Free Flip-Book Chemistry Class 11th by Study Innovations. 534 Pages

1 2 − 1 = 0.293 1 = 0.707 29.3 79.7
2 2 2
50
1 1 = 0.50 1 = 0.50 50 25
2 2 12.5
6.25
2 3 = 0.75 1 = 0.25 75 3.125
4 4 0

3 7 = 0.875 1 = 0.125 87.5
8 8

4 15 = 0.9375 1 = 0.0625 93.75
16 16

5 31 = 0.96875 1 = 0.03125 96.75
32 32

∞ Total 0 100

Let the initial amount of a radioactive substance be N0
After one half-life period (t1 / 2) it becomes = N0 / 2

After two half-life periods (2t1 / 2) it becomes = N0 / 4

After three half-life periods (3t1 / 2) it becomes = N0 / 8

After n half life periods (nt1 / 2) it shall becomes =  1 n N0
 2 

Half life periods of some isotopes

Radio isotope Half life Radio isotope Half life
14.3 days
238 U 4.5 × 109 years 32 P 8.0 days
92 15
1.5 × 10−4 seconds
230 Th 8.3 × 104 years 131 I 5 × 103 years
90 53 3.82 days

226 Ra 1.58 × 103 years 214 Po
88 24 days 84
44.3 days
234 Th 14 C
90 6

59 Fe 222 Rn
26 86

Thus, for the total disintegration of a radioactive substance an infinite time will be required.

Time (T) Amount of radioactive substance (N) Amount of radioactive substance decomposed (N0 – N)
0 (N0 ) 0

t1/ 2 1  1  1 1 N0 1 − 1  N
2  2  2 2 
N0 = N 0 = 0

2t1/ 2 1  1  2 3 N0 1 − 1  N
3t1/ 2 4  2  4 4 
4t1/ 2 N0 = N 0 = 0
nt / 2
1  1  3 7 N0 1 − 1  N
8  2  8 8 
N0 = N 0 = 0

1  1  4 15 N0 1 − 1  N
16  2  16 16 
N0 = N0 = 0

 1  n   1  n 
 2  1  2  
N 0  −  N 0

Amount of radioactive substance left after n half-life periods

 1  n
 2 
N = N0

and Total time T = n × t1/ 2

where n is a whole number.

(2) Average-life period (T) : Since total decay period of any element is infinity, it is meaningless to use the
term total decay period (total life period) for radioelements. Thus the term average life is used which the following
relation determines.

Average life (T) = Sum of lives of the nuclei
Total number of nuclei

Relation between average life and half-life : Average life (T) of an element is the inverse of its decay
constant, i.e., T = 1 , Substituting the value of λ in the above equation,

λ

T = t1/ 2 = 1.44 t1/ 2
0.693

Thus, Average life (T) = 1.44 × Half life(T1/ 2 ) = 2 × t1/ 2
Thus, the average life period of a radioisotope is approximately under-root two times of its half life period.

Note :  This is because greater the value of λ , i.e., faster is the disintegration, the smaller is the

average life (T).

Example : 12 The isotopes 238 U and 235U occur in nature in the ratio of 140 : 1. Assuming that at the time of earth

formation, they were present in equal ratio, make an estimation of the age of earth. The half life period of

238 U and 235U are 4.5 × 109 and 7.13 × 108 years respectively [Roorkee 1983]

(a) 6.04 × 109 years (b) 5.69 × 1010 years (c) 6.69 × 1011 years (d) 6.69 × 109 years

Solution: (a) Let the age of the earth be t years

For 238 U λ1 × t = 2.303 log N0U238 ......(i)
N U238 ......(ii)

For 235 U λ2 × t = 2.303 log N0U235
N U235

Subtracting eq. (ii) from eq. (i)

t(λ1 − λ2) = 2.303log N 0U 238 − log N 0U 235  = 2.303 log N 0U 238 ⋅ NU 235
 N U 238 N U 235  N U 235 N U 238


t  0.693 9 − 0.693  = 2.303 log 1
 4.5 × 10 7.13 × 108  140

= −(2.303)(2.1461)

= −4.9425

t = 6.04 × 109 years

Example : 13 Calculate the mass of 14 C (half life period = 5720 years) atoms which give 3.7 × 107 disintegrations per

second [Roorkee 1987]

(a) 2.34 × 10 −4 g (b) 2.24 × 10 −4 g (c) 2.64 × 10 −4 g (d) 2.64 × 10 −2 g

Solution: (a) Let the mass of 14 C atoms be m g

Number of atoms in m g of 14 C = m × 6.02 × 10 23
14

λ = 0.693 = 0.693 = 3.84 × 10 −12 sec −1
Half life 5720 × 365 × 24 × 60 × 60

We know that − dN t = λ ⋅ Nt
dt

i.e. Rate of disintegration = λ × No. of atoms

3.7 × 107 = 0.693 × m × 6.02 × 10 23 = 3.84 × 10 −12 × m × 6.02 × 10 23 = 2.24 × 10 −4 g
365 × 24 14 14
5720 × × 60 × 60

Example : 14 Prove that time required for 99.9% decay of a radioactive species is almost ten times to its half life period

Solution: We know that t = 2.303 log N0
λ Nt

N 0 = 100 , N t = (100 − 99.9) = 0.1

So, Time required for 99.9% decay t = 2.303 log 100
λ 0.1

= 2.303 × 3 ......(i)
λ

Half life period = 0.693 ......(ii)
λ

So Time requird for 99.9% decay = 2.303 × 3 × λ = 10
Half life period λ 0.693

Example : 15 1.0 g of 198 Au (t1 / 2 = 65 hours) decay by β-emission to produce mercury [Roorkee 1984]
79

(a) Write the nuclear reaction for the process.

(b) How much mercury will be present after 260 hours ?

Solution: (a) 198 Au → 198 Hg + 0 e
(b) 79 80 −1

No. of half-lives in 260 hours = 260 = 4
65

 1  n 1
 2  16
Amount of gold left after 4 half-lives = =

Amount of gold disintegrated = 1 − 1 = 15 g
16 16

So Amount of mercury formed = 15 = 0.9375g
16

Example : 16 A certain radio-isotope A X (Half life = 10 days) decays to A−4 Y . If 1 g of atoms of A X is kept in sealed
Z Z−2 Z

vessel, how much helium will accumulate in 20 days [Roorkee 1986]

(a) 16800 mL (b) 17800 mL (c) 18800 mL (d) 15800 mL

Solution: (a) A X → A−4 Y + 4 He
Z Z−2 2

In two half lives, 3 of the isotope A X has disintegrated, i.e., 3 g atom of helium has been formed from 3 g
4 Z 4 4

atom of A X
Z

Volume of 1 g atom of helium = 22400 mL

Thus, Volume of 3 g atom of helium = 3 × 22400 mL = 16800 mL
4 4

Example : 17 210 Po decays with α-particle to 206 Pb with a half life period of 138.4 days. If 1g of 210 Po is placed in a
84 82 84

sealed tube, how much helium will be accumulated in 69.2 days ? Express the answer in cm3 at STP.

[Roorkee 1991]

Solution : 210 Po → 206 Pb + 4 He
84 82 2

Amount of polonium left after 69.2 days can be calculated by applying,

No 1  n 69.2 1
 2  138.4 2
N = where n = =

1 1 1 / 2
 2 
= = 0.7072g

Amount of polonium disintegrated = (1 – 0.7072) = 0.2928g

No. of atoms of polonium in 0.2928g = 6.023 × 10 23 × 0.2928
210

Thus, No. of atoms of helium formed = 6.023 × 10 23 × 0.2928
210

Volume of helium collected = 22400 × No. of helium atoms = 22400 × 0.2928 = 31.23 cm3
6.023 × 10 23 210

Example : 18 In nature a decay chain series starts with 232 Th and finally terminates at 208 Pb. A thorium ore sample was
90 82

found to contain 8 × 10 −5 mL of helium at STP and 5 × 10 −7 g of 232 Th. Find the age of the ore sample

assuming the source of helium to be only decay of 232 Th . Also assume complete retention of helium within

the ore. (Half life of 232 Th = 1.39 × 1010 years) [Roorkee 1992]

(a) 6.89 × 109 years (b) 4.89 × 109 years (c) 3.69 × 109 years (d) 6.893 × 1010 years

Solution: (b) No. of moles of helium = 8 × 10 −5
22400

232 Th → 208 Pb + 6 4 He
90 82 2

No. of 232 Th moles which have disintegrated = 8 × 10 −5
90 6 × 22400

Mass of 232 Th which have disintegrated = 8 × 10 −5 × 232 = 1.3809 × 10−7 g
90 6 × 22400

Mass of 232Th left, ' N t ' = 5 × 10 −7 g

' N 0 ' = (5 × 10 −7 + 1.3809 × 10 −7 ) = 6.3809 × 10 −7 g

Applying t= 2.303 log N0 = 2.303 × 1.39 × 1010 log 6.3809 × 10 −7 = 4.89 × 109 years
λ Nt 0.693 5 × 10 −7

Radioactive disintegration series.

The phenomenon of natural radioactivity continues till stable nuclei are formed. All the nuclei from the initial
element to the final stable element constitute a series known as disintegration series. Further we know that mass
numbers change only when α-particles are emitted (and not when β-particles are emitted) causing the change in
mass of 4 units at each step. Hence the mass numbers of all elements in a series will fit into one of the formulae.

4n, 4n + 1 , 4n + 2 and 4n + 3

Hence there can be only four disintegration series

Series 4n 4n + 1 4n + 2 4n + 3
n 58 59 59 58
Parent element 90 Th232 94 Pu 241 92 U 238 92 U 235
1.39 ×1010 years 10 years 4.5 ×109 years
Half life 90 Th232 92 U 238 7.07 ×108 years
Prominent element 1.39 ×1010 years 93 Np237 4.5 ×109 years 89 Ac 227
Thorium (Natural) 2.2 ×106 years Uranium (Natural) 13.5 years
Half life Neptunium (Artificial) Actinium
Name of series (Natural)
82 Pb207
End product 82 Pb208 83 Bi 209 82 Pb206

n 52 52 51 51
lost α = 6 α =8 α =8 α =7
Number of β =5 β =6 β =4
particles β =4

The numbers indicate that in a particular series the mass numbers of all the members are either divisible by 4
(in case of 4n) or divisible by 4 with remainder of 1, 2 or 3 (in the rest three series), n being an integer. In other
words, the mass numbers of the members of 4n, 4n + 1 , 4n + 2 and 4n + 3 series are exactly divisible by 4, 4 + 1,
4 + 2 and 4 + 3 respectively.

Note :  4n + 1 series is an artificial series while the rest three are natural.

 The end product in the 4n + 1 series is bismuth, while in the rest three, a stable isotope of lead is the end
product.

 The 4n + 1 series starts from plutonium 94 Pu241 but commonly known as neptunium series because
neptunium is the longest-lived member of the series.

 The 4n + 3 series actually starts from 92 U 235 .

Activity of population, Radioactive equilibrium and Units of radioactivity.

(1) Activity of population or specific activity : It is the measure of radioactivity of a radioactive
substance. It is defined as ' the number of radioactive nuclei, which decay per second per gram of radioactive
isotope.' Mathematically, if 'm' is the mass of radioactive isotope, then

Specific activity = Rate of decay = λN = λ × Avogadro number
m m Atomic mass in g

where N is the number of radioactive nuclei which undergoes disintegration.

(2) Radioactive equilibrium : Suppose a radioactive element A disintegrates to form another radioactive
element B which in turn disintegrates to still another element C.

A → B → C

In the starting, the amount of A (in term of atoms) is large while that of B is very small. Hence the rate of
disintegration of A into B is high while that of B into C is low. With the passage of time, A go on disintegrating while
more and more of B is formed. As a result, the rate of disintegration of A to B goes on decreasing while that of B to
C goes on increasing. Ultimately, a stage is reached when the rate of disintegration of A to B is equal to that of B to
C with the result the amount of B remains constant. Under these conditions B is said to be in equilibrium with A.
For a radioactive equilibrium to be established half-life of the parent must be much more than half-life of the
daughter.

It is important to note that the term equilibrium is used for reversible reactions but the radioactive reactions are
irreversible, hence it is preferred to say that B is in a steady state rather than in equilibrium state.

At a steady state,

NA = λB = TA T = 1  , Where λA and λB are the radioactive constants for the
NB λA TB  λ 

processes A → B and B → C respectively. Where TA and TB are the average life periods of A and B respectively.

In terms of half-life periods, NA = (t1 / 2 )A
NB (t1 / 2 )B

Thus at a steady state (at radioactive equilibrium), the amounts (number of atoms) of the different
radioelements present in the reaction series are inversely proportional to their radioactive constants or directly
proportional to their half-life and also average life periods.

It is important to note that the radioactive equilibrium differs from ordinary chemical equilibrium because in
the former the amounts of the different substances involved are not constant and the changes are not reversible.

(3) Units of radioactivity : The standard unit in radioactivity is curie (c) which is defined as that amount of
any radioactive material which gives 3.7 × 1010 disintegration’s per second (dps), i.e.,

1c = Activity of 1g of Ra 226 = 3.7 × 1010 dps

The millicurie (mc) and microcurie (µc) are equal to 10−3 and 10−6 curies i.e. 3.7 × 107 and 3.7 × 104 dps
respectively.

1c = 103 mc = 106 µc ; 1c = 3.7 × 1010 dps ; 1mc = 3.7 × 107 dps ; 1µc = 3.7 × 104 dps

But now a day, the unit curie is replaced by rutherford (rd) which is defined as the amount of a radioactive
substance which undergoes 106 dps. i.e., 1rd = 106 dps . The millicurie and microcurie correspondingly
rutherford units are millirutherford (mrd) and microrutherford (µrd) respectively.

1c = 3.7 × 1010 dps = 37 × 103 rd ; 1mc = 3.7 × 107 dps = 37 rd ; 1 µc = 3.7 × 104 dps = 37 mrd

However, is SI system the unit of radioactivity is Becquerel (Bq)
1 Bq = 1 disintegration per second = 1 dps = 1µrd

106 Bq = 1rd

3.7 × 1010 Bq = 1c

(4) The Geiger-Nuttal relationship : It gives the relationship between decay constant of an α- radioactive
substance and the range of the α-particle emitted.

log λ = A + B log R

Where R is the range or the distance which an α-particle travels from source before it ceases to have ionizing
power. A is a constant which varies from one series to another and B is a constant for all series. It is obvious that the
greater the value of λ the greater the range of the α-particle.

Artificial transmutation, Synthetic or Transuranic elements and Artificial radioactivity.

(1) Artificial transmutation or Nuclear transformation or Nuclear transmutation : The conversion
of one element into another by artificial means, i.e., by means of bombarding with some fundamental particles, is

known as artificial transmutation. The phenomenon was first applied on nitrogen whose nucleus was bombarded
with α-particles to produce oxygen.

7 N 14 + 2 He 4 → 8 O17 + 1 H 1

Nitrogen isotope Alpha particle Oxygen isotope Proton

The element, which is produced, shows radioactivity, the phenomenon is known as Induced radioactivity.
The fundamental particles which have been used in the bombardment of different elements are as follows :

(i) α-particle : Helium nucleus, represented as 2 He 4 .

(ii) Proton : Hydrogen nucleus, represented as 1 H 1 .

(iii)Deutron : Deuterium nucleus, represented as 1 H 2 or 1 D 2 .

(iv)Neutron : A particle of mass number 1 but no change, represented as 0 n1 .

Since α-particles, protons and deutrons carry positive charge, they are repelled by the positively charged
nucleus and hence these are not good projectiles. On the other hand, neutrons, which carry no charge at all, are
the best projectiles. Further among α-particles, protons and deutrons; the latter two carrying single positive
charge are better projectiles than the α-particles. However, the positively charged α-particles, protons and deutrons
can be made much more effective if they are imparted with high velocity. Cyclotron is the most commonly used
instrument for accelerating these particles. The particles leave the instrument with a velocity of about 25,000 miles
per second. A more recent accelerating instrument is called the synchrotron or bevatron. It is important to note
that this instrument cannot accelerate the neutrons, being neutral.

When a target element is bombarded with neutrons, product depends upon the speed of neutrons. Slow
neutrons penetrate the nucleus while a high-speed neutron passes through the nucleus.

92 U 238 + 0 n1 → 92U 239 ; 92 U 238 + 0 n1 → 92U 237 + 20 n1
slow speed high speed

Thus slow neutrons, also called thermal neutrons are more effective in producing nuclear reactions than
high-speed neutrons.

Alchemy : The process of transforming one element into other is known as alchemy and the person involved
in such experiments is called alchemist. Although, gold can be prepared from lead by alchemy, the gold obtained is
radioactive and costs very high than natural gold.

Some examples are given below for different nuclear reactions :
(i) Transmutation by α-particles :
(a) α, n type

4 Be 9 (α,n) 6C12 i.e. 4 Be 9 + 2 He 4 → 6C12 + 0n1 ; 94 Pu239 (α,n) 96Cm242 i.e. 94 Pu 239 + 2 He 4 → 94 Cm242 + 0n1
(b) α, p type

9 F 19 (α, p) 10 Ne 22 i.e. 9 F 19 + 2 He 4 → 10 Ne 22 + 1 H 1 ; 7 N 14 (α, p) 8 O17 i.e., 7 N 14 + 2 He 4 → 8 O17 + 1 H 1
(c) α, β type

26 Fe 59 (α, β ) 29 Cu63 i.e., 26 Fe 59 + 2 He 4 → 29Cu63 + −1e 0
(ii) Transmutation by protons :

(a) p, n type

15 P 31 (p,n) 16 S 31 i.e., 15 P 31 + 1 H 1 → 16 S 31 + 0n1

(b) p, γ type

6 C12 (p,γ ) 7 N 13 i.e., 6 C12 + 1 H 1 → N 13 + γ
(c) p, d type

4 Be 9 (p, d) 4 Be 8 i.e., 4 Be 9 + 1 H 1 → 4 Be 8 + 1 H 2

(d) p, α type

8 O16 (p,α) 7 N 31 i.e., 8 O16 + 1 H 1 → 7 N 13 + 2 He 4
(iii)Transmutation by neutrons :

(a) n,p type

13 Al 27 (n, p) 12 Mg 27 i.e., 13 Al 27 + 0n1 → 12 Mg 27 + 1 H 1
(b) n,α type

8 O16 (n,α) 12Mg27 i.e., 8 O16 + 0n1 → 6C13 + 2 He 4
(c) n, γ type

92 U 238 (n, λ) 92U 239 i.e., U 238 + 0n1 → 92U 238 +λ

92

(d) n,β type

8 O18 (n, β ) 9 F 19 i.e., 8 O18 + 0n1 → 9 F 19 + −1e 0
(iv)Transmutation by deutrons :

(a) d,p type

3 Li6 (d, p) 3 Li7 i.e., 3 Li6 + 1 H 2 → 3 Li7 + 1 H 1 ; 32 As 75 (d, p) 32 As 76 i.e., 32 As75 + 1 H 2 → 32 As76 + 1 H 1
(v) Transmutation by γ-radiations :

(a) γ, n type

4 Be 9 (γ ,n) 4 Be 8 i.e., 4 Be 9 + γ → 4 Be 8 + 0n1

(2) Synthetic elements : Elements with atomic number greater than 92 i.e. the elements beyond uranium
in the periodic table are not found in nature like other elements. All these elements are prepared by artificial
transmutation technique and are therefore known as transuranic elements or synthetic elements. The nuclear
reactions for the preparation of some transuranic elements are cited below.

Elements 93 (neptunium) and 94 (plutonium) were first discovered in 1940. Bombarding uranium-238
with neutrons produced them.

92 U 238 + 0n1 → 92U 239 → 93 Np 239 + −1e 0 ; 94 Np 239 → 94 Pu 239 → −1e 0
(Uranium) (Neptunium)

Elements with larger atomic numbers are normally formed in small quantities in particle accelerators. For
example, curium-242 is formed when a plutonium-239 target is struck with alpha particles.

94 Pu 239 + 2 He 4 → 96 Cm242 + 0n1
Plutonium curium

(3) Artificial radioactivity or induced radioactivity : In 1934, Irene Curie and F. Joliot observed

that when boron and aluminium were bombarded by α-particles, neutrons, protons and positrons were emitted.

Curie and Joliot explained this observation by saying that during bombardment, a metastable isotope is

formed which behaves as a radioactive element. This process was termed as artificial radioactivty.

“The process in which a stable isotope is converted into radioactive element by artificial transmutation is called
artificial radioactivity.”

When 27 Al is bombarded by α-particles, radioactive isotope 30 P is formed.
13 15

27 Al + 4 He 30 Si + 1H1 (95% of total conversion)
13 2 14

30 P* + 0n1(5% of total conversion half life period of P30 is 3.2 minutes)
15
15

30 Si + +1 e 0
14
Positron

In a similar manner, the artificial radioactivity was observed when 10 B was bombarded by α-particles.
5

10 B + 4 He 13 C + 1 H
5 2 6 1

13 N* + 0 n1
7

13 C + +1e 0
6

The following are some of the nuclear reactions in which radioactive isotope are fomed.

23 Na + 2 H → 24 Na * + 1 H ( 24 Na − β radioactive) ; 238 U + 1 n → 239 U * +γ ( 239 U − β radioactive)
11 1 11 1 11 92 0 92 92

12 C + 1 H → 13 N * +γ ( 13 N − positron radioactive) ; 25 Mg + 4 He → 28 Al * + 1 H ( 28 Al − β radioactive)
6 1 7 7 12 2 13 1 13

Nuclear fission and Nuclear fusion.

(1) Nuclear fission : The splitting of a heavier atom like that of uranium – 235 into a number of fragments
of much smaller mass, by suitable bombardment with sub-atomic particles with liberation of huge amount of energy
is called Nuclear fission. Hahn and Startsman discovered that when uranium-235 is bombarded with neutrons,
it splits up into two relatively lighter elements.

92 U 235 + 0n1 → 56 Ba140 + 36 Kr 93 + 2 − 3 0n1 + Huge amount of energy

Spallation reactions are similar to nuclear fission. However, they differ by the fact that they are brought by
high energy bombarding particles or photons.

Elements capable of undergoing nuclear fission and their fission products. Among elements capable of
undergoing nuclear fission, uranium is the most common. The natural uranium consists of three isotopes, namely
U 234 (0.006%) , U 235 (0.7%) and U 238 (99.3%) . Of the three isomers of uranium, nuclear fission of U 235 and U 238

are more important. Uranium-238 undergoes fission by fast moving neutrons while U 235 undergoes fission by slow
moving neutrons; of these two, U 235 fission is of much significance. Other examples are Pu239 and U 233 .

Uranium-238, the more abundant (99.3%) isotope of uranium, although itself does not undergo nuclear
fission, is converted into plutonium-239.

92 U 238 + 0n1 → 92U 239 ; 92 U 239 → 93 Np 239 + −1e 0 ; 93 Np 238 → 94 Pu 239 + −1e 0

Which when bombarded with neutrons undergo fission to emit three neutrons per plutonium nucleus. Such
material like U-238 which themselves are non-fissible but can be converted into fissible material (Pu-239) are
known as fertile materials.

Release of tremendous amount of energy : The importance of nuclear fission lies in the release of
tremendous amount of energy during this process. During the U 235 fission nearly 0.215 mass unit per uranium
nucleus is found to be converted into energy.

2U352.13254+10.0n019 → X13e8.193595 +S94r.99455+202n×11.00+9E
236.133 235.918

The released energy is due to difference in the total sum of masses of the reactants and products, in according

to the Einsten's mass energy relation i.e. E = mc 2 .

Alternatively, ∆m = 236.133 − 235.918 = 0.215 amu

 1amu = 931 MeV

0.215 amu = 931× 0.215 MeV = 198 MeV = 198 × 2.3 × 107 kcal

∴ Energy released by the fission of 1 g of U 235 = 198 × 2.3 × 107 = 1.9 × 107 kcal
235

Recall that the combustion of 1 g of carbon releases only 94.0 / 12 = 7.83 kcal of energy while the fission of 1

g of U 235 releases 1.9 × 107 kcal . Hence nuclear fission releases several million times higher energy than the

ordinary chemical combustion.

Release of neutrons : During U 235 fission it is obvious that Uranium
2-3 neutrons per uranium molecule are emitted. Some neutrons are Neutron
ejected within an extremely short interval and are called prompt
neutrons; fission products for an appreciable time fraction of a E
second to several seconds emit the rest after the fission. These are
called delayed neutrons. 3E

Note :  Each fission yields 3 neutrons each of which 9E

can cause further fission to give 3 neutrons goes on increasing in A fission chain reactor of Uranium.

geometric progression 1, 3, 9, 27, 81, 243,.... and many geometric

progression take place in a very small fraction of a second.

Chain reaction : With a small lump of U 235 , most of the neutrons emitted during fission escape but if the
amount of U 235 exceeds a few kilograms (critical mass), neutrons emitted during fission are absorbed by adjacent

nuclei causing further fission and so producing more neutrons. Now since each fission releases a considerable
amount of energy, vast quantities of energy will be released during the chain reaction caused by U 235 fission.

Atomic bomb : An atomic bomb is based upon the process of that nuclear fission in which no secondary
neutron escapes the lump of a fissile material for which the size of the fissile material should not be less than a
minimum size called the critical size. There is accordingly a sudden release of a tremendous amount of energy,
which represents an explosive force much greater than that of the most powerful TNT bomb. In the world war II in
1945 two atom bombs were used against the Japanese cities of Hiroshima and Nagasaki, the former contained U-
235 and the latter contained Pu-239.

Atomic pile or Nuclear reactor : It is a device to obtain the nuclear energy in a controlled way to be used
for peaceful purposes. The most common reactor consists of a large assembly of graphite (an allotropic form of

carbon) blocks having rods of uranium metal (fuel). Many of the neutrons formed by the fission of nuclei of 92 U 235
escape into the graphite, where they are very much slow down (from a speed of about 6000 or more miles/sec to a
mile/sec) and now when these low speed neutrons come back into the uranium metal they are more likely to cause
additional fissions. Such a substance likes graphite, which slow down the neutrons without absorbing them is known
as a moderator. Heavy water, D2O is another important moderator where the nuclear reactor consists of rods of
uranium metal suspended in a big tank of heavy water (swimming pool type reactor). Cadmium or boron are used
as control rods for absorbing excess neutrons.

Plutonium from a nuclear reactor : For such purposes the fissile material used in nuclear reactors is the
natural uranium which consists mainly (99.3%) of U-238. In a nuclear reactor some of the neutrons produced in U-
235 (present in natural uranium) fission converts U-238 to a long-lived plutonium isotope, Pu-239 (another
fissionable material). Plutonium is an important nuclear fuel. Such reactors in which neutrons produced from fission
are partly used to carry out further fission and partly used to produce some other fissionable material are called
Breeder reactors.

Production of radioactive isotopes by bombarding with neutrons from a nuclear reactor : These
radioactive isotopes are used in medicine, industry and hospitals.

Nuclear reactors in India : India is equipped with the five nuclear reactors, namely Apsara (1952), Cirus
(1960), Zerlina (1961), Purnima (1972) and R-5. Purnima uses plutonium fuel while the others utilize uranium as
fuel.

Apsara the first nuclear reactor was completed on 14th August 1952 at Trombay under the guidance of the late
Dr. H.J. Bhabha. It is the swimming pool reactor, which consists of a lattice of enriched uranium (fuel) immersed in
a large pool of water. Water acts as a moderator, coolant and shield. This reactor is simple, safe, flexible, easily
accessible and cheap.

(2) Nuclear fusion : “ Oposite to nuclear fission, nuclear fusion is defined as a process in which lighter nuclei
fuse together to form a heavier nuclei. However, such processes can take place at reasonable rates only at very high
temperatures of the order of several million degrees, which exist only in the interior of stars. Such processes are,
therefore, called Thermonuclear reactions (temperature dependent reactions). Once a fusion reaction initiates,
the energy released in the process is sufficient to maintain the temperature and to keep the process going on.

4 1H 1 → 2 He 4 + 2 +1e 0 + Energy
Hydrogen Helium Positron

This is not a simple reaction but involves a set of the thermonuclear reactions, which take place in stars
including sun. In other words, energy of sun is derived due to nuclear fission.

Calculation of energy released in nuclear fusion : Let us write the reaction involving the fusion of four
hydrogen nuclei to form helium nucleus.

4 1H1 → 2 He 4 + 2 +1e 0

Hydrogen Helium Positron
4.003873 2×0.000558=0.001116
Mass: 4×1.008144 4.004989

or =4.032576

∴ Loss is mass, ∆m = 4.032576 − 4.004989 = 0.027587amu

∴ Energy released = 0.027587 × 931 MeV = 26.7 MeV

∴ Energy released/gm of hydrogen consumed = 26.7 = 6.7 MeV = 6.7 × 2.3 × 107 kcal = 1.54 × 108 kcal
4

Controlled nuclear fusion : Unlike the fission process, the fusion process could not be controlled. Since

there are estimated to be some 1017 pounds of deuterium ( 1H 2 ) in the water of the earth, and since each pound is
equivalent in energy to 2500 tonnes of coal, a controlled fusion reactor would provide a virtually inexhaustible
supply of energy.

Comparision of nuclear fission and nuclear fusion : Now let us compare the efficiency of the energy
conversion of the two processes, i.e. nuclear fission and nuclear fusion

Nuclear fission reaction, 92 U 235 + 0n1 → 56 Ba141 + 36 Kr 92 + 2 − 3 0n1 + 200 MeV

If one atom of uranium is fissioned by one neutron, the percent efficiency in terms of mass converted into

energy (where 1 mass unit = 931 MeV) will be : (235 + 200 MeV × 931 × 100 = 0.09%
1) mass units

Nuclear fusion reaction, 1H 2 + 1H 3 → 2 He 4 + 0n1 + 17.8 MeV

The percent efficiency of the reaction = (2 + 3 17.8 MeV × 931 × 100 = 0.35%
mass units)

Thus it indicates that for these two fission and fusion reactions the percent efficiency is approximately four
times greater for the fusion reaction.

Hydrogen bomb : Hydrogen bomb is based on the fusion of hydrogen nuclei into heavier ones by the
thermonuclear reactions with release of enormous energy.

As mentioned earlier the above nuclear reactions can take place only at very high temperatures. Therefore, it
is necessary to have an external source of energy to provide the required high temperature. For this purpose, the
atom bomb, (i.e., fission bomb) is used as a primer, which by exploding provides the high temperature necessary
for successful working of hydrogen bomb (i.e., fusion bomb). In the preparation of a hydrogen bomb, a suitable
quantity of deuterium or tritium or a mixture of both is enclosed in a space surrounding an ordinary atomic bomb.
The first hydrogen bomb was exploded in November 1952 in Marshall Islands; in 1953 Russia exploded a powerful
hydrogen bomb having power of 1 million tonnes of TNT

A hydrogen bomb is far more powerful than an atom bomb. Thus if it were possible to have sufficiently high
temperatures required for nuclear fusion, the deuterium present in sea (as D2O) sufficient to provide all energy
requirements of the world for millions of years.

Note :  The first nuclear reactor was assembled by Fermi in 1942.

Difference between Nuclear fission and fusion

Nuclear fission Nuclear fusion

The process occurs only in the nuclei of heavy elements. The process occurs only in the nuclei of light elements.
The process involves the fission of the lighter nuclei to
The process involves the fission of the heavy nucleus to heavy nucleus.
the lighter nuclei of comparable masses.
The process takes place at higher temperature 108 oC) .
The process can take place at ordinary temperature.

The energy liberated during this process is high (200 MeV The energy liberated during the process is comparatively
per fission) low (3 to 24 MeV per fusion)

Percentage efficiency of the energy conversion is Percentage efficiency of the energy conversion is high (four
comparatively less. times to that of the fission process).

The process can be controlled for useful purposes. The process cannot be controlled.

Isotopes, Isobars, Isotones, Isodiaphers, Isoelectronic species, Isosters and Nuclear isomers.

(1) Isotopes : Atoms of a given element which have same atomic number (nuclear charge) but different mass
number are called isotopes. In other words, isotopes are the atoms of the same element differing in mass number.
Thus isotopes have same number of protons and electrons but different number of neutrons. They have same
position in the periodic table, same chemical properties and same atomic charge. The term was first coined by

Soddy. However, Aston using mass spectrometer first separated isotopes (Ne 20 and Ne 22 ) .

Examples : (i) 1 H1 , 1 H 2 , 1 H 3 (ii) 6 C12 , 6 C13 and 6 C14

Hydrogen (Protium) Deuterium Tritium
(p =1, e =1, n= 0) (p =1, e =1, n= 0) (p =1, e =1, n= 0)

(iii) 8 O16 , 8 O17 , 8 O18 (iv) 17 Cl 35 and 17Cl 37

Of all the elements, tin has maximum number of stable isotopes (ten).

The fractional atomic weight (35.5) of chlorine is due to the fact that in the ordinary chlorine atom, Cl 35 and

Cl 37 are present in the ratio of 3 : 1.

∴ Average atomic weight of Cl = 3 × 35 + 1× 37 = 35.5 amu
4

The percentage of a given isotope in the naturally occurring sample of an element is called Isotopic
abundance. As the isotopic abundance of an element is constant irrespective of its source, atomic weight of an
element is constant.

(2) Isobars : Isobars are the atoms of different elements with the same mass number but different atomic
numbers. In other words, isobars have different number of protons, neutrons and electrons but the sum of protons
and neutrons (i.e., number of nucleons) is same.

Examples : (i) 18 Ar 40 , 19 K 40 and 20 Ca 40 (ii) 52 Te130 , 54 Xe130 and 56 Ba130 .

Since isobars are the atoms of different elements, they will have different physical and chemical properties.

(3) Isotones : Isotones are the atoms of different elements with the same number of neutrons but different
mass numbers, e.g. 14 Si30 , 15 P 31 and 16 S32 . Since the variable factor in isotones is the number of protons (atomic
number), they must have different physical and chemical properties.

Examples : (i) 14 Si30 , 14 P 31 and 16 S 32 (ii) 19 K 39 and 20Ca 40

(iii) 1 H 3 and 2 He 4 (iv) 6 C13 and 7 N 14

(4) Isodiaphers : Atoms having same isotopic number are called isodiaphers.

Mathematically, isotopic number (isotopic excess) = (N – Z) or (A – 2Z)

Where, N = Number of neutrons; Z = Number of protons

Examples : (i) 92 U 235 and 90Th231 (ii) 19 K 39 and 9 F 19 (iii) 29 Cu65 and 24 Cr 55

(5) Isoelectronic species : Species (atoms, molecules or ions) having same number of electrons are called
isoelectronic.

Examples : (i) N 3− , O 2− , F − , Ne, Na + , Mg 2+ , Al 3+ , CH 4 , NH 3 , H 2O and HF have 10 electrons each.

(ii) P 3− , S 2− , Cl − , Ar, K + and Ca 2+ have 18 electrons each.

(iii) H − , He, Li+ and Be 2+ have 2 electrons each.

(iv) CO, CN − and N 2 have 14 electrons each.

(v) N 2O, CO2 and CNO− have 22 electrons each.
(6) Isosters : Molecules having same number of atoms and also same number of electrons are called isosters.

Examples : (i) N 2 and CO (ii) CO2 and N 2O (iii) HCl and F2

(iv) CaO nad MgS (v) C6 H6 (benzene) and inorganic benzene B6 N6 .

(7) Nuclear isomers : Nuclear isomers (isomeric nuclei) are the atoms with the same atomic number and
same mass number but with different radioactive properties. They have same number of electrons, protons and
neutrons. An example of nuclear isomers is uranium-X (half-life 1.4 min) and uranium-Z (half-life 6.7 hours). Otto
Hahn discovered nuclear isomers.

The reason for nuclear isomerism is the different energy states of the two isomeric nuclei. One may be in the
ground state whereas the other should be in an excited state. The nucleus in the excited state will evidently have a
different half-life.

Now-a-days as much as more than 70 pairs of nuclear isomers have been found. Few examples areas follows

(i) 69 Zn and 69 Zn (ii) 80 Br and 80 Br

(T1 / 2 =13.8 hour) (T1 / 2 =57 min) (T1 / 2 =4.4 hour) (T1 / 2 =18 min)

Example : 19 Naturally occurring boron consists of two isotopes whose atomic weights are 10.01 and 11.01. The atomic
weight of natural boron is 10.81. Calculate the percentage of each isotope in natural boron

[IIT 1978, 82; MLNR 1994]

(a) 80 (b) 90 (c) 70 (d) 20

Solution: (a) Let the % of isotope with at. wt. 10.01 = x

∴ % of isotope with at. wt. 11.01 = (100 – x)

Now since, At. wt. = x × 10.01 + (100 − x) × 11.01
100

10.81 = x × 10.01 + (100 − x) × 11.01
100

x = 20

Hence, % of isotope with at. wt. 10.01 = 20

∴ % of isotope with at. wt. 11.01 = 100 – 20 = 80

Application of radioactivity and Hazards of radiations.

Radioisotopes find numerous applications in a variety of areas such as medicine, agriculture, biology,
chemistry, archeology, engineering and industry. Some of the are given below :

(1) Age determination (carbon dating) : Radioactive decay follows a very exact law, and is virtually
unaffected by heat, pressure, or the state of chemical combination of the decaying nuclei, it can be used as a very
precise clock for dating past events. For instance, the age of earth has been determined by uranium dating
technique as follows. Samples of uranium ores are found to contain Pb 206 as a result of long series of α- and β-
decays. Now if it is assumed that the ore sample contained no lead at the moment of its formation, and if none of
the lead formed from U 238 decay has been lost then the measurement of the Pb 206 / U 238 ratio will give the value
of time t of the mineral.

No. of atoms of Pb 206 = e −λt−1 , where λ is the decay constant of uranium-238
No. of atoms of U 238 left

Alternatively, t = 2.303 log Amount of Initial amount of U 238 till date
λ U 238 in the mineral
present

Similarly, the less abundant isotope of uranium, U 235 eventually decays to Pb 207 ;Th232 decays to Pb 208 and

thus the ratios of Pb 207 / U 235 and Pb 208 / Th232 can be used to determine the age of rocks and minerals. Many
ages have been determined this way to give result from hundreds to thousands of million of years,.

Besides the above long-lived radioactive substances viz. U 238 ,U 235 and Th232 (which have been present on

earth since the elements were formed), several short-lived radioactive species have been used to determine the age

of wood or animal fossils. One of the most interesting substances is 6 C14 (half-life 5760 years) which was used by
Willard Libby (Nobel lauret) in determining the age of carbon-bearing materials (e.g. wood, animal fossils, etc.)
Carbon-14 is produced by the bombardment of nitrogen atoms present in the upper atmosphere with neutrons
(from cosmic rays).

7 N 14 + 0n1 → 6 C14 + 1 H 1

Thus carbon-14 is oxidised to CO2 and eventually ingested by plants and animals. The death of plants or
animals puts an end to the intake of C14 from the atmosphere. After this the amount of C14 in the dead tissues
starts decreasing due to its disintegration.

6 C14 → 7 N 14 + −1e 0

It has been observed that on an average, one gram of radioactive carbon emits about 12 β-particles per
minute. Thus by knowing either the amount of C-14 or the number of β-particles emitted per minute per gram of
carbon at the initial and final (present) stages, the age of carbon material can be determined by using the following
formulae.

λ = 2.303 log N0 or t = 2.303 log N0
t Nt λ Nt

where t = Age of the fossil, λ = Decay constant, N0 = Initial radioactivity (in the fresh wood), N t =
Radioactivity in the fossil

The above formula can be modified as

t = 2.303 log Initial ratio of C14 / C12 (in fresh wood)
λ C14 / C12 ratio in the old wood

= 2.303 log Initial amount of C14 / C12 (in fresh wood) = 2.303 log Radioactivity in fresh wood due to C14
λ Amount of C14 in the old wood λ Radioactivity in old wood due to C14

= 2.303 × T1/ 2 of C14 log counts min−1 g −1 of C14 in fresh wood
0.693 counts min−1 g −1 of C14 in old wood

Similarly, tritium 1 H 3 has been used for dating purposes.

(2) Radioactive tracers (use of radio–isotopes) : A radioactive isotope can be easily identified by its
radioactivity. Because of similar physical and chemical properties of radioisotopes and non-radioisotopes of an
element, if a small quantity of the former is mixed with normal isotope, then chemical reactions can be studied by
determining the radioactivity of the radioisotope. The radioactivity can, therefore act as a tag or label that allows
studying the behaviour of the element or compounding which contains this isotope. An isotope added for this
purpose is known as isotopic tracer. The radioactive tracer is also known as an isotopic tracer. The radioactive tracer
is also known as an indicator because it indicates the reaction. Radioisotopes of moderate half-life periods are used
for tracer work. The activity of radioisotopes can be detected by means of electroscope, the electrometer or the
Geiger-Muller counter. Tracers have been used in the following fields :

(i) In medicine : Radioisotopes are used to diagnose many diseases. For example, Arsenic – 74 tracer is used
to detect the presence of tumours, Sodium –24 tracer is used to detect the presence of blood clots and Iodine –131
tracer is used to study the activity of the thyroid gland. It should be noted that the radioactive isotopes used in
medicine have very short half-life periods.

(ii) In agriculture : The use of radioactive phosphorus 32 P in fertilizers has revealed how phosphorus is
absorbed by plants. This study has led to an improvement in the preparation of fertilizers. 14 C is used to study the
kinetics of photo synthesis.

(iii) In industry : Radioisotopes are used in industry to detect the leakage in underground oil pipelines, gas
pipelines and water pipes. Radioactive isotopes are used to measure the thickness of materials, to test the wear and
tear inside a car engine and the effectiveness of various lubricants. Radioactive carbon has been used as a tracer in
studying mechanisms involved in many reactions of industrial importance such as alkylation, polymerization,
catalytic synthesis etc.

(iv) Analytical Studies : Several analytical procedures can be used employing radioisotopes as tracers.

(a) Adsorption and occlusion studies : A small amount of radioactive isotope is mixed with the inactive
substance and the activity is studied before and after adsorption. Fall in activity gives the amount of substance
adsorbed.

(b) Solubility of sparingly soluble salt : The solubility of lead sulphate in water may be estimated by
mixing a known amount of radioactive lead with ordinary lead. This is dissolved in nitric acid and precipitate as
lead sulphate by adding sulphuric acid. Insoluble lead sulphate is filtered and the activity of the water is measured.
From this, the amount of PbSO4 still present in water can be estimated.

(c) Ion-exchange technique : Ion exchange process of separation is readily followed by measuring activity
of successive fractions eluted from the column.

(d) Reaction mechanism : By labelling oxygen of the water, mechanism of ester hydrolysis has been
studied.

R−C O O
*
+HOH → R−C + R′OH

OR *

OH

(e) Study of efficiency of analytical separations : The efficiency of analytical procedures may be
measured by adding a known amount of radio-isotopes to the sample before analysis begins. After the completion,
the activity is again determined. The comparison of activity tells about the efficiency of separation.

(3) Use of γ − rays : γ − rays are used for disinfecting food grains and for preserving food stuffs. Onions,

potatoes, fruits and fish etc., when irradiated with γ − rays, can be preserved for long periods. High yielding disease

resistant varieties of wheat, rice, groundnut, jute etc., can be developed by the application of nuclear radiations. The
γ − rays radiations are used in the treatment of cancer. The γ − radiations emitted by cobalt –60 can burn

cancerous cells. The γ − radiations are used to sterilize medical instruments like syringes, blood transfusion sets. etc.

These radiations make the rubber and plastics objects heat resistant.

Hazards of radiations : The increased pace of synthesis and use of radio isotopes has led to increased
concern about the effect of radiations on matter, particularly in biological systems. Although the radioisotopes have
found wide spread uses to mankind such as atomic power generation, dating, tracer technique, medicinal treatment,
the use of nuclear energy is an extremely controversial social and political issue. You should ask yourself, how you
would feel about having a nuclear power plant in your town. The accident of Chernobyl occurred in 1986 in USSR
is no older when radioisotopes caused a hazard there. The nuclear radiations (alpha, beta, gamma as well as X-
rays) possess energies far in excess of ordinary bond energies and ionisation energies. Consequently, these
radiations are able to break up and ionise the molecules present in living organisms if they are exposed to such
radiations. This disrupts the normal functions of living organisms. The damage caused by the radiations, however,
depends upon the radiations received. We, therefore, conclude this chapter by examining the health hazards
associated with radioisotopes.

The resultant radiation damage to living system can be classified as :

(i) Somatic or pathological damage : This affects the organism during its own life time. It is a permanent
damage to living civilization produced in body. Larger dose of radiations cause immediate death whereas smaller
doses can cause the development of many diseases such as paralysis, cancer, leukaemia, burns, fatigue, nausea,
diarrhoea, gastrointestinal problems etc. some of these diseases are fatal.

Many scientists presently believe that the effect of radiations is proportional to exposure, even down to low
exposures. This means that any amount of radiation causes some finite risk to living civilization.

(ii) Genetic damage : As the term implies, radiations may develop genetic effect. This type of damage is
developed when radiations affect genes and chromosomes, the body's reproductive material. Genetic effects are
more difficult to study than somatic ones because they may not become apparent for several generations.

Chemical analysis

Analytical chemistry deals with qualitative and quantitative analysis of substances.

Qualitative analysis : In qualitative inorganic analysis, the given compound is analysed for the basic and acid
radicals (i.e., the cations and the anions), that it contains. For example zinc blend is analysed for the Zn2+ and S2−
ions that it contains.

Test for Different Gases.

(1) Colourless gases
(i) Tests for CO2 : It is colourless and odourless gas. It gives white ppt. with lime water which dissolves on

passing excess of CO2 . Ca(OH)2 + CO2 → CaCO3 ↓ +H 2O ; CaCO3 + CO2 + H 2O → Ca(HCO3 )2
Lime water White ppt. White ppt. Excess So lub le

(ii) Test for CO : It is colourless and odourless gas. It burns with a blue flame. 2CO + O2 → 2CO2

Note :  CO is highly poisonous gas.

(iii) Test for O2 : It is colourless and odourless gas. It rekindles a glowing splinter.

(iv) Tests for H2 S : It is a colourless gas with a smell of rotten eggs. It turns moist lead acetate paper black.

(CH 3COO)2 Pb + H2S → 2CH 3COOH + PbS

Black

(v) Tests for SO2 : It is a colourless gas with a suffocating odour of burning sulphur. It turns acidified
K 2Cr2O7 solution green. 3SO2 + K2Cr2O7 + H2SO4 → K2SO4 + Cr2(SO4 )3 + H2O

Green

(vi) Tests for NH3 : It is a colourless gas with a characteristic ammonical smell. It gives white fumes of
NH4Cl with HCl , NH3 + HCl → NH4Cl . With Nessler’s reagents, it gives brown ppt.

White fumes

2K2 [HgI 4 ] + NH 3 + KOH → NH 2 HgOHgI + 7KI + 2H 2O
Nessler 's reagent Iodine of Millon's base

( Brown ppt )

It gives deep blue colour with CuSO4 solution, CuSO4 + 4 NH3 → [Cu(NH3 )4 ]SO4 . NH3 dissolves in water
Deep blue

to give NH 4 OH, which being basic, turns red litmus blue, NH 3 + H2O → NH 4 OH ⇌ NH + + OH − .
4

(vii) Tests for HCl gas : It is colourless gas with a pungent irritating smell. It turns moist blue litmus paper

red i.e., it is acidic in nature. It gives white ppt. with AgNO3 solution. This white ppt. is soluble in NH4OH.

HCl + AgNO3 → AgCl + HNO3 ; AgCl + 2NH 4 OH → [Ag(NH 3 )2 ] + 2H 2O .
White ppt. Soluble

(viii) Test for CH3COOH vapours : These vapours are colourless with a vinegar like smell.

(2) Coloured gases

(i) Tests for Cl2 : It is a greenish yellow gas with a pungent smell. In small quantity it appears almost

colourless. It bleaches a moist litmus paper, Cl2 + H 2O → 2HCl + [O]; Colour + [O] → Colourless. Blue litmus

paper first turns red and then becomes colourless.

(ii) Tests for Br2 : Brown vapours with a pungent smell. It turns moist starch paper yellow.
(iii) Tests for I2 : Violet vapours with a pungent smell. It turns moist starch paper blue.

(iv) Tests for NO2 : Brown coloured pungent smelling gas. It turns moist starch KI paper blue
2KI + 2NO2 → 2KNO2 + I2 ; I2 + Starch → Blue colour.

It turns ferrous sulphate solution black, 3FeSO4 + NO2 + H2SO4 → Fe2(SO4 )3 + FeSO4 . NO+ H2O
Black brown

Systematic Procedure for Qualitative Analysis of Inorganic Salts.

It involves the following steps : (1) Preliminary tests (2) Wet tests for acid radicals and (3) Wet tests for basic
radicals.

(1) Preliminary tests
(i) Physical examination : It involves the study of colour, smell, density etc.
(ii) Dry heating : Substance is heated in a dry test tube.

Observation Result

(a) A gas or vapour is evolved. Compounds with water of crystallisation

Vapour, evolved, test with litmus paper. Ammonium salts, acid salts, and hydroxides.
(usually accompanied by change of colour)

The vapour is alkaline. Ammonium salts.

The vapour is acidic. Readily decomposable salts of strong acids.

Oxygen is evolved Nitrates,chlorates and certain oxides.

Dinitrogen oxide Ammonium nitrate or nitrate mixed with an
ammonium salt.

Dark-brown or reddish fumes (oxides of Nitrates and nitrites of heavy metals.
nitrogen), acidic in reaction.

CO2 is evolved, lime water becomes turbid. Carbonates or hydrogen carbonates.

NH3 is evolved which turns red litmus blue. Ammonium salts.

SO2 is evolved, which turns acidified Sulphates and thiosulphates.
K2Cr2O7 green, decolourises fuschin colour.

H2S is evolved, turns lead acetate paper Hydrates, sulphides or sulphides in the

black, or cadmium acetate yellow. prescnce of water.

Cl2 is evolved, yellowish green gas, bleaches Unstable chlorides e.g., copper chlorides in the
litmus paper, turns KI – starch blue, presence of oxidising agents.
poisonous.

Br2 is evolved (reddish brown, turns Bromides in the presence of oxidising agents.
fluorescent paper red).

I2 is evolved, violet vapours condensing to Free iodine and certain iodides

black crystals Ammonium and mercury salts.
(b) A sublimate is formed As2O3 , Sb2O3
White sublimate Hg

Grey sublimate

Steel grey, garlic odour As

Yellow sublimate S, As2S3 HgI 2(Re d)

Action of heat on different compounds : Many inorganic salts decomposes on heating, liberating characteristic
gases. A few such reactions are as follows,

2HgO ∆→ 2Hg + O2 2Pb3O4 ∆ → 6PbO+ O2 2PbO2 ∆ → 2PbO+ O2
(Re d ) (Silvery deposit ) (Re d) (Yellow) (Brown)

CuCO3 ∆ → CuO + CO2 ZnO3 ∆ → ZnO + CO2 CuSO4 .5H2O ∆ → CuSO4 + 5H2O
(Green) Yellow (hot ) (Blue) (White)
(White) White (cold)

CuSO4 ∆ → CuO + SO3 2FeSO4 ∆ → Fe 2O3 + SO2 + SO3 2Ag 2O → 4 Ag + O2
2AgNO3 450°C → 2Ag + 2NO2 + O2 2Cu(NO3 )2 → 2CuO + 4 NO2 + O2
2Zn(NO3 )2 ∆ → 2ZnO + 4 NO2 + O2
(White) (Brown)

2Ag 2CO3 → 4 Ag + 2CO2 + O2 2Pb(NO3)2 → 2PbO + 4 NO2 + O2 (NH 4 )2 Cr2O7 → N 2 + Cr2O3 + 4 H 2O
(Orange) (Green)

2NaHCO3 → Na2CO3 + CO2 + H2O NH4 HCO3 → NH3 + CO2 + H2O CaCO3 → CaO + CO2
2NaNO3 → 2NaNO2 + O2 MgCO3 → MgO + CO2 2NH3 Redhot → N2 + 3H2
2Mg(NO3 )2 → 2MgO + 4 NO2 + O2 2Ca(NO3 )2 → 2CaO + 4 NO2 + O2 Al 2 (SO4 )3 Redhot → Al 2O3 + 3SO3
2CaSO4.2H2O → 2CaSO4.H2O+ 2H2O 2BeSO4 ∆ → 2BeO + 2SO2 + O2
2AlCl3.6H2O → Al2O3 + 6HCl + 9H2O
(Plaster of Paris)

2AgNO3 350oC → 2AgNO2 + O2 2MgSO4 ∆ → 2MgO + 2SO2 + O2 2ZnSO4 ∆ → 2ZnO + 2SO2 + O2

(COO)2 Sn ∆ → SnO + CO2 + CO CaC2O4 → CaCO3 + CO NH 4 NO2 → N 2 + 2H 2O

NH 4 NO3 → N 2O + 2H 2O 2KClO3 → 2KCl + 3O2 2FeCl 3 → 2FeCl 2 + Cl 2

Li2CO3 → Li2O + CO2 (COO)2 Fe → FeO + CO + CO2 2KMnO4 → K 2 MnO4 + MnO2 + O2

MgCl 2 . 6H 2O → HgCl 2 + Hg NH 4 Cl → NH 3 + HCl 2LiNO3 → Li2O + 2NO2 + 1 O2
2

Hg(NO3 )2 → Hg + 2NO2 + O2 2CuCl 2 ∆ → Cu2Cl 2 + Cl 2 2Co(NO3 )2 ∆ → 2CoO + 4 NO2 + O2

4K 2Cr2O7 → 4 K 2CrO4 + 2Cr2O3 + 3O2 2Mg(NH 4 )PO4 ∆ → Mg 2 P2O7 + H 2O + 2NH 3

2Zn(NH 4 )PO4 ∆ → Zn2 P2O7 + H 2O + 2NH 3 K 4 Fe(CN)6 ∆ → 4KCN + Fe + 2C + N 2

ZnCl 2 . 2H 2O ∆ → Zn(OH)Cl + HCl + H 2O 2(ZnCl 2 .H 2O) ∆ → Zn2OCl 2 + 2HCl + H 2O

2[FeCl 3 .6H 2O] ∆ → Fe 2O3 + 9H 2O + 6HCl 2ZnSO4 800oC → 2ZnO + 2SO2 + O2

Na 2 B4 O7 .10H 2O ∆ → Na 2 B4 O7 ∆ → 2NaBO2 + B2O3
(Glassy bead)

H 3 BO3 100oC → HBO2 160oC → H 2 B4 O7 Redhot → B2O3

ZnSO4 .7 H 2 O 70oC → ZnSO4 .6 H 2O 100oC → ZnSO4 .H 2O 450oC → ZnSO4
−H2O −5 H 2O

(iii) Flame test

Characteristic flame colour : Certain metals and their salts impart specific colours to Bunsen burner flame.

• Pb imparts pale greenish colour to the flame.
• Cu and Cu salts impart blue or green colour to the flame.

• Borates also impart green colour to the flame.
• Ba and its salts impart apple green colour to the flame.

• Sr imparts crimsen red colour to the flame.
• Ca imparts brick red colour to the flame.
• Na imparts yellow colour to the flame.
• K imparts pink-violet (Lilac) colour to the flame.
• Li imparts crimsen-red, Rb imparts violet and Cs imparts violet colours to the flame.
• Livid- blue flame is given by As, Sb and Bi.
(iv) Borax bead test : The transparent glassy bead (NaBO2 + B2O3 ) when heated with inorganic salt and
the colour produced gives some idea of cation present in it.

Colour of bead in oxidising flame Colour of bead in Basic radical present
reducing flame

Greenish when hot, blue in cold. Red and opaque Cu

Dark green in hot and cold Same Cr

Deep – blue Deep blue Co

Yellow when hot Green Fe

Violet in hot and cold Colourless Mn

Brown in cold Grey or black or Ni
opaque

Microcosmic salt bead test : Microcosmic salt, Na(NH4 )HPO4 .4H 2O is also used to identify certain
cations just like borax. When microcosmic salt is heated in a loop of platinum wire, a colourless transparent bead of
sodium metaphosphate is formed.

Na (NH 4 )HPO4 .4H 2O → Na(NH 4 )HPO4 + 4H 2O ; Na(NH 4 )HPO4 → NaPO3 + NH 3 + H 2O
Now NaPO3 reacts with metallic oxides to give coloured orthophosphates.

NaPO3 + CuO → NaCuPO4 (Blue); NaPO3 + CoO → NaCoPO4 (Blue); NaPO3 + Cr2O3 → NaPO3 .Cr2O3 (Green)
(v) Charcoal cavity test

(a) Compound fused in cavity directly

Nature and colour of bead Cation

Yellow, brittle bead Bi 3+
Yellow, soft bead which marks on paper Pb 2+
White, brittle Sb 3+
White yellow when hot ZnO
White garlic odour As 2 O3
Brown CdO
Grey metallic particles attracted by magnet Fe, Ni,CO

Maleable beads Ag and Sn (White),Cu (Red flakes)

(b) Compound mixed with Na2CO3 Crystalline

Sustance Salts, NaCl, KCl ; Substance Oxidising agents like ; Substance infusible, perform test (a)
Decrepitates deflagrates
NO3− , NO2− chlorates

(vi) Cobalt Nitrate test

Colour Composition Result
Blue residue CoO. Al2O3 Al
Green residue CoO. ZnO ZnO
Pink dirty residue CoO. MgO
Blue residue NaCoPO4 MgO
PO43− in absence of Al.

(2) Wet tests for acid radicals : Salt or mixture is treated with dil. H 2SO4 and also with conc. H 2SO4
separately and by observing the types of gases evolved. Confirmatory tests of anions are performed.

Observations with Dilute H 2SO4

Observations Acid Radical Confirmatory test

Brisk effervescence with CO32− (carbonate) Gas turns lime water milky but milkyness disappears on passing gas inexcess,
evolution of colourless
and odourless gas. Na2CO3 + H 2SO4 → Na2SO4 + H 2O + CO2 ;

Ca(OH)2 + CO2 → CaCO3 + H 2O ;
lime water milky

CaCO3 + H 2O + CO2 → Ca(HCO3 )2
soluble

Brown fumes NO − (Nitrite) Add KI and starch solution blue colour
2
2NaNO3 + H 2 SO4 → Na2 SO4 + 2HNO2 ;

HNO2 → NO (colourless);
2NO + O2 (air) → 2NO2 (brown);

2KI + H 2SO4 + 2NHO2 → K 2SO4 + 2H 2O + 2NO + I 2 ;
I 2 + starch → blue colour

Smell of rotten eggs S 2− (sulphide) Gas turn lead acetate paper black

(H 2 S smell) on heating Sodium carbonate extract (SE)* + sodium nitroprusside – purple colour,

Na2 S + H 2 SO4 → H 2 S + Na2 SO4 ;

H 2S + (CH 3 COO) 2 Pb → PbS + 2CH 3 COOH ;

(black)

Na2S + Na2[Fe(CN)5 NO] → Na4[Fe(CN)5 NOS]
sodium nitroprusside (purple)

Colourless gas with SO32− (sulphite) Gas turns acidified K 2Cr2O7 solution green [different from CO32− ] since
pungent smell of burning
sulphur gas also turns lime water milky

Na2SO3 + H 2SO4 ∆ → Na2SO4 + H 2O + SO2 ;

Cr2O72− + 3SO2 + 2H + → 2Cr 3+ + 3SO42 + H 2O ;
(green)

Ca(OH)2 + SO2 → CaSO3
(milky)

Solution gives smell of CH 3COO − Aq. Solution + neutral FeCl 3 → blood red colour
vinegar
(acetate) 3CH 3COONa + FeCl3 → Fe(CH 3COO)3 + 3NaCl

neutral (red)

White or yellowish white S 2 O 32− Aq. Solution + AgNO3 → white ppt. changing to black (viii) on warming ,
turbidity on warming Na2S2O3 + 2AgNO3 → Ag 2S2O3 + 2NaNO3 ;
(thiosulphate)
white ppt.

Ag 2S2O3 + H 2O → Ag 2S + H 2SO4
black ppt.

Observation with concentrated H 2SO4

Observation Acid Radical Confiramatory Test

Colourless pungent Cl − (chloride) Add MnO2 in the same test tube and heat–pale green Cl2
gas giving white fumes gas (i)
with aq. NH 4 OH
S.E.+ HNO3 + AgNO3 solution –white ppt. soluble in aq.
NH 3 (ii)

Reddish brown fumes Br − (bromide) Chromyl chloride test (iii)
Add Mn2O and heat –yellowish brown Br2 gas (iv)
Violet pungent vapours I − (iodide) S.E.+ HNO3 + AgNO3 solution –pale yellow ppt. partially
turning starch paper soluble aq. NH 3 (v)
blue. Layer test (vi)
S.E.+ HNO3 + AgNO3 → yellow ppt. insoluble in aq. NH 3
Brown pungent fumes NO − (nitrate) (vii)
intensified by the 3 Layer test (viii)
addition of Cu- Ring test (viii)
turnigs.
Acidified KMnO4 solution is decolorised (ix)
Colourless gases C 2 O 42− S.E. + CH 3COOH + CaCl 2 solution–white ppt. decolorising
acidified KMnO4 solution (x)
turning lime water (oxalate)

milky and burning with

blue flame.

Reactions

Chloride : (i) KCl + conc.H 2SO4 → KHSO4 + HCl ; HCl + NH 3 → NH 4 Cl
(white fumes)

4HCl + MnO2 ∆ → MnCl2 + Cl 2 + 2H 2O

(ii) KCl + AgNO3 → AgCl ↓ + KNO3 ; AgCl + aq. 2NH 3 →[Ag(NH 3 )2 ]Cl
white ppt. soluble

(iii) Chromyl- chloride test : Chloride + K2Cr2O7 (solid) + conc. H 2SO4 heat → reddish brown vapours of
chromyl-chloride (CrO2Cl2 ). Pass these vapours into NaOH, when yellow Na2CrO4 solution is formed. On adding
CH 3COOH and (CH 3COO)2 Pb, yellow ppt. of lead chromate (PbCrO4 ) is formed.

KCl + conc.H 2SO4 ∆ → KHSO4 + HCl ; K2Cr2O7 + 2H 2SO4 ∆ → 2KHSO4 + 2CrO3 + H 2O
conc.

CrO3 + 2HCl → CrO2Cl 2 + 2H 2O ; CrO2Cl 2 + 4 NaOH → Na2CrO4 + 2NaCl + 2H 2O

Na2CrO4 + (CH 3COO)2 Pb → PbCrO4 ↓+ 2CH 3COONa
yellow ppt.

Bromide : (iv) KBr + conc. H 2SO4 ∆ → KHSO4 + H ; 4HBr + MnO2 ∆ → Br2 + 2H 2O + MnBr2

(v) NaBr + AgNO3 → AgBr ↓ + NaNO3 ; AgBr + aq. 2NH 3 →[Ag(NH 3 )2 ]Br
pale yellow ppt. partially soluble

(vi) Layer Test : S.E. + Cl2 water + CHCl3 shake → yellowish orange colour in CHCl3 layer (CS2 or
CCl4 can be taken instead of CHCl3 ); 2NaBr + Cl2 → 2NaCl + Br2

orange yellow
(soluble in CHCl3 )

In case of I − , violet colour of I 2 in CHCl3 layer, 2NaI + Cl2 → 2NaCl + I2 (violet)

Iodide : (vii) KI + conc.H 2SO4 ∆ → KHSO4 + HI ; 2HI + H 2SO4 → I 2 + 2H 2O + SO2
( violet )

Nitrate : NaNO3 + H 2SO4 → NaHSO4 + HNO3

4HNO3 → 4 NO2 + O2 2H 2O ; Cu + 4HNO3 → Cu (NO3 )2 + 2NO2 + 2H 2O
brown fumes

(viii) Ring test : To water extract (all NO3− are water soluble) add freshly prepared FeSO4 solution and then

conc. H 2SO4 carefully by the side of the test- tube. A dark brown ring of [Fe (H 2O)5 NO]2+ SO42− at the interface
between the two liquids is formed.

2NaNO3 + H 2SO4 → 2NaHSO4 + 2HNO3 ;

2HNO3 + 6Fe SO4 + 3H 2SO4 → 3Fe 2 (SO4 )3 + 2NO + 4H 2O
[Fe (H 2O)6 ] SO4 + NO →[Fe(H 2O)5 NO]2+ SO42− + H 2O

Oxalate : Na2C2O4 + H 2SO4 → Na2SO4 + H 2O + CO + CO2

CO burns with blue flame and CO2 turns lime water milky.

(ix) 5C2O42− + 2MnO4− + 16H + →10 CO2 + 2Mn2+ + 8H 2O
(violet) colourless

(x) CaCl2 + NaC2O4 → CaC2O4 ↓+ 2NaCl CaC2O2 decolourises acidified KMnO4 .
white ppt.

Specific test in solution :

(i) Sulphate : S.E. add dil. (to decompose CO32− until reaction ceases). Add BaCl 2 solution. White ppt.
insoluble in conc. HNO3 , BaCl2 + NaSO4 → BaSO4 ↓+ 2NaCl

white ppt.

(ii) Borate : lgnite the mixture containing borate, conc. H 2SO4 . And ethanol in a china-dish with a burning
splinter –green edged flame of ethyl borate.

2Na3 BO3 + 3H 2 SO4 → 2H 3 BO3 + 3Na 2 SO4 ; H 3 BO3 + 3C2 H5OH ∆ → (C2 H5O)3 B + 3H 2O
(conc.) ethanol burns with green flame
(volatile)

In presence of Cu2+ , perform this test in a test tube since Cu2+ salts are not volatile.

(iii) S.E. + HNO3 + ammonium molybdate solution. Heat, yellow crystalline ppt. confirms

Na3PO4 + 12(NH4 )2 MoO4 + 24HNO3 ∆ → (NH4 )3 PO4 . 12MoO3 + 21NH4 NH3 + NaNO3 + 12H2O
yellow ppt.

Arsenic also gives this test. Hence presence of phosphate should also be checked after group II.

(iv) Fluoride : Sand +salt (F − ) +conc. H 2SO4 ; heat and bring a water wetted rod in contact with vapours

at the mouth of the test tube. A white deposit on the rod shows the presence to F −

NaF + H 2SO4 ∆ → NaHSO4 + HF ; SiO2 + 4HF ∆ → SiF4 + 2H 2O ; 3SiF4 4H 2O → 2H 2SiF6 + H 4 SiO4
white

(3) Wet tests for basic radicals : Analysis of Basic Radicals

Group Group reagent Basic radical Composition and colour of the precipitate

I Dilute HCl Ag + AgCl : white  Chloride

Pb 2+ PbCl 2 : white  insouble
Hg 2+
HgCl 2 : white  cold dilute HCl

II H 2S in presence of Hg 2+ HgS : black 

dilute HCl Pb 2+ PbS : black 

Bi 3+ Bi 2S3 : black 
CuS : black 
Cu 2+  Sulphides
CdS : yellow  insoluble in
Cd 2+
As 2 S3 : yellow  dilute HCl

As 3+ Sb 2S3 : orange

Sb 3+ SnS : brown 

Sn 2+ SnS2 : yellow 

III NH 4 OH in presence of Fe 3+ Fe(OH)3 : 
reddish brown  Hydroxides are insoluble in
NH 4 Cl Cr 3+ Cr(OH)3 : green NH 4 OH

Al 3+ Al(OH3 : white 

IV H 2S in presence of Zn 2+ ZnS : 
NH 4 OH Mn 2+
Co 2+ greenish white  Sulphides are insoluble in
Ni 2+ MnS : buff  NH 4 OH
CoS : black 


NiS : black 

V (NH 4 )2 CO3 in Ba 2+ BaCO3 : white
presence of Sr 2+ 
Ca 2+ SrCO 3 : white  Carbonates are insoluble
NH 4 OH
CaCO3 : white

VI NaHPO4 Mg 2+ Mg (NH 4 )PO4 : White

VII NaOH NH + Ammonia gas is evolved
4

Chemical reactions involved in the tests of basic radicals

Group I : When dil. HCl is added to original solution, insoluble chlorides of lead, silver mercurous mercury
are precipitated.

Pb(NH 3 )2 + 2HCl → PbCl 2 + 2HNO3 ; AgNO3 + HCl → AgCl + HNO3

Hg(NO3 )2 + 2HCl → HgCl 2 + 2HNO3
Pb2+ (lead)
(i) PbCl2 is soluble in hot water and on cooling white crystals are again formed.

(ii) The solution of PbCl2 gives a yellow precipitate with potassium chromate solution which is insoluble in
acetic acid but soluble in sodium hydroxide.

PbCl2 + K2CrO4 → PbCrO4 + 2KCl ; PbCrO4 + 4 NaOH → Na2 PbO2 + Na2CrO4 + 2H 2O
yellow ppt.

(iii) The solution of PbCl2 forms a yellow precipitate with potassium iodide solution.

PbCl2 + 2KI → PbI 2 + 2KCl
Yellow ppt.

(iv) White precipitate of lead sulphate is formed with dilute H 2SO4 . The precipitate is soluble in ammonium
acetate, PbCl 2 + H 2SO4 → PbSO4 + 2HCl ; PbSO4 + 2CH 3COONH 4 → Pb(CH 3COO)2 + (NH 4 )2 SO4

Ag+(silver)

(i) AgCl dissolves in ammonium hydroxide, AgCl + 2NH 4 OH → Ag(NH 3 )2 Cl + 2H 2O
Diammine silver (I)
chloride

(ii) On adding dilute HNO3 to the above solution, white precipitate is again obtained

Ag(NH 3 )2 Cl + 2HNO3 → AgCl + 2NH 4 NO3
White ppt.

(iii) On adding KI to the complex solution, yellow precipitate is obtained.

Ag(NH 3 )2 Cl + KI → AgI + KCl + 2NH 3

Hg 2+ (mercurous)
2

(i) Hg 2Cl 2 turns black with NH 4 OH , Hg 2Cl 2 + 2NH 4 OH →Hg+Hg(NH2)Cl + NH 4 Cl + 2H 2O

Black

(ii) The black residue dissolves in aqua-regia forming mercuric chloride.

3HCl + HNO3 → NOCl + 2H 2O + 2Cl ; 2Hg(NH 2 )Cl + 6Cl → 2HgCl 2 + 4HCl + N 2

Hg + 2Cl → HgCl2
(iii) The solution of HgCl2` forms white or slate-coloured precipitate with stannous chloride.

2HgCl2 + SnCl2 → Hg 2Cl2 + SnCl4 ; Hg 2Cl2 + SnCl2 → 2Hg + SnCl4
white ppt. Grey ppt.

(iv) The solution of HgCl2 with copper turning forms a grey deposit.

HgCl2 + Cu → Hg + CuCl2
Grey ppt.

Group II : When hydrogen sulphide is passed in acidified solution, the radicals of second group are
precipitated as sulphides. The precipitate is treated with yellow ammonium sulphide. The sulphides of IIB are first
oxidised to higher sulphides which then dissolve to form thio-compounds.

Ag 2S3 + 2(NH 4 )2 S2 → 2(NH 4 )2 S + As 2S5 ; Sb2S3 + 2(NH 4 )2 S2 → 2(NH 4 )2 S + Sb2S5

SnS + (NH 4 )2 S2 →(NH 4 )2 S + SnS2

As2S5 + 3(NH 4 )S → 2(NH 4 )3 AsS4 ; Sb2S5 + 3(NH 4 )2 S → 2(NH 4 )2 SbS4 ; SnS2 + (NH 4 )2 S →(NH 4 )2 SnS3

Ammonium Ammonium Ammonium

thioarsenate thioantimonate thiostannate

All the three are soluble.

In case, the precipitate does not dissolve in yellow ammonium sulphide, it may be either HgS or PbS or
Bi2S3 or CuS or CdS. The precipitate is heated with dilute HNO3 . Except HgS , all other sulphides of IIA are
soluble.

3PbS + 8HNO3 → 3Pb(NO3 )2 + 2NO + 3S + 4H 2O ; Bi2 S3 + 8HNO3 → 2Bi(NO3 )3 + 2NO + 3S + 4H 2O
3CuS + 8HNO3 → 3Cu(NO3 )2 + 2NO + 3S + 4H 2O ;
3CdS + 8HNO3 → 3Cd(NO3 )2 + 2NO + 3S + 4H 2O
Hg2+ (mercuric)

HgS is dissolved in aqua-regia, 3HgS + 2HNO3 + 6HCl → 3HgCl 2 + 3S + 2NO + 4H 2O
The solution is divided into two parts:
Part I : Stannous chloride solution reduces HgCl 2 first into white Hg 2Cl2 and then to grey metallic mercury.
Part II : Copper displaces Hg from HgCl 2 which gets coated on copper turnings as a shining deposit.
Pb2+ (lead)
In case the sulphide dissolves in dilute HNO3 , a small part of the solution is taken. Dilute H 2SO4 is added. If
lead is present, a white precipitate of lead sulphate appears, Pb(NO3 )2 + H 2SO4 → PbSO4 + 2HNO3

(White ppt.)

In absence of lead, the remaining solution is made alkaline by the addition of excess of NH 4 OH.Bismuth
forms a white precipitat of Bi(OH)3 , copper forms a deep blue coloured solution while cadmium forms a colourless
soluble complex,

Bi(NO3 )3 + 3NH 4 OH → Bi(OH)3 + 3NH 4 NO3 ;
White ppt.

Cu(NO3 )2 + 4 NH 4 OH →[Cu(NH 3 )4 ](NO3 )2 + 4 H 2O; Cd(NO3 )2 + 4 NH 4 OH →[Cd (NH 3 )4 ](NO3 )2 + 4H 2O

Tetrammine cupric nitrate Tetrammine cadmium nitrate

(deep blue solution) (colourless solution)

Bi3+ (bismuth) : The precipitate dissolves in dilute HCl, Bi (OH)3 + 3HCl → BiCl3 + 3H 2O
Part I : Addition of excess of water to BiCl3 solution gives a white precipitate due to hydrolysis.

BiCl3 + H 2O → BiOCl + 2HCl
Bismuth Oxychloride (White ppt.)

Part II : The solution of BiCl3 is treated with sodium stannite when a black precipitate of metallic bismuth is

formed, 2BiCl3 + 3Na2SnO2 + 6NaOH → 3Na2SnO3 + 2Bi + 6NaCl + 3H 2O
Sod. stannite Sod. stannate

Cu2+ (copper) : Blue coloured solution is acidified with acetic acid. When potassium ferrocyanide is added a

chocolate coloured precipitate is formed, Cu(NH 3 )4 (NO3 )2 + 4CH 3COOH → Cu(NO3 )2 + 4CH 4 COONH 4

2Cu(NO3 )2 + K4 [Fe(CN)6 ] → Cu2[Fe(CN)6 + 4KNO3
Chocolate ppt.

Cu2+ (cadmium) : H 2S is passed through colourless solution. The appearance of yellow precipitate
confirms the presence of cadmium, Cd (NH 3 )4 (NO3 )2 + H 2S → CdS + 2NH 4 NO3 + NH 3

Yellow ppt.

Group IIB : In case the precipitate dissolves in yellow ammonium sulphide, the tests of the radicals arsenic,
antimony and tin are performed. The sulphide is treated with concentrated hydrochloric acid. Antimony and tin
sulphide dissolve while arsenic sulphide remains insoluble.

As3+ (arsenic) : The insoluble sulphide is treated with concentrated nitric acid which is then heated with
ammonium molybdate. Yellow precipitate of ammonium arsenomolybdate is formed.

As2S5 + 10HNO3 → 2H 3 AsO4 + 10NO2 + 2H 2O + 5S
Arsenic acid

H 3 AsO4 + 12(NH 4 )2 MoO4 + 21HNO3 →(NH 4 )3 AsO4 . 12MoO3 + 21NH 4 NO3 + 12H 2O
Yellow ppt.

Sn2+ or Sn4+ (tin) : Solution of sulphide in concentrated HCl is reduced with iron fillings or granulated zinc.

SnS2 + 4HCl → SnCl4 2H 2S ; SnCl4 + Fe → SnCl2 + FeCl 4
White ppt. Grey

HgCl2 solution is added to above solution which gives first a white precipitate that turns to grey.

2HgCl 2 + SnCl2 → HgCl 2 + SnCl4 ; Hg 2Cl2 + SnCl 2 → 2Hg + SnCl4
White ppt. Grey

Sb2+ (antimony) : Filtrate of sulphide in concentrated HCl is divided into two parts.

Part I : On dilution with excess of water, a white precipitate of antimony oxychloride is obtained.

SbCl3 + H 2O → SbOCl + 2HCl
White ppt.

Part II : H 2S is circulated. Orange precipitate is formed, 2SbCl3 + 3H 2S → Sb2S3 + 6HCl
Orange ppt.

Group III : Hydroxides are precipitated on addition of excess of ammonium hydroxide in presence of
ammonium chloride.

AlCl3 + 3NH 4 OH → Al(OH)3 + 3NH 4 Cl ; CrCl3 + 3NH 4 OH → Cr(OH)3 + 3NH 4 Cl
Gelatinous ppt. Green ppt.

FeCl3 + 3NH 4 OH → Fe(OH)3 + 3NH 4 Cl
Brownish red ppt.

Fe3+ (iron) : The brownish red precipitate dissolves in dilute HCl. The solution is divided into two parts.
Part I : K4 [Fe(CN)6 ] solution is added which forms deep blue solution or precipitate.

Fe(OH)3 + 3HCl → FeCl 3 + 3H 2O ; 4FeCl3 + 3K4 [Fe(CN)6 ] → Fe4 [Fe(CN)6 ]3 + 12KCl
Prussian blue

Part II : Addition of potassium thiocyanate solution gives a blood red colouration.

FeCl3 + 3KCNS → Fe(CNS)3 + 3KCl
Blood red colour

Cr3+(chromium) : The green precipitate is fused with fusion mixture (Na2CO3 + KNO3 ). The fused product
is extracted with water or the precipitate is heated with NaOH and bromine water.

2Cr(OH)3 + 3KNO + 2Na2CO3 → 2Na2CrO4 + 3KNO2 + 2CO2 + 3H 2O

or 2NaOH + Br2 → NaBrO4 + NaBr + H 2O ; NaBrO → NaBr + [O]

2Cr(OH)3 + 4 NaOH + 3[O] → 2NaCrO4 + 5H 2O
The solution thus obtained contains sodium chromate. The solution is acidified with acetic acid and treated
with lead acetate solution. A yellow precipitate appears.

Na2CrO4 + Pb(CH 3COO)2 → PbCrO4 + 2CH 3COONa
Yellow ppt.

Al3+(aluminium) : The gelatinous precipitate dissolves in NaOH , Al(OH)3 + NaOH → NaAlO2 + 2H 2O
Soluble

The solution is boiled with ammonium chloride when Al(OH)3 is again formed.

NaAl 2 + NH 4 Cl + H 2O → Al(OH)3 + NaCl + NH 3
Group IV : On passing H 2S through the filtrate of the third group, sulphides of fourth group are
precipitated. NiS and CoS are black and insoluble in concentrated HCl while MnS (buff coloured), ZnS (colourless)
are soluble in conc. HCl.

Zn2+ (zinc) : The sulphide dissolves in HCl. ZnS + 2HCl → ZnCl2 + H 2S
When the solution is treated with NaOH, first a white precipitate appears which dissolves in excess of NaOH

ZnCl2 + 2NaOH → Zn(OH)2 + 2NaCl ; Zn(OH)2 + 2NaOH → Na2 ZnO2 + 2H 2O
White ppt. (Soluble)

On passing H 2S , white precipitate of zinc sulphide is formed Na2 ZnO2 + H2S → ZnS + 2NaOH
White ppt.

Mn2+ (manganese) : Manganese sulphide dissolves in HCl MnS + 2HCl → MnCl2 + H 2S
On heating the solution with NaOH and Br2 -water, manganese dissolne gets precipitated.

MnCl2 + 2NaOH → Mn(OH)2 + 2NaCl ; Mn(OH)2 + O → MnO2 H 2O
The precipitate is treated with excess of nitric acid and PbO2 or Pb3O4 (red lead). The contents are heated.
The formation of permanganic acid imparts pink colour to the supernatant liquid.

2MnO2 + 4HNO3 → 2Mn(NO3 )2 + 2H 2O + O2

2Mn(NO3 )2 + 5Pb3O4 + 26HNO3 → 2HMnO4 + 15 Pb (NO3 )2 + 12 H 2O
Permanganic acid (pink)

Note :  The above test fails in presence of HCl.

Ni2+ (nickel) and Co2+ (cobalt)
The black precipitate is dissolved in aqua- regia.

3Nis + 6HCl + 2HNO3 → 2NiCl2 + 2NO + 3S + 2H 2O

3CoS + 6HCl + 2HNO3 → 3CoCl2 + 2NO + 3S + 4H 2O

The solution is evaporated to dryness and residue extracted with dilute HCl. It is divided into three parts.

Part I : Add NH 4 OH (excess) and dimethyl glyoxime. A rosy red precipitate appears, if nickel is present,

OH O

| ↑

CH 3 − C = NOH CH 3 − C = N N = C − CH 3
Ni | + 2NH 4 Cl + 2H 2O
NiCl2 + 2 | + 2NH 4 OH → |
N = C − CH 3
CH 3 − C = NOH CH 3 − C = N |

↓ OH

O

Part II : Add CH 3COOH in excess and KNO2 . The appearance of yellow precipite confirms the presence of
cobalt.

KNO2 + CH 3COOH → CH 3COOK + HNO2 ; CoCl 2 + 2KNO2 → Co(NO2 )2 + 2KCl

Co(NO2 )2 + 2HNO2 → Co(NO2 )3 + NO + H 2O ; Co(NO2 )3 + 3KNO2 → K 3 [Co(NO2 )6 ]

Part III : Solution containing either nickel or cobalt is treated with NaHCO3 and bromine water. Appearance
of apple green colour is observed, the solution is heated when black precipited is formed, which shows the presence
of nickel, CoCl2 + 2NaHCO3 → Co(HCO3 )2 + 2NaCl

Co(HCO3 )2 + 4 NaHCO3 → Na4 Co(CO3 )3 + 3H 2O + 3CO2 ; Br2 + H 2O → 2HBr + O

2Na4 Co(CO3 )3 + H 2O + O → 2Na3Co(CO3 )3 + 2NaOH
sod. cobalti carbonate
(Green colouration)

NiCl2 + 2NaHCO3 → NiCO3 + 2NaCl + H 2O + CO2 ; 2NiCO3 + O → Ni2O3 + 2CO2
(Black)

Group V : Ammonium carbonate precipitates V group radicals in the form of carbonates are soluble in acetic
acid.

BaCO3 + 2CH 3COOH →(CH 3COO)2 Ba + CO2 + H 2O

SrCO3 + 2CH 3COOH →(CH 3COO)2 Sr + CO2 + H 2O

CaCO3 + 2CH 3COOH →(CH 3COO)2 Ca + CO2 + H 2O
Ba2+ (barium) : Barium chromate is insoluble and precipitated by the addition of potassium chromate
solution, Ba(CH 3COO)2 + K 2CrO4 → BaCrO4 + 2CH 3COOK
Sr2+ (Strontium) : Strontium sulphate is insoluble and precipitated by the addition of ammonium sulphate
solution, Sr(CH 3COO)2 + (NH 4 )2 SO4 → SrSO4 + 2CH 3COONH4

White ppt.

Ca2+ (calcium) : Calcium oxalate is insoluble and precipitated by the addition of ammonium oxalate.

Ca(CH 3COO)2 + (NH 4 )2 C2O4 → CaC2O4 + 2CH 3COONH4
White ppt.

Group VI : In the filtrate of V group, some quantity of ammonium oxalate is added as to remove Ba, Ca and
Sr completely from the solution. The clear solution is concentrated and made alkaline with NH 4 OH. Disodium
hydrogen phosphate is now added, a white precipitate is formed.

MgCl2 + Na2 HPO4 + NH 4 OH → Mg(NH 4 )PO4 + 2NaCl + H 2O
Megnesium ammonium phosphate

(White ppt.)

NH + (ammonium) : The substance (salt or mixture) when heated with NaOH solution evolves ammonia.
4

NH 2Cl + NaOH → NaCl + NH 3 + H 2O

When a rod dipped in HCl is brought on the mouth of the test tube, white fumes of ammonia chloride are
formed, NH 3 + HCl → NH 4 Cl

White fumes

To the aqueous solution of ammonium salt when Nessler’s reagents is added, brown coloured precipitate is
formed.

NH 2
Hg

2K2 HgI 4 + NH 4 Cl + 4KOH → O + 7KI + KCl + 3H 2O
Hg

I

Iodide of Millon's base
(Brown ppt.)