Vidyamandir Classes
Supplementary Booklet - 8/Chemistry [CBSE]
d & f-Block Elements
8.1
Write down the electronic configuration of:
(i)
Cr3+
(iii)
Cu+
3+
(ii)
Pm
(iv)
Ce4+
NCERT Questions
(v)
(vi)
Co2 +
Lu2+
(vii)
(viii)
Mn2+
Th4+
8.2
Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their + 3 state?
8.3
Explain briefly how + 2 state becomes more stable in the first half of the first row transition elements with increasing
atomic number?
8.4
To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition
elements? Illustrate your answer with examples.
8.5
What may be the stable oxidation state of the transition element with the following d electron configurations in the
ground state of their atoms : 3d3. 3d5 . 3d8 and 3d4 ?
8.6
Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal
to its group number.
8.7
What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
8.8
What are the characteristics of the transition elements and why are they called transition elements? Which of the
d-block elements may not be regarded as the transition elements?
8.9
In what way is the electronic configuration of the transition elements different from that of the non transition elements?
8.10
What are the different oxidation states exhibited by the lanthanoids?
8.11
Explain giving reasons:
(i)
Transition metals and many of their compounds show paramagnetic behaviour.
(ii)
The enthalpies of atomization of the transition metals are high.
(iii)
The transition metals generally form coloured compounds.
(iv)
Transition metals and their many compounds act as good catalyst.
8.12
What are interstitial compounds? Why are such compounds well known for transition metals?
8.13
How is the variability in oxidation states of transition metals different from that of the non transition metals? Illustrate
with examples.
8.14
Describe the preparation of potassium dichromate from iron chromite ore.
What is the effect of increasing pH on a solution of potassium dichromate?
8.15
Describe the oxidising action of potassium dichromate and write the ionic equation for its reaction with:
(i)
iodide
(ii)
iron(II) solution and
(iii)
H 2S
Supplementary Booklet – 8
1
Chemistry
Vidyamandir Classes
8.16
Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i)
iron(II) ions (ii) SO2 and (iii) oxalic acid? Write the ionic equations for the reactions.
8.17
For M2+/M and M3+/M2+ system the E values for some metals are as follows:
Cr2+/Cr
0.9V
Cr3+/Cr2+
0.4 V
Mn 2  / Mn
2+
Fe /Fe
8.18
1.2 V
0.4 V
Mn3+/Mn2+
3+
Fe /Fe
2+
+ 1.5 V
+ 0.8 V
Use this data to comment upon:
(i)
the stability of Fe3+ in acid solution as compared to that of Cr3+ or Mn3+ and
(ii)
the ease with which iron can be oxidized as compared to a similar process for either chromium or manganese
metal.
Predict which of the following will be coloured in aqueous solution? Ti3+, V3+, Cu+, Sc3+, Mn2+, Fe3+ and Co2+. Give
reasons for each.
8.19
Compare the stability of +2 oxidation state for the elements of the first transition series.
8.20
Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
8.21
(i)
electronic configuration
(ii)
oxidation state
(iii)
atomic and ionic sizes and
(iv)
chemical reactivity.
How would you account for the following:
(i)
Of the d4 species, Cr2+ is strongly reducing while manganese (III) is strongly oxidising.
(ii)
Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.
(iii)
The d1 configuration is very unstable in ions.
8.22
What is meant by ‘disproportionation’? Give two examples of disproportionation reaction in aqueous solution.
8.23.
Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?
8.24
Calculate the number of unpaired electrons in the following gaseous ions: Mn3+, Cr3+, V3+ and Ti3+. Which one of these
is the most stable in aqueous solution?
8.25
Give examples and suggest reasons for the following features of the transition metal chemistry:
8.26
(i)
The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.
(ii)
A transition metal exhibits highest oxidation state in oxides and fluorides.
(iii)
The highest oxidation state is exhibited in oxoanions of a metal.
Indicate the steps in the preparation of :
(i)
K2Cr2O7 from chromite ore
(ii)
KMnO4 from pyrolusite ore
8.27
What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.
8.28
What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner
transition elements: 29, 59, 74, 95, 102, 104.
8.29
The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some
examples from the oxidation state of these elements.
8.30
Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on
the possible oxidation state of this element.
Supplementary Booklet – 8
2
Chemistry
Vidyamandir Classes
8.31
Use Hund’s rule to drive the electronic configuration of Ce3+ ion, and calculate its magnetic moment on the basis of
‘spin-only’ formula.
8.32
Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation
states. Try to correlate this type of behaviour with the electronic configurations of these elements.
8.33
Compare the chemistry of the actinoids with that of lanthanoids with reference to:
(i)
electronic configuration
(ii)
oxidation states and
(iii)
chemical reactivity
8.34
Write the electronic configurations of the elements with the atomic numbers 61, 91, 101 and 109.
8.35
Compare the general characteristics of the first series of the transition metals with those of the second and third series
metals in the respective vertical columns. Give special emphasis on the following points:
(i)
electronic configurations (ii)
oxidation states
(iii)
ionisation enthalpies
(iv)
atomic sizes
8.36
Write down the number of 3d electrons in each of the following ions: Ti2+, V2+, Cr3+, Mn2+, Fe2+, Fe3+, Co2+, Ni2+ and
Cu2+. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).
8.37
Comment on the statement that elements of the first transition series possess many properties different from those of
heavier transition elements.
8.38
What can be inferred from the magnetic moment values of the following complex species?
Example
Magnetic Moment (BM)
K4[Mn (CN)6]
2.2
2+
[Fe (H2O)6]
5.3
K2[MnCl4]
5.9
Hints and Answers :
8.1
(i)
Cr 3 
 3d3
(ii)
Pm3 
 4f 4
(iii)
Cu  
 3d10
(iv)
Ce 4  
 4f 0 (Ce  4f 3 5d 2 6s 2 )
(v)
Co 2 
 3d 7
(vi)
Lu 2 
 4f 14 5d1
(vii)
Mn 2 
 3d5
(viii)
Th 4  
 6d 0 7s0 (Th 
 6d 2 7s 2 )
8.2
Mn2+ (3d5) having half – filled stable configuration than Fe2+(3d6)
8.5
The stable oxidation states are predicted on the basis of electronic configuration on that up to Mn, the max. O.S. of
stability correspond to sum of S & and d electrons. After Mn, there is decrease in the stability of higher oxidation states.
8.6
(i)
3d3 : Stable oxidation state will be +5 due to outer electronic configuration 3d 34s2
(ii)
3d5 : Stable oxidation state will be +7 due to outer electronic configuration 3d 54s2
(iii)
3d8 : Stable oxidation state will be +2 due to outer electronic configuration 3d 84s2
(iv)
3d4 : Stable oxidation state will be +6 due to outer electronic configuration 3d 44s2

Though d4 species does not exist but as it is given it must be chromium (Cr)
MnO 4 : oxidation state of Mn   7 ; Group No.  7
CrO 4 2  : oxidation state of Cr   6 ; Group No.  6
Supplementary Booklet – 8
3
Chemistry
Vidyamandir Classes
8.7
The gradual and steady decrease in atomic/ionic radii along the lanthanoid series i.e. from La to Lu is called Lanthanoid
contraction. It can be explained on the basis of poor shielding effect of f-orbitals due to which the effective nuclear
charge increases regularly and hence size decreases regularly.
Consequences of Lanthanoid Contraction :
Due to lanthanoid contraction 2nd, 3rd T.S exhibit similar atomic radius
2nd
Zr
3
Hf
Series
rd
Serise
Nb 

 samesize
Ta 

8.8
A transition element is defined as the one which has incompletely filled d-orbitals in its ground state or in any one of its
oxidation state.

Zn, Cd, Hg are not regarded as transition elements because they have fully – filled d10 configuration in ground
state as well as in their most stable oxidation state.
8.9
E.C. of transition elements
(n – 1)d1–10 ns1–2

In transition element ultimate as well as penultimate orbit is incompletely filled whereas in representative
element only ultimate orbit is incompletely filled.
8.10
In lanthanoids the most predominant and stable O.S is +3. For certain elements +2, +4 also exist.
8.11
(i)
(iii)
(iv)
Due to presence of unpaired electrons
(ii)
Due to high interatomic attraction
Transition elements are coloured because when an electron from a lower energy d-orbital is excited to a higher
energy d-orbitals with the d sub-shell then the energy of excitation lies in the visible region and the
corresponding complementary colour is observed.
The transition element shows catalytic activity which can be attributed to their ability to show variable
oxidation state and their tendency to form complexes.
8.12
Transition elements are able to form interstitial compounds when small atoms like H, C, or N are trapped inside the
crystal lattice of metals. Due to similar size with these elements transition elements form interstitial compounds.
8.13
The transition elements have a tendency to show variable oxidation state because of penultimate d-electrons to enter
into chemical bond formation in addition to ns electrons as the energy of (n-1)d and ns orbitals are comparable.
8.14
Preparation of potassium dichromate
Step I : 4FeCrO4  8Na 2 CO3  7O 2 
 8Na 2 CrO 4  2Fe 2 O3  8CO 2
(yellow)
Step II : 2Na 2 CrO 4  2H  
 Na 2 Cr2 O7  2Na   H 2 O
chromate
(yellow)
Dichromate
(Orange)
Step III : Na 2 Cr2 O7  2KCl 
 K 2 Cr2 O7  2NaCl
(Orange crystals)

Chromate and dichromate ions are inter convertible depending upon the pH of the solution
(a)
In acidic solution 2CrO 24  2H  
 Cr2 O72   H 2 O
(yellow)
(b)
(Orange)
In basic solution Cr2 O72   2OH  
 2CrO 42  H 2 O
(Orange)
Supplementary Booklet – 8
(yellow)
4
Chemistry
Vidyamandir Classes
8.15
K2Cr2O7 is a very good oxidizing agent in acidic medium and is reduced to Cr3+
Cr2 O72   14H   6e  
 Cr 3  7H 2 O
It oxidizes iodides to iodine, sulphides to sulphur, ferrous to ferric
6I 
 3I2  6e
3H 2S 
 6H  3S  6e 
6Fe2 
 6Fe3  6e 
8.16
Preparation of KMnO4.
Step – I :
2MnO 2  4KOH  O 2 
 2K 2 MnO4  2H 2 O
Step – II :
3MnO 24  4H  
 2MnO 4  MnO 2  2H 2 O
 potassiumgreenmanganate
 permagnate
purple 
Oxidizing property
(i)
Iodide is oxidized to iodine 10I   2MnO 4  6H  
 2Mn 2   8H 2 O  5I 2
(ii)
Sulpher dioxide to sulphate 2H 2 O  5SO 2  2MnO 4 
 5SO 42   2Mn 2  4H 
(iii)
Oxalate or oxalic acid is oxidized to CO2.
5 C2 O 24   2MnO 24   16H  
 2Mn 2   8H 2 O  10CO 2
8.18
The ions which have incompletely filled d-orbitals will be coloured and those with fully filled or empty d-orbitals will
be colourless.
Ti3+, V3+, Mn2+, Fe3+, Co2+ are coloured because they have incompletely filled d-orbitals.
d0 d10 will be colourless.
8.21
(a)
Ti3 
 3d1
;
v3 
 3d 2
Cu  
 3d10
;
Sc3 
 3d 0
Mn 2 
 3d 5 ;
Fe3 
 3d5
;
Co 2 
 3d 7
E values for Cr3+/Cr2+ is negative  0.41v  and for Mn3+/Mn2+ is positive (+ 1.57v). Thus, Cr2+ can undergo
oxidation and therefore, is reducing agent. On the other hand, Mn (III) can undergo reduction, and therefore,
acts as an oxidsing agent.
(b)
In the presence of complexing agents, cobalt gets oxidised from +2 to +3 state because Co (III) is more stable
than Co(II)
(c)
After loss of ns e.s, d1 electron can easily be lost to give a stable configuration. Therefore, the elements having
d1 configuration are either reducing or undergo disproportionation.
8.22
A reaction in which same atom is oxidised and reduced at the same time is called as the disproportionation reaction.
Example :
(i)
3CrO34  8H  
 2CrO 24   Cr 3  4H 2 O
(ii)
3MnO24  4H 
 2 MnO4  MnO2  2H 2 O
Supplementary Booklet – 8
5
Chemistry
Vidyamandir Classes
8.23
Electronic configuration : [Ar] 3d10 4s1
Copper: Atomic No. 29
It can easily lose one (4s1) electron to attain a stable 3d10 configuration.
8.24
Mn3+ : {Ar]3d4
3+
V : [Ar]3d
2
No. of unpaired electrons = 4 ;
No. of unpaired electrons = 2 ;
Cr3+ : [Ar]3d3
3+
Ti : [Ar]3d
1
No. of unpaired electrons = 3
No. of unpaired electrons = 1
3+

Out of these, Cr is most stable in aqueous solution.
+
Cu and Sc3+ are colourless because Cu+ has completely filled 3d10 orbitals while Sc3+ has empty d-orbital (3d0).
8.25
(a)
The lower oxide of transition metal having low oxidation state is basic whereas the higher oxide of transition
metal having high oxidation state is amphoteric or acidic.
Example : MnO (II) is basic while Mn2O7 (VII) is acidic. Similarly, CrO (II) is basic while Cr2O3 (III) is amphoteric
Note : Higher is the oxidation state of the metal , more is the acidic character of the oxide.
(b)
A transition metal exhibits highest oxidation state in oxides and fluorides because oxygen and F are of small
size and high electronegativity and hence can readily oxidise the metals.
(c)
The highest oxidation state is exhibited in oxoanions of a metal because of high electronegativity of oxygen
and its high oxidising property.
Example :
8.27
Cr in Cr2 O72  : +6 and
Mn is MnO 4 : +7
The transition elements are able to form alloy i.e. homogeneous mixture of two or more metal atoms because of their
similar radii. Due to similar radii they are able to mutually substitute one another in their crystalline lattice.
MISCH METAL : It is an alloys of lanthanoid metal ( 95%) and iron (5%) with traces of S, C, Ca and Al
It is used in Mg-based alloy to produce bullets, shell and lighter flint.
8.28
The inner transition elements are those in which the last electron enter f-subshells. The f-block elements are known as
the inner transition elements in which the 4f and 5f orbitals are progressively filled.
Elements having atomic number 59, 95 and 102 are inner transition elements.
8.29
The actinoids show a wider range of O.S. which is partly attributed to the fact that the 5f, 6d and 7s levels are of
comparable energies. The actinoids show in general +3 oxidation state.
8.30
Lawrencium [Lr] is last element of actinoids
Lr = 5f 14 6d1 7S2
103
8.31
8.34
Ce (Z = 58) : [Xe] 4f15d16s2
Ce3+ : 4f1
No. of unpaired electrons = 1
µs  n(n  2 
11  2   3  1.73 BM
Promethium (Pm)
Z = 61 :
[Xe] 4f56s2
Protactinium(Pa)
Z = 91
[Rn] 5f26d17s2
:
Mandelevium (Md) Z = 101 :
Z = 109
Supplementary Booklet – 8
:
[Rn] 5f137s2
[Rn] 5f146c77s2
6
Chemistry
Vidyamandir Classes
8.36
Note : Water (H2O) forms high spin or outer orbital complexes.
8.38
Mn2+ has 4s0 3d5 electronic configuration  
n  n  2   1  3  1.732 B.M.
K4[Mn (CN)6] is low spin complex with one unpaired electron.
[either there can be 1 unpaired electron or 3 or 5
if
n = 1,   1.73 , n = 3,   3.9
Supplementary Booklet – 8
7
Chemistry
Vidyamandir Classes
So, the nearest possible value is 1.73]
[Fe (H 2 O) 6 ]2 ; Fe 2 : 4s0 3d 6 ,  
4 6 
24  4.9 B.M.
It has four unpaired electrons because H2O is weak field ligand. It is high spin complex.
In K2[MnCl4], Mn2+ has 5 unpaired electrons,   5  7  35  5.9 B.M.
It is because Cl is weak field ligand therefore, it is high spin complex.
Definitions :
1.
Minerals: Minerals are naturally occurring chemical substances in the earth’s crust obtainable by mining.
2.
Ores: Out of many minerals in which a metal may be found, only a few are viable to be used as sources for the
extraction of that metal. Such minerals are known as ores.
3.
Gangue: An ore is usually contaminated with earthly or undesired materials known as gangue.
4.
Metallurgy: The entire scientific and technological process used for isolation of the metal from its ores is known as
metallurgy.
5.
Pyrometallurgy: It is a branch of extractive metallurgy which consists of the thermal treatment of minerals and
metallurgical ores and concentrates to bring about physical and chemical transformations in the materials to enable
recovery of valuable metals. Examples of elements extracted by pyrometallurgical processes include the oxides of less
reactive elements like Fe, Cu, Zn, Chromium, Tin, Manganese.
6.
Hydrometallurgy: It is a method for obtaining metals from their ores. It is a technique within the field of
extractive metallurgy involving the use of aqueous chemistry for the recovery of metals from ores, concentrates, and
recycled or residual materials. Hydrometallurgy is typically divided into three general areas:

Leaching

Solution concentration and purification

Metal recovery

The extraction and isolation of metals from ores involve the following major steps:
1. Concentration of the ore: removal of impurities
2. Isolation of the metal from its concentrated ore:
 Conversion of ore to its oxides
 Reduction of metal oxide to metal
3.
Purification of the Metal :
Occurrence of Metals :
Principal Ores of Some Important Metals
Supplementary Booklet – 8
8
Chemistry
Vidyamandir Classes
Metal
Ores
Composition
Aluminium
Bauxite
Kaolinite (a form of clay)
AlOx(OH)3-2x [where 0 < x < 1]
[Al2 (OH)4 Si2O5]
Iron
Haematite
Magnetite
Siderite
Iron pyrites
Fe2O3
Fe3O4
FeCO3
FeS2
Copper
Copper pyrites
Malachite
Cuprite
Copper glance
CuFeS2
CuCO3 · Cu (OH)2
Cu2O
Cu2S
Zinc
Zinc blende or Sphalerite
Calamine
Zincite
ZnS
ZnCO3
ZnO
Concentration of Ores :

Removal of the unwanted materials (eg., sand, clays, etc.) from the ore is known as concentration, dressing or
benefaction.

It involves several steps and selection of these steps depends upon the differences in physical properties of the
compound of the metal present and that of the gangue.
The type of the metal, the available facilities and the environmental factors are also taken into consideration.
1.
Hydraulic Washing : Based on the differences in gravities of the ore and the gangue particles. It is therefore, a type of
gravity separation. In one such process, and upward stream of running water is used to wash the powdered ore. The
lighter gangue particles are washed away and the heavier ores are left behind.
2.
Magnetic Separation :
Based on differences in magnetic properties of the ore components
If either the ore or the gangue (one of these two) is capable of being attracted by a magnetic field, then such
separations are carried out (e.g., in case of iron ores). The ground ore is carried on a conveyer belt which passes
over a magnetic roller.
Supplementary Booklet – 8
9
Chemistry
Vidyamandir Classes
3.
Froth Floatation Method : For Sulphide Ores :
It is based on the difference in the wettability of the ore and gangue particles.
The mineral particles becomes wet by oils while the gangue particles by water. A rotating paddle agitates the
mixture and draws air in it. As a result, froth is formed which carries the mineral particles. The froth is light and is
skimmed off. It is then dried for recovery of the ore particles.
Practice Question :
What is the role of depressant in froth floatation method? Explain with the help of an example.
Ans :
Depressants are the chemical substances which help us to separate two sulphide ores.
For e.g., In case of an ore containing ZnS and PbS, NaCN is used as an depressant which selectively prevents ZnS from
coming to the froth but allows PbS to come with the froth.
Note :
Collectors and froth stabilities are also added.
Collectors (e.g., Pine oils, fatty acids, xanthates, etc.) enhance non-wettability of the mineral particles.
Froth stabilisers (e.g., Cresols, aniline) stabilize the froth.
4.
Leaching : It is used if the ore is more soluble in a suitable solvent.
(a) Leaching of Alumina from Bauxite :
Al2 O3  Bauxite  principle ore of aluminium 
It usually contains SiO2, iron oxides and TiO2 as impurities.
Step 1 : It is done by digesting the powdered ore with conc. NaOH [at 473 – 523 K and 35 – 36 bar pressure] when
Al2O3 is leached out as sodium aluminate leaving behind the impurities.
Al2 O3 (s)  2NaOH (aq)  3H 2 O () 
 2Na[Al (OH) 4 ](aq) .
The solution also contains SiO 2 as sodium silicate.
Supplementary Booklet – 8
10
Chemistry
Vidyamandir Classes
Step 2 : The aluminate solution is neutralised by passing CO2 gas when hydrated Al2O3 is precipitated out leaving
behind sodium silicate in the solution.
2Na[Al (OH) 4 ](aq)  CO 2 (g) 
 Al2 O3 ·xH 2 O(s)  2NaHCO3 (aq)
Step 3 : Hydrated alumina is then filtered, dried and heated to give pure Al2O3.
1470 K
Al2 O3 · xH 2 O 
 Al2 O3 (s)  xH 2 O (g)
(b) Other Examples : In the metallurgy of silver and gold, the metal is leached with a dilute solution of NaCN on KCN
in the presence of air (for O2) as follows:
Step 1 : 4M(s)  8 CN  (aq)  2H 2 O (aq)  O 2 (g) 
 4[M (CN) 2 ] (aq)  4 OH  (aq) (M  Ag or Au)
Step 2 : 2  M  CN 2 

 aq 
 Zn  s  
  Zn  CN 4 
2
 aq   2M s 
Extraction of Crude Metal from Concentrated Ore :
It involves two major steps
(a)
Conversion to oxide
(b)
Reduction of the oxide to metal.
Roasting
Calcination
 Heating the ore in presence of air at a temperature
below the melting point of the metal.
 Heating of ore in absence (limited supply) of air to remove
volatile impurities.


Generally done for carbonates

ZnCO3 (s) 
 ZnO(s)  CO 2 (g)
Generally done for sulphide ores.
 CO2 is released.
 SO2 is released
Example :
2ZnS
(Zinc blende)
 3O 2 
 2ZnO  2SO 2


CaCO3 · MgCO3 (s) 
 CaO (s)  MgO (s)  2CO 2 (g)
2PbS  3O 2 
 2PbO  2SO 2
(Galena)
2Cu 2S
(Copper Glance)
 3O 2 
 2Cu 2O  2SO 2

The SO2 produced is utilised for manufacturing H2SO4.
Note :
Sulphide ore of Cu sometimes contain iron oxide as impurity (obtained from oxidation of FeS). It is removed in the
form of slag by adding SiO2 as flux.
FeO  SiO 2 
 FeSiO3
flux
(b)
slag
Iron silicate
Reduction of Oxide to the Metal : Reduction of the metal oxide usually involves heating it with some reducing
agent like carbon (C), carbon monoxide (CO) or even another metal. The reducing agent
(e.g.
carbon) combines with the oxygen of the metal oxide
MxO y  yC 
 xM  yCO .
Supplementary Booklet – 8
11
Chemistry
Vidyamandir Classes
To choose a suitable reducing agent for a given metal oxide (MxOy), Gibbs energy interpretations are made.
Thermodynamic Principles of Metallurgy :
The change in Gibbs energy, G for any process at any specified temperature is given by equation:
G 

H

T

enthalpy change

S

entropy change
G  G   RT n Q
& For any reaction,
G    RT n K
Where K  equilibrium constant of the ‘reactant-product’ system at temperature T.
G  0  spontaneous reaction

G  0  non  spontaneous reaction
For a substance to be used as a reducing agent, G for the combined reaction should be negative i.e., from the
equation G  H  TS .
If S is +ve, on increasing temp. TS would increase and G will become negative.

A non-spontaneous reaction can be made to occur by coupling it with highly spontaneous reaction. The reduction of
metal oxide with the help of a reducing agent is also an example of coupling of reactions.
A 
B
G1 is  ve
C 
D
Coupling of reaction: A  C 
BD
G 2 is  ve
G  G1  G 2
For spontaneity of coupled reaction, G has to be  ve
 | G 2 |  | G1 |
Example :
M x O 
 xM 
(i)
C
(ii)
G1 is  ve
1
O2 
 CO
2
G1 is  ve
M x O  C 
 xM  CO
(i) + (ii)
Note :
1
O2
2
G  G1  G 2
If G of formation of oxide is more  ve , it means it has greater tendency to undergo oxid and hence act as reducing
for those oxide which have lesser negative G of formation.
All this information is represented in the form of Ellingham Diagram where  f G of various oxides ore plotted
against T.
Ellingham Diagram :


Ellingham Diagram  f G  vs. T helps in predicting the feasibility of thermal reduction of an ore and formation of oxides of
elements. For example in reaction :
2xM(s)  O 2 (g) 
 2M x O(s)
Dioxygen - a gas has been used up and metal oxide - a solid is formed. Since gases have higher entropy than liquids and solids,
therefore, during this reaction, S becomes negative. Thus, if temperature is increased, TS becomes more and more negative
and G becomes less and less negative.
Supplementary Booklet – 8
12
Chemistry
Vidyamandir Classes

G increases or becomes less  ve with increase in temperature.  f G  vs. T is straight line with +ve slope for most of the
reactions involving the formation of metal oxides.

Each plot is a straight line except when some phase change  s  liq or liq  g  takes place. The temperature at which such
a change occurs is indicated by an increase on the + ve side.
Gibbs energy (G  ) vs T plots (schematic) for formation of some oxides (Ellingham diagram).
Limitations of Ellingham Diagram :

Ellingham diagrams are based on thermodynamic concepts. It does not say about the kinetics of the reduction
process

The interpretation of G  is based upon K (i.e., G   RT ln K ). Thus it is presumed that reactants and products are
in equilibrium.
This is not always true because the reactant/product may be solid.
Note :

In Ellingham Diagram, the metal which lies lower will
reduce the oxide of the metal lying above it

i.e. from the curve, we can infer that at T < T1, Mg can
reduce Al2O3.

At T > T1, Al can reduce MgO.

At T = T1, G  0 and reaction will be at equilibrium.
O
Mg
Al
Al
f G
Mg
(  ve values)
T
T1
T2
Practice Questions :
1.
The value of  f G  Cr2O3    540 KJ mol 1 and  f G   Al2O3    827 KJ mol 1
Supplementary Booklet – 8
13
Chemistry
Vidyamandir Classes
Is reduction of Cr2O3 possible with Al?
Given Data:
(i)
2 Cr 
3
O2 
 Cr2O3
2
G1  540 kJ mol1
(ii)
2 Al 
3
O 2 
 Al2 O3
2
G 2  827 kJ mol1
Required:
(iii)
Cr2O3  2Al 
 2Cr  Al2 O3
G
So, (iii) = (ii) – (i)
G  G 2  G1 =  827    540   KJ mol1  287 KJ mol1
Since G for coupled reaction comes out to be  ve ,

reduction is possible.
Try out yourself
2.
3.
4.
In the blast surface, reduction of iron oxide takes place in different temp zones. Write the reactions and reducing agents
used at each temp. zone in the blast surface.
[For answer Refer NCERT Page : 154-156]
Explain the different forms of iron?
Sol.
The iron obtained from Blast furnace contains about 4% carbon and many impurities in smaller amount (e.g.,
S, P, Si, Mn). This is known as pig iron and cast into variety of shapes. Cast iron is different from pig iron and
is made by melting pig iron with scrap iron and coke using hot air blast. It has slightly lower carbon content
(about 3%) and is extremely hard and brittle. Further Reductions Wrought iron or malleable iron is the purest
form of commercial iron and is prepared from cast iron by oxidising impurities in a reverberatory furnace lined
with haematite. This haematite oxidises carbon to carbon monoxide: Fe2O3 + 3C → 2 Fe + 3CO (6.33)
Limestone is added as a flux and sulphur, silicon and phosphorus are oxidised and passed into the slag. The
metal is removed and freed from the slag by passing through rollers.
Explain the extraction of copper from cuprous oxide (Cu2O).
Sol.
(1)
Copper exists in the form of its ore as Cu2S which is subjected to roasting to obtain Cu2O.
2Cu 2S  3O 2 
 2Cu 2O  2SO 2
(2)
Cu2O is then reduced with the help of carbon as reducing agent
Cu 2 O  C 
 2Cu  CO
(3)
Sometimes copper is obtained from Cu2O by following oxidation – reduction reaction.
2Cu 2 O  Cu 2S 
 6Cu  SO2
The solidified copper obtained thus has a blistered appearance due to evolution of SO2 and hence is
called blister copper.
5.
Write chemical reactions taking place in extraction of zinc from zinc blende.
Sol.
(1)
Roasting :
2ZnS  3O 2 
 2ZnO  2SO 2
(2)
Reduction :
ZnO  C 
Zn  CO
673 K
coke
The metal is distilled off and collected by rapid chilling.
Supplementary Booklet – 8
14
Chemistry
Vidyamandir Classes
Electrochemical Principles of Metallurgy :
 G    n F E

More reactive metals have large negative values of the electrode potential. So, their reduction is
difficult.

If the difference of two E values corresponds to a positive E and consequently negative G ,
then the less reactive metal will come out of the solution and the more reactive metal will go to the
solution.
E.g.
6.
Cu 2 (aq)  Fe(s) 
 Cu(s)  Fe 2 (aq) .
Explain the Electrolytic reduction of pure alumina to give aluminium.
Sol.
(1)
In the metallurgy of aluminium, purified Al2O3 is mixed with Na3AlF6 (cryolite) or CaF2 [ to lower
the melting pt of the MIXTURE and increase the conductivity]
(2)
The electrolysis of the molten mass is carried out in an electrolytic cell using carbon electrodes.
The oxygen liberated at anode reacts with the carbon of anode producing CO and CO 2.
This way for each kg of aluminium produced , about 0.5 kg of carbon anode is burnt away.
The electrolytic reactions are:
Cathode : Al3  melt   3e 
 Al   
Anode:
C (s)  O 2  (melt) 
 CO (g)  2e
C (s)  2O 2  (melt) 
 CO 2 (g)  4 e
The overall reaction is:
2Al2 O3  3C 
 4Al  3CO 2
This process of electrolysis is widely known as Hall-Heroult process.
7.
How is leaching carried out in case of low grade copper ores?
Sol.
Copper is extracted by hydrometallurgy from low grade ores. It is leached out using acid or bacteria.
The solution containing Cu2+ is treated with scrap iron or H2
Cu 2  (aq)  H 2 (g) 
 Cu(s)  2H  (aq)
Oxidation - Reduction :
Besides reductions, some extractions are based on oxidation particularly for non-metals.
Supplementary Booklet – 8
15
Chemistry
Vidyamandir Classes
Examples :
(i)
EXTRACTION OF CHILORINE FROM BRINE (conc. NaCl solution)
2Cl (aq)  2H 2 O() 
 2OH  (aq)  H 2 (g)  Cl2 (g)
H2 is collected at cathode and Cl2 at anode.
(ii)
8..
Leaching process in the extraction of gold and silver is also on example of oxidation – reduction.
What are the products when (i) aq. NaCl (ii) molten NaCl are subjected to electrolysis?
Sol.
S.N.
aq.NaCl
(i)
.
Name
cathode  H 2
anode  Cl2
(ii)
Principle
molten NaCl
anode  Cl2
cathode  Na
Description
Type of
Metal/Example
1.
Distillation
The impure metal is evaporated to obtain the pure metal
as distillate.
Low boiling metals like
zinc and mercury.
2.
Liquation
A low melting metal (e.g., tin) is heated on a sloping
hearth of a reverberatory furnace. Metal flows down
into the receiver leaving the higher melting impurities
on the hearth.
Metals having lower
melting points than the
impurities.

Metals such as copper,
silver, gold, lead,
nickel, chromium, zinc,
aluminium, etc.
3.
Electrolysis
The more basic metal
remains in the solution
and the less basic ones
go to the anode mud
Impure metal acts as anode.
A strip of the same metal in pure form is used as
cathode.

The electrodes are put in a suitable electrolytic
bath containing soluble salt of the same metal.
The reactions are:
Example: Tin (Sn)
Anode: M  Mn+ + ne–
Cathode: Mn+ + ne–  M
Copper obtained by this method is
99.95- 99.99% pure
4.
5.
Zone refining
Vapour phase refining
Impurities are more
soluble in the molten
state (melt) than in the
solid state of the metal.


Supplementary Booklet – 8
The metal should
form a volatile
compound with a
suitable reagent,
the volatile
compound should
be easily
decomposable, so
that the recovery is

A circular mobile heater is fixed at one end of a
rod of the impure metal.

The molten zone moves along with the heater
which is moved forward. As the heater moves
forward, the pure metal crystallises out of the melt
and the impurities pass on into adjacent molten
zone.

The process is repeated several times and the
heater is moved in the same direction. At one end,
impurities get concentrated. This end is cut off.
Mond Process for Refining Nickel :

Nickel is heated in a stream of carbon monoxide
forming a volatile complex, nickel tetracarbonyl:
330 – 350 K
Ni  4CO 
Useful for producing
semiconductor and
other metals of very
high purity, e.g.
germanium, silicon,
boron, gallium and
indium.
Zirconium and titanium
is purified by this
method
Ni(CO4 )
T
Nickel tetracarbonyl
he nickel carbonyl on heating at higher
temperature is decomposed to give the pure metal:
16
Chemistry
Vidyamandir Classes
easy
450 – 470 K
Ni(CO) 4 
Ni
Pure nickel
 4CO
van Arkel Method :
Oxygen and nitrogen present as impurities are removed.
The crude metal is heated in an evacuated vessel with
iodine. The metal iodide being more covalent,
volatilises:
The metal iodide is decomposed on a tungsten filament,
electrically heated to about 1800 K. The pure metal is
thus deposited on the filament.
870 K
Zr(s)  2I 2 (g) 
 ZrI 4 (g)


1800 K, tungsten filament


Zr
 2I 2 
 ZrI4 
Pure zirconium


523 K
Ti(s)
 2I 2 
 TiI 4 (g)
impure titanium
1800 K
TiI4 

6.
Chromatographic
methods such as paper
chromatography,
column
chromatography, gas
chromatography, etc.
The method is based
upon the principle that
the different
components of a
mixture are adsorbed to
different extents on an
adsorbent.
Ti
 2I2 (g)
Pure titanium

The mixture is put in a liquid or gaseous medium
which is moved through the adsorbent. Different
components are adsorbed at different levels on the
column.

Later the adsorbed components are removed
(eluted) by using suitable solvents (eluent).
In column chromatography the column of Al2O3
is prepared in a glass tube.
Refining of Copper using Electrolytic Method :
Copper is refined using an electrolytic method. Anodes are of impure copper and pure copper strips are taken as cathodes.
The electrolyte is acidified solution of copper sulphate and the net result of electrolysis is the transfer of copper in pure form the
anode to the cathode:
Anode:
Cu  Cu 2   2e
Cathode:
Cu 2  2e  Cu
Impurities from the blister copper deposit as anode mud which contains antimony, selenium, tellurium, silver, gold and platinum
; recovery of these elements may meet the cost of refining.
Practice Question :
What criteria should be followed for selection of stationary phase in chromatography??
Sol.
(1)
Stationary phase should have high and selective adsorption properties.
(2)
It should not react with the solvent or the constituents of the mixture under consideration.
Uses :
Uses of Aluminium

Al foils are used
wrapping food items.
Uses of Copper
for
Supplementary Booklet – 8
Uses of Zinc
 Extensively used for making
electric
cables,
electrical
17

For galvanising iron to
protect it from corrosion.
Uses of Iron

Cast iron is used for casting
stoves, railway sleepers,
Chemistry
Vidyamandir Classes

The fine dust of Al is used in
paints and lacquers.

As reducing agent in the
extraction of chromium and
manganese from their oxides.

Al wires are used
electricity conductors.

Alloys containing
aluminium, being light, are
very useful.

as
appliances and water pipes.

 Used for making utensils,
steam
pipes,
kettles,
evaporating
pans,
calorimeters, etc.
 Important alloys of copper are
brass (Cu 60%, Zn 40%),
bronze (Cu 80%, Zn 10%,
Sn-10%) and German silver
(Cu 25-30%, Zn 25-30%,
Ni 40–50%).

Zinc plates and rods are
used in batteries and dry
cells.
Zn dust is used as a
reducing agent in the
manufacture of dye-stuffs,
paints, extraction of gold
and silver by the cyanide
process, etc.
Alloys are used in making
aircrafts and other articles of
daily use.
gutter pipes, toys, etc.

It
is
used
in
the
manufacture of wrought
iron and steel.

Wrought iron is used in
making anchors, wires,
bolts,
chains
and
agricultural implements.

Nickel steel is used
for
making
cables,
automobiles and aeroplane
parts,
chrome
steel
for cutting tools and
crushing machines, and
stainless steel for cycles,
automobiles, utensils, pens,
etc.
A summary of the Occurrence and Extraction of Some Metals
Metal
Aluminium
Occurrence
1. Bauxite, Al2O3 . x H2O
2. Cryolite, Na3AlF6
Iron
1. Haematite, Fe2O3
2. Magnetite, Fe3O4
Copper
1. Copper pyrites, CuFeS2
2. Copper glance, Cu2S
Common Method of Extraction
Remarks
Electrolysis of Al2O3 dissolved in
molten Na3AlF6
For the extraction, a good source of
electricity is required.
Reduction of the oxide with CO and
coke in Blast furnace
Temperature approaching 2170 K is
required.
Roasting of sulphide partially and
reduction
It is self reduction in a specially
designed converter. The reduction
takes place easily. Sulphuric acid
leaching
is
also
used
in
hydrometallurgy from low grade ores.
Roasting followed by reduction with
coke
The metal may be purified by fractional
distillation.
3. Malachite, CuCO3 .Cu(OH)2
4. Cuprite, Cu2O
Zinc
1. Zinc blende or Sphalerite, ZnS
2. Calamine, ZnCO3
3. Zincite, ZnO
d & f-Block Elements
Questions Based on CBSE Pattern
1.
Why does a transition series contain 10 elements?
2.
How many elements are present in the d-block of the periodic table?
3.
How many transition series of elements are there in the periodic table. Name them?
4.
On what basis can you say that scandium (Z = 21) is a transition element but zinc (Z = 30) is not?
5.
Name the elements which are not really transition elements but are discussed with them. Why is it so?
Supplementary Booklet – 8
18
Chemistry
Vidyamandir Classes
6.
Which of the transition metals do not show metallic property?
7.
Why maxima of enthalpy of atomisation occur at about middle of each transition series?
8.
Why is the Eº value for the Mn3+/Mn2+ couple much more positive than that for Cr3+/Cr2+ or Fe3+/ Fe2+. Explain.
9.
Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
10.
Calculate the ‘spin only’ magnetic moment of M2+ (aq) ion (Z = 27).
11.
Write all possible oxidation states of an element having atomic number 25.
12.
Why Cr, Mn, and Fe have nearly the same atomic radii?
13.
Why is Cr2+ reducing and Mn3+ oxidising when both have d4 configuration?
14.
Name a transition element which does not show variable oxidation state?
15.
Which of the 3d series of transition metals exhibits the largest number of oxidation states and why?
16.
Copper (I) compounds are white and diamagnetic while copper (II) compounds are coloured and paramagnetic.
Explain?
17.
Give examples of catalytic activity and formation of coloured ions.
18.
What are interstitial compounds? Why are such compounds well known for transition metals?
19.
Write the chemical equations when : (i)
20.
What happens when K2Cr2O7 is heated with sodium chloride and concentrated H2SO4 ?
21.
The +3 oxidation states of lanthanum (Z = 57), gadolinium (Z = 64) and lutetium (Z = 71) are especially stable. Why.
22.
Briefly explain why electronic configurations of lanthanoids are not known with certainty.
23.
Which out of two, La(OH)3 and Lu(OH)3, is more basic and why?
24.
Why is the separation of lanthanide elements difficult?
25.
Among the ionic species, Sc3+, Ce4+ and Eu2+, which one is good oxidising agent. Give a suitable reason for your answer.
(Atomic numbers: Sc = 21, Ce = 58, Eu = 63)
26.
Explain why lanthanoids are paramagnetic in nature.
27.
Give reasons for the following:
(i)
Size of trivalent lanthanoid cations decrease with increase in atomic number.
(ii)
Chemistry of all the lanthanoids is quite similar.
28.
Compare the chemistry of actinoids with that of lanthanoids with special reference to
(i)
electronic configuration
(ii)
oxidation state
(iii)
atomic and ionic sizes
(iv)
chemical reactivity
29.
Which is the last element in the series of the actinoids ? Write the electronic configuration of this element. Comment on
the possible oxidation state of this element.
30.
What is the importance of the complexes formed by transition elements?
31.
Name few important catalysts mentioning their use in the related reactions.
32.
Why does Mn(II) show maximum paramagnetic character amongst the bivalent ions of the first transition series?
33.
Why Zn2+ salts are white while Ni2+ salts are blue?
34.
Why do transition elements show variable oxidation state?
35.
The melting and boiling points of Zn, Cd and Hg are low. Why?
Supplementary Booklet – 8
KMnO4 is heated
19
(ii)
K2Cr2O7 is heated.
Chemistry
Vidyamandir Classes
36.
Explain why transition elements have many irregularities in their electronic configuration?
37.
Which out of the following is(are) transition element/s and why?
Zn, Ag, Cd, Au.
38.
(a)
Of the ions Ag+, Co2+ and Ti4+ , which one will be coloured in aqueous solutions?
(b)
(Atomic numbers Ag = 47, Co = 27, Ti = 22).
If each one of the above ionic species is in turn placed in a magnetic field, how will it respond and why?
39.
In the titration of Fe2+ ions with KMnO4 in acidic medium, why dilute H2SO4 is used and not dilute HCl?
40.
Give reasons for each of the following:
(i)
Size of trivalent lanthanoid cations decreases with increase in the atomic number.
(ii)
Transition metal fluorides are ionic in nature whereas bromides and chlorides are usually covalent in nature.
(iii)
Chemistry of all the lanthanoids is quite similar.
41.
The outer electronic configurations of two members of the lanthanoid series are as follows:
4f 1 5d1 6s 2 and 4f 7 5d 0 6s 2
What are their atomic numbers? Predict the oxidation states exhibited by these elements in their compounds.
42.
Write the electronic configuration of Gadolinium (Z = 64) and its most stable oxidation states.
43.
Explain the following observations:
(i)
Generally there is an increase in density of elements from titanium (Z = 22) to copper (Z = 29) in the first
series of transition elements.
(ii)
Transition elements and their compound are generally found to be good catalysts in chemical reactions.
44.
How would you account for the following?

The atomic radii of the metals of the third(5d) series of transition elements are virtually the same as those of
the corresponding members of the second (4d) series.
d & f-Block Elements
Solved Questions Based on CBSE Pattern
1.
Why do transition metals exhibit higher enthalpies of atomisation?
2.
Why is Cr2+ reducing and Mn3+ oxidising when both have d4 configuration?
3.
Why is the Eº values for Mn3+/Mn2+ couple much more positive than for Cr3+/Cr2+ or Fe3+/Fe2+ ?
4.
Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?
5.
Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with
increasing atomic number.
6.
In what way is the electronic configuration of transition elements different from that of the non-transition elements?
7.
Compare the chemistry of actinoids with that of lanthanoids with special reference to
(i)
Electronic configuration (ii)
oxidation state (iii)
atomic and ionic sizes
8.
The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some
examples from the oxidation state of these elements.
Supplementary Booklet – 8
20
Chemistry
Vidyamandir Classes
SOLUTIONS - QUESTIONS BASED ON CBSE PATTERN
1.
Enthalpy of atomisation is the amount of heat required to break the metal lattice to get free atoms. As transition metals
contain a large number of unpaired electrons, they have strong interatomic attractions (metallic bonds). Hence, they
have high enthalpies of atomisation.
2.
Cr2+ has the configuration 3d4. It can lose electron to form Cr3+ which has the stable 3d3 configuration (as it has halffilled t2g level. Hence, it is reducing. On the other hand, Mn3+ also has 3d4 configuration but it can gain electron to form
Mn2+ which has stable 3d5 configuration (as it is exactly half-filled). Hence, it is oxidizing.
Alternatively, Eº value for Cr3+/Cr2+ is negative (–0.41 V) whereas Eº value for Mn3+/Mn2+ is positive (+1.57 V).
Hence, Cr2+ ion can easily undergo oxidation to give Cr3+ ion and, therefore, acts as strong reducing agent whereas Mn3+
can easily undergo reduction to give Mn2+ and hence acts as oxidizing agent.
3.
The large positive Eº value for Mn3+/Mn2+ shows that Mn2+ is much more stable than Mn3+. This is due to the fact that
Mn2+ has the configuration 3d5 which is half-filled and hence is very stable. Thus, the third ionization energy of Mn
will be very high. In fact, this is the reason that +3 state of Mn is of little importance.
4.
Manganese (Z = 25) shows maximum number of oxidation states. This is because its electronic configuration is 3d 5 4s2.
As 3d and 4s are close in energy, it has maximum number of electrons to lose or share (as all the 3d electrons are
unpaired). Hence, it shows oxidation states from +2 to +7 (+2, +3, +4, +5, +6 and +7) which is the maximum number.
5.
Except scandium (which shows an oxidation state of +3), all other first row transition elements show an oxidation state
of +2. This is due to loss of two 4s electrons. In the first half, as we move from Ti2+ to Mn2+, the electronic
configuration changes from 3d2 to 3d5, i.e., more and more of d-orbitals are half-filled imparting greater and greater
stability to +2 state. In the second half, i.e., Fe2+ to Zn2+, the electronic configuration changes from 3d6 to 3d10,
i.e., electrons in the 3d orbitals pair up and the number of half-filled orbital decreases. Hence, the stability of
+2 state decreases.
6.
Transition elements contain incompletely filled d-subshell, i.e., their electronic configuration is (n – 1) d 1–10 ns0–2
whereas non-transition elements have no d-subshell or their d-subshell is completely filled and have ns1–2 or ns2 p1–6 in
their outermost shell.
7.
(i)
Electronic configuration. The general electronic configuration of lanthanoids is [Xe]54 4f1–14 5d0–1 6s2 whereas
that of actinoids is [Rn]86 5f1–14 6d0–1 7s2. Thus, lanthanoids belong to 4f-series whereas actinoids
belong to 5f-series.
(ii)
Oxidation states. Lanthanoids show limited oxidation states (+2, +3, +4) out of which +3 is most common.
This is because of large energy gap between 4 f and 5d subshells. On the other hand, actinoids show a large
number of oxidation states because of small energy gap between 5f, 6d and 7s subshells.
(iii)
Atomic and ionic sizes. Both show decrease in size of their atoms or ions in + 3 oxidation state.
In lanthanoids, the decrease is called lanthanoid contraction whereas in actinoids, it is called actinoid
contraction. However, the contraction is greater from element to element in actionoids due to poorer shielding
by 5f electrons than that by 4f electrons in lanthanoids.
8.
Lanthanoids show limited number of oxidation state, viz., +2, +3 and +4 (out of which +3 is most common). This is
because of large energy gap between 4f and 5d subshells. The dominant oxidation state of actinoids is also +3 but they
show a number of other oxidation states also, e.g., uranium (Z = 92) and plutonium (Z = 94), show +3, +4, +5 and +7
etc. This is due to small energy difference between 5f, 6d and 7s subshells of the actinoids.
Supplementary Booklet – 8
21
Chemistry