Download THE d- AND f-BLOCK ELEMENTS

The d- and f-block elements
CHAPTER
9
THE d- AND f-BLOCK ELEMENTS
LEARNING OBJECTIVES
(i) Know the electronic configurations of the transition (d-block) and the inner transition (f-block) elements.
(ii) Appreciate the relative stability of various oxidation states in terms of electrode potential values.
(iii) Describe the preparation, properties, structures and uses of some important compounds such as K2Cr2O7 and KMnO4.
(iv) Understand the general characteristics of the d– and f–block elements and the general horizontal and group trends in them.
(v) Describe the properties of the f-block elements and give a comparative account of the lanthanoids and actinoids with respect
to their electronic configurations, oxidation states and chemical behaviour.
INTRODUCTION
Many of the most important elements of modern society are transition and inner transition elements. Iron, copper, silver and gold
are among the transition elements that have played important roles in the development of human civilisation. The inner transition
elements such as Th, Pa and U are proving excellent sources of nuclear energy in modern times. The transition elements are those
which as elements or as ions have partly filled (n – 1) d-orbitals. Since in these the differentiating electron enters the (n – 1) dorbitals, these are also called d-block elements. The d-block of the periodic table contains the elements of the groups 3-12 in
which the d orbitals are progressively filled in each of the four long periods.
The elements which as free elements or as ions have partly filled (n – 2) f-orbitals are called f-block or inner-transition elements.
Since in these elements the differentiating electron enters the f-orbitals [viz. (n – 2) orbitals], these are also called f-block
elements. Elements of f-blocks are formal members of group 3 from which they have been taken out to form a separate f-block of
the periodic table.
THE TRANSITION ELEMENTS (d-BLOCK) :
In the long form of periodic table elements are classified based on electronic configuration. The elements which are classified
between’s’ and ‘p’ block are ‘d’ block elements or Transition elements. In these elements the differentiating electron enters in the
‘d’ orbitals of penultimate shell. General configuration for ‘d’ block elements is ns1-2 (n-1)d1-10. i.e., in ‘d’ block elements the
valence shell has constant number of electrons whereas the number of electron in penultimate shell go on increasing. Elements
which have atleast one unpaired electron in their ‘d’ orbital in atomic or any oxidation state are called as Transition elements.
Thus all transition elements are ‘d’ block elements but all ‘d’ block elements may not be transition elements. Transition elements
are classified between’s’ and ‘p’ blocks from fourth period onwards.
Series of transition elements are four –
1st Series - They are classified in fourth period and are called as ‘3d’ series of elements. Their atomic numbers are 21(Sc) to 30 (Zn).
2nd Series- They are classified in fifth period and are called as ‘4d’ series of elements. Their atomic numbers are 39(Y) to 48(Cd).
3rd series- They are classified in sixth period and are called as ‘5d’ series of elements. Their atomic numbers are 57(La), 72(Hf) to 80(Hg)
4th Series - They are classified in the seventh period and are called as ‘6d’ series of elements. This is an incomplete series. Their
atomic numbers are 89(Ac), 104(Ku) to 112 (Uub).
Electronic configuration of first series of transition elements.
Atomic number
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
Element
Scandium
Titanium
Vanadium
Chromium
Manganese
Iron
Cobalt
Nickel
Copper
Zinc
Symbol
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Electronic configuration
[Ar) 4s23d1
[Ar] 4s23d2
[Ar) 4s23d3
[Ar) 4s13d5
[Ar] 4s23d5
[Ar) 4s23d6
[Ar) 4s23d7
[Ar) 4s23d8
[Ar) 4s13d10
[Ar] 4s23d10
Gyaan Sankalp
1
The d- and f-block elements
Chromium and copper are elements having exceptional electronic configurations of [Ar] 4s13d5 and [Ar] 4s13d10 instead of [Ar]
4s23d4 and [Ar] 4s23d9.
Zn (30) is [Ar] 4s23d10 , Cd (48) is [Kr] 5s24d10, Hg (80) is [Xe] 6s2 4f14 5d10
These three elements do not have any unpaired electrons in their ‘d’ orbitals in atomic as well as in ionic states. Therefore they
are only classified as ‘d’ block elements and not as transition elements. Copper, Silver and Gold, the elements of IB group i.e.,
coinage metals, have ns1 (n-1)d10 configuration. They are transition elements as in their higher oxidation state they have an
unpaired electron in their ‘d’ orbitals.
GENERAL PROPERTIES OF d-BLOCK ELEMENTS
All the elements of ‘3d’ series are good reducing agents except copper. In general the reactivity of transition elements is less.
Their reactivity decreases with increase in atomic number. The atomic radii of transition elements decrease by negligible amounts
due to the shielding of valence shell electrons, provided by electrons of ‘d’ orbitals of penultimate shell. The ionisation potential
of transition elements increases by negligible amounts due to shielding effect.
Metallic character : All the d-block elements are metals as the numbers of electrons in the outer most shell are one or two. These
elements occur in three types face centered cubic (fcc), hexagonal closed packed (hcp) and body-centred cubic (bcc). They are
hard malleable and ductile. IB (11) group elements Cu, Ag and Au are most ductile and soft. These are good conductor of heat and
electricity (due to free e–). They are most conductive in nature. Their order of conductivity is Ag > Cu > Au > AI.
Atomic size : Atomic and ionic radii of d-block elements is smaller than s-block elements. Atomic radii depends on effective
nuclear charge (Zeff) and screening effect (SE).
In 3d reries :
Sc  Cr (Zeff > SE)  radius decreases
Mn  Ni (Zeff = SE)  radius remains constant
Cu  Zn (Zeff < SE)  radius increases
Decrease in the radii with increase in atomic number is not regular. Atomic radii tend to reach minimum near at the middle of the
series and increase slightly towards the end of the series.
In dipositive ions of 3d series Cu+2 is the smallest in size. The elements of 4d and 5d series belongs to a particular group have
almost same atomic radii. This is due to Lanthanide contraction. e.g. Zr  Hf, Tc  Re, Nb  Ta, Ru  Os etc.
In d-block elements : Largest atomic radii - La, Smallest atomic radii - Ni
In IIIB (13) group order of atomic radii is : Sc < Y < La (No lanthanide contraction)
Density : The atomic volume of the transition elements are low compared with s-block, so their density is comparatively high.
Except Sc, Y and Ti, all the d-block elements have density greater than 5gm cm–3. Os (22.57 gm cm–3) and Ir (22.61 gm cm–3) have
highest density. In all the groups there is normal increase in density from 3d to 4d series, and from 4d to 5d it increases just double
it is due to lanthanide contraction. Ti < Zr << Hf
In 3d series : Sc  Cr density increases, Mn, Fe, Co, Ni  almost constant, Cu  Zn decreases
In 3d series highest density – Cu, lowest density – Sc
Some important orders of density : Fe < Ni < Cu ; Fe < Cu < Au ; Fe < Hg < Au
Melting and boiling points : M.P. and B.P. of d-block is greater than s-block (the reason is stronger metallic bond and presence
of covalent bond formed by unpaired d-electrons.) In Zn, Cd, and Hg there is no unpaired electron present in d-orbital, hence due
to absence of covalent bond, their m.p. and b.p. are very low. (Volatile metals Zn, Cd, Hg). In 3d series: Sc Cr; m.p. b.p increases,
Mn  Zn m.p. b.p decreases. Mn and Tc possess comparatively low m.p., it due to stable configuration (Half filled). Lowest mp
Hg - 38°C, Highest mp. W  3400°C.
Ionisation potentials : The ionisation potential values of most of the d-block elements lie in between those of s- and p-block
elements. This indicates elements lie in between those of s- and p-block elements. This indicates that the d-block elements are
less electropositive than those of the elements of groups IA and IIA (s-block elements). Thus d-block elements do not form ionic
compounds so readily as the alkali and alkaline earth metals do.
The ionisation potentials of d-block elements increase as we move across each series from left to right, although the increase is
not quite regular, e.g., in the first series, the values for Sc, Ti, V and Cr differ very slightly. Similarly the values Fo, Co, Ni and Cu
are fairly close to one another. The value for Zn is appreciably higher due to the additional stability associated with completely
filled 3d-level in Zn (Zn 3d104s2).
Slight variations that occur in ionisation potentials across the series are mainly due to the slight changes in atomic radii which are
an account of the fact that the screening effect caused by the addition of extra electrons to 3d-level almost compensates the
effect in increase in nuclear charge, as we move from left to right in the series.
(a) Variable valency or variable oxidation states : They exhibit variable valency due to involvement of (ns) and (n-1)d electrons in
bonding. This is due to less energy difference between these electrons.
The oxidation states of all transition elements of ‘3d’ series are :
2
Gyaan Sankalp
The d- and f-block elements
4s23d1
Sc
[Ar]
+3
Ti
[Ar] 4s23d2
+2
+3
(+4)
V
[Ar] 4s23d3
+2
+3
+4
(+5)
Cr
[Ar] 4s13d5
+1
+2
(+3)
+4
+5
(+6)
Mn
[Ar] 4s23d5
(+2)
+3
+4
+5
+6
(+7)
Fe
[Ar] 4s23d6
(+2)
(+3)
+4
+5
+6
Co
[Ar] 4s23d7
(+2)
(+3)
+4
Ni
[Ar] 4s23d8
(+2)
+3
+4
Cu
[Ar] 4s13d10
+1
(+2)
The most common oxidation states are in bracket ( ).
Most common oxidation state among the transition elements is +2. Highest oxidation state shown by transition elements of ‘4d’
and ‘5d’ series is +8. The elements showing this oxidation state are Ruthenium (44) and Osmium (76). The common oxidation state
shown by elements of IIIB i.e., Sc, Y, La and Ac is +3 as their divalent compounds are highly unstable. In lower oxidation state
transition elements form ionic compounds and in higher oxidation state their compounds are covalent.
For example, In chromate ion CrO4–2, the bonds between Cr and O are covalent.
Generally higher oxidation states are exhibited in the compounds which are formed with highly electronegative elements like O
and F. They also shows zero oxidation state in their carbonyl compounds like Ni(CO)4. Usually transition metal ions in their lower
oxidation state act as reducing agents and in higher oxidation state they are oxidising agents.
Highest oxidation state of transition elements can be calculated by = n + 2 (n = no. of unpaired e-) (It is not applied for Cr and Cu)
The transition metal ions having stable configuration are stable. Metal ions of ‘3d’ series having 3dº configuration Sc+3, Ti+4 and
V+5 etc. are stable. Transition metal ions having 3d5 configuration are stable like Mn+2, Fe+3. In aqueous medium Cr+3 is stable.
Co+2 and Ni+2 are stable. Transition metal ion with 3d10 configuration which is stable is Cu+1. In aqueous medium Cu+2 is more
stable than Cu+1.
Ex. – Ti+2, V+2, Fe+2, Co+2 etc are reducing agents
Cr+6, Mn+7, Mn+4 Mn+5, Mn+6 etc are oxidising agents.
Standard oxidation potentials and reducing properties :
The oxidation potential value of the member of first transition series are as follows :
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Oxidation
potential value
–
1.63
1.18
0.90
1.18
0.44
0.28
0.25
0.34
0.76
(v) for M 
M2+ + 2e–
M  M3+ +
3e–
2.10
–
0.74
–
–
–
–
–
–
The irregular trend is due to the variation in ionization energies, sublimation energies and the hydration energies of the divalent
ion of the member of the transition series. The value of standard oxidation potential reveal that the value for any element except
Cu is higher than that of standard hydrogen electrode (taken as zero). It is therefore expected that these metals, except Cu, would
evolve H2 gas from acid solution. In actual practice the rate at which most of the metal react with acid is very slow. Some of the
metals get protected from further attack of the acid due to the formation of a thin protective layer of their oxide, which prevent the
acid to come in further contact with metals.
Ex. Although E 0ox of Cr is high, it get coated with its non-reactive oxide (Cr2O3) which make the Cr metal so unreactive that it can
be used as a protective non-oxidizing metals.
Inspite of the fact that metals of first transition series, except Cu, have high value of E 0ox , these are not good reducing agent as
the metal of IA, IIA and IIIA groups. The poor reducing capacity of the transition metal is due to high heats of vapourization, high
ionization potential and low heat of hydration of their ions.
Magnetic properties : Matter, in general is associated with magnetic properties. Majority of substances are either paramagnetic
or diamagnetic. A paramagnetic substance is one which is attracted into a magnetic field. Paramagnetism is mainly due to the
presence of unpaired electrons in atoms or ions or molecules. Diamagnetic substance is one which is slightly repelled by a
magnetic field. Ti+2 [Ar]3d2, Ti+3 [Ar]3d1. V+2[Ar]3d3, Cr+3[Ar]3d3
As is evident most of the transition metal ions have unpaired electrons in their ‘d’ orbitals. Hence most of the transition metal ions
are paramagnetic in nature. Transition metal ions having 3d0 and 3d10 configuration exhibit diamagnetic nature. An unpaired
electron spins and as it is a charged particle, magnetic field is created due to its spinning. Each electron may, in fact, be considered
as a micro magnet having a certain value of magnetic moment. The total magnetic moment of a substance is the resultant of the
magnetic moments of all the individual electrons. Thus substances containing unpaired electrons get attracted towards the
magnets exhibiting paramagnetic nature. The magnetic moment () created due to spinning of unpaired electrons can be calcuGyaan Sankalp
3
The d- and f-block elements
lated by using  =
n(n  2) , where ‘n’ is the number of unpaired electrons in the metal ion,  = Magnetic moment in Bohr
Magnetons (B.M.)
The magnetic moment of diamagnetic substances will be zero. As the number of unpaired electrons increase the magnetic
moment created goes on increasing and hence the paramagnetic nature also increases. Transition metal ions having d5 configuration will have maximum number of unpaired electrons therefore they will be maximum paramagnetic in nature.
Colour Property:Most of the transition metal ions exhibit colour property. This is due to the presence of unpaired electrons in
their ‘d’ orbitals. They require less amount of energy to undergo excitation of electrons. Hence they absorb visible region of light
exhibiting colour. Ti+2[Ar]3d2, V+2[Ar]3d3 etc.
These are having unpaired electrons in their ‘d’ orbitals therefore they are coloured. Transition metal ions which do not have any
unpaired electrons in their ‘d’ orbitals like 3d0 and 3d10 configurations, do not exhibit any colour property.
Ex.., Sc+3 [Ar]3d0, Cu+1[Ar]3d10, Ti+4[Ar]3d0 etc are colourless ions.
A transition metal ion absorbs a part of visible region of light and emits rest of the six colours, the combination of which is the
colour of emitted light. The colour of metal ion is the colour of the emitted light. In transition metal ion the ‘d’ orbitals split into
lower energy set t2g orbitals and higher energy set eg orbitals. The electrons from t2g set get excited to higher energy set i.e., eg
set. This excitation of electrons is called as ‘d-d’ transition. As d-d transition requires less amount of energy they absorb visible
region of light. Due to this ‘d -d’ transition the transition metal ions exhibit colour property.
2 2 d 2
dx–y
z
(eg)
presence of ligand
light
d-d transsition
d-orbitals
(degenerate)
dxy dyz dxz
t2g
KMnO4 (dark pink), K2Cr2O7 (orange) having dº configuration are coloured due to charge transfer spectrum.
Some of the coloured metal ions are as follows :
Ti+3 Purple
Cr+3
Green
Mn+2 Light pink
Fe+2 - Pale green
Fe+3 yellow
Co+2 - Blue
Ni+2
green
Cu+2 - Blue
Complex formation : Transition metal ions have maximum tendency to form complexes. In the formation of complexes they form
coordinate covalent bonds and act as electron pair acceptors. Transition metal ions have smaller size and have high positive
charge i.e., high charge density. Due to this they have maximum tendency to accept electrons. They have vacant ‘d’ orbitals
available on them hence they can accept lone pairs of electrons forming coordinate covalent bond. The greater the charge
density on the transition metal ion, the greater they have tendency to form complexes. Thus Ti+2 to Ni+2 the stability of
complexes formed goes on increasing. Compounds like NaCl, Al2(SO4)3 and K2SO4 provide only two types of ions in aqueous
solution i.e., cations and anions, are called as simple salts. When two or more simple salt solutions are mixed and then subjected
for crystallisation, the crystals which are obtained are of addition compounds. The addition compound formed may behave in
one of the following ways
H O
2  2K+ + 2Al+3 + 4SO –2
(a) K2SO4, Al2(SO4)3. 24H2O 
4
Common alum
The addition compound which undergoes complete ionisation to form three types of ions i.e., two types of cations and one type
of anions or vice versa, is called as double salt. These double salts loose their identity in aqueous solution.
H O
2 

(b) 4KCN + Fe(CN)2
4K+ + [Fe(CN)6]–4
potassium ferrocyanide
Ferrocyanide ion
Addition compound of this type which undergoes partial ionisation to form a complex ion in aqueous solution is called as
complex compound or coordination compound. Complexes retain their identity in aqueous solution
Catalytic property : Transition elements and their compounds exhibit catalytic properties. This is due to their variable valency
as well as due to the free valencies on their surface. When transition elements and their compounds are in powdered state their
catalytic properties exhibited will be to a greater extent. This is due to greater surface area available in the powdered state.
Transition metals and their compounds exhibiting catalytic properties in various processes are
4
Gyaan Sankalp
The d- and f-block elements
(i) Fe is used in Haber’s process for manufacture of NH3.
(ii) V2O5 is used in contact process for H2SO4 manufacture.
(iii) Pt is used in Ostwald’s process of nitric acid.
(iv) Ni is used in hydrogenation of oils.
(v) FeSO4 is used in oxidation of Benzene with H2O2.
(vi) Cu is used in dehydrogenation of alcohols.
(vii) TiCl4 is used as catalyst in Vinyl polymerisation.
Formation of alloy : Transition elements have maximum tendency to form alloys. The reactivity of transition elements is very less
and their sizes are almost similar. Due to this a transition metal atom in the Lattice can be easily replaced by other transition metal
atom and hence they have maximum tendency to form alloys. In the alloys ratio of component metals is fixed. These are extremely
hard and have high M.P.
Some important alloy :
1. Bronze - Cu (75 - 90 %) +Sn ( 10 - 25 %)
2. Brass - Cu ( 60 - 80 %) +Zn (20 - 40 %)
3. Gun metal - (Cu + Zn + Sn) (87 : 3 : 10)
4. German Silver - Cu + Zn + Ni ( 2 : 1 : 1)
5. Bell metal - Cu (80%) + Sn (20%)
6. Nichrome - (Ni + Cr + Fe)
7. Alnico - (Al, Ni, Co)
8. Type Metal - Pb + Sn + Sb
9. Alloys of steel : (a) Vanadium steel - V (0.2 – 1%) , (b) Chromium steel - Cr (2 – 4%), (c) Nickel steet - Ni (3 – 5%)
(d) Manganese steel - Mn (10 – 18%), (e) Stainless steel - Cr (12 – 14%) & Ni (2–4%), (f) Tungsten steel - W (10 – 20%),
(g) Invar - Ni (36%)
10. 14 Carat Gold - 54% Au + Ag(14 to 30%) + Cu (12 – 28%)
11. 24 Carat Gold - 100% Au
12. Solder - Pb + Sn
13. Magnallium - Mg (10%) + Al (90%)
14. Duralumin - (Al + Mn + Cu)
15. Artificial Gold - Cu (90%) + Al (10%)
16. Constantan - Cu (60%) + Ni (40%)
Formation of interstitial compounds :
Transition elements form interstitial compounds with smaller sized non metal elements like hydrogen, carbon, boron, nitrogen
etc. The smaller sized atoms get entrapped in between the interstitial spaces of the metal lattices. These interstitial compounds
are non stoichiometric in nature and hence cannot be given any definite formula. The smaller sized elements are held in interstitial
spaces of transition elements by weak Vander Waals forces of attractions. The interstitial compounds have essentially the same
chemical properties as the parent metals but they differ in physical properties such as density and hardness.
Some important compounds of transition elements :
Oxides and Oxoanions of Metals : These oxides are generally formed by the reaction of metals with oxygen at high temperatures.
All the metals except scandium form MO oxides which are ionic. The highest oxidation number in the oxides, coincides with the
group number and is attained in Sc2O3 to Mn2O7. Beyond group 7, no higher oxides of iron above Fe2O3 are known.
As the oxidation number of a metal increases, ionic character decreases. In the case of Mn, Mn2O7 is a explosive covalent green
oil. The oxides of metals in high oxidation states are acidic. For example, V2O5, CrO3, Mn2O7 are acidic.
Thus, Mn2O7 gives HMnO4 and CrO3 gives H2CrO4 and H2Cr2O7. V2O5 is, however, amphoteric though mainly acidic and it
gives VO43– as well as VO2+ salts. In vanadium there is gradual change from the basic V2O3 to less basic V2O4 and to amphoteric
V2O5. The well characterised CrO is basic but Cr2O3 is amphoteric. The basic and amphoteric oxides dissolve in one-oxidising
acids forming hexaquo ions [MCH2O)6]n+. A few of these oxides dissolve in acids and bases to form important oxometallic salts.
Potassium dichromate, chromate and permanganate are few of the examples.
POTASSIUM DICHROMATE (K2Cr2O7) :
Potassium dichromate is one of the most important compound of chromium, and also among dichromates. In this compound, Cr
is in the hexavalent (+6) state.
Preparation : It is manufactured from chromite ore, FeO.Cr2O3. The various steps are :
(a) Preparation of sodium dichromate, Na2Cr2O7 : For its preparation on a commercial scale, chrome iron ore is the starting point
which is roasted, ground and mixed with sodium carbonateand quick lime. The mixture is dried and then strongly heated to
redness in an ordinary flame in a reverberatory furnace. Thus the atmospheric oxygen is utilised for oxidation.
[2(FeO.Cr2O3) + O Fe2O3 + 2Cr2O3] × 2
[2Na2CO3 + Cr2O3 + 3O 2Na2CrO4 + 2CO2] × 4
––––––––––––––––––––––––––––––––––––––––––––––––––
4(FeO.Cr2O3) + 8Na2CO3 + 7O2 2Fe2O3 + 8Na2CrO4 + 8CO2
The function of quick-lime is probably to keep the mass porous and prevent fusion. The sodium chromate thus formed is
extracted with water. To this, the requisite amount of strong sulphuric acid is added which converts chromate to dichromate.
Sodium sulphate separates out and is removed. The solution is then concentrated to sp. gr. 1.7, when the dichromate comes down
in the crystalline form.
2Na2CrO4 + H2SO4 Na2Cr2O7 + Na2SO4 + H2O
5
Gyaan Sankalp
The d- and f-block elements
(b)
Conversion of sodium dichromate to potassium dichromate, K2Cr2O7 : A hot concentrated solution of sodium, dichromate is
treated with the calculated quantity of potassium chloride.
Na2Cr2O7 + 2KCl 2NaCl + K2Cr2O7
Potassium dichromate being much has soluble separates out from the solution on crystallisation.
Properties : Potassium dichromate gives out garnet red prismatic crystals (m.pt.398°C) which decompose only at a white heat
4K2Cr2O7 4K2CrO4 + 2Cr2O3 + 3O2
to oxygen and chromic oxide. Heated with alkalies, it forms potassium chromate.
K2Cr2O7 + 2KOH 2K2CrO4 + H2O
It is moderately soluble in water but more so in concentrated sulphuric acid. When conc. sulphuric acid is added to a cold
solution of potassium dichromate, red crystals of chromic anhydride separate out.
K2Cr2O7 + 2H2SO4 2KHSO4 + 2CrO3 + H2O
Potassium dichromate is a powerful oxidising agent and is generally used for this purpose in the presence of acids.
K2Cr2O7 + 4H2SO4 K2SO4 + Cr2(SO4)3 + 4H2O + 3O
The Cr2O72– ion takes up electrons and hence acts as an oxidising agent.
Cr2O72– + 14H+ + 6e– 2Cr3+ + 7H2O
It is quite clear that in this reaction Cr(+6) is reduced to Cr(+3) salt.
It is on this account used in titrations for which one-sixth of the molecular weight of K2Cr2O7 dissolved in a litre of water would
give the normal solution. In presence of H2SO4 it liberates I2 from KI, oxidises ferrous salts to ferric. H2S to S, SO2 to H2SO4,
halogen acids to halogens, SO32 to SO42–, primary alcohols to aldehydes and then to acids.
(a) K2Cr2O7 + 4H2SO4 K2SO4 + Cr2(SO4)3 + 4H2O + 3O
6KI + 3H2SO4 + 3O 3K2SO4 + 3I2 + 3H2O
––––––––––––––––––––––––––––––––––––––––––––––––
K2Cr2O7 + 7H2SO4 + 6KI 4K2SO4 + Cr2(SO4)3 + 3I2 + 7H2O
or Cr2O72– + 14H+ + 6I– 2Cr3+ + 7H2O + 3I2
(b) K2Cr2O7 + 4H2SO4 K2SO4 + Cr2(SO4)3 + 4H2O + 3O
[2FeSO4 + H2SO4 + O Fe2(SO4)3 + H2O] × 3
––––––––––––––––––––––––––––––––––––––––––––––––
K2Cr2O7 + 7H2SO4 + 6FeSO4 K2SO4 + Cr2(SO4)3 + 3Fe2(SO4)3 + 7H2O
or Cr2O72– + 6Fe2+ + 14H+ 2Cr3+ + 6Fe3+ + 7H2O
(c) K2Cr2O7 + 4H2SO4 K2SO4 + Cr2(SO4)3 + 4H2O + 3O
[H2S + O H2O + S] × 3
––––––––––––––––––––––––––––––––––––––––––––––––
K2Cr2O7 + 4H2SO4 + 3H2S K2SO4 + Cr2(SO4)3 + 7H2O + 3S
(d) SO2 + H2O + O H2SO4
(e) K2Cr2O7 + 14HCl 2KCl + 2CrCl3 + 7H2O + 3Cl2
(f) Cr2O72– + 3SO32– + 8H+ 2Cr3+ + 3SO42– + 4H2O
(g) K2Cr2O7 + 4H2SO4 K2SO4 + Cr2(SO4)3 + 4H2O + 3O
C2H5OH + O CH3CHO + H2O
CH3CHO + O CH3COOH
When treated with concentrated hydrochloric acid or with a chloride and strong sulphuric acid, reddish brown vapours of
chromyl chloride are obtained.
K2Cr2O7 + 2H2SO4 2KHSO4 + 2CrO3 + H2O
[KCl + H2SO4 KHSO4 + HCl] × 4
2CrO3 + 4HCl 2CrO2Cl2 + 2H2O
––––––––––––––––––––––––––––––––––––––––––––––––
K2Cr2O7 + 4KCl + 6H2SO4 2CrO2Cl2 + 6KHSO4 + 3H2O
Chromyl chloride vapour when passed through water give yellow-coloured solution containing chromic acid. Solution containing chromic acid. Chromyl chloride test can be used for the detection of chloride ion in any mixture.
Uses : Potassium dichromate finds extensive use in dyeing, calico-printing and chrome-tanning (leather industry). It is used as
an oxidising agent and in the preparation of various chromium compounds like Cr2O3, CrO3, K2SO4.Cr2(SO4)3.24H2O. CrO2Cl2,
K2CrO4, CrCl3, Cr(CH3COO)2 etc. It is also used in photography and in volumetric analysis for the estimation of ferrous salts. A
mixture of K2Cr2O7 and conc. H2SO4, usually known as chromic acid mixture, has strong oxidising properties and is used as a
cleaning agent for glass ware etc.
6
Gyaan Sankalp
The d- and f-block elements
POTASSIUM PERMANGANATE, KMnO4 :
It is salt of an unstable acid HMNO4 (Permanganic acid). The Mn is in +7 state in this compound.
Various steps involved in its preparation are :
(i) Preparation of potassium manganate, K2MnO4. It is prepared from manganese dioxide and potassium carbonate or KOH in
presence of atmospheric oxygen or other oxidising agents.
2MnO2 + 4KOH + O2 2K2MnO4 + 2H2O
2MnO2 + 2K2CO3 + O2 2K2MnO4 + 2CO2
(ii) Oxidation of K2MnO4 to KMnO4
(a) To potassium manganate solution is either added sulphuric acid or better the alkali neutralised by passing carbon dioxide
through it. As soon as the alkalinity of the solution is removed, the manganate decomposes to form permanganate.
3K2MnO4 + 2H2SO4 2K2SO4 + 2KMnO4 + MnO2 + 2H2O
3K2MnO4 + 2H2O + 4CO2 2KMnO4 + MnO2 + 2KHCO3
Due to the precipitation of manganese dioxide, there is a considerable loss of manganese in the process.
(b) In the Stadeler’s process, the K2MnO4 is converted to permanganate without the loss of manganese, by passing chlorine
through it.
2K2MnO4 + Cl2 2KMnO4 + 2KCl
(c) But now, for manufacturing potassium permanganate, the method of electrolytic oxidation, is preferred. The manganate
solution is electrolysed between iron electrodes separated by diaphragm. The oxygen evolved at the anode converts manganate
to permanganate :
2K2MnO4 + H2O + O 2KMnO4 + 2KOH
or MnO42– – e– MnO4
Properties : Potassium permanganate forms dark purple lustrous crystals giving deep pink colour in solution. It is isomorphous
with potassium perchlorate. Its solubility in water at 20° is only about 7%, whilst it dissolves more at higher temperatures (25%
at 63°).
(i) Action of heat : It gives off O2 at 200°C
2KMnO4  K2MnO4 + MnO2 + O2
At red heat K2MnO4 formed as above is decomposed into permanganite, K2MnO3 and O2.
2K2MnO4  2K2MnO3 + O2
(ii) Action of conc. H2SO4
(a) With well cooled conc. H2SO4, it gives Mn2O7 which decomposes on warming.
2KMnO4 + 2H2SO4 Mn2O7 + 2KHSO4 + H2O
2Mn2O7 4MnO2 + 3O2
(b) With warmed conc. H2SO4 oxygen gas is evolved.
4KMnO4 + 6H2SO4 2K2SO4 + 4MnSO4 + 6H2O + 5O2
(iii) When heated in a current of H2, solid KMnO4 gives KOH, MnO and steam.
2KMnO4 + 5H2 2KOH + 2MnO + 4H2O
(iv) Oxidation reactions of potassium permanganate : Potassium permanganate is a powerful oxidising reagent and is used for this
purpose in acidic, neutral or alkaline solutions.
(a) In acidic medium : In acidic medium, the potential oxidation equation for potassium permanganate is :
2KMnO4 + 3H2SO4 K2SO4 + 2MnSO4 + 3H2O + 5O
or ionically :
MnO4– + 8H+ + 5e– Mn2+ + 4H2O ; E° = 1.5 volts.
Since in the above reaction five atoms of oxygen (i.e. 10 equivalents) are made available from two molecules of KMnO4, the
equivalent weight of KMnO4 in this medium is one fifth of its molecular weight, i.e. 158/5 = 31.6.
In acidic medium KMnO4 oxidises sulphides H2S to S, SO2 to H2SO4, SO32– to SO42–, NO2– to NO3–, AsO32– to AsO43–,
oxalates or oxalic acid to CO2, Fe(ous) salts to Fe(ic) salts, H2O2 to H2O + O2, KI to I2, HX to X2, ethyl elcohol to acetaldehyde
etc.
(i) 2KMnO4 + 3H2SO4 K2SO4 + 2MnSO4 + 3H2O + 5O
[H2S + O H2O + S] × 5
––––––––––––––––––––––––––––––––––––––––––––––––
2KMnO4 + 3H2SO4 + 5H2S K2SO4 + 2MnSO4 + 8H2O + 5S
or 2MnO42– +16H+ + 5S2– 2Mn2+ + 8H2O + 5S
––––––––––––––––––––––––––––––––––––––––––––––––
(ii) 2KMnO4 + 3H2SO4 K2SO4 + 2MnSO4 + 3H2O + 5O
[SO2 + H2O + O H2SO4] × 5
––––––––––––––––––––––––––––––––––––––––––––––––
2KMnO4 + 5SO2 + 2H2O K2SO4 + 2MnSO4 + 2H2SO4
7
Gyaan Sankalp
The d- and f-block elements
or 2MnO4– + 5SO2 + 2H2O 5SO42– + 2Mn2+ + 4H+
––––––––––––––––––––––––––––––––––––––––––––––––
(iii)2MnO42– + 6H+ + 5SO2 2Mn2+ + 5SO42– + 3H2O
Similar equations can be written for the oxidation of NO2– to NO3– , AsO33– to AsO43– and thiosulphates to sulphates and
sulphur.
(iv) 2KMnO4 + 3H2SO4 K2SO4 + 2MnSO4 + 3H2O + 5O
[C2H2O4 + O H2O + 2CO2] × 5
––––––––––––––––––––––––––––––––––––––––––––––––
2KMnO4 + 3H2SO4 + 5C2H2O4 K2SO4 + 2MnSO4 + 8H2O + 10CO2
or 2MnO42– + 16H+ + 5C2O42– Mn2+ + 8H2O + 10CO2
––––––––––––––––––––––––––––––––––––––––––––––––
(v) 2KMnO4 + 3H2SO4 K2SO4 + 2MnSO4 + 3H2O + 5O
[2FeSO4 + H2SO4 + O Fe2(SO4)3 + H2O] × 5
––––––––––––––––––––––––––––––––––––––––––––––––
2KMnO4 + 8H2SO4 + 10FeSO4 K2SO4 + 2MnSO4 + 5Fe2(SO4)2 + 8H2O
or 2MnO42– + 16H+ + 10Fe2+ 2Mn2+ + 8H2O + 10Fe+3
––––––––––––––––––––––––––––––––––––––––––––––––
(vi) 2KMnO4 + 3H2SO4 K2SO4 + 2MnSO4 + 3H2O + 5O
[H2O2 H2O + O] × 5
O + O O2] × 5
––––––––––––––––––––––––––––––––––––––––––––––––
2KMnO4 + 3H2SO4 + 5H2O2 K2SO4 + 2MnSO4 + 8H2O + 5O2
or 2MnO42– + 6H+ + 5H2O2 2Mn2+ + 8H2O + 5O2
(vii) 2KMnO4 + 3H2SO4 K2SO4 + 2MnSO4 + 3H2O + 5O
[2KI + H2SO4 K2SO4 + 2HI] × 5
[2HI + O H2O + I2] × 5
––––––––––––––––––––––––––––––––––––––––––––––––
2KMnO4 + 8H2SO4 + KI K2SO4 + 2MnSO4 + 8H2O + 5I2
or 2MnO42– + 16H+ + 10I– 2Mn2+ + 8H2O + 5I2
(viii) 2KMnO4 + 3H2SO4 K2SO4 + 2MnSO4 + 3H2O + 5O
[2HX + O H2O + X2] × 5
––––––––––––––––––––––––––––––––––––––––––––––––
2KMnO4 + 3H2SO4 + 10HX K2SO4 + 2MnSO4 + 8H2O + 5X2
or 2MnO42– + 16H+ + 10X– 2Mn2+ + 8H2O + 5X2
[X = Cl, Br, I]
––––––––––––––––––––––––––––––––––––––––––––––––
KMnO 4
(ix) CH3CH2OH + O 
 CH3CHO + H2O
Ethyl alcohol
Acetyldehyde
(b) In alkaline and neutral media : Potassium permanganate changes to green manganate in presence of alkali.
2KMnO4 + 2KOH 2K2MnO4 + H2O + O
............. (i)
and the manganate is also further reduced to MnO2 when a reducing substance is present.
2K2MnO4 + 2H2O 2MnO2 + 4KOH + 2O
............. (ii)
So the complete potential reaction is :
2KMnO4 + H2O 2MnO2 + 2KOH + 3O
............. (iii)
or MnO4– + 2H2O + 3e– MnO2 + 4OH– ; E° = 1.23 volts
In all these cases manganese dioxide is precipitated. The potential equation for the neutral media is also the same, in which from
two molecules of permanganate, three atoms of oxygen are available. Infact, during the course of reaction, the alkali generated
makes the medium alkaline even when we start with neutral solutions.
Quite evidently the equivalent weight of KMnO4 in alkaline and neutral media is one-third of its molecular weight.
In strongly alkaline solution MnO42– ion (instead of MnO2) is produced.
2KMnO4 + 2KOH 2K2MnO4 + H2O + O
or MnO4– + e– MnO42– ; E° = 0.56 volts
Alkaline permanganate is very much used in the oxidation of organic compounds. Thus it will oxidise nitro-toluene to nitrobenzoic
acid :
8
Gyaan Sankalp
The d- and f-block elements
NO2C6H4CH3 + 3O  NO2C6H4COOH + H2O
The oxidation of ammonia to nitrogen proceeds thus :
2KMnO4 + H2O 2KOH + 2MnO2 + 3O
2NH3 + 3O  N2 + 3H2O
––––––––––––––––––––––––––––––––––––––
2KMnO4 + 2NH3 2KOH + 2MnO2 + N2 + 2H2O
In alkaline medium, potassium iodide is oxidised to potassium iodate by potassium permanganate :
2KMnO4 + H2O + KI 2MnO2 + 2KOH + KIO3
and in this case, iodine is not liberated as in acid medium.
In neutral medium, (i) hot manganous sulphate is oxidises to manganese dioxide by KMnO4 :
2KMnO4 + H2O 2KOH + 2MnO2 + 3O
2MnSO4 + 3H2O + 3O 3MnO2 + 3H2SO4
2KOH + H2SO4 K2SO4 + 2H2O
––––––––––––––––––––––––––––––––––––––
3MnSO4 + 2KMnO4 + 2H2O 5MnO2 + K2SO4 + 2H2SO4
(ii) KMnO4 oxidises Na2S2O3 to Na2SO4
3Na2S2O3 + 8KMnO4 + H2O 3Na2SO4 + 8MnO2 + 3K2SO4 + 2KOH
(ii) 2KMnO4 + 4H2S  2MnS + S + K2SO4 + 4H2O
(iv) The reaction with hydrogen peroxide is characteristic. Potassium permanganate is reduced to manganese dioxide and oxygen
is evolved :
5H2O2 + 2KMnO4 2MnO2 + 2KOH + 4H2O + 4O2
or in acidic solution, manganeous sulphate is formed :
5H2O2 + 2KMnO4 + 3H2SO4 K2SO4 + 2MnSO4 + 8H2O + 5O2
Uses : Potassium permanganate is used as a disinfectant and as an oxidising agent. It is valuable volumetric reagent and is used
for the estimation of ferrous salts, oxalic acid, hydrogen peroxide, etc. Alkaline KMnO4 is used in organic chemistry under the
name of Baeyer’s reagent.
Example 1 :
On what ground can you say that scandium (Z = 21) is a transition element but zinc (Z = 30) is not?
Sol. On the basis of incompletely filled 3d orbitals in case of scandium atom in its ground state (3d1), it is regarded as a transition
element. On the other hand, zinc atom has completely filled d orbitals (3d10) in its ground state as well as in its oxidised state,
hence it is not regarded as a transition element.
Example 2 :
In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 kJ mol–1. Why?
Sol. In the formation of metallic bonds, no electrons from 3d-orbitals are involved in case of zinc, while in all other metals of the 3d
series, electrons from the d-orbitals are always involved in the formation of metallic bonds.
Example 3 :
How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition
elements?
Sol. Irregular variation of ionisation enthalpies is mainly attributed to varying degree of stability of different 3d-configurations
(e.g., d0, d5, d10 are exceptionally stable).
Example 4 :
For the first row transition metals the E values are:
V
Cr
Mn
Fe
Co
Ni
E
(M2+/M) –1.18
– 0.91 –1.18 – 0.44
– 0.28 – 0.25
Explain the irregularity in the above values.
Cu
+0.34
Sol. The E (M2+/M) values are not regular which can be explained from the irregular variation of ionisation enthalpies
(1H1+ 1H2) and also the sublimation enthalpies which are relatively much less for manganese and vanadium.
Example 5 :
Explain why Cu+ ion is not stable in aqueous solutions?
Sol. Because of small size and high electronegativity oxygen or fluorine can oxidise the metal to its highest oxidation state.
TRY IT YOURSELF
Q.1 Which of the following will not related with first transition series –
(A) Fe
(B) V
(C) Cu
Q.2 Which of the following pair of ion having 2.83 magnetic moment (A) V+3, Ni+2
(B) V+2, Cr+3
(C) Co+2, Zn+2
(D) Ag
(D) Sc+3, Mn+2
Gyaan Sankalp
9
The d- and f-block elements
Q.3 Transition elements are coloured –
(A) Due to small size
(B) Due to metallic nature
(C) Due to unpaired d–electrons
(D) All the above
Q.4 The colour of KMnO4 solution is decolourised by Fe2+ solution, one mole of Fe2+ reacts with --- Moles of KMnO4 (A) 5
(B) 1/5
(C) 4
(D) 1/4
Q.5 Which is most dark in colour (A) Mn+2
(B) Cu+1
(C) Ti+3
(D) Zn+2
Q.6 Elements which generally exhibit multiple oxidation states and whose ions are usually coloured are –
(A) Metalloids
(B) Transition elements (C) Non–metals
(D) Gases
Q.7 In which of the following ionic radii of chromium would be smallest –
(A) K2CrO4
(B) CrO2
(C) CrCl3
(D) CrF2
Q.8 The placement of Zn, Cd and Hg along with 'd' block elements is not proper because –
(A) Their 'd' orbitals are completely filled
(B) Their 'd' orbitals are empty
(3) They do not form complex compounds
(D) They do not form coloured compounds
Q.9 Which of the following electronic configuration belong to those of transition elements –
(A) KL 3s2p6d54s1
(B) KL 3s2p6d104s2p3
(C) KL 3s23p63d104s1
(D) KLM 4s24p64d105p1
Q.10 Oxidation number of Mn in KMnO4 –
(A) +4
(B) +6
(C) –6
(D) +7
ANSWERS
(1) (D)
(6) (B)
(2) (A)
(7) (A)
(3) (C)
(8) (A)
(4) (B)
(9) (A)
(5) (C)
(10) (D)
THE INNER TRANSITION ELEMENTS (f-BLOCK)
They were earlier called as rare earth metals as it was believed that they exist in earth’s crust to a very less extent. But this
terminology is now not applicable as they exist in earth’s crust to a sufficient extent.
The elements in which the additional electron enters in (n - 2)f orbitals are called inner transition elements or f-block elements.
Position in the periodic table : The lanthanides resemble ytterbium in most of their properties. So it became necessary to
accommodate all the fifteen elements together at one place. This has been done by placing the first element, lanthanum below
ytterbium and placing the remaining fourteen elements separately in the lower part of the periodic table.
Lanthanide series (Z = 58 - 71)
(Ce – Lu)
Actinide series
(Z = 90 - 103)
(Th – Lw)
Lanthanides (Lanthanones) : Lanthanides are reactive elements so do not found in free state in nature. Most important minerals
for lighter Lanthanides are - Monazite, cerites and orthite and for heavier lanthanides - Gadolinite and Xenotime
Electronic configuration : The general configuration of lanthanides may be given as 4f2-145s25p65d0/16s2.They have outer
three shells incomplete.
Atomic Number
58
59
60.
61.
62.
63.
64.
65.
66.
67.
68.
69.
70.
71.
Element
Atomic
Cerium
Praseodymium
Neodymium
Promethium
Samarium
Europium
Gadolinium
Terbium
Dysprosium
Holmium
Erbium
Thulium
Ytterbium
Lutecium
Symbol
+3 ion
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
Outer electronic configuration
4f2 6s2 4f1
4f3 6s2 4f2
4f4 6s2 4f3
4f5 6s2 4f4
4f6 6s2 4f5
4f7 6s2 4f6
4f7 5d1 6s2 4f7
4f9 6s2 4f8
4f10 6s2 4f9
4f11 6s2 4f10
4f12 6s2 4f11
4f13 6s2 4f12
4f14 6s2 4f13
4f14 5d1 6s2 4f14
It is to be noted here that filling of 4f orbitals in the atoms is not regular. A 5d electron appears in gadolinium (z = 64) with an outer
electronic configuration of 4f75d16s2 (and not 4f86s2). This is because the 4f and 5d electrons are at about the same potential
energy and that the atoms have a tendency to retain stable half filled configuration.
10
Gyaan Sankalp
The d- and f-block elements
On the other hand, the filling of f orbitals is regular in tripositive ions.
After losing outer electrons, the f orbitals shrink in size and became more stable. Pm is the only synthetic radioactive lanthanide.
Atomic and ionic sizes : In the lanthanide series with increasing atomic number, there is a progressive decrease in the size from
lanthanum to lutecium (La+3 to Lu+3). This contraction in size is known as lanthanide contraction. The general electronic
configuration of these elements is 4f0–145s2p6d0-16s2. In these elements the added electron enters the deep seated f-orbitals
and therefore experiences considerable pull by the nucleus. Such an electron cannot add to the size of the element and also
because the intervening 5s2p6d1 electronic shells, it is very little screening effect on the outermost 6s2 electrons. Hence with
increasing atomic number, the enhanced nuclear charge leads to contraction in the size of atoms and ions. The atomic volumes
of europium and ytterbium are unexpectedly large. The large atomic size of Eu and Yb suggest weaker bonding in the solid
elements. Both these elements have only two electrons extra than the stable configurations (half filled, f7, and completely filled,
f14), hence they utilise two electrons in metallic bonding as in the case with barium.
Effects of Lanthanide Contraction :
Close resembalace of Lanthanides :- The general decrease in the sizes of the lanthanides with an increase in their nuclear
charges result in a small increase in their ionisation energies. Hence their basic and ionic nature gradually decreases from La to
Lu. This also explains the variations in properties such as increased tendency for hydrolysis and formation of complex salts and
decreased thermal stability. solubility of their salts.
(ii) Similarity of ytterbium with lanthanides :- The properties of ytterbium are so similar to the lanthanides that it is considered
more a member of the lanthanide series than a congener of scandium.
(iii) Anomalous behaviour of post-lanthanides :- The following anomalies may be observed in the behaviour of post-lanthanide
elements.
(a) Atomic size : The ionic radii of Zr+4 is about 9% more than Ti+4. Similar trend is not maintained on passing from the second
to third transition series. The ionic radius of Hf+4, instead of increasing (because of inclusion of one more electronic shell)
decreases (or is virtually equal to Zr+4) as a consequence of the lanthanide contraction. This explains the close similarities
between the members of the second and third transition series than between the elements of the first and second series.
(b) lonisation potential and electronegativity : The effect of lanthanide contraction is also seen in the increase in the ionisation
potential values and electronegativities of the elements of the third transition series, contrary to the general trend.
Because of the lanthanide contraction, the post-lanthanide elements have stronger positive field and thus the electrons are held
more tightly. The greater effective nuclear charge of the former make them more electronegative than the latter.
(c) High density : Because of lanthanide contraction the atomic sizes of the post lanthanide elements become very small
consequently, the packing of atoms in their metallic crystals become so much compact that their densities are very high.
The densities of the third transition series elements are almost double to those of the second series elements.
(i)
Oxidation states : The lanthanides contains two s electrons in the outermost shell, they are therefore expected to exhibit a
characteristic oxidation state of +2. But for the lanthanides, the +3 oxidation is common. This corresponds to the use of two
outermost electrons (6s2) alongwith one inner electron. The inner electron used is a 5d electron (in La, Gd and Lu), or one of the
4f electron if no 5d electrons present. All the lanthanides attains +3 oxidation state and only cerium, Praseodymium, and terbium
exhibit higher oxidation state (+4).
Oxidation states + 2 and +4 occur particularly when they lead to
(i) A noble gas configuration e.g. Ce4+ (f0)
(ii) A half filled ‘f ‘ orbital e.g. Eu2+, Tb4+, (f7)
(iii) A completely filled ‘f ‘ orbital e.g. Yb2+ (f14)
Therefore, in higher oxidation state, they act as oxidising while in lower state as reducing agents.
Lanthanides
Ce58
Nd60
Sm62
Gd64
Dy66
Er68
Yb70
Oxidation States
+3,+4
+3
(+2), +3
+3
+3; (+4)
(+2), +3
+2,+3
Lanthanides
Pr59
Pm61
Eu63
Tb65
Ho67
Tm69
Lu71
Oxidation States
+3, (+4)
+3
+2,+3
+3,+4
+3
(+2), +3
+3
Oxidation states in brackets are unstable states.
Gyaan Sankalp
11
The d- and f-block elements
Magnetic properties : In tripositive lanthanide ions the number of unpaired electrons regularly increases from lanthanum to
Gadolinium (0 to 7) and then continuously decreases upto lutecium (7 to 0). So lanthanum and lutecium ions which are diamagnetic, all other tripositive lanthanide ions are paramagnetic.
Colour - The lanthanide ions have unpaired electrons in their 4f orbitals. Thus these ions absorbs visible region of light and
undergo f-f transition and hence exhibit colour. The colour exhibited depends on the number of unpaired electrons in the 4f
orbitals. The ions often with 4fn configuration have similar colour to those ions having 4f14–n configuration.
Lanthanide ions having 4f0, 4f7,4f14 are colourless. Lanthanide ions 4f1 and 4d13 are also colourless.
Other Properties :
(a) All the lanthanoids are silvery white soft metals and tarnish rapidly in air.
(b) Highly dense metals with high m.pts. (do not show any regular trend).
(c) lonisation Energies - Lanthanides have fairly low ionisation energies comparable to alkaline earth metals.
(d) Electro positive character - High due to low I.P.
(e) Complex formation - Do not have much tendency to form complexes due to low charge density because of their large size.
Lu+3 is smallest in size can only form complex.
(e) Reducing Agent - They readily lose electrones so are good reducing agent.
In +3 oxidation states, nitrates, perchlorates and sulphates of lanthanides and actinides are water soluble, while their hydroxides,
fluorides and carbonates are water insoluble. Alloys of lanthanides with Fe are called misch metals. La(OH)3 is most basic in
nature while Lu(OH)3 least basic. Lanthanides form MC2 type carbide with carbon, which on hydrolysis gives C2H2.
General reactions of lanthanoids are :
heat gently
With hydrogen :
 LnH 2 (s)
Ln + H2(g) 
With oxygen :
burn in O 2
4Ln + 3O2(g) 
 2LnO 2 (s)
With sulphur :
heat
2Ln + 3S(s) 
 Ln 2S3 (s)
With carbon :
2773K
Ln + 2C 

 LnC2 (s)
With nitrogen :
heat
2Ln + N2(g) 
 2LnN
With water :
2Ln + 6H2O(  ) 
 2Ln(OH)3  3H 2 (g)
With halogens :
heat
2Ln + 3X2(g) 
 2LnX3
With acids :
normal temp.
2Ln + 6H+(aq) 
 2Ln 3 (aq)  3H 2 (g)
from acid
573 673K
ACTINIDES (5f - BLOCK ELEMENTS)
The elements in which the extra electron enters 5f-orbitals of (n - 2)th main shell are known as actinides. The man-made eleven
elements Np93 - Lr103 are placed beyond uranium in the periodic table and are collectively called trans-uranic elements. Th, Pa
and U first three actinides are natural elements.
Electronic configuration :The general configuration of actinides may be given as 5f1–14 6d0/1, 7s2.
Atomic No.
Elements
Symbol
Electronic Configuration
90
Thorium
Th
6d27s2
91
Protactinium
Pa
5f26d17s2
92
Uranium
U
5f36d17s2
93
Neptunium
Np
5f46d17s2
94
Plutonium
Pu
5f67s2
95
Americium
Am
5f77s2
96
Curium
Cm
5f76d17s2
97
Berkellium
Bk
5f97s2
98
Californium
Cf
5f10 7s2
99
Einstenium
Es
5f117s2
100
Fermium
Fm
5f12 7s2
101
Mandelevnium
Md
5f137s2
102
Nobellium
No
5f147s2
103
Lawrencium
Lw
5f14 6d17s2
12
Gyaan Sankalp
The d- and f-block elements
Ionic sizes : The general trend in lanthanoids is observable in the actinoids as well. There is a gradual decrease in the size of
atoms or M3+ ions across the series. This may be referred to as the actinoid contraction (like lanthanoid contraction). The
contraction is, however, greater from element to element in this series resulting from poor shielding by 5f electrons.
Oxidation states : In lanthanides-and actinides +3 oxidation is the most common for both of the series of elements. This oxidation
state becomes increasingly more stable as the atomic number increases in the actinide series.
Actinides
Th 90
Pa 91
U92
Np93
Pu94
Am95
Cm96
Oxidation state
+4
(+4), +5
(+3), (+4), (+5), +6
(+3), (+4), +5, (+6), (+7)
(+3), +4, (+5), (+6), (+ 7)
+2,(+3),(+4),(+5),(+6)
+3, (+4)
Actinides
Bk97
Cf98
Es99
Fm100
Md 101
No102
Lw103
Oxidation state
+3, (+4)
+3
+3
+3
+3
+3
+3
General characteristics and comparison with Lanthanoids : The actinoid metals are all silvery in appearance but display a
variety of structures. The structural variability is obtained due to irregularities in metallic radii which are far greater than in
lanthanoids.
The actinoids are highly reactive metals, especially when finely divided. The action of boiling water on them, for example, gives
a mixture of oxide and hydride and combination with most non metals takes place at moderate temperatures. Hydrochloric acid
attacks all metals but most are slightly affected by nitric acid owing to the formation of protective oxide layers; alkalies have no
action.
The magnetic properties of the actinoids are more complex than those of the lanthanoids. Although the variation in the magnetic
susceptibility of the actinoids with the number of unpaired 5 f electrons is roughly parallel to the corresponding results for the
lanthanoids, the latter have higher values.
It is evident from the behaviour of the actinoids that the ionisation enthalpies of the early actinoids, though not accurately
known, but are lower than for the early lanthanoids. This is quite reasonable since it is to be expected that when 5f orbitals are
beginning to be occupied, they will penetrate less into the inner core of electrons. The 5f electrons, will therefore, be more
effectively shielded from the nuclear charge than the 4f electrons of the corresponding lanthanoids. Because the outer electrons
are less firmly held, they are available for bonding in the actinoids.
APPLICATIONS OF D- AND F-BLOCK ELEMENTS :
Iron and steels are the most important construction materials. Their production is based on the reduction of iron oxides, the
removal of impurities and the addition of carbon and alloying metals such as Cr, Mn and Ni. Some compounds are manufactured
for special purposes such as TiO for the pigment industry and MnO2 for use in dry battery cells. The battery industry also
requires Zn and Ni/Cd. The elements of Group 11 are still worthy of being called the coinage metals, although Ag and Au are
restricted to collection items and the contemporary UK ‘copper’ coins are copper-coated steel. The ‘silver’ UK coins are a Cu/Ni
alloy. Many of the metals and/or their compounds are essential catalysts in the chemical industry. V2O5 catalyses the oxidation
of SO2 in the manufacture of sulphuric acid. TiCl4 with A1(CH3)3 forms the basis of the Ziegler catalysts used to manufacture
polyethylene (polythene). Iron catalysts are used in the Haber process for the production of ammonia from N2/H2 mixtures.
Nickel catalysts enable the hydrogenation of fats to proceed. In the Wacker process the oxidation of ethyne to ethanal is
catalysed by PdCl2. Nickel complexes are useful in the polymerisation of alkynes and other organic compounds such as benzene.
The photographic industry relies on the special light-sensitive properties of AgBr.
Example 6 :
Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.
Sol. Cerium (Z = 58)
Example 7 :
Actinoid contraction is greater from element to element than lanthanoid contraction. Why?
Sol. The 5f electrons are more effectively shielded from nuclear charge. In other words the 5f electrons themselves provide poor
shielding from element to element in the series.
Gyaan Sankalp
13
The d- and f-block elements
TRY IT YOURSELF
Q.1 The radius of La3+ is 1.06 Å which of the following given values will be closest to the radius of Lu3+ (At. no. Lu = 71, La = 57)
(A) 1.6 Å
(B) 1.4 Å
(C) 1.06 Å
(D) 0.85 Å
Q.2 Which property is less affected by “Lanthanide contraction”.
(A) Basic strength
(B) Ionisation energy
(C) Complex formation
(D) All of above
Q.3 The lanthanide contraction is responsible for the fact that(A) Zr and Y have about the same radius
(B) Zr and Nb have similar oxidation state
(C) Zr and Hf have about the same radius
(D) Zr and Zn have the same oxidation state
Q.4 Which of the following transition metals shows only + 3 oxidation state :–
(A) Ce
(B) Pt
(C) Nd
(D) Gd
Q.5 As the increase in atomic number of lanthanide element, the ionic radius.
(A) Increases
(B) Decreases
(C) Almost unchanged
(D) Irregular change
Q.6 Which of the following statements is not correct (A) La(OH)3 is less basic than Lu(OH)3
(B) La is actually an element of transition series rather than lanthanides
(C) Atomic radii of Zr and Hf are same because of Lanthanide contraction
(D) In lanthanide series the ionic radius of Lu3+ is smallest
Q.7 Zr and Hf have almost equal atomic and ionic radii because (A) of diagonal relationship
(B) of lanthanide contraction
(C) of actinide contraction
(D) both belong to f-block of elements
Q.8 Lanthanide ion, which is diamagnetic in nature.
(A) La3+
(B) Lu3+
(C) Gd3+
(D) (A) and (B) both
Q.9 The lanthanide element which is obtained by synthesis (A) Lu
(B) Pr
(C) Pm
(D) Gd
Q.10 Which of the following is the strongest base (A) Sc(OH)3
(B) La(OH)3
(C) Ln(OH)3
(D) Yb(OH)3
ANSWERS
(1) (D)
(6) (AD)
(2) (D)
(7) (B)
(3) (C)
(8) (D)
(4) (D)
(9) (C)
(5) (B)
(10) (B)
USEFUL TIPS
1.
11.
The highest oxidation states are found in fluorides and oxides because fluorine and oxygen are the most electronegative
elements. The highest oxidation state shown by any transition metal is eight. The oxidation state of eight is shown by Ru and Os.
Most of the transition elements and their compounds show paramagnetism due to the presence of one or more unpaired
electrons in them. Paramagnetism in any transition series first increases, reaches a maximum value of d5 cases and then decreases
thereafter.
The elements of first row transition metals (3d-series) from oxides of the type, MO, M2O3, M3O4, MO2, M2O5 and MO3.
The oxides of metals in the intermediate oxidation states, are generally amphoteric. For example, CuO, Cr2O3, MnO2 etc. are
amphoteric.
Halides of transition metals in the higher oxidation states exhibit a greater tendency towards hydrolysis.
Potassium permanganate (KMnO4) is the salt of an unstable acid HMnO4 (permanganic acid).
When a solid chloride salt is heated with potassium dichromate and conc. H2SO4, orange coloured vapour of chromyl chloride
is obtained.
Lanthanoids in aqueous solution and in the solid state exhibit oxidation states of +2, +3 and +4. The trivalent state (+3) being
more stable.
Trivalent ions of lanthanoids (except La3+ and Lu3+) are paramagnetic.
Direct consequences of the lanthanoid contraction are (a) almost equal size of Hf4+ and Zr4+. (b) Decrease in the basic character
of lanthanoid hydroxides with increase in the atomic number. La(OH)3 is the most basic, while Lu(OH)3 is least basic.
Actinoids show variable oxidation states. The oxidation state of +3 is the most common.
14
Gyaan Sankalp
2.
3.
4.
5.
6.
7.
8.
9.
10.
The d- and f-block elements
MISCELLANEOUS SOLVED EXAMPLES
Example 1 :
When Mn(OH)2 is made by adding an alkali to a solution containing Mn2+ ions, the precipitate quickly darkens, and eventually
goes black. What might be the chemical giving the black colour, and how is it made ?
Sol. The black colour is due to the manganese (IV) oxide, MnO2. It is made by the Mn(OH)2 being oxidised by oxygen in the air.
Mn(OH)2  MnO + H2O
1
MnO + O  MnO2
2
air
black
Example 2 :
What is the electronic difference between lanthanoids and actinoids ? Why is Eu(II) more stable than Ce(II)
Sol. In Lanthanoid last electron enter in 4f subshell where as in actinoid last electron enter in 5f subshell
Eu+2 = [Xe]4f7
Ce+2 = [Xe]4f2
Eu+ has half filled 4f subshell
Example 3 :
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 configuration of these elements.
Sol. (i) Ce = [Xe]4f25d06s2 ; Ce+4 = [Xe]
Stable inert gas configuration
(ii) Tb = [Xe]4f95dº6s2 ; Tb+4 = [Xe]4f7 ; Half filled ‘f’ subshell.
(iii) Eu = [Xe]4f75d06s2 ; Eu+2 = [Xe]4f7 ; Half filled ‘f’ subshell is more stable
(iv) Yb = [Xe]4f145d06s2 ; Yb+2 = [Xe]4f14 ; Fully filled ‘f’ subshell.
Example 4 :
Describe the two uses of each of the following :
(a) Copper sulphate (b) Silver bromide
Sol. (a) (i) Anhydrous CuSO4 is used for detections of moisture (water) in organic liquid such as alcohol and ether.
(ii) Used as a fungicide under the name bordeaux mixture [CuSO4 + Ca(OH)2].
(iii) Used as a electrolyte in electroplating
(b) (i) In qualitative and quantitative analysis (ii) Preparing silver halide
(iii) AgBr is photosensative and hence are widely used in photography.
Example 5 :
Indicate the steps in the preparation of :–
(a) K2Cr2O7 from chromite ore
(b) KMnO4 from pyrolusite ore
(c) Copper sulphate from metallic copper
Sol. (a) (i) Conversion of chromite ore to sodium chromate
4FeCr2O4 + 16NaOH + 7O2  8Na2CrO4 + 2Fe2O3 + 8H2O
(ii) Conversion of sodium chromate to sodium dichromate
2Na2CrO4 + H2SO4  Na2Cr2O7 + 2Na2SO4 + H2O
(iii) Conversion of sodium dichromate to potassium dichromate
Na2Cr2O7 + 2KCl  K2Cr2O7 + 2NaCl
(b) (i) Convertion of pyrolusite ore to potassium manganate.

2MnO2 + 4KOH + O2  2K2MnO4 + 2H2O
(ii) Oxidation of potassium manganate to potassium permanganate.
2K2MnO4 + Cl2  2KCl + 2KMnO4
(c) 2Cu + 2H2SO4 + O2(air) 2CuSO4 + 2H2O
Example 6 :
Compare the periodic trends in the transition elements with reference to the following
(i) electronic configuration
(ii) oxidation states
(iii) colour
(iv) atomic radii
(v) covalent radii
(vi) ionization energies
Sol. Trends in the periodic table
Electronic configuration : They are known as d-block elements. they have partially filled d-shells. The number of electrons in the
outermost shell remains the same. Elements with atomic number Z = 21 to Z = 30 are known as first series to transition elements.
All elements have 2 electrons in the 4s shell except Cr and Cu which have exactly half filled or fully filled d orbitals and so they
have one electron in the 4s shell. The second set of transition elements come in the fifth period. They have atomic number from
Z = 37 to Z = 54.
15
Gyaan Sankalp
The d- and f-block elements
Oxidation states : The transition elements show a side variation oxidation states depending on the electronic structure.
All the elements exhibit +2 and +3 oxidation state. Chromium and copper exhibit +1 valency state.
Ti, V, Cr, Mn, Fe, Co, Ni, exhibit tetra valent states.
+5 oxidation states is exhibited by V, Cr, Mn and Fe, +6 by Cr, Mn and Fe, +7 by Mn only
Colour : Transition metal compound are often coloured in the solid or solution states. The transition metal ions have incomplete
‘d’ shells. The energy needed for transitions with in the d shell is small and therefore they appear coloured by absorbing light in
the visible region.
Atomic radii : The atomic radii are seen to decrease with the increase in atomic number. Increase in the nuclear charge tend to
draw the electronic cloud inside. The radii for elements from chromium to copper is very close to each other.
Ionic radii : The ionic radii follow the same order as atomic radii
Ionization energies : The ionization energies are fairly high, these elements are less electropositive than s and p block elements.
The ionization energy along a given period increases with the atomic number. The d-electrons provide a screening effect. Thus
the effect of increasing nuclear charge and the expansion of d-orbitals oppose each other.
Melting and boiling points : The melting and boiling points of the transition elements are generally very high. Zn Cd, Hg are
notable exceptions with low melting points.
Example 7 :
Identify R, S, T ....... X.
KOH
K 2 Cr2 O7  R 
S
Lead Acetate
 NH 4 Cl

Al
T 
 OU  (V)  W 

(Gas)
X
(Oxide)
(Metal)
(CH3COO) 2 Pb
KOH
Sol. K 2 Cr2 O7 
 K 2 CrO 4 
 PbCrO 4  2CH3COOK
(R )
(S)
 NH 4 Cl

Al
(NH 4 )2 Cr2 O7 
 N 2  Cr2 O3  Al2 O3  Cr
 H 2O (U)
(Gas)
(T)
(V)
(Gas)

(X)
Oxide
(W)
Metal
Example 8 :
MnO42– change to MnO2 & MnO4– in acidic medium. Calculate equivalent weight of MnO42–.
+6
Oxidation
–
–e
+7
Sol. MnO 42– + 2H +  MnO 2 + 2MnO 4– + H 2O
–
+2e
Reduction
+4
Thus in such disproportion reaction
Equivalent Wt. = Eq. wt. (oxidation) + Eq. Wt. (Reduction)
 Eq. wt. of MnO42– =
Eq. wt. of MnO42– =
Mol. Weight
Mol. Weight M
M
+
=
+
1
2
1
2
3
Mol. Wt.
2
Example 9 :
What is meant by ‘disproportionation’ ? Give two examples of disproportionation reaction in aqueous solution.
Sol. The reaction in which a particular oxidation state of an element becomes less stable relative to the two other oxidation states, one
lower and one higher, is called disproportionation. Thus, the species undergoing disproportionation acts both as self-oxidizing
and self reducing agent. Examples :
(i) In acidic medium, Mn6+ disproportionates into Mn4+ and Mn7+.
(ii)
Cu+
 2MnO42– + MnO2 + 2H2O
3MnO42– + 4H+ 
disproportionates into Cu2+ and Cu0
 Cu2+ + Cu0
2Cu+ 
16
Gyaan Sankalp
The d- and f-block elements
Example 10 :
Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why ?
Sol. Copper exhibits +1 oxidation state because of the extra stability of completely filled 3d orbitals, (3d10).
Example 11 :
Determine the equivalent weight of potassium permanganate from the reaction given
(a) 2MnO4– + 2H2O  MnO4 + 4OH–
(b) 2MnO4 + 8H+  MnSO4 + 4H2O
Sol. (a) As the reaction is a king place in alkaline solution.
2MnO4– + 2H2O + 3e–  MnO4 + 4OH–
Mol. weight 158
=
= 52.6
3
3
(b) 2MnO4– + 8H+ + 5e–  MnSO4 + 4H2O
+7
+2
As the reaction is taking place acidic medium
 Equivalent weight =
Eq. Wt. =
Mol. Weight 158
=
= 31.6
5
5
Example 12 :
Dimercury (I) iodide, Hg2I2 is a greenish colour and is precipitated if iodide ions are added to a solution of dimercury (I) sulphate.
Hg2SO4. Likewise the red mercury (II) iodide, HgI2, is precipitated from a solution of mercury (II) sulphate, HgSO4. However,
both precipitates dissolves in excess iodide solution. What might be the reason for this ?
Sol. It is due to formation of HgI42– (a soluble complex) in both the cases with HgI2 :
 HgI42–
HgI2 + 2I– 
But in Hg2I2, first there is oxidation of Hg(I) to Hg(II) and then complex formation takes place, it is by following disproportion
reaction :
 Hg42– + Hg
Hg22+ + 4I– 
+1
+2
0
Example 13 :
What happens when(i) Ferric chloride is added to potassium ferricyanide
(ii) Potassium ferricyanide is added to ferrous sulphate
(iii) Excess of potassium iodide is added to mercuric chloride
(iv) Ammonium thiocyanate is added to ferric chloride solution.
Sol. (i) Prussian blue is formed
4FeCl3 + 3K4Fe(CN)6  Fe4[Fe(CN)6]3 + 12KCl
Prussian blue
(Ferric ferrocyanide)
(ii) Ferrous ion is first oxidised to ferric ion while ferricyanide ion is reduced to ferrocyanide ion. Then, ferric ions react with
ferrocyanide ions to form potassium ferric ferrocyanide (Turnbull’s blue).
Fe2+ + [Fe(CN)6]3–  Fe3+ + [Fe(CN)6]4–
K+ + Fe3+ + [Fe(CN)6]4–  K+ Fe3+ [Fe(CN)6]4–
Pot. Ferricferrocyanide (Turnbull’s blue)
(iii)First scarlet precipitate is formed which then dissolves in excess of potassium iodide forming a complex.
HgCl2 + 2KI  HgI2 + 2KCl
HgI2 + 2KI  K2HgI4
Pot. tetraiodo mercurate (colourless)
(iv) Deep red colouration due to the formation of a complex is developed.
FeCl3 + NH4CNS  Fe(CNS)Cl2 + NH4Cl
or FeCl3 + 3NH4CNS  Fe(CNS)3 + 3NH4Cl
Example 14 :
(a) Fill in the blanks as directed.
(i) Cu(I) is . . . . . . . . . . and Cu (II) is . . . . . . . . . . (paramagnetic/diamagnetic)
(ii) Zn unlike other transition elements show only one oxidation state of . . . . . . .
(iii)The covalent radius of cadmium is . . . . . . . . . than that of Ag (more/less)
(iv) Transition metals compounds are coloured. This is due to the electronic transitions . . . . . . . . .
Gyaan Sankalp
17
The d- and f-block elements
(v) The maximum paramagnetic character in the 3d series of elements is shown by . . . . . . .atom.
(b) Write balanced equations for the following reactions described
(i) Potassium ferricyanide reacts with H2O2 in basic medium.
(ii) Cobalt (II) reacts with KNO2 in acetic acid solution.
Sol. (a) (i) CuI diamagnetic, CuII paramagnetic
(ii) Zn = +2
(iii) More (iv) within the same d orbital
(v) Chromium atom.
(b) (i) 2K3Fe(CN)6 + H2O2 + 2KOH 
 2K4Fe(CN)6 + O2 + 2H2O
(ii) Co2+ + 3KNO2 + 4NO2– + 2H+ 
 K3[Co(NO2)6] + NO + H2O
Example 15 :
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 ?
Sol. The extra stability of empty, half-filled and completely filled d-orbitals. For the 3d series, the oxidation states involving both 4s
and 3d electrons are stable. In the second half of the series, +2 and +3 are the stable oxidation states.
Electronic configuration
Stable oxidation states
3d3 4s2
+3, +5
3d5 4s2
+2, +7
3d8 4s2
+2
3d4 4s2
+3, +6
Example 16 :
How would you account for the following :
(a) Of the d4 species, Cr2+ is strongly reducing while manganese (III) is strongly oxidising.
(b) Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.
(c) The d1 configuration is very unstable in ions.
o
Sol. (a) The E Cr3 /Cr 2  is –0.41 V. Therefore, the tendency for Cr2+ to form Cr3+ by releasing one electron is positive. Thus Cr2+ is
strongly reducing.
The Mn3+ (d4 configuration) acts as a strongly oxidising agent as it can revert to Mn2+ (d5 configuration) in the process. Mn2+
2+
has half-filled d-orbitals, which impart greater stability to Mn2+ ion. The Eº value for Mn3+ + e– 
 Mn is +1.57 V..
(b) Cobalt (II) is a d7 case, and cobalt (III) is a d6 case. In the presence of complexing agent, the crystal field stabilization energy
of Co3+ (having d6 configuration) is higher than that for Co2+ (having a d7 configuration). Therefore, in the presence of
complexing agent, Co2+ gets easily oxidised to Co3+.
Co2+
(d7)
Co3+
6
(d )
(c) The d1 configuration in ions is unstable because it can lose its only electrons in the d orbital to attain extra stability of the
empty d-level.
Example 17 :
In qualitative analysis, a confirmatory test for chromium involves the formation of a blue transient peroxo species, CrO5, which
is better represented as CrO(O2)2. It is formed when H2O2 is added to an acidic solution containing Cr2O72–.
What is oxidation state of Cr is CrO (O2) ? Write the equation of formation of CrO5.
Sol. Cr2O72– + 2H+ + 4H2O2 
 2CrO(O2)2 + 5H2O
Oxidation state of oxygen in peroxide = – 1
Oxidation sate of oxygen in oxide = – 2
Hence in [Cr O (O2)2] = four oxygen of peroxide.
  
x–2–4=0
x=+6
18
Gyaan Sankalp
O
O
O
Cr
O
O
The d- and f-block elements
Example 18 :
Write down the electronic configuration of
(a) Cr3+
(b) Cu+
(c) Co2+
(d) Mn2+
3+
4+
2+
(e) Pm
(f) Ce
(g) Lu
(h) Th4+
3+
2
2
6
2
6
3
Sol. (a) Cr (24)
1s 2s 2p
3s 3p 3d
(b) Cu+ (29)
1s2 2s2 2p6
3s2 3p6 3d10
2+
2
2
6
(c) Co (27)
1s 2s 2p
3s2 3p6 3d7
2+
2
2
6
(d) Mn (25)
1s 2s 2p
3s23p6 3d5
2+
2
2
6
(e) Pm (61)
1s 2s 2p
3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p 6s0 4f5
(f) Ce4+ (58)
2
2
6
1s 2s 2p
3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6
2+
2
2
6
(g) Lu (71)
1s 2s 2p
3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s0 4f14 5d1
4+
2
2
6
(h) Th (90)
1s 2s 2p
3s23p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d10 5f0
Example 19 :
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.
Sol. Oxometal anion
Oxidation state of metal
Group number
VO43–
+5
5
Cr2O73–, CrO42–
+6
6
MnO4 –
+7
7
Example 20 :
How is the variability in oxidation states of transition metals different from that of the non-transition metals ? Illustrate with
examples.
Sol. In the transition elements, the oxidation states differ from each other by unity (by 1), whereas the oxidation states of the
nontransition elements usually differ from each other by two units.
For example, Vanadium shows the oxidation states of +2, +3, +4 and +5
Sulphur shows the oxidation states of +2, +4 and +6.
Example 21 :
For M2+/M and M3+/M2+ systems the Eº values for some metals are as follows :
Cr2+/Cr
– 0.9 V
Cr3+/Cr2+
–0.4 V
2+
Mn /Mn
– 1.2 V
Mn 3+/Mn 2+
+ 1.5 V
Fe2+/Fe
– 0.4 V
Fe3+ /Fe2+
+ 0.8 V
Use this data to comment upon :
(a) The Stability of Fe3+ in acid solution as compared to that of Cr3+ or Mn3+.
(b) The case with which iron can be oxidised as compared to the similar process for either chromium or mangnese metal.
Sol. (a) The Eº values for Fe3+/Fe2+, Cr3+/Cr2+ and Mn3+/Mn2+ systems are +0.80V, –0.4V and +1.5 V.
Thus, the reduction tendency of Fe3+, Cr3+ and Mn3+ follows the order
Mn3+(+1.l5 V) > Fe3+ (+0.80 V) > Cr3+ (– 0.4V)
3+
Thus Fe is more stable than Mn3+ and less stable than Cr3+.
(b) The Eº values for Fe2+/Fe, Cr2+/Cr and Mn2+/Mn systems are –0.4V, –0.9V and –1.2V. Thus the oxidation tendency of these
metals follows the order, Mn > Cr > Fe
Thus, the case of oxidation is the least for iron, followed by chromium and then manganese.
Example 22 :
Comment on the statement that elements of the first transition series possess many properties different from those of heavier
transition elements.
Sol. It is due to lanthanoid contraction that the elements of second and third series show similar properties and differ considerably
from the elements of the first series.
Example 23 :
Calculate the number of unpaired electrons in following gaseous ions : Mn3+, Cr3+, V3+ and Ti3+. Which one of these is the most
stable in aqueous solution ?
Sol. Species
Mn 3+
Electronic configuration (outer)
3d4
No. of unpaired electrons
Eo
M3 /M 2 
/V
4
+ 1.57
Cr3+
3d3
3
– 0.41
V3+
3d2
2
– 0.26
Ti3+
3d1
1
– 0.37
Only Mn3+ has positive Eº value which corresponds to a strong tendency to get reduced to Mn2+. This may be due to the extra
stability of d5 configuration for Mn2+.
19
Gyaan Sankalp
The d- and f-block elements
Example 24 :
Give example and suggest reasons for the following features of the transition metal chemistry :
(a) The lowest oxide of transition metal is basic, the highest is acidic.
(b) A transition metal exhibits higher oxidation states in oxides and fluorides.
(c) The highest oxidation state is exhibited in oxoanions of a metal.
Sol. (a) The oxide of a transition metal in low oxidation state is basic. Examples, TiO, VO, MnO, FeO, Cu2O, NiO.
The oxides of transition elements in higher oxidation states are acidic. Examples, V2O5, CrO3 and Mn2O7.
(b) This is because oxygen and fluorine are highly electronegative elements.
(c) In highest oxidation state, the covalent character increases. As a result, the metal in higher oxidation state is stabilized by
forming oxoions.
Example 25 :
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.
Sol. The chemistry of actinoids is not so smooth as that of the lanthanoids. Such a behaviour of actinoids is due to a wider range of
oxidation states exhibited by these elements. This can be attributed to very small energy difference between the 5f, 6d and 7s
levels.
The most common oxidation state in lanthanoids is +3. The oxidation states of +2 and +4 are also observed.
Actinoids also show in general the +3 oxidation state. However, the actinoids in the first half of the series also exhibit higher
oxidation states. For example, the highest oxidation state increases from +4 in Th to +5, +6 and +7 in Pa, U and Np respectively.
In the second half of the series, the maximum oxidation state decreases to +3.
This uneven distribution of oxidation states in the case of actinoids results in not so smooth chemistry of actinoids.
Example 26 :
Use Hund’s rule to derive the electronic configuration of Ce3+ ion, and calculate its magnetic moment on the basis of ‘spin-only’
formula.
Sol. The electronic configuration of Ce3+ (atomic no. of Ce = 58) is
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s1
There is only one unpaired electron in Ce3+ ion. The spin-only formula for calculating the magnetic moment is
=
n (n  2) BM
where n is the number of unpaired electrons in the species. Here, n = 1,
So,
 = 1 3 BM = 1.73 BM
Example 27 :
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 configuration of these elements.
Sol.
Lanthanoids showing
+4 oxidation state
Outer electronic configuration
+2 oxidation state
Outer electronic configuration
Cerium (Ce)
4f0
Samarium (Sm)
4f6
1
Praseodymium (Pr)
4f
Europium (Eu)
4f7
Thulium (Tm)
4f13
Ytterbium (Yb)
4f14
This behaviour shows that the empty, half-filled and completely filled f-orbitals exhibit more stability. However, this trend is not
followed by certain lanthanoids.
Example 28 :
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).
Sol. Water (H2O) is a weak field ligand. So it gives high spin hydrated ions.
Ion
No. of 3d electrons
Distribution of electrons
t2 g
eg
d 2 2 d 3
d xy
dyz
d zx
x y
z
2+
Ti
2


V2+
3



Cr3+
5





Mn 2+
5





Fe2+
4




Fe3+
5





Co2+
8





Ni2+
2


Cu2+
1

20
Gyaan Sankalp
The d- and f-block elements
QUESTION BANK
EXERCISE - 1
ONLY ONE OPTION IS CORRECT
Q.1
Q.2
Q.3
Q.4
Q.5
Q.6
Which one of the following statements is not correct(A) Nickel forms Ni(CO)4
(B) All the transition metals form mono metallic carbonyls
(C) Carbonyls are formed by transition metals
(D) Transition metals form complexes
Which of the following pairs of compounds are more stable(A) K2[NiCl6],[Pt(CN)4]2– (B) NiCl4, PtCl4
(C) [Ni(CN)4]2–, K2[Pt(Cl6)] (D) PtCl2, NiCl2
Which of the following compound is used as the starting
material for the preparation of potassium dichromates:
(A) K2SO4Cr2 (SO4)3.24H2O (Chrome alum)
(B) PbCrO4 (Chrome yellow)
(C) FeCr2O4 (Chromite)
(D) PbCrO4PbO(Chrome red)
When permanganate is reduced in alkaline medium, the
resulting species will be(A) MnO3–(aq)
(B) MnO2(s)
(C) Mn+2(aq)
(D) MnO4–2(aq)
When acidified solution of K2Cr2O7 is shaken with aqueous solution of FeSO4 then :
(A) Cr2O72– ion reduced to Cr+3 ions
(B) Cr2O72– ion reduced to CrO72– ions
(C) Cr2O72– ion oxidised to Cr
(D) Cr2O72– ion is oxidised to CrO3
The radius of La3+ (atomic number of La = 57) is 1.06 Å.
Which one of the following given values will be closest to
the radius of Lu3+ (atomic number of Lu = 71)
(A) 1.60 Å
(B) 1.40 Å
(C) 1.96 Å
(D) 0.85 Å
Heat
K2Cr2O7 
 4K2Cr2O4+ 3O2 + X In the above reaction X is :
(A) Cr2O3
(B) CrO3
(C) Cr2O7
(D) CrO5
Q.8 Give the ions having 4f145do6so outer electronic configuration:
(A) Yb2+
(B) Lu3+
3+
(C) Yb
(D) Both (A) and (B)
Q.9 Compounds of transition elements are coloured because
(A) They have high oxidation states
(B) They have low oxidation states
(C) They energy levels of electron differ by a value corresponding to the part of the visible spectrum
(D) The excited electron re-emit the absorbed light
Q.10 What would happen when a solution of potassium
dichromate is treated with an excess of dilute nitric acid –
(A) Cr3+ and Cr2O72– are formed
(B) Cr2O72– and H2O are formed
(C) Cr2O72– is reduced to + 3 state of Cr
(D) Cr2O72– is oxidised to + 7 state of Cr
Q.7
Q.11 Wilkinson’s catalyst is
(A) [Rh (CO)2I2]–
(B) (Ph3P3) RhCl
(C) Co(CO)8
(D) (Ph3P)2 Rh(CO)Cl
Q.12 Which is most readily decomposed
(A) V(CO)6
(B) Cr(CO)6
(C) Fe(CO)5
(D) Ni(CO)4
Q.13 For the same transition metal ion, the colour of its compound will depend upon the
(A) Temperature of the reaction
(B) Pressure of the reaction
(C) Nature of ligands or Lewis bases attached to the
metal ion.
(D) None of these
Q.14 Identify the wrong statement about ferrocene .
(A) It is paramagnetic
(B) The compound has a sandwich structure .
(C) The ring structures exhibit aromatic character.
(D) The symmetry of the space group requires the five
membered rings to be staggered.
Q.15 With increase in atomic number the ionic radii or actinides:
(A) contract slightly
(B) increase gradually
(C) show no change
(D) change irregularly
Q.16 The net bonding in a metal carbonyl is M = C = O. This is
due to (A) The overlap of px orbital of the metal with the 2py
orbital of carbon
(B) Back bonding
(C) The original M  C  bond is very strong
(D) The C – O bond order remaining unaltered
Q.17 Match the underlined atom in column A with oxidation
number in column B.
A
B

[Fe(CN)5 N O ]2–
0(I)
[Ag(CN)2]–
3(II)
[Cr(en)3]3+
1(III)
Ni(CO)4
2(IV)
a
b
c
d
(A)
I
II
III
IV
(B)
II
III
IV
I
(C)
IV
III
II
I
(D)
II
III
I
IV
Q.18 The correct calculate value for H for
M(s)  M2+(aq)
From following arbitrary value
M(s)  M(g) ; H = 1000 KJ mol–1
M(g)  M+(g) ; H = 750 KJ mol–1
M+(g)  M2+(g) ; H = 1200 KJ mol–1
M2+(g) + aq  M2+(aq.) ; H = – 1800 KJ mol–1
(A) 1950 KJ mol–1
(B) 1150 KJ mol–1
(C) 2300 KJ mol–1
(D) None of these
(a)
(b)
(c)
(d)
Gyaan Sankalp
21
The d- and f-block elements
Q.19 By calculating Eº value of the following reaction, which
one of the following reaction are spontaneous.
= – 0.76 V,, E0 
Ag / Ag
= 0.80 V.
(a) 2Ag + Zn2+  2Ag+ + Zn(b) Cu + Zn2+  Zn + Cu2+
(c) Zn + Cu2+  Zn2+ + Cu (d) 2Ag+ + Cu  2Ag + Cu2+
(A) Only d
(B) Only b
(C) Both c & d
(D) Both a & c
Which of the following has maximum number of unpaired
electrons ?
(A) Fe2+
(B) Fe3+
3+
(C) Co
(D) Co2+
The atomic number of potassium is 19 & that of Manganeseis 25. Although thecolour of MnO4– is dark violet yet
the K+ is colourless. This is due to the fact that(A) Mn is a transition element; while K+ is not
(B) [MnO4]– is negatively charged; while K+ has a positive charge.
(C) The effective atomic number of Mn in [MnO4]– is 26;
while for K+, the atomic number is 18.
(D) The Mn in a high positive oxidation state allow charge
transfer transitions.
Heating of Ag with dil. HNO3 give(A) NO
(B) NO2
(C) N2O
(D) N2O3
Zinc dissolve in an excess of NaOH, because of the formation of(A) ZnO
(B) Zn(OH)2
(C) NaZn(OH)3
(D) Na2ZnO2
TiCl4(l) is used to make smoke screens by spraying the
liquid in air. The product(s) of the reaction of TiCl4 with
water vapour are(A) TiO2 and HCl
(B) TiCl4(g) and HCl
(C) Ti(s) and Cl2
(D) TiO2, O2 and Cl–(aq)
When excess of SnCl2 is added to HgCl2, the substance
formed is(A) Hg2Cl2
(B) Sn
(C) Hg
(D) Cl2
Which of the following statement is wrong :
(A) An acidified solution of K2Cr2O7 liberates iodine from
iodides.
(B) In acidic solution dichromates ions are converted to
Chromate ions.
(C) Ammonium dichromate on heating undergo exothermic
decomposition to give Cr2O3
(D) Potassium dichromate is used as a titrant for Fe+2 ions.
The oxidation number is changed in which of the following
case(A) SO2 gas is passed into Cr2O72–/H+
(B) Aqueous solution of CrO42– is acidified
(C) CrO2Cl2 is dissolved in NaOH
(D) Cr2O72– solution is made alkaline
A magnetic moment of 1.73 BM will be shown by one among
of the following compounds.
(A) [Cu(NH3)4]2+
(B) [Ni(CN)4]2–
(C) TiCl4
(D) [CoCl6]–
Given E 0
Cu 2  /Cu
Q.20
Q.21
Q.22
Q.23
Q.24
Q.25
Q.26
Q.27
Q.28
22
Gyaan Sankalp
= 0.34 V, E 0
Zn 2  / Zn
Q.29 Which of the following elements form stable dinuclear ions
(A) Zn
(B) Cd
(C) Hg
(D) Fe
Q.30 The formula of blue perchromate is :
(A) CrO3
(B) Cr2O3
(C) CrO5
(D) None
Q.31 A mixture of Mn2+ & Zn2+ can be separated by using an
excess of(A) NH4OH
(B) NaOH
(C) H2SO4
(D) HNO3
Q.32 The correct electronic configuration of Ti (Z = 22 atom) is
(A) 1s22s22p63s23p63d2
(B) 1s22s22p63s23p63d24p6
2
2
6
2
6
4
(C) 1s 2s 2p 3s 3p 3d
(D) 1s22s22p63s23p64s23d2
Q.33 The blue colour produced on adding H2O2 to acidified
K2Cr2O7 is due to the formation of ;
(A) CrO5
(B) Cr2O3
(C) CrO72–
(D) CrO3
Q.34 To which of the following series the transition elements
from Z = 39 to Z = 48 belong?
(A) 3d series
(B) 4d series
(C) 5d series
(D) 6d series
Q.35 Acidified K2Cr2O7 is treated with H2S. In the reaction the
oxidation number of chromium :
(A) Increases from + 3 to + 6 (B) Decreases from + 6 to + 3
(C) Remains unchanged
(D) Decreased from + 6 to + 2
Q.36 Which oxide of manganese is acidic in nature.
(A) MnO
(B) Mn2O7
(C) Mn2O3
(D) MnO2
Q.37 Considering ionic radii of dipositive ions from Ca2+ to Zn2+
in an octahedral environment. Account for this
(A) The ionic radii steadily decreases along a smooth curve.
(B) There is practically no change in ionic radii.
(C) Except at Mn2+ the trend of ionic radii show minima at
V 2 and Ni 2  .
(D) The ionic radius reaches a maximum at Ti2  , then decreases from Ti2  through Mn2+ to Co2+ and then increases.
Q.38 Of the ions, Zn2+, Ni2+ and Cr3+ (atomic number of Zn=30,
Ni = 28 and Cr = 24).
(A) Only Zn2+ is colourless and Ni2+ and Cr3+ are coloured
(B) All these are colourless
(C) All these are coloured
(D) Only Ni2+ is coloured and Zn2+ and Cr3+ are colouless
Q.39 Which of the following ion in aqueous medium has orange
colour :
(A) Cr2O7–2
(B) Cr+3
(C) MnO4–
(D) MnO42–
Q.40 In which of the following does not have valence electron
in 3d-subshell –
(A) Fe(III)
(B) Mn(II)
(C) Cr(I)
(D) P(0)
Q.41 Which is not true in case of transition metals –
(A) They are malleable and ductile
(B) They have high melting and boiling points
(C) They crystallise with body centered cubic and hexagonal close packed structure only
(D) They show variable oxidation states although not
always
The d- and f-block elements
Q.42 Among the following series of transition metal ions, the
one where all metal ions have 3d2 electronic configuration
is –
(A) Ti3+, V2+,Cr3+, Mn4+
(B) Ti+, V4+, Cr6+, Mn4+
4+
3+
2+
3+
(C) Ti , V , Cr , Mn
(D) Ti2+, V3+, Cr4+, Mn5+
Q.43 Transition metals are less reactive because of their –
(A) High ionisation potential and low melting point
(B) High ionisation potential and high melting point
(C) Low ionisation potential and low melting point
(D) Low ionisation potential and high melting point
Q.44 Transition elements exhibits positive oxidation states only.
This is because of –
(A) Their large size of the atoms
(B) Their electropositive nature
(C) Their electronegative nature
(D) Their paramagnetic nature
Q.45 Transition metals, despite high E° oxidation, are poor
reducing agent due to –
(A) High heat of vapourization
(B) High ionization energies
(C) Low heats of hydration
(D) All of these
Q.46 Transition elements are frequently used as catalyst because.
(A) Of paired d-electrons
(B) Of high ionic charge
(C) Free valency on the surface
(D) Of their specific nature
Q.47 Cerium (Z = 58) is an important member of lanthanoids.
Which of the following statements about cerium is incorrect –
(A) The common oxidation states of cerium are +3 and +4.
(B) Cerium (IV) acts as an oxidising agent
(C) The +4 oxidation state of cerium is not known in solutions
(D) The +3 oxidation state of cerium is more stable than the
+4 oxidation state
Q.48 Which of the following oxide of chromium is amphoteric in
nature :
(A) CrO3
(B) Cr2O3
(C) CrO3
(D) CrO5
Q.49 Chromium has most stable oxidation state is :
(A) + 5
(B) + 3
(C) + 4
(D) + 2
Q.50 The element with the electronic configuration
[Xe]54 4f14 5d16s2 is a
(A) representative element (B) transition element
(C) lanthanide
(D) actinide
Q.51 The most characteristic oxidation state of lanthanides is :
(A) +2
(B) +3
(C) +4
(D) None of these
Q.52 Of the following outer electronic configurations of atoms,
the highest oxidation state is achieved by which one of
them –
(A) (n – 1)d8ns2
(B) (n – 1)d5ns1
3
2
(C) (n – 1)d ns
(D) (n – 1)d5ns2
Q.53 When a mixture of K2Cr 2O7 and KCl is heated with
conc.H2SO4,which of the following is produced
in the form of red vapours :- .
(A) CrO3
(B) CrO2Cl2
(C) CrCI3
(D) Cr2O3
8
Q.54 When d arrangement in nickel ion changes from a weak
octahedral field in [Ni (H2O)6 ]2 to a strong field in
[Ni(CN)4 ]2 , which orbitals get lowered in energy ? Assume tetragonal distortion along the Z-axis.
(A) d
x 2  y2
(C) d xy , d
z2
(B) d
z2
(D) d xz , d yz , d
z2
Q.55 When pink [Co(H2 O)6 ]2 is dehydrated the colour
changes to blue. Account for this :
(A) The octahedral complex becomes square planar.
(B) A tetrahedral complex is formed.
(C) Distorted octahedral structure is obtained.
(D) Dehydration results in the formation of polymeric species.
Q.56 Acidified solution of chromic acid on treatment with H2O2
yields :
(A) CrO3 + H2O + O2
(B) Cr2O3 + H2O + O2
(C) CrOs + H2O
(D) H2Cr2O7 + H2O + O2
Q.57 Which of the following is diamagnetic ?
(A) d5 arrangement in Fe3 is strong octahedral field.
(B) d7 arrangement in Co 2 is weak octahedral field.
Q.58
Q.59
Q.60
Q.61
(C) d5 arrangement in Mn 2 is weak octahedral field.
(D) d10 arrangement of Ni is tetrahedral Ni(CO)4.
The mineral perovskite is CaTiO3. Its unit cell has the ccp
structure. (i) Where are the calcium ions located ? The
ilmenite structure is FeTiO3 (ii) Where are the oxide ions
located in the latter ?
(A) (i) Corners are occupied by Ca++ ions.
(ii) Oxide ions form a hcp structure.
(B) (i) Ca++ ions at the body centre.
(ii) Oxide ions at face centres of the unit cell.
(C) (i) Ti4+ ions at the body centre.
(ii) Oxide ions at the corners.
(D) (i) Ti4+ ions at the corners.
(ii) Oxide ions at face centres and corners of the unit cell.
Crystals of potassium dichromate heated to give –
(A) Oxygen
(B) Hydrogen
(C) Sulphur dioxide
(D) Hydrogen sulphide
The atomic numbers of vanadium (V), chromium (Cr),
manganese (Mn) and iron (Fe) are 23, 24, 25 and 26
respectively. Which one of these may be expected to have
the highest second ionisation enthalpy –
(A) V
(B) Cr
(C) Mn
(D) Fe
The colourless species is –
(A) VCl3
(B) VOSO4
(C) Na3VO4
(D) (V(H2O)6)SO4.H2O
Gyaan Sankalp
23
The d- and f-block elements
Q.62 Which of the following is the greatest paramagnetic (A) Cu+
(B) Cu2+
2+
(C) Fe
(D) Fe3+
Q.63 Which of the following transition element shows the
highest oxidation state –
(A) Mn
(B) Fe
(C) V
(D) Cr
Q.64 Which of the following compounds is not coloured (A) Na2[CuCl4]
(B) Na2[CdCl4]
(C) K4[Fe(CN)6]
(D) K3[Fe(CN)6]
Q.65 Which of the following metal acts as the most efficient
catalyst (A) Alkali
(B) Transition
(C) Alkaline earth metals
(D) Coloured metals
EXERCISE - 2
(C) Chlorine gas is evolved
(D) Chlorine is formed
ONE OR MORE THAN ONE CHOICE MAY
BE CORRECT
Q.1
Q.2
Q.3
Q.4
Q.5
Q.6
Q.7
Pair of coloured compound would be (A) PbCl2
(B) ZnCO3
(C) (CH3COO)2Cu
(D) Co(NO3)2
In the acidic medium oxidation state of Cr in Cr 2O72–
change to X, In such case dichromate is acting as an :
Cr2O72– + 14H+ + A  B + 7H2O then choose the correct
options
(A) A is 6e–
(B) B is 2Cr3+
3+
(C) A is 2Cr
(D) B is 6e–
A reduction in atomic size with increase in atomic number
is a characteristic of elements of –
(A) high atomic masses
(B) d-block
(C) f-block
(D) radioactive series
The hydrated cupric chloride is strongly heated. Which of
the following statement(s) is (are) correct for this –
(A) It is reduced to Cu2Cl2
(B) Cupric oxide is formed along with Cu2Cl2
(C) Only Cl2 is liberated
(D) Cl2 and HCl both are liberated
When CO2 is passed into aqueous –
(A) Na2CrO4 solution, its yellow colour changes to orange
(B) K2MnO4 solution, it disproportionates to KMnO4 and
MnO2
(C) Na2Cr2O7 solution, its orange colour changes to green
(D) KMnO4 solution, its pink colour changes to green
Which of the following statement(s) is/are correct with
reference to the ferrous and ferric ions –
(A) Fe3+ gives brown colour with potassium ferricyanide
(B) Fe2+ gives blue precipitate with potassium ferricyanide
(C) Fe3+ gives red colour with potassium thiocyanate
(D) Fe3+ gives brown colour with ammonium thiocyanate
Which of the following chemical reaction(s) is(are) involved
in the developing of photographic plate
(A) C6 H 4 (OH)2  2AgBr  2Ag  C6 H 4 O2  2HBr
ASSERTION AND REASON QUESTIONS
(Q.9-Q.16)
Q.9
Q.10
Q.11
Q.12
Q.13
(B) AgBr  2Na 2S3O3  Na 3[Ag(S2O3 ) 2  NaBr
(C) AgBr  2NH3 (aq)  [Ag(NH3 )2 ]Br
Q.8
24
(D) 2AgBr  Na 2S2 O3  Ag 2S2 O3  NaBr
Which of the following statement(s) is (are) correct when a
mixture of NaCl and K2Cr2O7 is gently warmed with conc.
H2SO4 ?
(A) A deep red vapour is evolved
(B) The vapour when passed into NaOH solution gives a
yellow solution of Na2CrO4
Gyaan Sankalp
Q.14
Q.15
Note : Each question contains STATEMENT-1 (Assertion)
and STATEMENT-2 (Reason). Each question has 5 choices
(A), (B), (C), (D) and (E) out of which ONLY ONE is correct.
(A) Statement-1 is True, Statement-2 is True; Statement-2
is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2
is NOT a correct explanation for Statement-1.
(C) Statement -1 is True, Statement-2 is False.
(D) Statement -1 is False, Statement-2 is True.
(E) Statement -1 is False, Statement-2 is False.
Statement 1 : Transition metals have high boiling points
and high enthalpy of atomization.
Statement 2 : The transition metals occur in bcc, hcp or
ccp structures characteristic of metals.
Statement 1 : Zn, Cd and Hg are normally not considered
transition metals.
Statement 2 : d-orbitals in above elements are completely
filled, hence above metals do not show the general
characteristic properties of the transition elements.
Statement 1 : A transition metal forms alloy with other
transition metal easily.
Statement 2 : The transition metals have similar electronic
configuration, nearly equal radii and exhibit similar packing
in their lattices.
Statement 1 : Anhydrous copper (II) chloride is a covalent
while anhydrous copper (II) fluoride is ionic in nature.
Statement 2 : In halides of transition metals, the ionic
character decreases with increase in atomic mass of the
halogen.
Statement 1 : Ce3+ cannot be easily oxidised to Ce4+.
Statement 2 : Ce3+ tends to lose its only electron from 4f
orbitals.
Statement 1 : E° for Mn3+/Mn2+ couple is more positive
than for Fe3+/Fe2+.
Statement 2 : Mn3+ has a d4 configuration and Fe3+ has a
d5 configuration.
Statement 1 : Transition metals exhibits higher enthalpies
of atomization.
Statement 2 : The atoms of transition metals are held
together by strong metallic bonds.
The d- and f-block elements
Q.16 Statement 1 : The transition elements form interstitial
compounds.
Statement 2 : Steel and cast iron are hard due to the
formation of an interstitial compound with carbon.
MATCH THE COLUMN TYPE QUESTIONS
(Q.17-Q.19)
Each question contains statements given in two columns
which have to be matched. Statements
(A, B, C, D) in column I have to be matched with statements (p, q, r, s) in column II.
Q.17 Match the column correctly –
Column I
Column II
(A)Pt/PtO
(p) Decomposition of KClO3
(B) MnO2
(q) conversion of SO2 to SO3
(C) PdCl2
(r) Deacon’s process
(D) V2O5
(s) Wacker’s process
(E) CuCl2
(t) Adam’s catalyst for reductions
Q.18 Column II gives stable oxidation states corresponding to
electronic configuration in column I match them correctly.
Column I
Column II
(A) 3d34s2
(p) 2
(B) 3d54s2
(q) 3
(C) 3d84s2
(r) 5
(D) 3d44s2
(s) 6
Q.19 Column II gives oxidation state for oxide of 3d series given
in column I match them correctly.
Column I
Column II
(A) Mn2O7
(p) +6
(B) CrO3
(q) +5
(C) V2O5
(r) +2
(D) TiO
(s) +7
PASSAGE BASED QUESTIONS
Passage 1- (Q.20-Q.22)
A scarlet compound A is treated with conc. HNO3 to give
a chocolate brown precipitate B. The precipitate is filtered
and the filtrate is neutralised with NaOH. Addition of KI to
the resulting solution gives a yellow precipitate C. The
precipitate B on warming with conc. HNO3 in the presence
of Mn(NO3)2 produces a pink-coloured solution due to
the formation of D..
Q.20 A is –
(A) Pb3O4
(B) PbO2
(C) PbI2
(D) HMnO4
Q.21 B is –
(A) Pb3O4
(B) PbO2
(C) PbI2
(D) HMnO4
Q.22 C is –
(A) Pb3O4
(B) PbO2
(C) PbI2
(D) HMnO4
Passage 2- (Q.23-Q.25)
A white substance A reacts with dilute H2SO4 to produce
a colourless gas B and a colourless solution C. The reaction
between B and acidified K2Cr2O7 solution produces a
green solution and a slightly coloured precipitate D. The
substance D burns in air to produce a gas E which reacts
with B to yield D and a colourless liquid. Anhydrous copper
sulphate is turned blue on addition of this colourless liquid.
Addition of aqueous NH3 or NaOH to C produces first a
precipitate, which dissolves in the excess of the respective
reagent to produce a clear solution in each case.
Q.23 A is –
(A) ZnS
(B) H2S
(C) ZnSO4
(D) S
Q.24 B is –
(A) ZnS
(B) S
(C) ZnSO4
(D) H2S
Q.25 C is –
(A) ZnS
(B) H2S
(C) ZnSO4
(D) S
EXERCISE - 3
Q.1
Q.2
Q.3
Q.4
Q.5
(a) Find the oxidation number of transition metal in each of
the following compounds
(i) FeCO3 (ii) Cu2(OH)3Cl (iii) CrO2Cl2
(iv) CuCl
(b) Describe by means of equations, the sequence of reactions that occur when chromite is converted to a solution
of dichromate.
What happen when FeSO4(X) is subjected to heating, compound R, S, T are obtained, R is red-brown solid, S can be
oxidized to (T), Identify R, S, T.
What happen when FeSO4 react with K2Cr2O7/H+ & then
K4Fe(CN)6 is added.
What happen when H2S gas is passed into FeCl3 solution.
Gas (A) and Gas (B) both turns K2Cr2O7/H+ green. Gas (A)
Q.6
also turns lead acetate paper black. When gas (A) is passed
into gas (B) in aqueous solution, yellowish white turbidity
appears. Identify gas (A) and (B) and explain reactions.
Why are Mn2+ compounds more stable than Fe2+ towards
oxidation to their +3 state ?
100ºC
230ºC
750ºC
CuSO4.5H2O 
A 
B 
C
Identify A, B, C ?
Q.8 Explain briefly how +2 state becomes more and more stable
in the second half of the first row transition elements with
increasing atomic number ?
Q.9 Name the product formed when Fe2(SO4)3 react with Fe ?
Q.10 To what extent do the electronic configurations decide the
stability of oxidation states in the first series of the transition elements?
Q.7
Gyaan Sankalp
25
The d- and f-block elements
EXERCISE - 4
PREVIOUS YEAR IIT-JEE QUESTIONS
Q.1
Q.2
When MnO2 is fused with KOH, a coloured compound is
formed, the product and its colour is –
[2003]
(A) K2MnO4, purple green (B) KMnO4, purple
(C) Mn2O3, brown
(D) Mn3O4, black
In the process of extraction of gold,
Q.3
Q.4
O2
Roasted gold ore + CN– + H2O 

 [X] + OH–
[X] + Zn 
 [Y] + Au
Identify the complexes [X] and [Y]
(A) X = [Au(CN)2]–, y = [Zn(CN)4]2–
(B) X = [Au(CN)4]3–, Y = [Zn(CN)4]2–
(C) X = [Au(CN)2]–, Y = [Zn(CN)6]4–
26
Gyaan Sankalp
[2003]
Q.5
(D) X = [Au(CN)4]–, Y = [Zn(CN)4]2–
The product of oxidation I– with MnO4– in alkaline medium
is –
[2004]
(A) IO3–
(B) I2
(C) IO–
(D) IO4–
(NH4)2Cr2O7 on heating liberates a gas. The same gas will
be obtained by –
[2004]
(A) heating NH4NO2
(B) heating NH4NO3
(C) treating H2O2 with NaNO2
(D) treating with Mg3N2 with H2O
Among the following the coloured compound is [2008]
(A) CuCl
(B) K3[Cu(CN)4]
(C) CuF2
(D) [Cu(CH3CN)4]BF4
The d- and f-block elements
HINTS & SOLUTIONS
EX ERCIS E - 1
(1)
(2)
(6)
(17)
(18)
(19)
Q
1
2
3
4
5
6
7
8
9
10
11
A
B
C
C
B
A
D
A
D
C
C
B
Q
12
13
14
15
16
17
18
19
20
21
22
A
D
C
A
A
B
C
B
C
B
D
A
Q
23
24
25
26
27
28
29
30
31
32
33
A
D
A
C
B
A
A
C
C
B
D
A
Q
34
35
36
37
38
39
40
41
42
43
44
A
B
B
B
C
A
A
D
C
D
B
B
Q
45
46
47
48
49
50
51
52
53
54
55
A
D
C
C
B
B
C
B
D
B
D
B
Q
56
57
58
59
60
61
62
63
64
65
A
C
D
A
A
B
C
(B). Always transition metals combines with more than
one carbonyl group.
(C). By comparing the value of ionization energies the thermodynamic stability of transition metal compound can be
compared. Ionization value shows that Ni(II) compound
are more stable than Pt (II) compound while Pt(IV) compound are relatively more stable than Ni(IV).
(D). The ionic radius of Lu3+ is 0.87 Å, which is very close
to the value 0.85Å. This decrease in the ionic radius is due
to lanthanoid contraction.
 

(C). (a)  Fe(CN)5 N O 


(c) Zn + Cu
–
–2e
2Ag + Zn
2+
Oxidation
Oxidation
–
–2e
+
2+
+2e
Reduction
(20)
Eºcell = 0.84 – (0.34) = 0.46
 Eºcell is positive
 Reaction is spontaneous.
(B).
4s
3d
Fe2+
Fe3+
Co2+
Co3+
+
 2Ag + Zn
(21)
+2e
Reduction
Thus Eºcell = – 0.76 – (0.80) = – 1.56 v
 Eºcell is negative, reaction is non-spontaneous
(22)
Reduction
+2e–
2+
+
(d) 2Ag + Cu – 2Ag + Cu
 Fe + (–1) × 5 + 1 = – 2
–
Zn –
(b) Cu + –2e
 Zn2+ + Cu
Eºcell = 0.34 – (–0.76) = + 1.1 v
 Eºcell is positive
 Reaction is spontaneous
Oxidation
–
–2e
(a)
B
Reduction
+2e–
2+
2
Fe = + 2
(b) [Ag(CN)2]–  Ag + (–1) × 2 = – 1 ; Ag = + 1
(c) [Cr(en)3]3+  Cr + 0 = + 3 ; Cr = + 3
(d) Ni(CO)4  Ni + 0 = 0 (Zero)
(B). H = Hsub + IE1 + IE2 + Hhydration
= 1000 + 750 + 12000 + (– 1800)
H = 1150 KJ mol–1
(C). The emf of the cell can be calculated as
Eºcell = Eº(Reduction) – Eº(Oxidation)
Cathode
Anode
D
A
B
 Reaction is non spontaneous.
 Zn + Cu
Thus Eºcell = – 0.76 – (0.34) = – 1.1 v
 Eºcell is negative
(A). 2HNO3(dil.) 
 2NO + H2O + 3O
2Ag + 2HNO3 + O 
 2AgNO3 + H2O] × 3
+
Oxidation
 Fe3+ has maximum number (= 5) of unpaired electrons.
(D). In MnO4– the oxidation state of Mn is +7 (dº configuration), expected to be colourless but MnO4– is coloured
due to transfer of charge from oxygen to manganese.
(23)
6Ag + 8HNO3 
 6AgNO3 + 2NO + 4H2O
(D). It react wih NaOH on moderate heating & form soluble
Zincate.
Gyaan Sankalp
27
The d- and f-block elements
(25)
(27)
(29)
Zn + 2NaOH 
 Na2ZnO2 + H2
(C). SnCl2 reduces HgCl2 to Hg2Cl2 and finally to Hg2HgCl2 + SnCl2 
 Hg2Cl2 + SnCl4
Hg2Cl2 + SnCl2 
 2Hg + SnCl4
(A).
(A) Cr2O72– + SO2 + 2H+  2Cr3+ + 3SO42– + H2O
+6
+4
+3
+6
(change)
(B) 2CrO42– + 2H+  Cr2O72– + H2O (No change)
+6
+6
(C)CrO2Cl2 + 2NaOH  NaCrO4
(No change)
+6
+6
(D)Cr2O72– + 2OH–  2CrO42– + H2O(No change)
+6
+6
(C). Hg(l) compound contain mercurous ions (Hg – Hg)2+
& not Hg+.
The two atoms are bonded together using 6s-orbital.
Zn & Cd also form dinuclear ion (Zn – Zn)2+ & (Cd –
Cd)2+; however these ions are unstable & have only been
detected in melts of Zn/ZnCl2 & Cd/CdCl2 respectively.
(31)
(32)
(38)
(41)
(47)
(54)
(B).
Mn2+ + 2NaOH 
 Mn(OH)2 + 2Na+
(ppt)
Zn2+ + 2NaOH 
 Zn(OH)2 + 2Na+
(ppt)
(Amphoteric in nature)
Zn(OH)2 + 2NaOH 
 Na2[Zn(OH)2]
Soluble complex.
(D). Ti (Z = 22) is 1s22s22p63s23p64s23d2
(A). Zn2+ is colourless due to absence of any unpaired
electron. Ni2+ and Cr3+ are coloured due to 2 and 3 unpaired electrons respectively.
(C).The transition elements exhibit hexagonal closed
packed, face centred cube and body centred cubic structures and co-ordination number range from 8 to 12.
(C). Ce (IV) is a stable species in solution.
(D). In a tetragonal distortion, the ligands along the z-axis
move away. Their repulsion significantly decreases for the
orbitals involving z-direction. These are lowered in energy.
EXERCISE - 2
(1)
(2)
(3)
(4)
(CD)
(AB). In the acidic medium Cr2O72– take up the electron
Cr2O72– + 14H+ + 6e–  2Cr3+ + 7H2O
+6
(A)
(B)
Thus acting as an oxidizing agent, the oxidation change
from +6 to +3.
(BC). Both the d-block and f-block elements show a
decrease in the atomic size with an increase in atomic
number.
(ABD)

3CuCl 2 .2H 2 O 
 CuO  Cu 2 Cl2  2HCl + Cl2 + H2O
(5)
(9)
(10)
(11)
(C) Only Cl2 is liberated is wrong as evident from chemical
equation.
(AB).

H
 Na 2 Cr2 O7 (orange colour)
(A) Na 2 CrO 4 
(12)
H
(B) MnO24  
 MnO 4  MnO 2
in neutral or acidic medium
(C) False = In acidic medium no colour change
MnO4
(6)
(7)
(8)

 MnO 4 2
e
(D)
in alkaline (strong) medium changes to green.
(BC)
(AB).
(A) In making negative/positive of photographic film
(B) In fixing of photographic film by removing unreduced
AgBr as soluble complex.
(ABD).
Heat
K 2 Cr2 O7  4NaCl  6H 2SO 4 

2KHSO4  4NaHSO4  2CrO2 Cl2  3H 2 O
chromyl chloride
(deep red vapour)
28
Gyaan Sankalp
(13)
OH 
(14)
(15)
(16)
CrO2 Cl2  4NaOH 
 Na 2 CrO4  2NaCl  2H 2 O
yellow solution
(B). Strong metallic bonding in transition metals is
responsible for their high boiling points and high enthalpy
of atomization.
(A). Zn, Cd and Hg are normally not considered transition
metals because d-orbitals are completely filled, hence they
do not show the general characteristic properties of the
transition elements. Therefore, Zn, Cd and Hg are normally
not considered transition metals.
(A). The transition metals have similar electronic
configuration, nearly equal radii and exhibit similar packing
in their lattices hence these metals dissolve in each other
easily even in the solid state, i.e. one metal can be easily
substituted by another. It is because of this reason that
the transition metals form alloys with each other easily.
(A). Anhydrous copper (II) chloride is a covalent while
anhydrous copper (II) fluoride is ionic in nature because in
halides of transition metals, the ionic character decreases
with increase in atomic mass of the halogen. Fluorine being
the most electronegative, forms ionic salt with copper.
(D). Ce3+ can be easily oxidised to Ce4+ because Ce3+
tends to lose its only electron from 4f orbitals.
(A). Mn3+ has a d4 configuration so it has greater tendency
to accept one electron to acquire d5 configuration.
On the other hand Fe3+ has a d5 configuration which is
more stable than the d6 configuration of Fe2+. As a result,
reduction of Fe3+ to Fe2+ is not favoured. Since, E° values
reflect the reduction tendency, therefore, E° value for Mn3+/
Mn2+ couple is more positive than for Fe3+/Fe2+.
(A). Transition metals exhibits higher enthalpies of
atomization because the atoms of transition metals are held
together by strong metallic bonds.
(B). The transition elements form interstitial compounds in
which small atoms of light elements such as, H, C and N
The d- and f-block elements
occupy the voids in their lattices.
(17) (A) t, (B)  p, (C)  s, (D)  q, (E)  r
(18) (A) q, r ; (B)  p ; (C)  p (D)  q, s
(19) (A) s, (B)  p, (C)  q, (D)  r
(20) (A), (21) (B), (22) (C).
PbO4 (s) 
ZnS  dil. H 2SO4 
 ZnSO4 (aq)  H 2S(g)
(A)
(C)
(B)
2Pb(NO3 )2  PbO 2 (s)  2H 2 O
filter | chocolate
 brown
2KI
neutralised
PbI2 (s)  2KNO3 
 Pb(NO3 )(aq) 
 Pb(NO3 )(aq)  PbO2 (s)
4HNO3 (conc.)
B
burn

 E  D  colourless liquid
air
(gas)
(gas)
 CuSO 4 (anhy.)
CuSO4 turns blue
A is ZnS, B is H2S, C is ZnSO4


K 2Cr2 O7  4H 2SO4  3H 2S
with NaOH
yellow ppt
filtrate
(residue)
warm
5PbO 2 (s)  6HNO3  2Mn(NO3 )2  2HMnO4 (aq)  5Pb(NO3 )2  2H 2 O
acting as
pink coloured sol.
an oxidising
agent
 K 2SO 4  Cr2 (SO 4 )3  7H 2 O  3S
(D)
2H S
2  3S  2H O
S  O 2 (g) 
 SO 2 (g) 
2
(D)
So, A is Pb3O4, B is PbO2, C is PbI2.
(23) (A), (24) (D), (25) (C).
(E)
CuSO 4 (anhyd.)  5H 2 O 
 CuSO4 .5H 2 O
white
blue
Precipitate to
dil. H 2SO4
in an
A 
 B 
C  dissolve
excess of the
(white
(colourless (colourless
substance)
gas)
solution)
reagent
| K 2 Cr2 O7 , H 

Green solution  D
coloured ppt
NH3 (aq)
or NaOH
EXERCISE - 3
(1)
(2)
(a) (i) Fe = +2
(ii) Cu = +2 (iii) Cr = +6 (iv) Cu = +1
(b) Chromite is FeCrO4 – the principle chromium are.
4FeCrO4 + 4NaCO3 + O2  4Na2CrO4 + 2Fe2O2 + 4CO2
2 Na2CrO4 + 2HCl  Na2Cr2O7 + H2O + 2NaCl
by losing one of the d-electrons. That is why Fe2+ is less
stable than Mn2+ towards oxidation to +3 state.
(7)
(R)
(S)
(T)
Solid
750ºC
CuSO4 
CuO + 5O2 + O2
White(B)
(8)
SO3
(T)
(5)
(6)
230ºC

Bluish white
(A)
Blue

(X)
(4)
100ºC
 CuSO4 .H 2 O
2FeSO4 
 Fe 2 O3  SO 2  SO3
Red - Brown (O)
(3)
CuSO 4 .5H 2 O
6Fe2+ + Cr2O72– + 14H+  6Fe3+ + 2Cr3+ + 7H2O
Fe3+ + [FeII(CN)6]4–  FeIII [FeII(CN)6]–
Prussian Blue
When H2S is passed through FeCl3.
2FeCl3 + H2S  FeCl2 + 2HCl + S
In this case Fe2+ ion are formed due to reduction. Hence
solution appear light green.
(A) : H2S, oxidised to S by Cr2O72–/H+
(B) : SO2, oxidised to SO42– by Cr2O72–/H+
(9)
(10)
2H2S + SO2 
 3S + 2H2O
(Yellowish white)
The outer electronic configuration of Mn2+ is 3d5, whereas
that of Fe2+ is 3d6. The extra stability of the half-filled (d5)
configuration is Mn2+ make it more stable.
The Fe2+ ion tends to achieve the half filled configuration
(C)
The first five elements (Sc to Mn) of the first row transition
elements show variable oxidation states depending upon
the number of 4s and 3d electrons involved.
After the d5 configuration, i.e., in the last five elements,
the tendency for the d-electrons to participate in bonding
decreases and the +2 state becomes more stable.
Fe2(SO4)3 + Fe  3FeSO4
Ferric Sulphate Ferrous Sulphate
The general electronic configuration of 3d-series is
3d1-10 4s1-2. Thus by using varying number of 3d electrons, various oxidation states become possible.
In the highest oxidation states of the first five elements of
this series, all the 4s and 3d electrons are used up in bonding. After the d5 configuration, i.e., in the last five elements
the tendency for all the d-electrons to participate in bonding decreases. As a result, the highest oxidation states
shown by these elements are lower than the group number.
EXERCISE - 4
(1)
(3)
(A)
(A)
(2) (A)
(4) (A)
(5)
(C). In the crystalline form CuF2 is blue coloured.
Gyaan Sankalp
29