to t = t. This is the area under the graph i.e. area of the trapezium OABC. The area of the trapezium is given by,

Area = 1

2 (Sum of parallel sides) (distance between parallel sides)

= 1

2 (OA + CB) OC =

1

2 (u + ) t

= 1

2 (u + u + at) t =

1

2 (2u + at) t = ut +

1

2 at2

This is the total displacement in time t.

S = ut + 1

2 at2

t t

C

B

A

O

(t)

u

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Also, we have = ds

dt = u + at

ds = (u + at) dt

or ds = udt + at dt

s = t

0ds =

t

0udt +

t

0at dt = ut +

1

2 at2

This shows that the value of the integral is equal to the area under the graph.

Solved Problems : (Integral Calculus)

1. Find the integral of y = x3 + x + 2.

Solution :

3ydx (x x 2)dx

= 4 2x x

2x C4 2

[Ans.]

C = constant. (Constant is added because the derivative of a constant is 0)

2. Find the integral of y = sinx + cosx

Solution :

ydx (sinx cosx)dx = sinxdx cosxdx

= cosx + sinx + C [Ans.]

C = constant.

3. Find the change in velocity if a is given by a = t2 + 2t in interval t = 2sec to t = 4 sec.

Solution :

v = 4

2

adt = 4

2

2

(t 2t)dt =

43

2

2

tt

3

= 3 3

2 24 24 2

3 3 =

64 816 4

3 3 =

5612

3 =

56 36

3

v = 92

.3

[Ans.]

4. If a = s2 + 5s then give the relation of v v/s s. When v = 0, s = 0.

Solution :

We know a = dv

dt

As we know dm dm du

.dn du dn

a = dv ds

.ds dt

Remember

But, ds

vdt

a = v. dv

ds Remember

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a = s2 + 5s

v. 2dvs 5s

ds

v.dv = (s2 + 5s)ds

Integrating both side

2 3 2v s 5s

C2 3 2

as v = 0, s = 0

C = 0

2 3 2v s 5s

2 3 2

v2 = 3 22s 15s

3. [Ans.]

5. Rate of change of acceleration w.r.t. time is given by da

2t 3dt

, then find a, v, s w.r.t. time, assuming

zero initial condition.

Solution :

a =

dadt

dt = (2t 3)dt

a = t2 + 3t + C as a = 0 when t = 0

C = 0

a = t2 + 3t.

v = adt = 2(t 3t)dt

v = 3 2t 3t

C3 2

as t = 0, v = 0

C = 0

v = 3 2t 3t

.3 2

s = v dt =

3 2t 3tdt

3 2 =

4 3t tC

12 2

At t = 0, s = 0. s = 4 3t t

.12 2

[Ans.]

6. Find (i) Displacement of particle during t = 0s to t = 5s (ii) Displacement of particle during t = 5s to t = 10s (iii) Displacement of particle during t = 5s to t = 12s.

Solution :

(i) s1 = 1 25

5 5 m2 2

[Ans.]

(ii) s2 = 1 25

5 5 m2 2

[Ans.]

(iii) 3

1 45s 5 5 5 2 m.

2 2 [Ans.]

m/s

v

5

5 10 12 t

sec Vidyala

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7. The momentum p of a particle changes with time t according to the relation

dp

(10N)dt

+ (2 N/s)t. If the momentum is zero at t = 0, what will the momentum be at

t = 10s ?

Solution :

dp

dt(10 N) + (2 N/s) t

Integration, P = 10.t + 22t

c2

But at t = 0, p = 0 c = 0

p = (t2 + 10t) kg m

s

at t = 10

p = (100 + 100) kg m

s

= 200 kg m

.s

[Ans.]

8. Find the area enclosed by the curve y = sin x and the X-axis between x = 0 and x = .

Solution : Required area

=

0

sinxdx =

0

cosx

= cos + cos 0 = 1 + 1 = 2 [Ans.]

Exercise 2

Classwork Problems

1. Integrate w.r.t. x

i. x5 ii. 5 x + 4 iii. x 1

x iv. x2 cos x +

1

x

v. sin x + cos x

2. Evaluate, 41

1dx

x

3. Evaluate, 2R

GmMdx

x

4. Evaluate, 2

2

cos x dx

5. Evaluate, 2

0sinx dx

Homework Problems


i. x4 ii. x3/2 iii. x7 iv. 9

6x


i. 34x ii. 1

6 5(x ) + 7

1

x iii. (x2 + 4x + 5) (x2 2)

3. i. Integrate 3x2 + 7x w.r.t. x ii. Evaluate 4

3

1

x dx iii. Evaluate

33

1x dx

iv. Evaluate 0(sinx cosx)

dx v. Evaluate 2

2(3x 2x 1)

dx

y

xy O

y

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Classwork Problems

Exercise 1

1. i. 3x2 ii. 1

2 x iii.

3

1

2 x iv. 6x v. 1

2. i. 3 x ii. 3x2 4

3

x iii.

2

3

6x

4x 5

3. i.

x

1 2x + 1 2x ii.

2

1

(4x 5)

4. i. 2 cos 2x ii. 3x2 cos x3 iii. 5 sin 5x iv. 2x sin x2

Exercise 2

1. i. 6x

c6 ii.

5

6

6

5x + 4x + c iii. 2

3

3

2x 1

22x + c

iv. 3x

3 sin x + x

elog + c v. cos x + sin x + c

2. 1

3 3.

GmM

R 4. 2 5. 1

Homework Problems

1. i. 4x3 ii. 3

x2

iii. 7x8 iv. 15

69

x6

2. i. 2

3

6x

4x 5 ii. 5

8

6 7x

5 x iii. 4x3 + 12x2 + 6x 8

3. i. x3 + 7

2x2 + c ii. loge

4

3 iii. 20 iv. 2 v. 25

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Measurements

Syllabus : Need for Measurements, Units for Measurements, System of Units, S.I. Units, Fundamental and Derived Units, Dimension Analysis, Order of Magnitude and Significant figures, Accuracy and Errors in measurements.

Albert Einstein (1879 - 1955) Albert Einstein, a German born American physicist, is regarded by many as one of the two great physicists the world has known (the other is Isaac Newton). His three research papers (on special relativity, Brownian motion and the photoelectric effect) which he published in 1905, while he was employed as a technical assistant in a Swiss patent office in. Berne have profoundly influenced the development of physics. He received the Nobel Prize in Physics in 1921 for his explanation of the photoelectric effect.

MODULE 1

MODULE SYNOPSIS 1.0 Introduction 1.1 Need for measurements 1.2 Units for measurements 1.3 System of units 1.4 S. I. Units 1.5.1 Fundamental and derived units 1.5.2 Method for measurement of length 1.5.3 Method for measurement of mass 1.5.4 Method for measurement of time

Introduction

Physics is a quantitative science. Physicists deal in numbers - but not just the numbers of the mathematician. This is an important point that is often missed by beginning physicists. Physicist's numbers

are often (or could be) measurements, not the pure numbers of the mathematician.

Therefore, physicists measure things. Measurement is very important in physics – physicists are serious about measurement. One of the major contributions of physics to others sciences and society are the many measuring devices and techniques that physics has developed. In everyday life, we pick up a ruler and measure something without giving it much thought. Physicists think about their measurements, and need to have a much more sophisticated understanding of the measurement process than "normal" people do.

1. What is the need for measurement of a physical quantity? (3 marks)

A. Knowledge of the surroundings is obtained through our senses of vision, hearing, touch, smell etc., but physical quantities like length, mass, temperature, electric current etc. cannot be measured by mental judgment. For that purpose, we have to use different instruments. Experiments & measurements form the basis of physics.

In general, the observation of a phenomenon is incomplete, unless it gives us some quantitative information. Physics being a quantitative science, accurate measurements of the physical quantities are needed at every stage.

In our study of physics, we come across a number of discoveries, which were possible because of accurate measurements. e.g. the discovery of isotopes in atomic physics. In some cases, we find that the scientists observed discrepancies in the theoretical and experimental values and that led them to the development of new theories and formulae e.g. Newton’s formula and Laplace’s correction for the speed of sound waves in air. Thus advances in science were possible because the scientists tried for the accurate measurements and they did not remain satisfied with only a qualitative explanation.

2. What is meant by unit of a physical quantity? What are the properties possessed by it? (3 marks)

A. The reference standard used for the measurement of a physical quantity is called a unit. A unit should have the following properties.

It should be easily available.

It should be invariable (should not change with space and time).

It should be universally accepted.

It should be reproducible and not perishable.

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Notes on Measurements (15)


3. What is meant by a system of units? Give some examples of different systems of units. (2 marks)

A. A set of fundamental and derived units is called a system of units. Following are some examples of system of units used :

C.G.S. means centimeter gram second system.

M.K.S. means meter kilogram second system

F.P.S. means foot pound second system

S. I. System International system of units

4. What is S.I. system of units? Explain its need. (2 marks)

A. Use of different system of units became very inconvenient for exchanging scientific information between different parts of the world. Hence a common system of units called “Systeme Internationale d’ Units” (S.I. Units) was accepted. This international system of units called S.I. units consists of seven fundamental units, two supplementary units and large number of derived units.

5. What is meant by derived quantities and derived units? Give two examples. (2 marks)

A. Physical quantities other than fundamental quantities which depend on one or more fundamental quantities or their measurements are called derived quantities. e.g., force, density, speed etc. The units of derived quantities are called derived units. e.g., newton, kg/m3, m/s etc.

6. What are the conventions to be followed while writing S.I. units? (2 marks)

A. Full name of unit always starts with small letter even if named after a person. e.g., newton, joule and not Newton, Joule.

Symbol for unit named after a person should be in capital letter e.g., ‘N” for Newton, ‘J’ for joule, ‘A’ for ampere etc.

Symbol for all other units are written in small letters, e.g., ‘m’ for meter, ‘s’ for second etc.

Symbols of units are not to be expressed in plural form e.g., 25 m and not 25 ms.

Full stop and any other punctuation mark should not be written after the symbols e.g., kg and not kg. or N and not N.

7. What is meant by fundamental quantities and fundamental units? State fundamental quantities

and their units in S.I. (3 marks)

A. Fundamental quantities : The physical quantities which do not depend on any other physical quantities for their measurements are known as fundamental quantities.

Fundamental units : The units used to measure fundamental quantities are called fundamental units.

No. Physical

Quantity

S.I.

Unit Symbol No.

Physical

Quantity

S.I.

Unit Symbol

1. Length Metre m 5. Thermodynamic

Temperature Kelvin K

2. Mass Kilogram kg 6. Luminous intensity Candela cd

3. Time Second s 7. Amount of substance

Mole mol

4. Electric current Ampere A

8. Explain Parallax Method for Measurement of length. (3 marks)

A. Large distances such as the distance of a planet or a star from the earth cannot be measured directly with a metre scale. An important method in such cases is the parallax method.

To measure the distance D of a far away planet S by the parallax method, we observe it from two different positions (observatories) A and B on the Earth, separated by distance AB = b at the same time as shown in figure. We measure the angle between the two directions, along which the planet is viewed

at these two points. The ASB represented by symbol is called the parallax angle or parallactic angle. (see figure).

As the planet is very far away, b

D << 1, and therefore, is very small. Then we approximately take AB

as an arc of length b of a circle with centre at S and the distance D as the radius, AS = BS so that AB = b

= D where is in radians. D =

b.

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We can employ a similar method to determine the size or angular diameter of the planet. If d is the

diameter of the planet and the angular size of the planet (the angle subtended by d at the earth), we

have = d/D. The angle can be measured from the same location on the earth. It is the angle between the two directions when two diametrically opposite points of the planet are viewed through the telescope.

Since D is known, the diameter d of the planet can be determined using = d/D.

9. Explain the method of measurement of mass. (3 marks)

A. Mass is a fundamental property of matter. It does not depend on the temperature, pressure or location of the object in space. The SI unit of mass is kilogram (kg).

But while dealing with atoms and molecules, the kilogram is an inconvenient unit. For atomic levels unified atomic mass is used.

1 unified atomic mass unit = 1 u = 0.8333 101 of the mass of an atom of carbon 12 in kg. Isotope

126 C including the mass of electrons = 1.66 1027 kg.

Mass of commonly available objects can be determined by a common balance like the one used in a grocery shop. Large masses in the universe like planets, stars, etc. based on Newton’s law of gravitation can be measured by using gravitational method.

For measurement of small masses of atomic/subatomic particles etc., we make use of mass spectrograph in which radius of the trajectory is proportional to the mass of a charged particle moving in uniform electric and magnetic field.

10. Explain the method of measurement of time. (3 marks)

A. The mean solar day on the earth is considered to be duration of 24 hours for which an hour is of 60 minutes and each minute is of 60 seconds.

A solar day is the interval from one noon to the next noon. Average of length of a solar day over a year is considered as a mean solar day.

To measure any time interval, we need a clock. We now use an atomic standard of time, which is based on the periodic vibration produced in a cesium atom. This is the basis of the cesium clock, sometimes called atomic clock, used in the national standards. Such standards are available in many laboratories. In the cesium atomic clock, the second is taken as the time needed for 9,192,631,770 vibrations of the

radiation corresponding to the transition between the two hyperfine states of cesium 133 atom.

MODULE 2

MODULE SYNOPSIS

1.6.1 Dimensional Analysis 1.7.1 Order of magnitude 1.6.2 Uses of dimensional analysis 1.7.2 Significant figures 1.8 Accuracy and errors in measurements 1.8.1 Different types of errors 1.8.2 Error, mean value, absolute error, mean absolute error, relative error, percentage error.

11. Define dimensions and dimensional equation. Give two examples. (2 marks)

A. The dimensions of a physical quantity are the powers to which fundamental units must be raised in order to obtain the unit of a given physical quantity. The units of fundamental quantities are represented by ‘L’ for length, ‘M’ for mass, ‘T’ for time, ‘K’ for temperature, ‘I’ for current, ‘C’ for luminous intensity and ‘mol’ for mole. An expression, which gives the relation between the derived units and fundamental units in terms of dimensions is called a dimensional equation. e.g.

i) Dimensional equation for speed = [L1M0T1]. ii) Dimensional equation for charge = [L0M0T1I1]

Measurement of length b

B A

D D

S

D

d

Measurement of Diameter of planet

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12. State the uses of dimensional analysis. Explain each use with the help of an example.

(3 marks for each use)

A. (i) To find the correctness of a physical equation : The dimensions of all the terms on the two sides of a physical equation must be same. This is called the principle of homogeneity of dimensions. e.g., consider the equation,

S = ut + 21at .

2By writing the dimensions,

we get, S = [L1M0T0] ut = [L1M0T0] at2 = [L1M0T0]

The number 1

2has no dimensions. Hence each term has the same dimensions. Thus, the equation

is dimensionally correct. (Though this does not mean that it is always a correct physical equation).

(ii) To find conversion factor between the units of the same physical quantity in two different

systems of units : e.g., To find the conversion factor between the units of Force. i.e., Newton in SI system to dyne in c.g.s. system. Let 1 Newton = x dyne …..… (1)

The dimensions of force are [L1M1T2].

Equation (1) in dimension form can be written as

1 1 21 1 1[L M T ] = 1 1 2

2 2 2x[L M T ]

x =

1 1 21 1 1

1 1 22 2 2

L M T

L M T

=

1 1 2

1 1 1

2 2 2

L M T

L M T ……… (2)

where suffix 1 indicated SI system and suffix 2 indicates C.G.S. system. In S.I. system, L, M and T are expressed in m, Kg, s and in C.G.S. system L, M and T are expressed in cm, g and s respectively.

Equation (2) becomes

x =

11 2kgm s

cm g s=

1122 3 gcm

10 10 1cm g

x = 105

1 Newton = 105 dyne

(iii) To establish relationship between related physical quantities : e.g., The period ‘T’ of oscillation of a simple pendulum depends on length ‘L’ and acceleration due to

gravity ‘g’. Let us derive the relation between T, and g.

Let us assume that,

T = x yK g …..… (3)

K = constant which is dimensionless. The dimensions of T = [L0M0T1].

The dimensions of = [L1M0T0]

and dimensions of g = [L1M0T2]

Equation (1.3) can be dimensionally written as

[L0M0T1] = K[L1M0T0]x [L1M0T2]y

[L0M0T1] = K[Lx+yM0T2y] …..… (4)

By comparing the powers of L, M, T on both sides of equation (4), we get

O = x + y and 1 = 2y

y = 1 1

and x2 2

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Substituting these values of x and y in (3), we get

T = 1/2 1/2K g , i.e. T = Kg

The value of ‘K’ which is a numerical constant, cannot be found by using dimensional analysis and can be found out experimentally.

13. Explain with example the term ‘order of magnitude of a physical quantity’. (3 marks)

A. Many times, we are not interested in the exact value of a physical quantity, but only the rough value of its magnitude. For this purpose, magnitude of these quantities is rounded off to the nearest integral power of 10. The power of 10 is called order of magnitude of that quantity.

Hence order of magnitude of a physical quantity is defined as the value of its magnitude rounded off to the nearest integral power of 10.

The magnitude of any physical quantity can be expressed as A 10n, where ‘A’ is a number such that

0.5 A < 5 and ‘n’ is an integer, called order of magnitude.

For example, mass of electron = 9.1 1031 kg = 0.91 1030 kg

Order of magnitude = 1030 kg.

Also charge of electron = 1.6 1019 C.

Order of magnitude = 1019C.

14. What is meant by Significant figures. State the rules for determining significant figures. (3 marks)

A. The significant figures are those number of digits in a measured quantity which are known reliably plus one additional digit that is uncertain.

Rules for determining significant figures : (i) Retain only one uncertain digit in the measurement of a physical quantity. (ii) When the value of the measurement is to be rounded off to given number of significant figures, then

the figures to be dropped is (1) less than 5, then the last significant figure is left unchanged (2) 5 or greater than 5, the last significant figure is increased by one.

(iii) The zeros on the right hand side of the number are significant because they indicate the accuracy of the instrument used for measurement.

(iv) The zeros on the left hand side of the number are not significant, e.g., the number 0.0753 has only three significant figures.

(v) If the number of digits is more than the number of significant figures, the number should be

expressed in the power of 10. e.g., the mass of the earth is written as 5.98 1024kg, as it is known only upto 3 significant figures.

15. Explain in brief, accuracy and errors in measurement. (2 marks)

A. Different instruments are used to measure different physical quantities in physics and accuracy of the measurement depends on the accuracy of the instrument used for measurement. Defects in measurement of physical quantities can lead to errors and mistakes. Mistakes are committed by an observer and can be totally avoided, errors which occur due to the limitations of the instruments can be reduced. Lesser the percentage error, more is the accuracy in the measurement of a physical quantity.

16. Explain different types of errors in measurements with remedies. (4 marks)

A. The errors can be classified into four groups.

(i) Instrumental (or constant) error : These errors are caused due to faulty construction of instruments e.g., if a thermometer is not graduated properly i.e., one degree on the thermometer

actually corresponds to 0.99, the temperature measured by such a thermometer will differ from its value by a constant amount. Hence it is also called as constant error. Such errors can be minimized by taking the same measurement with different accurate instruments.

(ii) Systematic Error (Persistent errors) : This is an error due to defective setting of an instrument. If the pointer of an ammeter is not pivoted exactly at the zero of the scale, it will not point to zero when no current is passing through it. Such errors can be minimized by detecting its causes.

(iii) Personal Error : These errors are introduced due to fault of an observer taking reading, referred to

as human errors. They vary from person to person e.g., error due to nonremoval of parallax between pointer and its image in case of a magnetic compass needle e.g., error made in counting number of oscillations while measuring period of simple pendulum.

(iv) Random Error (Accidental) : Even after minimizing above types of errors, errors may occur due to different factors like change in temperature, pressure or fluctuation in voltage while the experiment is being performed. Such errors cannot be eliminated but can be minimized.

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17. Define (i) most probable value (ii) Absolute error (iii) Relative error (iv) Percentage error.

(4 marks)

A. (i) Most probable value : Most probable value is the one which is observed maximum number of times when large number of measurement of a particular value are done.

(ii) Absolute error : The magnitude of the difference between mean value and each individual value is called absolute error.

(iii) Relative error : The ratio of the mean absolute error in the measurement of a physical quantity to its most probable value is called relative error.

Relative error = m

m

| a |

a

(iv) Percentage error : The relative error multiplied by 100 is called the percentage error.

Percentage error = m

m

| a |

a 100 %

18. How can the effect of errors be minimized ? (2 marks)

A. The effect of errors can be minimized by : (i) taking a large magnitude of the quantity to be measured.

(ii) taking larger number of readings and calculating their mean value. (iii) using an instrument whose least count is as small as possible.

19. Can you call a physical quantity large or small without specifying a standard for comparison?

(2 marks)

A. To call a physical quantity large or small without specifying a standard for comparison is meaningless. To specify a physical quantity clearly we have to specify it in terms of a numerical value. When we say a physical quantity is large or small we are comparing it.

20. If two physical quantities have the same dimensions, are they the same? (2 marks)

A. If two physical quantities have the same dimensions, then they are not necessarily the same. Two different physical quantities can have the same dimensions. For e.g., work and torque have the same dimensions but they are entirely different physical quantities. Work is a scalar quantity whereas torque is a vector quantity.

21. A dimensionally correct equation need not actually be a correct equation but dimensionally

incorrect equation is necessarily wrong. Justify. (2 marks)

A. An equation may be dimensionally correct but still it may not be a physically correct equation because there may be some dimensionless constants which may be missing or erroneously present in the equation.

For example we have the kinematical equation

21S ut at

2

If we write the equation as

S = ut + at2 or 21S ut at

2

They are dimensionally correct but physically not correct.

However if the equation is written as

2 1S ut at

2

Then the equation is dimensionally wrong and we can say with certainty that it is a wrong equation.

22. Are all constant dimensionless or unit less? (2 marks)

A. All constants are not dimensionless or unit less. There are some constants which have dimensions and

units. For example, universal gravitational constant G, permittivity of free space 0 have units and dimensions. On the other hand refractive index and dielectric constant are examples of dimensionless constants.

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23. How to determine the distance of different stars from the earth? (3 marks)

A. The distance of a star from the earth can be determined by parallax method. Since the stars are of very large distances from earth, the parallactic angle θ is very small. The star is viewed from two diametrically opposite positions of the earth’s orbit and the angle θ is determined. Thus the base of the triangle is the diameter of the earth’s orbit around the sum. The radius of the earth’s orbit around the sun is known as one

astronomical unit and has a value 1.496 1011 m. Knowing the value of θ the distance D can be determined.

Solved Examples :

1. Find the dimensions of (i) Power (ii) Force and (iii) Permittivity of Vacuum (0)

Solution : (i) Dimensions of P :

By definition, Power = work done

time

Work done = Force displacement

Power = Force displacement

time

Dimensions of [P] = [F] [S]

[t]

[P] = 1 1 2 1[M L T ] [L ]

[T]

[P] = [M1 L2 T3] (ii) Dimensions of F :

By definition, force = mass acceleration

dimensions of force [F] = [M] [a]

[F] = [M1] [L1 T2]

[F] = [M1 L1 T2]

(iii) Dimensions of Permitivity of Vacuum (0) : From Coulomb’s law electrostatic force

F = 1 2

20

q q1

4 r

where F is electrostatic force of attraction or repulsion q1 and q2 are electric charges, r is distance

between charges, 0 is permittivity of vacuum. Rearranging above equation

0 = 1 2

2

q q

Fr, where 4 = constant

dimension of permittivity of vacuum,

[0] = 2

[q] [q]

[F] [r ]

charge q = It

[0] = 1 1 1 1

1 1 2 1 2

[I T ] [I T ]

[M L T ] [L ]

[0] = [M1 L3 T+4 I2]

2. Using dimensions, show that 1 Joule = 107 erg.

Solution : In the S.I. system, unit of work is Joule and in c.g.s. system, unit of work is erg. Both quantities represent

work and hence there dimensional formula is same.

Let 1 joule = x erg

1[M1 L2 T2] = x [M1 L2 T2]

or 1[kg m2 s2] = x [g cm2 s2]

C

A B S

D

θ

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x = 2 2

2 2

kg m s

g cm s

x =

3 2 2

2

10 g (10 cm)

g cm =

3 4 2

2

10 10 g cm

g cm

x = 107

1 joule = 107 erg Hence proved.

3. If length ‘L’, force ‘F’ and time ‘T’ are taken as fundamental quantities. What would the dimensional equation of mass and density?

Solution: From definition, Force F = ma

m = F

a

dimensions of mass, [m] = [F]

[a]

[m] = 1

1 2

[F ]

[L T ] [ F is given as fundamental quantity]

[m] = [L1 F1 T2] …..… (1)

By definition density d = mass

volume

dimension of [d] = 3

[m]

[L]

= 1 1 2

3

[L F T ]

[L ]

{from eqn (1)}

[d] = [L4 F1 T2] …..… (2)

4. The hydrostatic pressure ‘p’ of a liquid column depends upon the density ‘d’, height ‘h’ of liquid column and also an acceleration ‘g’ due to gravity. Using dimensional analysis, derive formula for pressure P.

Solution : Let, pressure p = k hx dy gz …..… (1)

dimensions of p = [M1 L1 T2] dimensions of h = [L1]

dimensions of d = [M1 L3]

dimensions of g = [L1 T2] Substituting in eqn (1) we get,

[M1 L1 T2] = k[L1]x [M1 L3]y [L1 T2]z

[M1L1 T2] = k [Lx 3y + z My T2z] Comparing we get, y = 1

2z = 2

z = 1 and 1 = x 3y + z

1 = x 3(1) + 1

x = 1 Substituting values of x, y and z in eqn (1) we get P = kh1d1g1

p = hdg …..… (2) where k is dimensionless constant. egn (2) is a formula for hydrostatic pressure p of a liquid column.

5. Derive an expression of kinetic energy of a body of mass ‘m’ and moving with velocity ‘v’, using dimensional analysis.

Solution : let, kinetic energy K.E = k mx vy …..… (1)

dimension of K.E = [M1 L2 T2] dimension of mass = [M1]

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dimension of velocity = [L1 T1] Subsitiuting in eqn (1) we get,

[M1 L2 T2] = k [M1]x [L1 T1]y

[M1 L2 T2] = k [Mx Ly Ty] Comparing we get, x = 1 and y = 2 Substituting values of x and y in equation (1) we get, K.E = k m1v2

K.E = k mv2 …..… (2) (where k is dimensionless constant) Equation (2) is an expression for kinetic energy of a body.

6. State the order of magnitude of the following : (a) Acceleration due to gravity g = 9.81 m/s2

(b) The gravitation constant G = 6.67 1011 Nm2 / kg2 (c) The period of rotation of the earth about its own axis; in seconds.

Solution :

The order of magnitude of any physical quantity can be expressed as A 10n where ‘A’ is a number such

that 0.5 A < 5 and ‘n’ is an integer, called order of magnitude.

(a) order of magnitude of acceleration due to gravity g = 9.81 m/s2

= 0.981 101 m/s2

order of magnitude = 101 m/s2

(b) The gravitation constant G = 6.67 1011 Nm2/kg2

G = 0.667 1010 Nm2/kg

order of magnitude of G = 1010 Nm2/kg (c) The period of rotation of the earth about its own axis = 24 hrs

= 24 3600 s = 86400 s

= 0.86400 105 s

order of magnitude = 105 s

7. Determine the number of significant figures in the following measurement :

(a) 0.05718 (b) 93.26 (c) 2.35 1019 (d) 1.3725 109

Solution : (a) The zeros on the left hand side of the number are not significant

Significant figures in 0.05718 are 4. (b) Significant figures in 93.26 are 4

(c) The number of significiant figures in 2.35 1019 are 3

(d) The number of significant figures in 1.3725 109 are 5

8. An object was weighed by a physical balance and following readings were obtained 5.04g, 5.06 g, 4.97 g, 5.00 g; 4.93 g. Find (i) Mean value (ii) Absolute error (iii) Percentage error.

Solution :

(i) Mean value Wm = 1 2 3 4 5w w w w w

5

wm = 5.04 5.06 4.97 5.00 4.93 25

5 5

wm = 5.00 g (ii) Average absolute error =

1 m 2 m 3 m 4 m 5 5w w w w w w w w w w

5

= 5.04 5 5.06 5 4.97 5 5.00 5 4.93 5

5

= 0.04 0.06

5

=

0.2

5

Average absolute error = 0.04 g

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(iii) Percentage error = Average absolute error

Mean vlaue

=

0.04100

5

Percentage error = 0.8 %

9. Using the dimensional method, show that the velocity v of a body falling freely under gravity varies as

gh , where g is the acceleration due to gravity and h is the distance through which the body falls.

Solution : Assuming the velocity depends upon the mass m of the body, the acceleration due to gravity g and the

distance h through which the body falls, we write

v mx hy gz v = k mx hy gz (k is a pure number) Substituting the dimensions of the quantities involved.

0 1 x y 2 zM L T k [M] [L] [LT ] = k [M]x [L]y + z [T]2z

Comparing the dimensions of L, M and T on the two sides x = 0

y + z = 1 and 2z = 1

On Solving, z = 1

2 and y =

1

2

v = k h g = k hg v hg

10. The velocity V of sound in a gas depends upon its pressure and density . Show by dimensional analysis

that V

P.

Solution : Dimensions of pressure and density are found as follows :

Pressure = 2

2

Force [MLT ]

Area L

= [ML1 T2]

Density = 3

Mass [M]

Volume [L ] = [ML3]

Let V Px y

V = k Px y, where k is a dimensionless constant. Expressing this in dimensional terms,

[M0LT1] = k [ML1 T2]x [ML3]y

= k [Mx + y Lx3y T2x] Equating the dimensions of M, L and T on the two sides,

x + y = 0

x 3y = 1

2x = 1

On solving these, x = 1

2 and y =

1

2

V = 1 1

2 2k P

= k P

V P

Classwork Problems

1. Deduce the dimensional formula for the following quantities from their definitions : (a) Momentum (b) Moment of force (c) Work (d) Power.

2. Find the dimensions of (a) Pressure and (b) Potential difference (c) Electric resistance

3. Find the conversion factor between S.I. units and CGS units of density.

4. Using dimensional analysis, show that 1 N/m2 = 10 dynes / cm2

5. The distance s covered by a body in time t is given by the relation s = a + bt + ct2. What are the dimensions of a, b and c ?

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6. Show that the equations of rectilinear motion

(a) v = u + at and (b) s = ut + 21at

2 are dimensionally homogeneous.

7. Find if the formula v2 = u2 + 2as2 is dimensionally homogeneous.

8. The velocity v, the acceleration a and the displacement s of a body in motion are related by the

expression v2 ax sy. Find x and y.

9. If the force F acting on a body in uniform circular motion depends upon the mass m of body, its speed v

and the radius of path ‘r’ , show that, F 2mv

r.

10. State the order of magnitude of the following :

(a) Permittivity of free space = 8.85 1012 C2 / Nm2

(b) Permeability of free space = 4 107 wb / Am

(c) Radius of the earth = 6.4 106 m

11. Determine the number of significant figures in the following measurement :

(a) 0.04531 (b) 35.26 (c) 4.56 1019 (d) 6.2346 109

12. An object was weighed by a physical balance and following readings were obtained : 10.8g; 10.4g; 10.2g; 10.6g; 9.8g

Find (i) Mean value (ii) Mean Absolute error, (iii) Percentage error

13. The velocity (v) of transverse waves on a string depends on the tension (T) in the string and its mass per unit length (m). Using dimensional analysis obtain the expression for v in terms of these quantities.

14. Using dimensional analysis show that speed (v) of sound through a gas is proportional to p

d, where p is

the pressure of the gas and d is the density of the gas.

15. Find the percentage error in volume V of a block of length 20 cm, breadth 15 cm and height 10 cm if these are measured with a metre scale (having a least count of 1 mm).

Homework Problems

1. The frequency of vibration n of a string of length kept under tension F is given by n = 1 F

2 m, where m

is the mass per unit length of string. Test this relation dimensionally.

2. Find the dimensions of (a) Constant of gravitation G, (b) Coefficient of viscosity.

3. Using dimensional method, show that 1 N = 105 dynes.

4. The length of a metal plate was measured using a vernier calipers of least count 0.01 cm. The measurements made were 3.11 cm, 3.13 cm, 3.14 cm and 3.14 cm. Find the mean length, the mean absolute error and the percentage error in the measurement of length.

1 2 3 4a 3.11cm,a 3.13cm,a 3.14cm,a 3.14cm

5. Find the percentage error in kinetic energy of a body of mass m = 60.0 0.3g and moving with a velocity of

v = 25.0 0.1 cm/s.

m 60.0 0.3g, v 25.0 0.1 cm/s

6. Fill in the blanks :

(a) 1 kg m2 s2 = _________ g cm2 s2

(b) 3 m s2 = _________ km h2

(c) 1 kg m3 = _________ g cm3

7. State the number of significant figures in the following :

(a) 0.007 m2 (b) 2.64 1024 kg (c) 0.2370 g cm3

(d) 6.320 J (e) 6.032 Nm2 (f) 0.0006032 m2

8. The length, breadth and thickness of a rectangular sheet of metal are 4.234m, 1.005 m and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

9. Check whether the following equation is dimensionally correct. 21mv

2 = mgh, where m is the mass of the

body, v its velocity, g is acceleration due to gravity.

10. The radius of a nucleus of mass number ‘A’ is given by R = 1.3 1015 A1/3m. Find the order of magnitude of radius for a nucleus with A = 125.

Vidyala

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Classwork Problems

1. (a) MLT1 (b) ML2T2 (c) ML2T2 (d) ML2T3

2. (a) [M L1 T2] (b) [M L2 T3 I1] (c) [M L2 T3 I2]

3. 1kg/m3 = 103 g/cm3 5. [L], [LT1], [LT2] 7. No 8. x = 1 and y = 1

10. (a) 1011 C2 / Nm2 (b) 106 Wb/Am (c) 107m

11. (a) 4 (b) 4 (c) 3 (d) 5

12. (i) 10.4 g (ii) 0.28 g (iii) 2.692 % 15. 2.17%

Homework Problems

2. (a) [M1 L3 T2] (b) [M L1 T1] 3. 105 dynes

4. Mean length = 3.13 cm The mean absolute error = 0.01 cm The percentage error = 0.319 %

5. 1.3 % 6. (a) 107 (b) 3.9 104 (c) 103

7. (a) 1 (b) 3 (c) 4 (d) 4 (e) 4 (f) 4

8. 4.255 m2, 0.0855 m3 10. 1014 m

THINK IT OVER Is there a physical quantity which has units but no dimension?

Vidyala

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MODULE 3 : MULTIPLE CHOICE QUESTIONS

Select the correct answer from each of the following questions :

i) Each question is allocated 1 marks for each correct response.

ii) No marks will be deducted for incorrect response of each question or if no answer is given. iii) There is only one correct answer for each question.

Classwork Problems

1. An atomic clock is based upon the periodic vibrations produced in (A) Sodium atoms (B) Germanium atoms (C) Cesium atoms (D) Neon atoms

2. Which one of the following is not a unit of energy? (A) Calorie (B) Electron volt (C) Megawatt (D) Watt hour

3. One micron is related to centimeter as

(A) 1 micron = 108 cm (B) 1 micron = 106 cm (C) 1 micron = 105 cm (D) 1 micron = 104 cm

4. SI unit of temperature is (A) celsius (B) kelvin (C) degree celsius (D) degree kelvin

5. The wavelength of a spectral line is 480 nm. What is its value in mm?

(A) 480 107 mm (B) 48 105 mm (C) 48 106 mm (D) 4.8 105 mm

6. Which one of the following units is correctly expressed? (A) Metre (B) Newton (C) newton (D) Joule / sec

7. The unit of which one of the following physical quantities is not a derived unit? (A) Frequency (B) Charge

(C) Gravitational constant (D) Electric current

8. The unit 1 N / m is equivalent to (A) 1 erg / cm (B) 1 erg / cm2 (C) 1 J /m (D) 1 J / m2

9. What is the order of magnitude of the number 8245? (A) 3 (B) 4 (C) 2 (D) 5

10. A student measured the diameter of a wire using a micrometer screw gauge of least count 0.001 cm. He recorded the following measurements. The correct measurement is

(A) 5.3 cm (B) 5.320 cm (C) 5.32 cm (D) 5.3200 cm

11. In the following numbers, the one having four significant figures is (A) 0.0004 (B) 0.0078 (C) 0.530 (D) 5.038

12. Error committed by a student in removing the parallax is (A) an instrumental error (B) a personal error (C) systematic error (D) a random error

13. Light year is a unit of (A) Time (B) Mass (C) Distance (D) Energy

14. A suitable unit for gravitational constant is

(A) kg – m sec1 (B) N m1 sec (C) N m2 kg2 (D) kg m sec1

15. SI unit of pressure is (A) Pascal (B) dynes/cm2 (C) cm of Hg (D) Atmosphere

16. Which of the following is not a unit of energy

(A) W s (B) kg m/sec (C) N m (D) Joule

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17. Unit of power is

(A) Kilowatt (B) Kilowatt hour (C) Dyne (D) Joule

18. The dimensions of power are

(A) M1L2T3 (B) M2L1T2 (C) M1L2T1 (D) M1L1T2

19. The dimensional formula for impulse is same as the dimensional formula for (A) Momentum (B) Force (C) Rate of change of momentum (D) Torque

20. According to Joule’s law of heating, heat produced H = I2 Rt, where I is current, R is resistance and t is time. If the errors in the measurement of I, R and t are 3%, 4% and 6% respectively then error in the measurement of H is

(A) 17% (B) 16% (C) 19% (D) 25%

Homework Problems

1. The error in the measurement of the radius of a sphere is 0.2%. What is the percentage error in the measurement of its surface area? (A) 0.2% (B) 0.4% (C) 0.6% (D) 0.8%

2. Zero error is included in the category of (A) Personal errors (B) Instrumental errors (C) Accidental errors (D) Constant errors

3. When a metal sphere is heated, then the maximum percentage change will be observed in its (A) area (B) radius (C) volume (D) mass

4. Identify the pair whose dimensions are equal. (A) Stress and Energy (B) Torque and Work (C) Force and Work (D) Force and Stress

5. Which one of the following quantities has dimensions? (A) Relative Density (B) Relative Refractive Index (C) Relative Velocity (D) Relative Permittivity

6. Which group of the following has all quantities with the same dimensions? (A) mass, length, time

(B) momentum, force and impulse (C) Young’s modulus, stress and pressure (D) Surface tension, force constant and acceleration

7. Which one of the following is a dimensional constant? (A) Refractive index (B) Dielectric constant (C) Relative density (D) Gravitational constant

8. The expression [M L2 T2] represents (A) Power (B) Momentum (C) Pressure (D) Kinetic Energy

9. The dimensional formula for impulse is the same as the dimensional formula for (A) momentum (B) force (C) rate of change in momentum (D) acceleration

10. Dimensional formula for torque is

(A) M L2 T1 (B) M L T2 (C) M L T1 (D) M L2 T2

11. Hertz is the unit for (A) Frequency (B) Force (C) Electric charge (D) Magnetic flux

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12. The unit of e.m.f. is (A) Joule (B) Joule-coulomb (C) Volt-coulomb (D) Joule/coulomb

13. Unit of impulse is

(A) Newton (B) kgm (C) kgm/s (D) JC1 m1

14. Which of the following is a derived unit ? (A) Unit of mass (B) Unit of length (C) Unit of time (D) Unit of volume

15. Dimensional formula for force is

(A) [M1L2T2] (B) [M1L1T2] (C) [M1L1T2] (D) [M1L2T2]

16. Which one has the dimensions different from the remaining three ? (A) Power (B) Work (C) Torque (D) Energy

17. Dimensions of frequency are

(A) M0L1T0 (B) M0L0T1 (C) M0L0T (D) MT2

18. Density of a liquid in CGS system is 0.625 g/cm3. What is its magnitude in SI system? (A) 0.625 (B) 0.0625 (C) 0.00625 (D) 625

Classwork Problems

1. (C) 2. (C) 3. (D) 4. (B) 5. (B) 6. (C) 7. (D) 8. (D) 9. (B)

10. (B) 11. (D) 12. (B) 13. (C) 14. (C) 15. (A) 16. (B) 17. (A) 18. (A)

19. (A) 20. (B)

Homework Problems

1. (B) 2. (B) 3. (C) 4. (B) 5. (C) 6. (C) 7. (D) 8. (D) 9. (A)

10. (D) 11. (C) 12. (D) 13. (B) 14. (B) 15. (B) 16. (A) 17. (B) 18. (D)

Vidyala

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XI.23/Chem/FC/Ch.1/Pg.29

Structure of Atom (Electronic Configuration)

Q.1 Explain

*(i) Aufbau principle

*(ii) Pauli’s exclusion principle

*(iii) Hund’s rule of maximum multiplicity

A. (i) Aufbau principle : “According to Aufbau principle “In the ground state of the atoms the orbitals are filled with electrons in order of the increasing energies.”

Orbitals are filled in order of increasing value of (n + ) rules. e.g. consider 3d and 4s orbitals For 3d, n = 3 and = 2. Hence (n + ) = 3 + 2 = 5 For 4s, n = 4 and = 0. Hence (n + ) = 4 + 0 = 4

As value of (n + ) for 4s orbital is less than that of 3d orbital with lower value of n will be filled first.

If two orbitals have same value of (n + ), then the orbital with lower value of n will be filled first. e.g. consider 3d and 4p orbitals For 3d, n = 3 and = 2. Hence (n + ) = 3 + 2 = 5 For 4p, n = 4 and = 1. Hence (n + ) = 4 + 1 = 5

Therefore, 3d orbital is filled up before 4p orbital. The order of energy of different orbitals in an atom is given below. 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d and so on. The sequence given above can be represented by diagram as shown in figure.

(ii) Pauli’s exclusion principle :

“Two electrons in an atom cannot have the same set of all four quantum numbers”

Or

“Only two electrons may exist in the same orbital with three quantum numbers same i.e. the fourth

quantum number different and these electrons must have opposite spin.”

Consider helium atom which has two electrons in 1s orbital. The four quantum numbers for two electrons in 2s orbital are as follows (Table.)

Electron number Quantum number Set of values of

quantum nos. n m s

1st Electron 1 0 0 1

2 (1, 0, 0,

1

2 )

2nd Electron 1 0 0 1

2 (1, 0, 0,

1

2 )

Fig. : Order of filling of atomic orbitals

Vidyala

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(30) Vidyalankar : Foundation - Chemistry


(iii) Hund’s rule of maximum multiplicity :

“When several orbitals of equal energy (degenerate orbitals) are available, the electrons first fill all

the orbitals singly before pairing in any of these orbitals.

The Configuration of four electron occupying porbitals is

Completely filled or half filled orbitals give extra stability.

*Q.2 Explain and write electronic configuration of chromium and copper.

A. Atomic number Cr is 24. Expected configuration of 24Cr is 1s2, 2s2 2p6, 3s2, 3p6, 4s2, 3d4.

The 3d orbital is not half filled. Hence it has less stability. Inter-electronic repulsion forces one 4s electron to enter into empty 3d orbital to make 4s and 3d orbitals half filled, so Cr acquires extra stability and its configuration becomes, 1s2, 2s2 2p6, 3s2, 3p6, 4s1, 3d5.

In this configuration, both 4s and 3d sub shells are half filled. This gives extra stability to Cr atom.

Cu (Atomic number 29) : The expected configuration of Cu is 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d9.

The 3d orbital is neither half filled nor completely filled. Hence it has less stability. Interelectronic repulsion forces one 4s electron to enter into one half filled 3d orbital to make it completely filled and 4s orbital half filled. So Cu acquires configuration.

1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d10.

In this configuration 3d sub shell in completely filled and 4s is half filled. This gives extra stability to Cu atom.

*Q.3 Write electronic configurations of the elements (number in the brackets indicate atomic numbers)

(i) Helium (2) (ii) Boron (5) (iii) Nitrogen (7) (iv) Sodium (11)

(v) Silicon (14) (vi) Chlorine (17) (vii) Potassium (19) (viii )Oxygen (8)

A. (i) Helium, 2He 1s2 (ii) Boron, 5B 1s2 2s2 2p1 (iii) Nitrogen, 7N 1s2 2s2 2p3 (iv) Sodium, 11Na 1s2, 2s2, 2p6, 3s1 or [Ne] 3s1 (v) Silicon, 14Si 1s2, 2s2, 2p6, 3s2, 3p2 or [Ne] 3s2 3p2 (vi) Chlorine, 17Cl 1s2, 2s2, 2p6, 3s2, 3p5 or [Ne] 3s2 3p5 (vii) Potassium, 19K 1s2, 2s2, 2p6, 3s2, 3p6, 4s1 or [Ar] 4s1 (viii)Oxygen, 8O, 1s2, 2s2 2p4 or [He] 2s2 2p4

Q.4 Give electronic configuration for the elements from atomic numbers 1 to 30.

A. Electronic Configuration

Element Atomic

No.

K

(n=1)

L

( n=2)

M

(n=3)

N

(n=4)

Condensed outer

electronic configuration

Hydrogen (H) 1 1s1 1s1

Helium (He) 2 1s2 1s2

Lithium (Li) 3 1s2 2s1 [He] 2s1

Beryllium (Be) 4 1s2 2s2 [He] 2s2

Boron (B) 5 1s2 2s2 2p1 [He] 2s2 2p1

Carbon (C) 6 1s2 2s2 2p2 [He] 2s2 2p2

Nitrogen (N) 7 1s2 2s2 2p3 [He] 2s2 2p3

Oxygen (O) 8 1s2 2s2 2p4 [He] 2s2 2p4

Fluorine (F) 9 1s2 2s2 2p5 [He] 2s2 2p5

Neon (Ne) 10 1s2 2s2 2p6 [He] 2s2 2p6

Sodium (Na) 11 1s2 2s2 2p6 3s1 [Ne] 3s1

Magnesium (Mg) 12 1s2 2s2 2p6 3s2 [Ne] 3s2

Aluminium (Al) 13 1s2 2s2 2p6 3s1 3p1 [Ne] 3s2 3p1

Silicon (Si) 14 1s2 2s2 2p6 3s2 3p2 [Ne] 3s2 3p2

Phosphorus (P) 15 1s2 2s2 2p6 3s2 3p3 [Ne] 3s2 3p3

Sulphur (S) 16 1s2 2s2 2p6 3s2 3p4 [Ne] 3s2 3p4

Chlorine (Cl) 17 1s2 2s2 2p6 3s2 3p5 [Ne] 3s2 3p5

↑↓ ↑ ↑ ↑↓ and not as ↑↓

Vidyala

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Notes on Structure of Atom (31)


Element Atomic

No.

K

(n=1)

L

( n=2)

M

(n=3)

N

(n=4)

Condensed outer

electronic configuration

Argon (Ar) 18 1s2 2s2 2p6 3s2 3p6 [Ne] 3s2 3p6

Potassium (K) 19 1s2 2s2 2p6 3s2 3p6 4s1 [Ar] 4s1

Calcium (Ca) 20 1s2 2s2 2p6 3s2 3p6 4s2 [Ar] 4s2

Scandium (Sc) 21 1s2 2s2 2p6 3s2 3p6 3d1 4s2 [Ar] 3d1 4s2

Titanium (Ti) 22 1s2 2s2 2p6 3s2 3p6 3d2 4s2 [Ar] 3d2 4s2

Vandium (V) 23 1s2 2s2 2p6 3s2 3p6 3d3 4s2 [Ar] 3d3 4s2

Chromium (Cr) 24 1s2 2s2 2p6 3s2 3p6 3d5 4s1 [Ar] 3d5 4s1

Manganese(Mn) 25 1s2 2s2 2p6 3s2 3p6 3d5 4s2 [Ar] 3d5 4s2

Iron (Fe) 26 1s2 2s2 2p6 3s2 3p6 3d6 4s2 [Ar] 3d6 4s2

Cobalt (Co) 27 1s2 2s2 2p6 3s2 3p6 3d7 4s2 [Ar] 3d7 4s2

Nickel (Ni) 28 1s2 2s2 2p6 3s2 3p6 3d8 4s2 [Ar] 3d8 4s2

Copper (Cu) 29 1s2 2s2 2p6 3s2 3p6 3d10 4s1 [Ar] 3d10 4s1

Zinc (Zn) 30 1s2 2s2 2p6 3s2 3p6 3d10 4s2 [Ar] 3d10 4s2

*Q.5 Define electronic configuration and valence shell electronic configuration.

A. Electronic configuration : It is representation of the arrangement of electron in various orbits and orbitals in the atom. For example,

the electronic configuration of Boron is 5B : 1s2, 2s2, 2p1.

Valence shell electronic configuration : It is the representation of the arrangement of electrons in the valence shell (orbit) of an atom with valence

orbitals. For example for oxygen 8O : [He] 2s2, 2p4.

Q.6 Explain the stability of halffilled and completely filled orbitals.

A. The halffilled and completely filled orbitals acquire extra stability due to the following reasons :

1) Symmetrical distribution of electrons :

(a) The halffilled and completely filled orbitals have symmetrical distribution of electrons in them and due to this symmetry, these orbitals are more stable.

(b) Electrons in the same subshell have equal energy but different spatial distribution.

(c) Hence their shielding of oneanother is relatively small and the electrons are strongly attracted by the nucleus.

2) Exchange energy : (a) This is the energy released when electrons exchange their positions within the subshell. The

larger the exchange energy, more is the stability.

(b) The number of exchanges are maximum when the orbitals are halffilled or completely filled.

(c) Due to maximum exchanges, the exchange energy of halffilled and completely filled orbitals is maximum and hence the stability is maximum.

(d) Consider the exchanges in the halffilled d5 configuration. (e) Hence the total exchanges are 4 + 3 + 2 + 1 = 10. Due to this, d5 electronic configuration

acquires more stability.

4 exchange by electron 1 3 exchange by electron 4

2 exchange by electron 3 1 exchange by electron 4

Fig. : Possible exchange for a d5 configuration Vidyala

nkar



Solved Examples

1. An atom has 2K, 8L, 5M electrons. Write down its electronic configuration and indicate in it. (i) number of sub-shells (ii) number of orbitals (iii) number of unpaired electrons (iv) number of electrons having l = 1.

Solution : In the given atom there are (2 + 8 + 5) = 15 electrons. The electronic configuration of the atom is

2 2 6 2 1 1 1x y z1s , 2s , 2p , 3s , 3p , 3p , 3p

(i) Number of subshells is 5 (ii) Number of orbitals is 9 (iii) Number of unpaired electrons is 3 (iv) Number of electrons having l = 1 is 9.

2. Which of the following are isoelectronic species, i.e., those having the same number of electrons?

Na+, K+, Mg2+, Ca2+, S2, Ar

Solution :

Na+ has (11 1) = 10 electrons K+ has (19 1) = 18 electrons

Mg2+ has (12 2) = 10 electrons Ca2+ has (20 2) = 18 electrons Ar has 18 electrons Thus, Na+ and Mg2+ are isoelectronic with 10 electrons K+, Ca2+ and Ar are also isoelectronic with 18 electrons.

3. Write electronic configurations of Fe2+ and Fe3+ ions. Which of these has more number of unpaired electrons? Atomic no. of Fe is 26.

Solution : Fe atom has 26 electrons. Fe2+ and Fe3+ ions are formed by removal of two and three electrons respectively from Fe atom and hence contain 24 and 23 electrons respectively. Their electronic configurations are :

Fe2+ : 1s2, 2s2, 2p6, 3s2, 3p6, 3d6 Fe3+ : 1s2, 2s2, 2p6, 3s2, 3p6, 3d5

Fe2+ has 4 unpaired electrons while Fe3+ has 5 unpaired electrons. Therefore, Fe3+ has more number of unpaired electrons.

Homework Problems

1. How many electrons in an atom of neon (Z = 10) have clockwise spin?

2. Write down electronic configuration of chromium (Z = 24) and indicate in it (i) number of sub-shells (ii) number of orbitals (iii) number of electrons in M-shell.

3. What is the electronic configuration of Zn2+ ion?

4. Write the electronic configurations of the following ions :

(a) H (b) Na+ (c) O2 (d) F.

5. What atoms are indicated by the following configurations : (a) [He] 2s1 (b) [Ne] 3s2 3p3 (c) [Ar] 4s2 3d1.

6. Give the number of electrons in the species H2+, H2 and O2

+.

7. The quantum numbers of six electrons are given below. Arrange them in order of increasing energies.

1) n = 4, l = 2, ml = 2, ms = 1/2 2) n = 3, l = 2, ml = 1, ms = + 1/2

3) n = 4, l = 1, ml = 0, ms = + 1/2 4) n = 3, l = 2, ml = 2, ms = 1/2

5) n = 3, l = 1, ml = 1, ms = + 1/2 6) n = 4, l = 1, ml = +1, ms = + 1/2

Homework Problems

1. Five 2. (i) 7 (ii) 15 (iii) 13

3. 1s2, 2s2, 2p6, 3s2, 3p6, 3d10. During the formation of ion the two 4s electrons are lost.

4. H : 1s2 Na+ : 1s2, 2s2, 2p6

O2 : 1s2, 2s2, 2p6 F : 1s2, 2s2, 2p6

5. (a) Li (b) P (c) Sc

6. The number of electrons in He+, H2 and O2+ are 1, 2 and 15 respectively.

7. 5 < 2 = 4 < 3 = 6 < 1.

Vidyala

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Redox Reactions

Table : The oxidation numbers of some of the elements in their compounds

Element Oxidation

Number Element

Oxidation

Number Element

Oxidation

number

H +1, 1 Co +3, +2 Fe +3, +2

Li +1 Ni +2 Sn +4, +2

Be +2 Cu +2, +1 Sb +5, +3, 3

B +3 Zn +2 Te +6, +4, 2

C +4, +2, 4 Ga +3 I +7, +5, +1, 1

N 5, +4, +3

+2, +1, 3, 2

Ge +4, 4 Xe +6, +4, +2

O +2, 1/2, 1 As +5, +3, 3 Cs +1

F 1 Se +6, +4, 2 Ba +2

Na +1 Br +5, +3, +1, 1 La +3

Mg +2 Kr +4, +2 Hf +4

Al +3 Rb +1 Ta +5

Si +4, 4 Sr +2 W +6, +4

P +5, +3, 3 Y +3 Re +7, +6, +4

S +6, +4, +2, 2 Zr +4 Os +8, +4

Cl +7, +6, +5, +4,

+3, +1, 1

Nb +5, +4 Ir +4, +3

K +1 Mo +6, +4, +3 Pt +4, +2

Ca +2 Tc +7, +6, +4 Au +3, +1

Sc +3 Ru +8, +6, +4, +3 Hg +2, +1

Ti +4, +3, +2 Pd +4, +2 Tl +3, +1

V +5, +4, +3, +2 Ag +1 Pb +4, +2

Cr +6, +5, +4, +3 Cd +2 Bi +5, +3

Mn +7, +6, +4, +3, +2

In +3 Po +2

At 1

*Q.1 Define the terms oxidation, reduction, in terms of electron transfer.

A. Oxidation : It is defined as the process of loss of one or more electrons from a substance (an element, or a compound or an ion).

M M+ + e

Reduction : It is defined as the process of gain of one or more electrons by a substance (an element, or a compound or an ion).

A + e A

*Q.2 Define Oxidant or Oxidising agent and Reductant or reducing agent in terms of electron transfer.

Oxidant : A substance that causes oxidation, and it self under goes reduction is called oxidant or

oxidising agent. OR Oxidising agent is a species that accepts electron (s) and causes other substance to lose electrons (s).

Reductant : A substance that causes reduction and it self under goes oxidisdation is called reductant or

reducing agent. OR A reducing agent is a species that donates electron(s) and causes other substance to accept electron (s).

For example, consider following reaction :

(s) (aq) 2(aq) 2(g)Zn 2HCl ZnCl H

1+ 2+ 00

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In this reaction Zn atom loses two electrons and gets oxidised to Zn2+ while H+ in HCl gains electron and gets reduced from + 1 to 0.

Zn(s) Zn(aq)2+ + 2e (oxidation) 2H+ + 2e 0

2(g)H (reduction)

In the above reaction Zn reduces from H+ ions to 02H hence Zn acts as reluctants while H+ oxidises Zn to

Zn2+ hence H+ acts as oxidant.

Q.3 What are the characteristics of (1) oxidant and (2) reductant.

A. (1) Characteristics of an oxidant:

Consider a reaction: 2 2

s aq aq sZn Cu Zn Cu

(1) Oxidant or oxidizing agent is a species that gains one or more electrons.

For example, aq sCu 2e Cu

(2) It causes other substances (like Zn) to lose one or more electrons and thus oxidises. For example, Cu2+ accepts two electrons from Zn and oxidises it to Zn2+.

(3) Oxidant itself gets reduced (Cu2+ to Cu). Hence in the above reaction, Cu2+ is an oxidant.

(2) Characteristics of a reductant: (1) A reductant or a reducing agent is a species that donates (loses) one or more electrons.

For example, Zn loses two electrons, 2s aqZn Zn 2e , hence it is a reductant.

(2) A reductant causes other substance (e.g. Cu2+) to accept one or more electrons and thus reduces. For example, Cu2+ is reduced to Cu.

(3) A reductant itself gets oxidized (Zn0 to Zn2 +). Hence in the above reaction Zn is a reductant.

Q.4 What is a redox reaction?

A. 1) Redox reaction : A reaction in which oxidation and reduction reactions occur simultaneously is called redox reaction. For example,

2 2(s) (aq) (aq) (s)Zn Cu Zn Cu

2) In this reaction, one species (Zn) loses electrons and undergoes oxidation.

For example, Zn(s) Zn2 + (aq) + 2e (oxidation) 3) The other species (Cu2+) accepts the donated electrons and undergoes reduction.

For example, (aq) (s)Cu 2e Cu (reduction)

4) Hence the overall redox reaction involves simultaneous reactions of oxidation and reduction. In the above reaction, Zn gets oxidized and Cu2 + get reduced.

Arrhenius, Svante (1859-1927) Arrhenius, a Swedish physical chemist, proposed the theory of ionic dissociation. Arrhenius came from a family of farmers, and his father was an estate manager and surveyor. He attended Uppsala University and did very well in physical science. Then he moved to Stockholm to work for a higher studies on aqueous solutions of electrolytes. He concluded that such solutions conduct electricity because the electrolyte exists in the form of charged atoms or groups of atoms, which move through the solution under the application of electric current. In 1903 he was awarded the Nobel Prize in chemistry. He worked on different subjects including immunology, cosmic physics and also on ‘greenhouse effect’. He also studied the effect of temperature on the rates of chemical reactions.

Q.5 What is an oxidation number or oxidation state ? How is it useful to identify redox reactions,

oxidant and reductant ?

A. The oxidation number or oxidation state of an atom in a molecule or ion is defined as the number of charges it would carry if electrons were completely transferred.

(s) (aq) 2(aq) 2(g)Zn 2HCl ZnCl H

In this reaction, Zn atoms lose two electrons and are oxidized to Zn2+ ions.

2(s) (aq)Zn Zn 2e

In water, HCl dissociates into H+ and Cl ions.

(aq) (aq) (aq)2HCl 2H 2Cl

H+ ions accept one electorn each given by Zn atoms.

(aq) 2(g)2H 2e H

Vidyala

nkar

Notes on Redox Reactions (35)


Thus, H+ ions are reduced to H atoms which combine to form H2. It can be said that Zn atoms cause reduction of H+ ions and H+ ions cause oxidation of Zn atoms. Hence,

Zn is a reducing agent and H+ ion acts as an oxidizing agent. Zn atoms lose electrons and H+ ions accept electrons.

Q.6 Explain rules to assign oxidation numbers or oxidation states.

A. The following are the general rules formulated to assign an oxidation number to an atom in any molecule or ion.

Rule-1 : The oxidation number of an atom in free uncombined elemental state is zero. For example, each atom in the elements such as Ca, H2, Cl2, O3, S8, P4 and so on has oxidation number of zero.

Rule-2 : The oxidation number of an atom in a monoatomic ion is equal to its charge. For example,

Ba2+ Cr3+ K+ Br S2

+2 +3 +1 1 2

Rule-3 : The oxidation number of H atom is either +1 or 1. When H atom is bonded to

non-metals, its oxidation number is +1. When it is bonded to metals, it will have oxidation number of 1. For example,

[O H] H O H Li H H Ca H

2 +1 +1 2 +1 +1 1 1 +2 1

Rule-4 : The oxidation number of oxygen is usually 2 in all of its compounds except in peroxides and

peroxide ions where it has oxidation number of 1. For example,

Ca O H O O H [O O]2

+2 2 +1 1 1 +1 1 1

In OF2, oxidation number of oxygen is +2.

Rule-5 : The oxidation number of F is always 1 in all of its compounds. The other halogens Cl, Br and I

usually have oxidation number of 1 in their compounds. The exception is the compounds in which these halogens (Cl, Br and I) are bonded to oxygen. In such compounds oxidation number of Cl, Br or I is +1. For example,

H F KBr Cl O Cl H O Cl

+1 1 +1 1 +1 2 +1 +1 2 +1

Rule-6 : The algebraic sum of the oxidation states of all atoms in a neutral molecule is zero, For example, in CdS,

Oxidation number of Cd + oxidation number of S = + 2 2 = 0

Rule-7 : The algebraic sum of the oxidation numbers of all the atoms in a polyatomic ion is equal to the

net charge of the ion. For example, in 24SO ion,

Oxidation state of S + 4 oxidation state of oxygen

= + 6 + 4 (2)

= 6 8

= 2 Note that (a) oxidation numbers may be fractional (b) the metallic elements have only positive oxidation states in their compounds whereas the non-metallic elements may have either positive or negative oxidation states. (c) the alkali metals have only +1 oxidation state while the oxidation number of the alkaline earth metals is +2 in their compounds.

Q.7 Give characteristics of the following :

(i) Oxidation (ii) Reduction (iii) redox reaction

(iv) oxidizing agent (oxidant) (v) reducing agent (reductant).

A. (i) Oxidation : a) Loss of electrons. b) Increase in oxidation number of oxidized species.

(ii) Reduction : a) Gain of electrons. b) Decrease in oxidation number of reduced species.

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24 0O S S

2

O

2O

(iii) Redox reaction : a) Oxidation and reduction together. b) Simultaneous loss and gain of electrons. c) Simultaneous increase and decrease in oxidation numbers.

(iv) Oxidizing agent (Oxidant) : a) Causes oxidation. b) Accepts electron(s). c) Itself undergoes reduction. d) Undergoes decrease in oxidation number.

(v) Reducing agent (Reductant) : a) Causes reduction. b) Donates electron(s). c) Itself undergoes oxidation. d) Undergoes increase in oxidation number.

Q.8 How identification of redox reaction is explained on the basis of oxidation number concept

A. Oxidation is loss of electrons. When a species loses electrons, its oxidation number increases.

Therefore, increase in oxidation number of a substance by loss of electrons is called oxidation. Reduction is gain of electrons by a substance. The gain of electrons decreases the oxidation number of

the substance. Thus, decrease in oxidation number of a substance by gain of electrons is reduction. An oxidizing agent or oxidant is a substance that accepts electrons and thereby undergoes decrease in oxidation number. A reducing agent or reductant is a substance that donates electrons and thereby undergoes increase in oxidation number.

A redox reaction involves oxidation and reduction simultaneously. It is an electron transfer reaction. One substance donates electrons and goes to higher oxidation state. Whereas, the other substance gains electrons and lowers its oxidation number. Therefore, a reaction in which oxidation number of one species increases due to the loss of electrons and the oxidation number of another species decrease due to the gain of electrons is a redox reaction. For example, consider the unbalanced reaction.

In this reaction the oxidation number of Cu increases from 0 to +2. It is therefore oxidized by loss of electrons and acts as a reducing agent. On the other hand, oxidation number of N decreases from +5 to +2. Evidently, N is reduced by gain of electrons and acts as an oxidizing agent. Therefore, the reaction is a redox reaction as oxidation number of one species (Cu) increases and that of the species (N) decreases.

Solved Problems

* Example 1 : Assign oxidation numbers to each element in the following compounds or ions.

(a) 22 3S O (b) 3HNO (c) 4Cr (OH) (d) 2

4 6S O

Solution:

(a) 2-2 3S O

According to rule 4, the oxidation number of oxygen is 2 and according rule

7 the sum of the oxidation numbers of all atoms is 2. Therefore,

2 oxidation number of S + 3 oxidation number of O = 2 It follows that,

2 oxidation number of S + 3 (2) = 2

2 oxidation number of S 6 = 2

2 oxidation number of S = +6 2 = + 4 Thus, oxidation number of S = +2

22 3S O

+2 2

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2 5 5 20 0O S S S S O

2

O 2

O

2O 2

O

(b) HNO3

According to rule 3 oxidation number of H is +1. According to rule 4 oxidation number of O is 2. Rule 6 states that sum of the oxidation states of all atoms is zero.

Hence, oxidation number of H+ oxidation number of N + 3 oxidation number of O = 0.

+1 + oxidation number of N + 3 (2) = 0

+1 + oxidation number of N 6 = 0 Therefore, oxidation number of N = +5 H N O3

+1 +5 2

(c) -4Cr(OH)

According to rule 3 oxidation number of H is +1 and according to rule 4 oxidation number of O is 2.

The rule 7 states that the algebraic sum of the oxidation number of all atoms is 1. Therefore,

Oxidation number of Cr + 4 oxidation number of H + 4 oxidation number of O = 1

Oxidation number of Cr + 4 (+1) + 4 (2) = 1

Oxidation number of Cr + 4 8 = 1 Hence, oxidation number of Cr = +3

4Cr(OH)

+3 2 +1

(d) 2-4 6S O

Oxidation number of O is 2 and the algebraic sum of the oxidation states of all the atoms is 2.

Therefore,

4 oxidation state of S + 6 oxidation state of O = 2

4 oxidation state of S + 6 (2) = 2

4 oxidation state of S 12 = 2

4 oxidation state of S = + 12 2 = + 10

Hence, oxidation state of S = 10

2.54

2-4 6S O

+2.5 2

* Example 2 : Determine the oxidation number of

(a) As in H3AsO3 (b) Pt in 26PtCl (c) C in K2C2O4 (d) V in 4

2 7V O

Solution :

(a) As in H3AsO3 : Oxidation number of H = +1

Oxidation number of O = 2 Sum of the oxidation states of all atoms = 0

Hence, 3 oxidation number of H + oxidation number of As + 3 oxidation number of O = 0

3 (+1) + oxidation number of As + 3 (2) = 0

oxidation number of As + 3 6 = 0 oxidation number of As = +3

(b) Pt in 2-6PtCl :

Oxidation number of Cl = 1

Sum of the oxidation numbers of all atoms = 2

Hence, oxidation number of Pt + 6 oxidation number of Cl = 2

Oxidation number of Pt + 6 (1) = 2

Oxidation number of Pt 6 = 2

Oxidation number of Pt = +6 2 = +4

(c) C in K2C2O4 : Oxidation number of K = +1

Oxidation number of O = 2

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Sum of the oxidation states of all atoms = 0 Therefore,

2 oxidation number of K + 2 oxidation number of C + 4 oxidation number of O = 0

2 (+1) +2 oxidation number of C + 4 (2) = 0

+2 +2 oxidation number of C 8 = 0

2 oxidation number of C = +6

Oxidation number of 6

C 32

.

(d) V in 4-2 7V O :

Oxidation number of O = 2

Sum of the oxidation states of all atoms = 4

Therefore, 2 oxidation state of V + 7 oxidation state of O = 4

2 oxidation state of V + 7 (2) = 4

2 oxidation state of V = +14 4 = +10

Oxidation state of V = 10

52

* Example 3 : Identify whether the following reactions are the redox reactions. Identify oxidants and reductants in the reactions that are found to be redox reactions.

(a) 3 3(aq) 3(aq) (aq) 3 4(aq)3H AsO BrO Br 3H AsO (b) (aq)(aq) (aq) 2 ( )HF OH H O F

Solution: (a) Write the oxidation number of all the atoms of reactants and products.

3 3(aq) 3(aq) (aq) 3 4(aq)3H AsO BrO Br 3H AsO

+1 +3 2 +5 2 1 +1 +5 2

Identify the species that undergo change in oxidation numbers.

The oxidation number of As increases from +3 to +5 and that of Br decreases from +5 to 1. Because oxidation number of one species increases and that of other decreases, the reaction is a redox reaction. The oxidation number of As increases by loss of electrons. As is a reducing agent and it is itself oxidized. The oxidation number of Br decreases. Hence, it is reduced by the gain of electrons and acts as an oxidizing agent.

Results :

(i) The reaction is a redox reaction.

(ii) Oxidant : 3BrO

(iii) Reductant : H3AsO3

(b) Assign oxidation numbers to all the atoms.

(aq) (aq) 2 ( ) (aq)HF OH H O F

+11 2+1 +1 2 1

Identify the species undergoing change in oxidation number. It is observed that no species undergoes change in oxidation number. Therefore, the reaction is not a redox reaction.

* Example 4 : Identify oxidizing and reducing agents in the following reaction.

2 22(aq) 2 3(aq) 4 6(aq) (aq)I 2S O S O 2I

Solution : Assign oxidation numbers to all the atoms. The element I2 has oxidation number of zero. The oxidation

numbers of O and I are respectively 2 and 1.

3 3(aq) 3(aq) (aq) 3 4(aq)3H AsO BrO Br 3H AsO

+3 +5 +1 +5 Gain of e

Loss of e

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2 2 1-2(aq) 2 3(aq) 4 6(aq) (aq)I 2S O S O 2I

0 +2 2 +2.5 2 1

Now, identify the species that undergoes change in oxidation number.

The oxidation number of I decreases from 0 to 1. It is therefore, reduced and is an oxidizing agent. The oxidation number of S increases from +2 to +2.5. It is oxidized and is a reducing agent.

Hence, Oxidizing agent is I2 and reducing agent is 22 3S O

* Example 5 :

Assign oxidation numbers to all the atoms present in :

(i) 22 4C O (ii) Mn2O3 (iii)

4Al OH

(iv) KAuCl4.

Solution:

(i) C2 2-4O :

Oxidation number of O = 2 Let the oxidation number of C = x

22 4C O ion carries 2 charge, hence the

sum of the oxidation numbers of all the

atoms present must be 2.

2x (C) + 4(2 ) (O) = 2

2x 8 = 2 x = 3

Hence oxidation number of C in 22 4C O = 3 +.

(ii) Mn2O3 :

Oxidation number of O = 2 Let the oxidation number of Mn = x Since the molecule is neutral,

2x(Mn) + 3(2)(O) = 0

2x 6 = 0

x = 3 Hence the oxidation number of Mn = 3 +.

(iii) -

4Al OH :

Oxidation number of H = 1 +

Oxidation number of O = 2 Let the oxidation number of Al = x

Since the ion carries 1 charge,

x (A) + 4 (2 ) (O) + 4(1 +) (H) = 1

x 8 + 4 = 1 x = 3 Hence the oxidation number of Al = 3 +.

(iv) KAuCl4 : Oxidation number of alkali metal K = 1 +

Oxidation number of Cl = 1 Let the oxidation number of Au = x Since the molecule is neutral,

(1 +) (K) +x(Au) + 4(1 ) (Cl) = 0

1 + x 4 = 0

x = 3 Hence the oxidation number of Au = 3 +.

* Example 6 :

Find the oxidation numbers to the underlined species in the following compounds or ions: 6PF , 3Na I O ,

2 72K Cr O , 3NaHCO , 3CIF , 6SbF , 4NaBH , 2 6H PtCl , 5 3 10H P O , 4CuSO , 3BiO , 3CH OH , 2 2H O , 24 4 6C H O ,

2 4H AsO , 3

Mn OH , 3I , 2 5C H OH , 3 3H PO , 2 3Na CO , 4I O , 34VO , 32Ni O , 3 6

K Fe CN

Solution:

(1) -6PF

Oxidation number of F = 1 Let the oxidation number of P = x

Since the ion carries 1 charge,

x (P) + 6 (1 ) (F) = 1

x = 5 Hence the oxidation number of P = 5 +.

(2) 3Na I O :

Oxidation number of alkali metal, Na = 1+

Oxidation number of O = 2 Let the oxidation number of I = x Since the molecule is neutral,

1 + (Na) + x + 3 (2 ) (O) = 0

1 + x 6 = 0 x = 5 Hence the oxidation number of I = 5 +

(3) 2 72K Cr O :

Oxidation number of K = 1 +

Oxidation number of O = 2 Oxidation number of Cr = x

Since the molecule is neutral,

2 (1 + ) (K) + 2x (Cr) + 7 (2 ) (O) = 0

2 + 2x 14 = 0

x = 6 Hence the oxidation number of Cr = 6 +.

(4) 3NaHCO :

Oxidation number of Na = 1 + Oxidation number of H = 1 +

Oxidation number of O = 2 Let the oxidation number of C = x Since the molecule is neutral,

1 + (Na) + (1 +) (H) + x (C) + 3 (2 ) (O) = 0

1 + 1 + x 6 = 0 x = 4 Hence the oxidation number of C = 4 +

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(5) 3CI F :

Oxidation number of F = 1 Let the oxidation number of Cl = x

x (Cl) + 3 (1 ) (F) = 0

x 3 = 0

x = 3 Hence the oxidation number of Cl = 3 +.

(6) -6Sb F :

Oxidation number of F = 1 Let the oxidation number of Sb = x

Since the ion 6SbF carries 1 charge

x (Sb) + 6 (1 ) (F) = 1

x 6 = 1 x = 5 Hence the oxidation number of Sb = 5 +

(7) 4NaBH :

Oxidation number of Na = 1 +

Oxidation number of H (in hydride) = 1 Let the oxidation number of B = x

(1 +) (Na) + x(B) + 4(1 )(1 +) = 0

1 + x 4 = 0 x = 3 Hence the oxidation number of B = 3 +.

(8) 2 6H PtCl :


Oxidation number of Cl = 1 Let the oxidation number of Pt = x

2(1 +) H + x(Pt) + 6(1 ) (Cl) = 0

2 + x 6 = 0 x = 4 Hence the oxidation number of Pt = 4 +

(9) 5 3 10H P O :


Oxidation number of O = 2 Let the oxidation number of P = x

5(1 +)(H) + 3x(P) + 10(2 )(O) = 0

5 + 3x 20 = 0 x = 5 Hence the oxidation number of P = 5 +.

(10) 4CuSO :

Oxidation number of O = 2 Oxidation number of Cu = 2 + Let the oxidation number of S = x

(2+)(Cu) + x (S) + 4(2 )(O) = 0

2 + x 8 = 0 x = 6 Hence the oxidation number of S = 6 +.

(11) -3Bi O :

Oxidation number of O = 2 Let the oxidation number of Bi = x

Since the ion 3BiO carries 1 charge,

x(Bi) + 3(2 ) (O) = 1

x 6 = 1 x = 5 Hence the oxidation number of Bi = 5 +.

(12) 3CH OH :



x(C) + 4(1 +)(H) + (2 )(O) = 0

x + 4 2 = 0 x = 2

Hence the oxidation number of C = 2 .

(13) 2 2H O :

In hydrogen peroxide,

Oxidation number of O = 1 Let the oxidation number of H = x

2x (H) + 2(1 ) (O) = 0

2x 2 = 0 x = 1 Hence the oxidation number of H = 1 +

(14) 2-4 4 6C H O :



4x (C) + 4(+ 1) (H) + 6(2 ) (O) = 2

4x + 4 12 = 2 4x = 6 x = 1.5 Hence the oxidation number of C = 1.5 +.

(15) -2 4H AsO :


Oxidation number of O = 2 Let the oxidation number of As = x

Since the ion 2 4H AsO carries 1 charge,

2(1 +) (H) + x (As) + 4(2 ) (O) = 1

2 + x 8 = 1 x = 5 Hence, the oxidation number of As = 5 +

(16) 3

Mn OH :


Oxidation number of O = 2 Let the oxidation number of Mn = x

The molecule 3

Mn OH has 1 Mn atom,

3H atoms and 3O atoms.

x + 3(1 +) (H) + 3(2 ) (O) = 0

x 3 6 = 0 x = 3 Hence the oxidation number of Mn = 3 +.

(17) -3I :

Triiodide ion 3I has two different

oxidation numbers for I, namely zero

and 1 .

(18) 2 5C H OH :



The molecule 2 5C H OH has 2C atoms,

6H atoms and 1O atom.

2x (C) + 6 (1 +) (H) + (2 ) (O) = 0

2x + 6 2 = 0 x = 2

Hence the oxidation number of C = 2 .

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2 aq 3 aq g

1 3 2 5 2 2 2

HNO NO NO

(19) 3 3H PO :


Oxidation number of O = 2 Let the oxidation number of P = x

3 (1 +) (H) + x + 3 (2 ) (O) = 0

3 + x 6 = 0 x = 3

Hence the oxidation number of P = 3 +.

(20) 2 3Na CO :

Oxidation number of Na = 1 +


2 (1 +) (Na) + x (C) + 3 (2 ) (O) = 0

2 + x 6 = 0 x = 4 Hence the oxidation number of C in

2 3Na CO = 4 +.

(21) -4I O :

Oxidation number of Na = 1 + Let the oxidation number of I = x

Since the ion 4IO carries 1 charge,

x (I) + 4 (2 ) (O) = 1

x 8 = 1 x = 7 Hence the oxidation number of I = 7 +

(22) 3-4VO :

Oxidation number of O = 2 Let the oxidation number of V = x

Since the in 3-4VO carries 3 charge,

x (V) + 4(2 ) (O) = 3

x 8 = 3 x = 5 Hence the oxidation number of V = 5 +.

(23) 32Ni O :

Oxidation number of O = 2 Let the oxidation number of Ni = x

2x (Ni) + 3(2 ) (O) = 0

2x 6 = 0

x = 3 Hence the oxidation number of Ni = 3 +.

(24) 3 6

K Fe CN :

Oxidation number of K = 1 + Oxidation number of C = 2 +

Oxidation number of N = 3 Let the oxidation number of Fe = x

3 6K Fe CN

has 3K, 1Fe, 6C and

6N atoms

3 (1 +) (K) + x (Fe) + 6 (2 +) (C) + 6(3 ) (N) = 0

3 + x + 12 18 = 0 x = 3 Hence the oxidation number of Fe = 3 +.

* Example 7 : Assign oxidation number to each atom in (i) ZnNH4PO4 and (ii) HCN.

Solution :

(i) ZnNH4PO4 :

Oxidation numbers of the atoms are,

Zn = 2 +, N = 3 , H = 1 +, O = 2 . Since P shows variable oxidation numbers, let the oxidation number of P = x.

(2 +) (Zn) + (3 ) (N) + 4 (1 +) (H) + x (P) + 4 (2 ) (O) = 0

2 3 + 4 + x 8 = 0 x = 5 Hence the oxidation number of P = 5 +.

(ii) HCN:

Oxidation number of the atoms are, H = 1 +, C = 2 +, N = 3

* Example 8 : Identify, which atom is oxidized and which is reduced in the following unbalanced redox equations:

(i) - - - -3 aq aq aq 3 aqIO IO I I (ii) -

2 aq 3 aq gHNO NO NO

Solution :

(i) 3 aq aq aq 3 aqIO IO I I

5+2 1+ 2 1 0,1

In this reaction I(5 +) in 3IO undergoes reduction to I(1 ) and I(1 +) in IO undergoes oxidation to I2(0).

(ii)

In this reaction N (3 +) in HNO2 is oxidized to N(5 +) in 3NO and N (3 +) is reduced to N (2 +) in NO.

(This type of a reaction is called disproportionation reaction)

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* Example 9 : Divide the following redox equations into oxidation half equation and reduction half equation.

(a) 3 2

s 4 aq 3 aqP PO HPO (b) 2 aq 3 aq gHNO NO NO

(c) 3 aq s 4 aq aqIO Re ReO I (d) 2

4 aq 2 s 4 aqMnO MnO MnO

Solution: (a)

P(s) is oxidized from 0 to 3 + in 23HPO and P in 3

4PO is reduced from 5 + to 3 +.

Oxidation half equation : 2

s 3 aqP HPO

Reduction half equation : 3 24 aq 3 aqPO HPO

(b)

Oxidation half equation : 2 aq 3 aqHNO NO

Reduction half equaton : 2 aq gHNO NO

(This is a disproportionation reacton.)

(c)

5 10 7

aq3 aq s 4 aqI O R e ReO I

Oxidation half equation : s 4 aqRe ReO

Reduction half equation : 3 aq aqIO I

(d) 6 4 7

24 aq 2 s 4 aqMnO MnO MnO

Oxidation half equation : 24 aq 4 aqMnO MnO

Reduction half equation : 24 aq 2 sMnO MnO

Example 10 : Assign oxidation number to the underlined elements in each of the following species : (a) NaH2PO4 (b) NaHSO4 (c) H4P2O7 (d) K2MnO4 (e) CaO2 (f) H2S2O7 (g) KAI(SO4)212H2O.

Solution : (a) Let the oxidation number of P be x. The oxidation number of each atom is shown above its symbol,

1 1 x 2

2 4NaH PO

Sum of oxidation numbers of various atoms in NaH2PO4 is zero, hence

1 (+1) + 2 (+1) + 1 (x) + 4 (2) = 0

or x 5 = 0 or x = + 5

Thus, the oxidation number of P in NaH2PO4 = + 5.

oxidised reduced

oxidised

reduced

oxidised

reduced

3 2

s 4 aq 3 aq

0 5 3

P PO HPO

oxidised

reduced

2 aq 3 aq g

3 5 2

HNO NO NO

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(b) In the similar way as discussed, the sum of oxidation numbers in 1 1 x 2

4NaHSO

will be,

1 (+1) + 1 (+1) + x + 4(2) = 0 or x = + 6

Thus, the oxidation number of S in NaHSO4 = + 6.

(c) The sum of oxidation numbers in, 1 x 2

4 72H P O

, will be

4 (+1) + 2(x) + 7(2) = 0 or x = + 5

Thus, the oxidation number of P in H4P2O7 = + 5.

(d) The sum of oxidation numbers in 1 x 2

2 4K MnO

, will be

2 (+1) + 1 (x) + 4(2) = 0 or x = + 6.

Thus, the oxidation number of Mn in K2MnO4 = + 7.

(e) The oxidation number of Ca is +2. Thus, the sum of oxidation numbers in 2 x

2CaO

will be

+ 2 + 2x = 0 or x = 1

Thus, oxidation number of oxygen in CaO2 = 1.

(f) The sum of oxidation numbers in, 1 x 2

2 72Na S O

will be given as

2 (+1) + 2 (x) + 7 (2) = 0 or x = + 6

Thus, the oxidation number of S in Na2S2O7 = + 6.

(g) The sum of oxidation numbers in, 1 3 x 12 2

4 2 2K Al(SO ) 12(H O ) ,

is

+ 1 + 3 + 2x + 8 (2) + 12 (2 1 2) = 0 or x = + 6 Alternatively, since H2O is a neutral molecule, therefore, sum of oxidation numbers of all the atoms in H2O may be taken as zero. As such water molecules can be ignored while computing the oxidation number of S. Accordingly,

+ 1 + 3 + 2x 16 = 0 or x = + 6

Thus, the oxidation number of S in KAl (SO4)2. 12 H2O = + 6.

Example 11 : What are the oxidation number of the underlined elements in each of the following and how do you rationalize your results? (a) KI3 (b) H2S4O6 (c) Fe3O4 (d) CH3CH2OH (e) CH3COOH

Solution :

(a) In KI3, since the oxidation number of K is +1, therefore, the average oxidation number of iodine = 1/3. But the oxidation number cannot be fractional. Therefore, in such cases we must consider its

structure, 3(K I )

. In 3I ion, a coordinate bond is formed between I2 molecule and I ion. The oxidation

number of two iodine atoms forming the I2 molecule is zero while that of iodine forming the coordinate bond

is 1.

(b) Conventional method. The average oxidation number in 1 x 2

62 4H S O

is calculated as:

2 (+1) + 4x + 6 (2) = 0 or x = + 2.5. But it is wrong because all the four S atoms cannot be in the same oxidation state, in the compound.

Thus, in such cases the oxidation number is calculated by Chemical bonding

method, i.e., the oxidation number is computed on the basis of its structure. The structure of H2S4O6 is as follows :

0 05 5H O S S S S OH

O

O O

O

The O. N. of each of the S-atoms linked with each other in the middle is zero while that each of the remaining two S-atoms is + 5.

(c) Conventional method. The average oxidation number of Fe in x 2

43Fe O

will be

3x + 4 (2) = 0 or x = 8 / 3.

Fe3O4 is mixed oxide therefore its oxidation number is calculated by Stoichiometry method.

Accordingly, Fe3O4 shall be written as: 2 2 3 2

2 3FeO.Fe O

.

Fe has O.N. of + 2 and + 3.

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2 1H C C OH

H

H H

H

2 1H C C OH

H O

H

(d) Conventional method. The average oxidation number of carbon in C2H5OH or x 1 2

62C H O

will be

2x + 6(+1) + 1 (2) = 0 or x = 2.

On the basis of chemical bonding method as carbon C-2 is attached to three H-atoms

(less electronegative than carbon) and one CH2OH group

(more electronegative than carbon), therefore, O.N. of C-2 will be = 3 (+1) + x + 1 (1) = 0

or x = 2.

The carbon, C-1 is, however, attached to one OH group (O.N. 1) and one CH3 group (O. N. = + 1),

therefore, O.N. of C-1 = + 1 + 2 (+1) + x + 1 (1) = 0

or x = 2.

(e) Conventional method. The oxidation number of carbon in CH3COOH or x 1 2

42 2C H O

will be

2x + 4 4 = 0 or x = 0.

Chemical bonding method The carbon C-2 is attached to three H-atoms (less electronegative than carbon)

and one COOH group (more electronegative than carbon), therefore,

O.N. of C-2 = 3 (+1) + x + 1 (1) = 0 or x = 2

C-1 is, however, attached to one oxygen atom by a double bond, one OH group (O.N. = 1) and one

CH3 group (O.N. = + 1), therefore, O.N. of C-1 = + 1 + x + 1 (2) + 1 (1) = 0 or x = + 2.

Example 12 : Fluorine reacts with ice and results in the change :

1 2 0 1 1 0 1

2 2H O S F g HF HOF g

Justify that this reaction is a redox reaction. If oxygen of HOF disproportionates at room temperature, then what reaction is possible?

Solution :

In the given reaction O.N. of F2 changes from zero to 1 in HF and HOF whereas O.N. of oxygen change

from 2 in H2O to zero in HOF. Thus, F2 is reduced, whereas oxygen is oxidised and, therefore, it is a redox reaction.

HOF is an highly unstable molecule and hence decomposes to form O2 and HF

1 1 0

22HOF 2H F O

If oxygen of HOF disproportionates, then in the disproportionation reaction oxygen must have three oxidation states. In simple words, the oxidation state of oxygen in one of the products should be higher. The oxidation state of oxygen is zero in HOF.

It decrease its oxidation state to 2 if HOF gets reduced to H2O and also it can increase its oxidation state to + 2 if HOF gets oxidized to OF2. Therefore, the possible reaction is :

1 0 1 1 2 2 1

222H O F H O O F

Example 13 :

Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, CrO5 and NO3. Suggest

structures of these three compounds. Count for the fallacy.

Solution :

(i) O.N. of S in H2SO5 On the basis of conventional method, the O.N. of S in H2SO5 comes out to be + 8 as shown below.

2 (+1) + x + 5 (2) = 0 or x = + 8 This is impossible because the maximum O.N. of S cannot be more than six since it has only six electrons in the valence shell. This fallacy is overcome by calculating the O.N. of S by chemical bonding method.

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N

2O

2

O 2

O

-1

The structure of H2SO5 along with O.N. of different atoms is :

1 2 1 1 1

H O S O O H

2

O

2O

Hence, 2 (+1) + 3 (2) + x + 2(1) = 0

or x = + 6

(ii) O.N. of Cr in CrO5

According to conventional method, O.N. of Cr is : x + 5 (2) = 0 or x = + 10. This is impossible

because the maximum O.N. of Cr cannot be more than six since it has a maximum of six electrons

(3d5 4s2) in the outer orbital configuration which can participate in bonding. This fallacy is removed

by calculating O.N. of Cr by chemical bonding method. The structure of CrO5 is :

Cr

2

O

1

O

1

O

1

O

1

O

From the structure, the O.N. of Cr can be calculated as follows :

x + 4 (1) + 1 (2) = 0 or x = + 6

It should be noted that four O atoms have peroxy linkage (O.N. = 1) and O held by double bond has

O.N. = 2.

(iii) O.N. of N in -3NO

According to conventional method, O.N. of N in 3NO

= x + 3 (2) = 1 or x = + 5

According to chemical bonding method, the structure of nitrate ion is : Hence the O.N. of nitrogen is,

x + 3 (2) = 1 or x = + 5

Thus, the O.N. of N in NO3 whether one calculates by conventional method or by chemical bonding

method is same.

Classwork Problems

1. Calculate oxidation number of : i) S in H2S ii) C in CH2Cl2 iii) Cr in K2Cr2O7 iv) Pb in Pb3O4

v) P in Na3PO4 vi) C in 23CO vii) Cr in 2

2 7Cr O viii) Mn in 4MnO

ix) Mn in 2MnO

2. Classify giving reasons the following unbalanced half equations as oxidation and reduction.

Use oxidation number concept.

i) Fe3+ Fe2+ ii) NH3(g) NO(g) iii) 2

4 aq aqMnO Mn

iv) 3 aq 2 aqClO ClO v) 3 aq aqClO Cl vi) 3 aq 2 gNO NO

vii) 2aq 2 sMn MnO viii)

22 4 aq 2 gC O CO ix) aq 2 aqBr Br

3. Using oxidation number concept, indentify the redox reactions. Hence, identify oxidizing and reducing

agents in the redox reactions.

i) 3 3 24 aq aq 4 aqH PO 3KOH K PO 3H O

ii) s aq 2 aq 2 gZn 2HCl ZnCl H

iii) 2 3

2aq 3 aq aq aq aq6Fe BrO Fe Br 3H O 6H

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iv) 22 aq aq s aqMnCl Na S MnS 2NaCl

v) 2aq 4 aq 4 aqNaOH HClO NaClO H O

vi) s 2 g s2Zn O 2ZnO

vii) s aq aq sB D B D viii) 2 4aq 4 aq aq aqSn IO Sn I

ix) 2 aq aq aq2 sFeCl 2KOH Fe OH 2KCl

* 4. Arrange the following in order of increasing oxidation number of C atom.

CaC2,23CO , 2

2 4C O , C2H4, CH4

* 5. Identify which atoms are oxidized and which are reduced in the following redox reactions. Mention oxidants and reductants.

i) 2

aq 3 aq aq 2 aqCu IO Cu I

ii) 2 22 aq 7 aq 2 aq 2 gH O Cl O ClO O

iii) 2 2 2 3

2 2 43 aq 7 aq 6 aq aqS O Cr O S O Cr

iv) 2 3

4 26 aq s aq aqS O Al H S Al

6. Divide the following redox equations into oxidation half equation and reduction half equation.

i) 3 2

s 4 aq 3 aqP PO HPO ii) 32 aq gHNO NO NO

iii) 3 aq s 4 aq aqIO Re ReO I iv) 24 aq 2 s 4 aqMnO MnO MnO

Homework Problems

1. Calculate the oxidation number of (i) C in CO2, (ii) N in (NH4)2SO4,

2. Calculate the oxidation number of (i) P in 3 2 7H P O ; (ii) Cl in 4ClO

3. What is the the oxidation number of metals in (i) 4

6Fe CN

and (ii) 4MnO?

4. Find the oxidation number of the element in bold in the following species:

(i) SiH4,BH3, BF3, 23O

2S , Br 4O and HP 24O (ii) PbSO4, U2

47O , Cr 2

4O , K2MnO4.

5. Determine the oxidation number of C in the following :

C2H6, C4H10, CO, CO2, and HCO3

6. Determine the oxidation number of O in the following : OF2, Na2O2, Na2O, KO2, KO3 and O2F2.

7. Find out the oxidation number of Cl in HCl, HClO, CaOCl2 and ClO2.

8. Find out the oxidation number of sulphur in the following species:

24 4 2 4 2 42

NH SO , H SO ,S O , 22 7 3 4S O , HSO and HSO .

9. Determine the oxidation number of all the atoms in the following well known oxidants KMnO4, K2Cr2O7 and KClO4.

10. Determine the change in the oxidation number of S in H2S and SO2 in the following industrial reaction:

2H2S(g) + SO2(g) 3S (s) + 2H2O(g)

11. What is the oxidation number of S in (i) Na2S4O6 (ii) S2Cl2

12. Identify the oxidant and reductant in the following reactions:

(i) 23 4 210H aq 4Zn s NO aq 4Zn aq NH aq 3H O

(ii) I2 (g) + H2S(g) 2 HI (g) + S (s).

13. Try all possible approaches to justify that the following reactions are redox reactions.

(i) CuO (s) + H2(g) Cu (s) + H2O (g)

(ii) Fe2O3 (s) + 3 CO (g) 2 Fe (s) + 3 CO2 (g)

(iii) 4 BCl3 (g) + 3 LiAIH4 (s) 2 B2H6 (g) + 3 LiCl (s) + 3 AlCl3 (s)

(iv) 2 K (s) + F2 (g) 2 K+ F (s)

(v) 4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (g)

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14. Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions.

(i) 2 AgBr (s) + C6H6O2 (aq) 2 Ag (s) + 2 HBr (aq) + C6H4O2 (aq)

(ii) HCHO() + 2 [Ag (NH3)2]+(aq) + 3 OH (aq) 2 Ag (s) + HCOO (aq) + 4 NH3 (aq) + 2 H2O ()

(iii) HCHO () + 2 Cu2+ (aq) + 5 OH (aq) Cu2O (s) + HCOO (aq) + 3 H2O ()

(iv) N2H4 () + 2 H2O2 () N2 (g) + 4 H2O ()

(v) Pb (s) + PbO2 (s) + 2 H2SO4 (aq) 2 PbSO4 (s) + 2 H2O ()

15. Calculate the oxidation number of phosphorus in the following species.

(i) 23HPO and (ii) 3

4PO

16. Calculate the oxidation number of each sulphur atom in the following compounds: (i) Na2S2O3 (ii) Na2S4O6 (iii) Na2SO3 (iv) Na2SO4

17. Identify the redox reactions out of the following reactions and identify the oxidising and reducing agents in them.

(i) 3 HCl (aq) + HNO3 (aq) Cl2 (g) + NOCl (g) + 2 H2O ()

(ii) HgCl2 (aq) + 2 KI (aq) HgI2 (s) + 2 KCl (aq)

(iii) 2 3 2Fe O (s) 3CO(g) 2Fe(s) 3CO (g)

(iv) PCl3 (l) + 3 H2O () 3 HCl (aq) + H3PO3 (aq)

(v) 4 NH3 + 3 O2 (g) 2 N2 (g) + 6 H2O (g)

Classwork Problems

1. (i) 2 (ii) 0 (iii) +6 (iv) 8 / 3 (v) +5 (vi) +4 (vii) +6 (viii) +7 (ix) +4

2. (i) reduction (ii) oxidation (iii) reduction (iv) reduction (v) reduction (vi) reduction (vii) oxidation (viii) oxidation (ix) oxidation

3. (i) It is not redox but acidbase neutralization reaction, since oxidation numbers of all the atoms remain unchanged.

(ii) HCl (oxidant), Zn (reductant) (iii) 3BrO (oxidant), Fe2+ (reductant)

(iv) It is a double displacement reaction. (v) Acidbase neutralization reaction

(vi) O2 (oxidant), Zn (reductant) (vii) B (reductant), D+ (oxidant)

(viii) 4IO (oxidant), Sn2+ (reductant) (ix) It is a double displacement reaction.

4. 2 24 2 4 2 2 4 3CH C H CaC C O CO

5. (i) 3IO (oxidant), Cu+ (reductant) (ii) H2O2 (reductant), Cl2O7 (oxidant)

(iii) 22 7Cr O (aq) (oxidant), 2

2 3S O (reductant) (iv) 24 6S O (oxidant), Al(s) (reductant)

6. (i) 2 3 2

s 3 aq 4 aq 3 aqP HPO oxidation; PO HPO reduction

(ii) 2 3 2 gHNO aq NO aq oxidation; HNO aq NO reduction

(iii) 4 3sRe ReO aq oxidation; IO aq I aq reduction

(iv) 2 24 4 4 2 sMnO aq MnO aq oxidation; MnO aq MnO reduction

Homework Problems

1. (i) +4 (ii) 3 2. (i) +5 (ii) +7 3. (i) +2 (ii) +7

4. (i) Si = 4 in SiH4, B = 3 in BH3, B = + 3 in BF3, S = + 2 in S223O , Br = + 7 in Br 4O and P = + 5 in

HP 24O ,

(ii) S = + 6 in PbSO4, U = + 5 in 42 7U O , B = + 3 in 2

4 7B O , Cr = + 6 in Cr 24O and Mn = + 6 in K2MnO4.

5. O.N. of C = 3 in C2H6, 2.5 in C4H10, + 2 in CO, + 4 in CO2 and + 4 in HC 3O .

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6. O.N. of O = + 2 in OF2, 1 in Na2O2 2 in Na2O, 1/2 in KO2, 1/3 in KO3 and + 1 in O2F2.

7. O.N. of Cl = 1 in HCl, + 1 in HClO, 0 in CaOCl2 and + 4 in ClO2.

8. O.N. of S = + 6 in (NH4)2 SO4, H2SO4 and HS 4O and 22 7S O . + 4 in HS 3O and + 3 in S2

24O .

9. K = + 1, Mn = + 7, O = 2; K = + 1, Cr = + 6, O = 2; K = + 1, Cl = + 7, O = 2.

10. O.N. of S changes from 2 in H2S and + 4 in SO2 to zero in elemental sulphur.

11. (i) 2.5, (ii) The O.N. of S in S2Cl2 is + 1 as shown: 1 1 1 1

Cl S S Cl

12 (i) 1 5 20 3 1 2

23 4 210H aq 4Zn s NO aq 4Zn aq NH aq 3H O l

Zn acts as a reductant and 3NO acts as oxidant.

(ii) 0 1 2 1 1 0

2 2I g H S g 2H I g S s .

I2 acts as an oxidant, H2S acts as the reductant.

13. Hint :

(i) 2 2 0 0 1 2

2 2CuO s H g Cu s H O g

(ii) 3 2 2 0 4

2 3 2Fe O s 3CO g 2 Fe s 3 CO g

(iii) 3 1 1 3 1 3 1 1 1 3 1

4 2 63 34 B Cl g 3LiAIH s 2 B H g 3LiCl s 3AlCl s

(iv) 22 K s F g 2 K F s

In this K is oxidized to K+ while F2 is reduced

It is a redox reaction.

(v) 3 1 0 2 2 1 2

3 224NH g 5O g 4NO g 6H O g

14. Substance oxidized Substance reduced Oxidising agent Reducing agent (i) C6 H6O2 (aq) AgBr (s) AgBr (s) C6H6O2 (aq) (ii) HCHO(aq) [Ag (NH3)2]+ [Ag (NH3)2]+ HCHO(aq) (iii) HCHO(aq) Cu2+ (aq) Cu2+ (aq) HCHO(aq) (iv) N2H4 (l) H2O2 (l) H2O2 (l) N2H4 (l) (v) Pb (s) PbO2 (s) PbO2 (s) Pb (s)

15. (i) +3 (ii) +5

16. (i) 2 and +6 (ii) +5, 0, 0, +5 (iii) +4 (iv) +6

17. (i) 1 1 1 5 2 0 3 2 1 1 2

23 23HCl aq HNO aq Cl g NOCl g 2H O l

(ii) 1 12 1 1 1 2 1

22HgCl aq 2K I aq Hg I s 2K Cl aq

(iii) 3 2 2 2 0 4 2

2 3 2Fe O (s) 3 CO(g) 2Fe(s) 3 CO (g)

(iv) 3 1 1 2 1 1 1 3 2

2 33 3P Cl l 3 H O l 3HCl aq H P O aq

(v) 3 1 0 0 1 2

3 2 224NH 3O g 2N g 6H O g

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IUPAC - Nomenclature

MODULE 1

Q.1 Explain the tetravalency of carbon atom.

A. 1) Carbon has electronic configuration,

C(6) = 1s2 2s2 1 1x y2p 2p Ground state

1s2 2s1 1 1 1x y z2p 2p 2p Excited state

2) Due to four unpaired electrons, carbon has tetravalency.

Q.2 Write the geometry and the type of hybridization of carbon in

(1) Methane (2) Ethylene (3) Acetylene.

A. Compound Geometry

Type of

hybridization

1) Methane Tetrahedral sp3

2) Ethylene Trigonal planar sp2

3) Acetylene Linear sp

Q.3 How are organic compound represented?

A. Organic compounds are represented by :

1) Lewis structure (or dot structure) : Two electrons bonded to form a covalent bond are shown by dots.

2) Structural formula : The covalent bond is represented by a dash (). A double bond is

represented by double dash (=) and a triple bond by triple dash ().

C C

H

H

H

H

Ethylene

Acetylene H C C H.

3) Condensed structural formula : In this dashes which represent covalent bonds in the complete

structural formula are removed. The number of identical groups attached to an atom are indicated

by subscript.

Example : CH3CH2CH2CH3 as CH3(CH2)2 CH3

4) Bond line formula : In this carbon and hydrogen atoms are not written. The dashes representing

carbon-carbon (CC) are drawn in a zig-zag fashion. The terminals denote methyl groups and the

line junctions denote carbon atoms bonded to appropriate number of hydrogen atoms required to

satisfy the valency of carbon atom.

Example : Methane H

H

H

HC ....:

:

Example : Methane H C H

H

H

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CH3

CH2

CH2

CH2

CH2

CH3

Br

Example :

1) 3-Methylhexane

CH2

CH2

CH

CH2

CH3

CH3

CH3

:

2) Cyclobutane

CH2

CH2

CH2

CH2

:

*Q.4 Write the complete structural formula, condensed structural formula and bond line formula of

isooctane.

A. Isooctane :

Complete structural formula :

Condensed formula : (CH3)2CH(CH2)4CH3

Bond line formula :

Q.5 Represent the following by molecular formula, complete structural formula, condensed formula

and Bond line formula :

1) nHexane 2) 2Bromobutane

A. 1) nHexane : (a) Molecular formula : C6H14 (b) Complete structural formula :

CH3CH2CH2CH2CH2CH3

(c) Condensed formula : CH3(CH2)4CH3

(d) Bond line formula :

2) 2Bromobutane (a) Molecular formula : C4H8Br

(b) Complete structural formula : CH3 CH CH2 CH3

(c) Condensed formula : CH3CH(Br)CH2CH3

(d) Bond line formula :

Q.6 How are alkanes classified?

A. Alkanes are classified as follows :

(i) Straight chain alkanes : The alkanes in which each carbon atom is bonded to not more than two other carbon atoms are known as or normal or straight chain alkanes (n-alkanes). e.g.

n-Hexane (C6H14) where n means normal.

Or Hexane

CH3

1 2 3 4 5 6 73 2 2 2 2 3CH CH CH CH CH CH CH

Br

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Notes on IUPAC-Nomenclature (51)


CH2

CH3

CH2

CH2CH3

CH2CH

CH CH

CH2

CH3

CH3

CH3

(ii) Branched chain alkanes : The alkanes in which at least one carbon atom is bonded to more than two other carbon atoms are known as branched chain alkanes.

Note that the carbon atom marked by asterisk * is attached to more than two carbon atoms.

*Q.7 Write the condensed structural formula and complete structural formula

of following alkane. Give its IUPAC name.

A.

1) Condensed structural formula of alkane : CH3(CH2)2CH(CH3)CH(C2H5)CH2CH(CH3)CH2CH3

2) Complete structural formula of alkane :

3) IUPAC name of alkane :

5Ethyl3, 6dimethylnonane

Q.8 Explain what are (i) n alkanes (ii) isoalkanes (iii) neoalkanes (3 marks)

A. (i) n alkanes :

Normal or nalkanes are straight chain alkanes.

Example :

npentane CH3 CH2 CH2 CH2 CH3.

(ii) Isoalkanes : The alkanes in which one methyl group is attached to the second carbon atom of

the normal chain of carbon atoms are called isoalkanes.

Example :

Isopentane CH3 CH CH2 CH3

Isohexane CH3 CH CH2 CH2 CH3

In isoalkanes, CH (CH3)2, isopropyl group is present at one end of the normal chain of carbon atoms.

CH3

CH3

Isobutane (C4H10)

Neopentane (C5H12)

CH3

CH3 CH2 CH3

9 8 7 6 5 4 3 2 13 2 2 2 2 3CH CH CH CH CH CH CH CH CH

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(iii) Neoalkanes : The alkanes in which two methyl groups are attached to the second carbon atom of the normal chain

of carbon atoms are called neoalkanes.

Example :

In these alkanes : C (CH3)3, tertbutyl group is present at one end of the normal chain of carbon atoms.

Example :

*Q.9 Write the condensed structural formula of nalkane C25H52.

A. Condensed structural formula of alkane with molecular formula C25H52 is CH3(CH2)23CH3.

*Q.10 Write the structures of the following alkyl groups.

A. Alkyl group Structure

1) 2-Methylpropyl 3 2 1

3 2CH CH CH

CH3

2) 1,2-Dimethylbutyl 4 3 2 1

3 2 2CH CH CH CH

CH3

CH3

Q.11 Mention different types of carbon atoms.

A. Types of carbon atoms in a molecule : Depending upon the number of carbon atoms to which a particular carbon atom is bonded, there are four types of carbon atoms.

(i) A carbon atom which is bonded to one another carbon atom is called a primary (1) carbon atom.

(ii) A carbon atom which is bonded to two other carbon atoms is called a secondary carbon atom (2)

(iii) A carbon atom which is bonded to three other carbon atoms is called a tertiary carbon atom (3)

(iv) A carbon atom which is bonded to four other carbon atom is called a quaternary carbon atom (4).

Note : Carbon atom in methane is considered as a primary carbon atom.

In this molecule : 1) C1 C6 C7 C8 and C9 are all primary carbon atoms 2) C3 and C4 are secondary carbon atoms 3) C5 is a tertiary carbon atom 4) C2 is quaternary carbon atom

*Q.12 Consider the following alkane and answer the following questions. (CH3)2CHCH2CH(C2H5)CH(CH3)CH3. (a) Write the complete structural formula of the alkene. (b) What is its molecular formula? (c) Write its bond line formula and IUPAC name. (d) How many methyl, methylene and methine groups are present in this alkane? (e) How many ethyl and isopropyl groups are present in this alkane? (f) How many carbon atoms are 1°, 2°, 3° and 4°?

CH3C CH3

CH3

CH3

Neopentane

CH3C CH3

CH3

CH3

CH2Neohexane

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3 3CH CH CH

CH3

A. (a) Structural formula of alkane :

(b) Molecular formula of alkane : C10H22

(c) Bond line formula :

IUPAC name : 3Ethyl2, 5dimethylhexane.

(d) Number of methyl groups ( CH3) : 05

Number of methylene groups ( CH2 ) : 02

Number of methine groups

(e) Number of ethyl groups : 01 Number of isopropyl groups : 02

(f) Number of 1° carbon atoms : 05 Number of 2° carbon atoms : 02 Number of 3° carbon atoms : 03 Number of 4° carbon atoms : 00

* Q.13 Give the classification of organic compounds based on their structure. (4 marks)

A.

(i) Aliphatic or Acyclic Compounds : Organic compounds in which carbon atoms are joined to form an open chain are called aliphatic or acyclic compounds.They may be long straight chain compounds or branched chain compounds.

e.g.

3 2 2 3CH CH CH CH Butane (Straight chain)

Isobutane (branched chain)

Aliphatic compounds may be saturated or unsaturated organic compounds like alkenes

2 2CH CH , alkynes H C C H .

C H : 03

1°CH3 2°CH2

1°CH3

1°CH3

6 5 4 3 2 1

3 2 31 3 2 3 3 1CH CH CH CH CH CH

HeteroAromatic

Aliphatic or Acyclic Compounds Cyclic or Closed chain compounds

Straight chain Branched chain

Homocyclic

Alicyclic Aromatic

Heterocyclic

HeteroAlicyclic

Organic Compounds

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CH2

CH2

CH2

CH2

O

CH2

CH2

N

CH2

CH2

CH2

H

Tetrahydrofuran

Piperidine

CH

CHCH

CH

O

CH

CH

N

CH

CH

CH

Furan Pyridine

(ii) Cyclic or closed chain compounds : Organic compounds in which carbon atoms are joined to form a closed chain or ring are called cyclic compounds.

They are further classified as : (A) Homocyclic Compounds (B) Heterocyclic Compounds

(A) Homocyclic Compounds : The cyclic organic compounds which have a ring made up of carbon atoms only are called homocyclic or carbocyclic compounds.

They are further classified as : (i) Alicyclic compounds, (ii) Aromatic compounds

(i) Alicyclic compounds : These are cyclic organic compounds in which carbon atoms form a ring and exhibit properties similar to aliphatic compounds. e.g.

CH2

CH2

CH2 CH

2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

Cyclopropane Cyclobutane Cyclohexane (ii) Aromatic compounds :

These are cyclic organic compounds which contain one or more benzene rings in their molecules. Benzene ring is a hexagon of carbon atoms linked by single and double bonds at alternate positions. e.g.

CH

CH

CH

CH

CH

CH

(C6H

6)Benzene

C

CH

CH

CH

CH

CH

CH

C

CH

CH

(C10

H8)Naphthalene

(Bicyclic) (B) Heterocyclic compounds :

These are the cyclic organic compounds in which along with carbon atoms there are one or more other or non-carbon atoms present in the ring like O, N, S etc.

(i) Hetero alicyclic compounds : In these organic compounds, carbon and hetero atoms are linked by single bonds. e.g.

(ii) Hetero aromatic compounds : In these organic compounds, carbon and hetero atoms are linked by conjugated or alternate double bonds.

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*Q.14 What Homologous series. Explain with example.

A. Homologous Series : The organic compounds of the same class (with the same functional group) when arranged in increasing

order of their molecular weights the two successive members of the series constantly differ by CH2 group in their molecular formulae. Such series is called homologous series and each member is called homologue. e.g.

(i) Homologous series of alkanes :

(ii) Homologous series of alcohols

Q.15 Give characteristics of a homologous series.

A. The characteristics of a homologous series are as follows : (i) All the members of the homologous series have same general formula and same functional group. (ii) Hence their chemical properties are same. (iii) The homologous show gradation in their properties like melting points, boiling points, density etc.

with the gradual change in their molecular weights.

(iv) The consecutive members differ in their molecular formulae by CH2 group, hence differ in molecular weights by 14 units.

(v) All the members of a homologous series can be prepared by the same general method.

Q.16 Explain various rules to give IUPAC name to the alkanes. OR

Explain Nomenclature of alkanes.

A. (i) Common name system (Trivial system) : The early chemists named the organic compounds on the basis of their source or history or special properties shown by the compound. CH3OH was named as methyl alcohol since its parent alkane is CH4 i.e. methane. It is also known as ‘wood spirit’ since it was obtained from destructive distillation of wood. In Greek language, the word ‘methu’ means ‘wine’, and ‘hule’ means ‘wood’ and hence also named as ‘methyl alcohol’. Even though common names are simple, short and informative, all compounds cannot be named by this approach. The common names of first four alkanes are the same as in IUPAC nomenclature of system.

(ii) IUPAC System : The need for systematic nomenclature was felt as the number of organic compounds increased. The task was begun in 1892 at Geneva by international committee of chemists and system developed

for nomenclature was called as Geneva system of nomenclature. There are several revisions in the systems of nomenclature. The latest rules in use are framed by a committee on nomenclature

of organic compounds by the International Union of Pure and Applied Chemistry (IUPAC). The common names of the first four alkanes are retained, from fifth member onwards Latin or Greek numeral are used to indicate the number of carbon atoms in the molecule.

All normal chain alkanes are considered as parent hydrocarbons.

Formula Name of alkane Formula Name of alkane

CH4 Methane C9H20 Nonane

C2H6 Ethane C10H22 Decane

C3H8 Propane C11H24 Undecane

C4H10 Butane C12H26 Dodecane

C5H12 Pentane C13H28 Tridecane

C6H14 Hexane C14H30 Tetradecane

C7H16 Heptane C15H32 Pentadecane

C8H18 Octane C16H34 Hexadecane

CH2

CH2

CH2

4

2 6

Methane CH

Ethane C H

3 8

4 10

Propane C H

Butane C H

3

2 5

Methyl alcohol CH OH

Ethyl alcohol C H OH

3 7

4 9

Propyl alcohol C H OH

Butyl alcohol C H OH

CH2

CH2

CH2

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CH

H

H

Methyl

Alkyl Groups : The univalent groups obtained by removing one hydrogen atom from alkanes are called as alkyl

groups. Alkyl groups are named by replacing the ending ‘ane’ of the parent alkane by-‘yl’.

e.g. Methane

ane

yl methyl.

Parent alkane Alkyl group

(i) CH4 Methane,

3 2H C C H CH CH

H H

H H

C2H

6(ii)

Ethane Ethyl

3 8(iii) C H H C C C H CH3

CH2

CH2

H H H

H H H

CH3

CH CH3

n-PropylPropane

or

Iso-propyl

In general, alkane is represented by RH and the alkyl group is represented by R-. For highly branched alkanes, IUPAC system is used for their nomenclature. According to this system for naming alkanes, the following rules are taken into consideration :

(a) Longest chain rule : The longest continuous chain of carbon atoms in the molecule is selected. This chain is regarded as the parent chain and it gives the name of the parent hydrocarbon. It must be noted that the longest chain may or may not be straight, but it has to be a continuous chain. If two chains are of equal length then a chain which has the maximum number of side chain is selected as the parent alkane.

e.g.

CH3

CH2

CH2

CH3

CH3 CH

3

7 6 5 4 3 2 1

3 2 3H C H C HC CH CH CH CH

[longest chain with 4 Substituents, correct]

CH2

CH2

CH3

CH3 CH

3CH

3

4 3 2 1

3 2 3H C H C CH CH CH CH CH

[longest chain with 3 substituents, incorrect]

5

6

7

(b) Lowest number rule [position of the substituent] :

When numbering of carbon atoms in the parent chain (1, 2, 3,…etc.) is done, it is started from the end which gives lower number to the carbon atoms, carrying the substituents C the substituted carbon atoms have the lowest possible numbers).

e.g.

CH3

CH3 CH

3

6 5 4 3 2 1

3 2 3H C H C HC HC CH CH

[Numbering is correct]

CH3

CH3 CH

3

1 2 3 4 5 6

3 2 3H C CH CH CH CH CH

[Numbering is incorrect]

(c) Lowest set of locants rule : The number which indicates the position of the substituent (side chain) is called locant. When two or more substituents are present, then numbering of carbon atoms in the parent chain is preferred from the end, which gives the lowest set of the locants.

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e.g.

6 5 4 3 2 1

3 2 3H C HC CH CH CH CH

CH3 CH

3CH

3

(correct set of locants as 2,3,5)

1 2 3 4 5 6

3 2 3H C HC CH CH CH CH

CH3

CH3

CH3

(Incorrect set of locants as 2,4,5)

(d) Lowest sum rule : According to this rule, the numbering of the parent chain containing two or more substituents is done in such a way that the sum of the locants is the lowest.

e.g.

1 2 3 4 5

3 3H C C HC CH CH

CH3

CH3

C2H

5

CH3

(Sum of the locants= 2+ 2 + 3 + 4 = 11correct)

5 2 1

3 34 3

CH C CH CH CH

CH3

C2H

5

CH3

CH3

(Sum of the locants= 2 + 3 + 4 + 4 = 13, Incorrect)

(e) Prefixes rule for same kind of substituents : If more than one similar kind of substituents (alkyl groups) are present in molecule, then their

positions are indicated separately and an appropriate numerical prefix (di,tri,etc) is attached to the name of the substituent. The positions of the substituents are separate by comma.

e.g.

3 2 3H C CH CH CH CH

CH3

CH3

(2, 3-Dimethyl pentane)

3 2 2 3H C H C HC H C C CH

CH3

CH3

CH3

(2, 2, 4-Trimethyl hexane)

(f) Prefixes rule for different kinds of substituents : If two or more different kinds of substituents (alkayl groups) are present in the molecule, then they are named in the alphabetical order alongwith their appropriate positions. For example,

e.g.

1 2 3 4 5

3 2 3H C CH CH CH CH

C2H

5

CH3

CH3

(3-Ethyl -2, 3-dimethyl pentane)

(g) Prefixes rule for different substituents at equivalent positions :

It two different substituent are located at equivalent positions from the two ends of the parent chain, then the numbering of the chain is done in such a way that the alkyl group which comes first in alphabetic order gets the lower position (it is written first in the name).

e.g.

1 2 3 5 6

3 2 2 2 34

H C CH CH CH CH CH

CH3

C2H

5

3-Ethyl-4-methyl hexane

(h) Prefixes rule for complex substituents :

If the substituent is further branched it is named as substituted alkyl group, by numbering the

carbon atom of this group attached to the parent chain as 1. The name of such a substituent is

enclosed in brackets to avoid confusion with the numbering of parent chain.

e.g.

8 7 6 5 4 3 2 1

3 2 2 2 3CH CH CH CH CH CH C CH

CH3

CH3

CH3

CHCH3

CH2

CH3

2, 2, 6-Trimethyl-4-(1-methyl propyl) octane

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Q.17 Give IUPAC name of the following compound?

(i) CH

3CH

2CH CH

2CH

3

CH3

1 2 3 4 5

(ii) CH

3CH

2CH CH CH

2

CH3

CH3

CH3

1 2 3 4 5 6

A. 3-Methyl pentane A. 3,4-Dimethylhexane

(iii) CH

3CH CH CH CH

2

C2H

5

CH3

CH3CH

3

1 2 3 4 5 6

A. 3-Ethyl 2, 4-dimethyl hexane

Q.18 Give the structural formulae of the following compounds?

(i) 3-Ethyl5methyl heptane

A.

(ii) 3-Methylpentane

A.

(iii) 2,2,4, 5-Tetramethylheptane

A.

(iv) 2, 2-Dimethyl butane

A.

(v) 3, 3-Dimethyl5ethyl heptane

A.

Q.19 How are alkyl groups classified?

A. Alkyl groups are classified as follows :

1) Primary alkyl group :

In this alkyl group, valency on the primary carbon atom is available.

Example : methyl radical 1°CH3

2) Secondary alkyl group :

In this alkyl group, valency of a secondary carbon atom is available.

Example : Isopropyl radical,

2

3 3CH CH CH

3) Tertiary alkyl group : In this alkyl group, valency of a tertiary carbon atom is available.

Example : tertButyl radical,

3

3 3CH C .

CH3

CH2

CH

C2H

5

CH2

CH CH2

CH3

CH3

CH3

CH2

CH

C2H

5

CH3

CH3

C CH2

CH3

CH3

CH

CH3

CH CH2

CH3

CH3

C CH2

CH

C2H

5

CH2

CH3

CH2

CH3

CH3

CH3

C CH2

CH3

CH3

CH3

CH3

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* Q.20 Following are the alkyl groups derived from an alkane C5H12.

Give their IUPAC names and classify them as primary, secondary and tertiary.

A.

Structure of alkyl group IUPAC name of alkyl group Type of alkyl group

1) 5 4 3 2 1

3 2 2 2 2CH CH CH CH CH Pentyl Primary

2) 4 3 2 1

3 2 2 3CH CH CH CH CH

1-Methylbutyl Secondary

3) 3 2 1


1-Ethylpropyl Secondary

4) 4 3 2 1

3 2 2CH CH CH CH

CH3

3-Methylbutyl Primary

5) 3 2 1

3 3CH CH CH CH

CH3

1,2-Dimethylpropyl Secondary

6) 2 3

3 2 3CH C CH CH

CH3

1

1,1-Dimethylpropyl Tertiary

7) 2 3 4

3 2 3CH CH CH CH

12CH

2-Methylbutyl Primary

8) 3 1

23 2CH C CH

CH3

CH3

2,2-Dimethylpropyl Primary

Solved Examples

Example 1 :

Give the IUPAC names of the following compounds.

3 2 2 2 3(i) CH CH CH CH CH CH CH

CH3

C2H

5

3 2 2 2 2 3(ii) CH CH CH CH CH C CH CH

CH3

CH2CH

3CH

CH3

CH3

3 2(iii) CH CH C

3 3H C C CH

3 3H C C CH

2 3CH C CH

CH3

CH3

CH3

CH3

Solution : (i) 5-Ethyl-2-methylheptane (ii) 3-Ethyl-5-isopropyl-3-methyloctane (iii) 4-tert-butyl-4-ethyl-2, 2, 5, 5-tetramethylhexane.

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Example 2 : Write structures for each of the following compounds. Why are the given names incorrect? Write correct IUPAC names. (i) 2, 2-Diethylbutane (ii) 5-Ethyl-3-methylheptane (iii) 2-Methyl-3-ethylhexane.

Solution :

3 2 3(i) CH C CH CH

CH2CH

3

CH2CH

3

The longest chain in the molecule is of five carbon atoms and not that of four.

Hence, the correct IUPAC name is 3-Ethyl-3-methylpentane.

7 6 5 4 3 2 1

3 2 2 2 3(ii) CH CH CH CH CH CH CH

CH3

CH2CH

3

Since ethyl and methyl groups are at equivalent positions, ethyl group would get preference while

numbering. Hence, the correct name of the compound is 3-Ethyl-5-methylheptane.

3 2 2 3(iii) CH CH CH CH CH CH

CH3 CH

2CH

3

The prefixes are to be written in alphabetical order. The correct IUPAC name is

3-Ethyl-2-methylhexane.

Classwork Problems

1. Write formula, common name and IUPAC name of alkyl groups derived from methane, ethane, propane and butane.

2. Draw the structures of all eight isomeric alkyl groups derived from an alkane C5H12. Give their IUPAC names and classify them as primary, secondary and tertiary.

3. Write common names and IUPAC names of the following alkanes.

(i) CH4 (ii) CH3 CH3 (iii) CH3 CH2 CH3

(iv) CH3 CH2 CH2 CH3 (v)

(vi) CH3 CH2 CH2 CH2 CH3 (vii)

(viii) (ix) CH3 CH2 CH2 CH2 CH2 CH3

(x) (xi)

* 4. Write the structures and give IUPAC names of following alkyl groups. (i) Isopentyl (ii) tert-pentyl (iii) Isohexyl (iv) Neohexyl

* 5. Write IUPAC names of the following alkanes.

1 2 3 4 5 6 7

3 2 2 3(i) CH CH CH CH CH CH CH

CH3

CH3

13CH CH

CH2

CH3

1 23

3 2 2 3(ii) CH CH CH CH CH

4 5 6

3 2 3CH CH CH CH

3 3CH CH CH

CH3

3 2 3CH CH CH CH

CH3

3 3CH C CH

CH3

CH3


CH3

3 2 3CH C CH CH

CH3

CH3

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1 2 3

3 2 2 3(iii)CH CH CH CH CH

5 64

2 2 3CH CH CH

3 3(iv) CH CH CH CH

CH3 C

2H

5

3 3(v) CH CH CH CH CH

C2H

5 C2H

5

C2H

5

Homework Problems

1. Write IUPAC name of the following alkanes.

(i) (ii)

(iii) (iv)

(v) (vi)

(vii) (viii)

(ix) (x)

(xi) (xii)

2. Write structure of the following alkanes.

(i) 3Ethyl2, 2dimethylpentane (ii) 3Ethyl4methylhexane

(iii) 3Ethyl2methylhexane

(iv) 2, 2, 6Trimethyl4(1methylpropyl) octane

(v) 2, 7, 8 Trimethyldecane (vi) 2, 3Dimethylhexane

(vii) 4Ethyl 3, 5 dimethyloctane (viii) 3Ethyl-2, 2dimethyl pentane

(ix) 4, 5 Diethyl 2 methyloctane

(x) 4 Ethyl 2, 3, 7, 7 tetramethylnonane

(xi) 5 Butyl 4 ethyldecane

3 3CH CH CH CH

CH2

CH3

CH2

CH3

CH3

3 2 3CH CH CH CH

3 2 3CH CH CH CH CH

CH3 CH3

CH3

3 2 3CH C CH CH

CH3

CH3

3 3CH C CH

CH3

CH3

3 2 2 2 3CH CH CH CH C CH

CH3

3 3CH CH CH CH CH

CH2 CH3

CH3 CH3

3 2 3CH CH C C CH

CH3 CH3

CH3 CH3

CH2 CH2 CH3

CH3

3 2 2 34CH CH CH CH CH CH

CH3

3 2 35CH CH C CH

CH2 CH2 CH3

3 3CH CH C CH CH

CH3 CH3 CH2 CH3

CH3

3 2 3CH C CH CH CH CH

CH3 CH2 CH3

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3. Write structure of the following compounds. (i) Isohexane (ii) Neopentane

* 4. Write the structures of following alkanes. (i) 5-(2-Ethylbutyl)-3, 3-dimethyldecane (ii) 5-sec-Butyl-4-isopropyldecane (iii) 2, 3-Dimethyl-6-(2-methylpropyl) decane (iv) 6-Isopropyl-2, 3-dimethylnonane (v) 5-(1, 2-Dimethylpropyl)-2-methylnonane (vi) 4-tert-Butyl-3-methylheptane (vii) 3, 3-Diethyl-5-isopropyl-4-methyloctane

* 5. Write the complete structural formula and give IUPAC names to the following alkanes. (i) (C2H5)4C (ii) (CH3)3CCH2CH(CH3)2

(iii) (C2H5)2CHCH2C(CH3)3 (iv) CH3(CH2)8CH3

Classwork Problems

3.

Common Name IUPAC Name

(i) Methane Methane

(ii) Ethane Ethane

(iii) Propane Propane

(iv) nButane Butane

(v) Isobutane 2Methylpropane

(vi) nPentane Pentane

(vii) Isopentane 2Methylbutane

(viii) Neopentane 2,2Dimethylpropane

(ix) nHexane Hexane

(x) Isohexane 2Methylpentane

(xi) Neohexane 2, 2Dimethylbutane

4. 4 3 2 1

3 2 2(i) CH CH CH CH

CH3

3-Methylbutyl

13(ii) CH C

CH3

22CH

33CH

1,1-Dimethylpropyl

5 4 3 2 1

3 2 2 2(iii) CH CH CH CH CH

CH3

4-Methylpentyl

4 2 13

3 2 2CH C CH CH

CH3

CH3

3,3-Dimethylbutyl

5. (i) 2,3-Dimethyl-4-(1-methylpropyl)heptane (ii) 3-Ethyl-4-methylhexane (iii) 3-Ethylhexane (iv) 2,3-Dimethylpentane (v) 4-Ethyl-3, 5-dimethylheptane

Homework Problems

1. (i) 3, 4Dimethylhexane (ii) 2Methylbutane

(iii) 2, 3Dimethylpentane (iv) 2, 2Dimethylbutane

(v) 2, 2Dimethylpropane (vi) 2, 2Dimethyhexane

(vii) 2, 3, 4Trimethyhexane (viii)2, 2, 3, 3Tetramethylpentane

(ix) 2Methyl4propylnonane (x) 44Dimethyldecane

(xi) 2, 3, 3, 4Tetramethylhexane (xii) 3Ethyl2,2,5trimethylheptane

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6 5 4 3 2 1

3 2 2 3(ii) CH CH CH CH CH CH

CH3

C2H

5

6 5 4 3 2 1

3 2 2 2 3(iii) CH CH CH CH CH CH

2

3CH CH

1

3CH

8 7 5 4 3 16 2

3 2 2 2 3(iv) CH CH CH CH CH CH C CH

CH3

CH3

CH3

1

3CH CH 2 3

2 3CH CH

10 9 8 7 6 5 4 3 2 1

3 2 2 2 2 2 3(v) CH CH CH CH CH CH CH CH CH CH

CH3 CH

3CH

3

1 2 3 4 5 6

3 2 2 3(vi) CH CH CH CH CH CH

CH3

CH3

1 2 3 4 5 6 7 8

3 2 2 2 3(vii) H C CH CH CH CH CH CH CH

CH3 CH

2

CH3

CH3

1 4 52 3

3 2 3(viii) CH C CH CH CH

CH3

CH3 2 3CH CH

1 2 3 4 6 7 85

3 2 2 2 3(ix) CH CH CH CH CH CH CH CH

CH3 CH

2

CH3

2 3CH CH

1 2 3 4 5 6 7 8 9 10

3 2 2 2 2 2 2 3(xi) CH CH CH CH CH CH CH CH CH CH

CH2

CH3

CH2

CH2

CH2

CH3

1 2 3 4 5 6 7 8 9 10

3 2 2 2 2 2 2 3(ii) CH CH CH CH CH CH CH CH CH CH

3CH CH

2 3CH CH

13CH CH

23CH

2. 1 3 4 5

23 3 3(i) CH C CH CH CH

CH3

CH3 C

2H

5

4. 1 2 3 4 5 6 7 8 9 10

3 2 2 2 2 2 2 3(i) CH CH C CH CH CH CH CH CH CH

CH3

CH3

1

2CH3 42

2 3H C CH CH

2 3CH CH

1 2 3 4 5

3 2 2 3(i) CH CH CH CH CH

CH3

3. (ii) 1 3

23 3

CH C CH

CH3

CH3

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1 2 3 4 5 6 7 8 9 10

3 2 2 2 2 2 3(iii) CH CH CH CH CH CH CH CH CH CH

CH3

CH3 1

2CH

23CH CH

23CH

(iv)

1 2 3 4 5 6 7 8 9

3 2 2 2 2 2 3(v) CH CH CH CH CH CH CH CH CH

CH3 3HC CH

3 3CH CH CH

1 2 3 4 5 6 7

3 2 2 2 3(vi) CH CH CH CH CH CH CH

CH3

3 3CH C CH

CH3

1 2 3

3 2(vii) CH CH C 4 5 6 7 8

2 2 3CH CH CH CH CH

CH3

2 3CH CH

2 3CH CH3CH CH

CH3

1 2 3 4 5

3 2 3(ii) CH C CH CH CH

CH3

CH3

CH3

2,2,4-Trimethylpentane

6 5 3 14 2

3 2 2 3(iii) CH CH CH CH C CH

CH3

CH3

2 3CH CH

4-Ethyl-2,2-dimethyl-hexane

3 2 2 2 2 2 2 2 2 3iv CH CH CH CH CH CH CH CH CH CH

Decane

5. 1 2 3 4 5

3 2 2 3(i) CH CH C CH CH

2 3CH CH

2 3CH CH

3,3-Diethylpentane

1 2 3 4 5 6 7 8 9

3 2 2 2 2 3CH CH CH CH CH CH CH CH CH

CH3 CH

3 CH

CH3

CH3

1

2

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MODULE 2

IUPAC of Cyclic Compounds : A saturated monocyclic compound is named by prefixing ‘cyclo’ to the corresponding straight chain alkane. If side chains are present, then the rules given above are applied. Names of some cyclic compounds are given below.

Table : Some Functional Groups and Classes of Organic Compounds

Class of

compounds

Functional group

structure

IUPAC group

prefix

IUPAC group

suffix

Example

Alkanes ane Butane, CH3(CH2)2CH3

Alkenes >C = C< ene But-1-ene, CH2=CHCH2CH3

Alkynes CC yne But-1-yne

CH CCH2CH2

Arenes

Benzene,

Halides X

(X = F, Cl, Br, I) halo 1-Bromobutane,

CH3(CH2)2CH2Br

Alcohols OH hydroxy- ol Butan-2-ol, CH3CH2CHOHCH3

Aldehydes CHO formyl, or oxo al Butanal, CH3(CH2)2CHO

Ketones >C=O oxo one Butan-2-one, CH3CH2COCH3

Nitriles CN cyano nitrile Pentanenitrile, CH3CH2CH2CH2CN

Ethers ROR alkoxy Ethoxyethane, CH3CH2OCH2CH3

Carboxylic acids COOH carboxy oic acid Butanoic acid, CH3(CH2)2CO2H

Carboxylate ions COO oate Sodium butanoate, CH3(CH2)2CO2 Na+

Esters COOR alkoxycarbonyl oate Methyl propanoate, CH3CH2COOCH3

Acyl halides COX (X = F, Cl, Br, I)

halocarbonyl oyl halide Butanoyl chloride, CH3(CH2)2COCl

Amines NH2,

>NH, >N

amino amine Butan-2-amine, CH3CHNH2CH2CH3

Amides CONH2,

CONHR,

CONR2

carbamoyl amide Butanamide, CH3(CH2)2CONH2

Nitros NO2 nitro 1-Nitrobutane, CH3(CH2)3NO2

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| |

2| |

C O OH NH C C C C

Q.1 Explain the rules for the IUPAC nomenclature of organic compounds having functional

groups.

A. 1) The functional group in the molecule is first identified and accordingly appropriate suffix is used. 2) The longest continuous chain of carbon atoms having functional group is selected. 3) The longest carbon chain is numbered from that end which is nearest to the functional group. 4) Using the appropriate number of the functional group, the complete name is written.

5) The functional groups F (fluoro), Cl (chloro), Br (bromo), I (iodo), NO2 (nitro), OR (alkoxy),

R (alkyl), Ar (aryl) are always prefix substituents.

6) In case of polyfunctional compounds, one of the functional groups is chosen as the principal functional group and the compound is named on that basis.

7) The following priority order is used in choosing the principal functional groups for the base name.

COOH > COOR > CONH2 > CN > CHO >

8) If two functional groups of the same type are present they are indicated by using di, tri, etc before the class name suffix.

Full name of the parent alkane is written before the class name suffix,

9) In case compounds have more than one double or triple bond, the ending ne of the parent alkane is dropped.

Solved Examples

*Example 1 Give IUPAC names of the following compounds. (i)

Ans. 2,3,5Trimethylhexane (ii)

Ans. 1Ethyl1methyl2propylcyclohexane (iii)

Ans. Cyclohex2enone

(iv)

Ans. 4Chlorocyclopentan1ene

(v)

Ans. 2Bromo3methylbutane

(vi)

Ans. Butan2olal

(vii) 1 2 3 42 2CH CH CH CH

Ans. Buta1, 3diene

CH3

CH CH2

CH CH CH3

CH3

CH3

CH3

123456

CH2-CH

2-CH

3

C2H

5CH

3

12

34

56

O

Cl

1 2

34

5

CH3

CH

CH3

CH CH3

Br

1234

CH3

CH2

CH C

OH

H

O

1234Vidy

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ar



1 2 3 4 5 6 7 83 2 2 2 2 3

| |CH CH CH CH CH CH CH CH

CH CH CH3 2 3

OHCH

3C CH

2CH

3

CH3

OH

1 2 3 4

2 2 2|| | ||

H C CH CH CH CH C H

OHO O

CHO

OHC

OH

CH3

CH2

CH CH CH2I

CH3

CH3

13 2

I

HC

H2C

C - CH3

CH2

CH2

CH2

CH2

CH

C-H

C-CH3

CHCH

3

*Example 2

Write IUPAC names of the following organic compounds.

(i)

Ans. 1Bromo2methylcyclobutane

(ii)

Ans. Ethylethanoate

(iii)

Ans. 5, 5Dimethylheptan3ol

(iv)

Ans. 1Chloro4ethylhept3ene

*Example 3

Write complete structural formula and bond line formula of the following compounds :

(i) 4Ethyl3methyloctane

Ans.

(ii) 2Methylbutan2ol

Ans.

(iii) 3Hydroxyhexanedial

Ans.

(iv) 1Iodo2, 3dimethylpentane

Ans.

(v) Methylcyclopentene

Ans.

(vi) 2Methylcyclohexa1, 3diene

Ans.

Br

O

O

OH

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5 4 3 2 12

| ||Br CH CH CH CH C H

OH O

*Example 4 Name the following organic compounds. List the names of the functional groups in each compound.

(i) 5 4 3 2 13

| | |CH CH CH CH C OH

OH O

Ans. IUPAC name Name of the functional group

Pent4ol2enoic acid OR

4Hydroxy2pentenoic acid

1) Hydroxyl 2) Double bond 3) Carboxylic acid

(ii) | |

5 4 3 23 2

|

1CH CH3 3

CH C CH C O

NH2


4Amino4methylpentan2one 1) Keto 2) Amino

(iii)


5Bromopent2en4olal or

5Bromo4hydroxypent2enal

1) Aldehyde 2) Hydroxy 3) Bromo 4) Double bond

(iv)


2,3Dimethylbutan2, 3diol Two hydroxyl

*Example 5

Following compounds are wrongly named. Write their correct IUPAC names.

(i) 1,5Dimenthylcyclohexane

Ans. Structure Corrected name

CH3

CH3

54

3

21

6

1, 3Dimethylcyclohexane

(ii) 2Chlorobut3ene


1 2 3 42 3

|H C CH CH CH

Cl

3Chlorobut1ene

(iii) Butan2,3diol


1 2 3 43 3

| |CH CH CH CH

OH OH

Butane2, 3diol

(iv) Octane1, 3diyne


1 2 3 4 5 6 7 82 2 2 3H C C C C CH CH CH CH Octa1, 3diyne

CH3

C C CH3

CH3

OH OH

CH3

1 2 3 4

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(v) 2Methylbutan3ol


4 3 2 13 3

| |CH CH CH CH

CH OH3

3Methylbutan2ol

Classwork Problems

*1. Write the complete structural formula of the following :

(i) (ii)

*2. Identify the functional group present in the following compounds :

1) npropyl alcohol 2) Acetone 3) Acetylene 4) Ethyl alcohol

*3. What is the structure of functional group in the following classes of organic compounds. 1) Esters 2) Ethers 3) Carboxylic acids 4) Ketones

*4. Write the general representation (general formula) of the following classes of organic compounds : 1) Alcohols 2) Ethers 3) Esters 4) Amines

*5. Write name of the class of following organic compounds. 1) CH3(CH2)3CH2Cl 2) CH3CH2CH2NH2 3) CH3CH2COCH3 4) CH3CH2OCH3

*6. Write structure of an organic compound containing the following functional groups. 1) Keto 2) Carboxyl 3) Hydroxyl 4) Nitrile

Homework Problems

1. Write IUPAC names of the following compounds.

2. Write IUPAC names of the following compound :

3. Write IUPAC names of the following alicyclic compounds.

O(6)

O(7)

4. Write IUPAC names for the following compounds :

(1) (2) (3)

5. Expand each of the following condensed formulae into their complete structural formulae

(1) CH3CH2COCH2CH3 (2) CH3CH = CH(CH2)3CH3

CH3

Br

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6. For each of the following compounds, write a condensed formula and also their bond line formulae. (1) HOCH2CH2CH2CH(CH3)CH(CH3)CH3

(2) |OH

N C CH C N

7. Expand each of the following bondline formulae to show all the atoms including carbon and hydrogen. (a) (b) (c) (d)

8. Give the IUPAC names of the following compounds :

(i) (ii)

(iii) (iv)

(v) (vi) (vii)

9. Give the IUPAC names of the following compounds :

(i) (ii) (iii) (iv) (v) (vi)

(vii)

Classwork Problems

1. (i) (ii)

2. Name of compound Formula of compound

Functional group

Formula/Name

1) npropyl alcohol 3 2 2CH CH CH OH OH, hydroxyl

2) Acetone 3 3CH C CH

O

C , keto

O

3) Acetylene H C C H C C alkyne

4) Ehtyl alcohol CH3CH2OH OH hydroxyl

CH3CH2COCH2CH2COCH3

3 2 3CH CH CH CH CH CH

CH3

Br OH

CH3 CH = CH CHO

CH3 CH = CH COOH

2 5 2C H C CH OH

CH2

3 2 2 2 3CH CH C Cl CH CONHCH CH

CH3

3CH CH CHO

CH3O

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3. Class name Structure of functional group

1) Esters C OR

O 2) Ethers

C O C

3) Carboxylic acids C

O

OH

4) Ketones C

O

4. 1) ROH 2) ROR 3) 4) RNH2

5. 1) Alkyl chloride (Alkyl halide) 2) Amine 3) Ketone 4) Ether

6. Class name Structure of compound Name of compound

1) Keto

Acetone

2) Carboxyl

Acetic acid

3) Hydroxyl CH3CH2OH Ethanol

4) Nitrile CH3C N Methyl cyanide or Acetonitrile

Homework Problems

1. 1) pentane2, 4dione 2) Penta1, 3diene

2. 1) 3Methylbutan2ol 2) 4Oxopentanoic acid

3. 1) Cyclopentane 2) Methylcyclohexane

3) 1Ethyl2methylcyclopentane 4) 2Ethyl1, 1dimethylcyclohexane

5) 1Ethyl1methyl2propylcyclohexane 6) Cyclohexanone

7) Cyclohex2enone 8) 3Methylcyclohex1yne

9) 2,3Dimethylcyclopent1ene

4. 1) 3,3Dimethylpent1en4yne

2) Hexa1, 3dien5yne

3) 4Cyclopropylhex1ene

5. 1) 2)

6. Condensed formulae 1) HO(CH2)3CH(CH3)CH(CH3)2 2) HOCH(CN)2

Bondline formulae 1) 2)

R C OR

O

OH OH CN

CN

Vidyala

nkar



7. (a)

(b)

(c)

(d)

8. (i) 4Ethyl2,4dimethylhept1ene (ii) 3Methylbut1yne

(iii) 3Chloro1phenylprop1ene (iv) 1Bromobut2ene

(v) 2Chlorobuta1, 3diene (vi) 2Chloro3ethylpenta1, 4diene

(vii) 5Propyloct2yne

9. (i) 2Ethylprop2en1ol

(ii) But2en1al

(iii) But2en1oic acid

(iv) 3ChloroNethyl3methylpentan1amide

(v) 2Methoxypropan1al

(vi) Hexane2, 4dione

(vii) 3Bromo5methylhexan2ol

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ar

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Std.XI : Foundation Course