THE d- AND f-BLOCK ELEMENTS (C-8)
The 'd'-block elements are called transition elements because they represent a change or transition in
properties from most electropositive S-block elements to least electropositive P-block elements. The
transition elements are those elements which have partially filled d-sub shells in their elementary
form or in their commonly occurring oxidation state. Group 12 elements such as Zn, Cd & Hg are
not considered as transition elements because they never have partially filled d-orbitals either in their
elementary form or commonly occurring oxidation state. However they show similarities to other
transition elements in some of their chemical properties. They are characterised by the general outer
electronic configuration (n-1)d1-10 ns1-2 .
The first transition series(3d series) involves the filling of 3d orbitals.[Z=21(Sc) to Z=30(Zn)]. The
second transition series (4d series) involves the filling of 4d orbitals [Z=39 (Y) to Z=48(Cd)]. The third
transition series (5d series) involves the filling of 5d orbitals [Z=57(La) to Z=80(Hg)]. The first element
lanthanum is followed by 14 elements called lanthanoids which involves the filling of 4f orbitals. The
fourth transition series (6d series) involves the filling of 6d orbitals[Z=89(Ac) to Z=112(Cn)]. Actinium
is followed by 14 elements called actinoids.
Q1. On what ground can you say that Scandium(Z=21) is a transition element but Zinc (Z=30) is not?
Scandium atom has incompletely filled 3d orbitals in their ground state (3d1), ie, Sc is a
transition element. On the other hand, Zinc atom has completely filled 3d orbitals (3d 10) in its ground
state as well as in its common oxidation state (Zn2+), hence it is not regarded as a transition element.
Q2. Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it is a
transition element?
It can exist in +2 oxidation state (4d9) ie, in common oxidation state it has partially filled 4d orbital.
General Characteristic Properties of Transition Elements.
• Metallic character :- All the transition elements are metals due to their relatively low
ionization energies and number of vacant d-orbitals in the outermost shell, which enable them
to form metallic bond. All these metals have high density. Osmium has the highest density
among these elements.
• Melting and Boiling points:- The transition metals have very high melting and boiling points
due to strong bonds between the atoms of these elements.
• Enthalpies of Atomisation:- They have high enthalpies of atomisation(Maximum at the middle
of each series). Greater the number of valence electrons, stronger is the resultant bonding. The
metals of second and third series have greater enthalpies of atomisation than the elements of
first series.
• Variation in Atomic and Ionic Sizes:The atomic and ionic radii decreases with increasing atomic number. But the variation within a
series is quite small. The filling of 4f before 5d orbital results in a regular decrease in atomic radii called
Lanthanoid contraction. The decrease in metallic radius coupled with increase in atomic mass results
in a general increase in the density of transition elements.
Q3: Why do the transition elements exhibit higher enthalpies of atomisation?
Because of large number of unpaired electrons in their atoms they have stronger inter atomic interaction
and hence stronger bonding between atoms resulting in higher enthalpies of atomisation.
Q4: In the series Sc(Z=21) to Zn(Z=30), the enthalpy of atomisation of zinc is the lowest. ie, 126
KJ mol-1. Why?
In case of zinc, no electrons from 3d- orbitals are involved in the formation of metallic bonds. But in
all other metals of the 3d series, electrons from the d-orbitals are always involved in the formation of
metallic bonds.
• Variable oxidation state:- They show a variety of oxidation state in their compounds due to
the availability of both ns and (n-1)d electrons for bond formation. The general oxidation state
is +2 and other oxidation states are (+1,+3,+4,+5,+6 &+7) possible. The highest oxidation state
shown by Mn is +7 and Osmium and Rhuthenium shows +8. The elements which give the
greatest number of oxidation states occur in or near the middle of the series.
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•
Ionisation Enthalpies and Electrode Potential:- The first ionisation energies of d-block
elements are higher than those of S block elements and less than that of P block elements. Along
a transition series ionisation enthalpy increases gradually, though not quite regularly.
The thermodynamic stability of various oxidation state of transition metals can be evaluated in
terms of the magnitude of their ionization energies. Smaller the ionization energies of the metal, the
more stable will be its compounds. For example Ni(ll) compounds are stable than Pt(ll) compounds.
This is because the sum of the first two ionisation energies of Ni is smaller than the sum of the first two
ionisation energies of Pt. Similarly Pt(lV) compounds are more stable than Ni(lV) compounds, because
the sum of the first four ionisation energies of Pt is smaller than that of Nickel.
• Trends in the M2+/M standard electrode potentials:Due to high values of enthalpy of atomisation and also high values of ionisation enthalpies, the
electrode potentials, EƟ are relatively low.
• Trends in the M3+/M2+ EƟ values and Relative stabilities of oxidation states :- The values
show an irregular trend. Very low values of EƟ will be found for ions having noble gas d10 and
d5 configuration respectively.
• Chemical reactivity and Eɵ values:- Transition metals exhibit different chemical behaviour.
Some are quite electropositive while some are least reactive or noble. The reactivity of 3d
transition metals in acid solution can be seen from their EƟ red values for the couple M2+/M. EƟ
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values increases from V to Cu and so the reactivity decreases.( EƟ α reactivity )
EƟ values depend upon ΔaH, Δhyd H and ΔiH(IE). If ΔhydH> ΔaH & IE, M2+ is stable (M2+ /M) and M3+
is stable (M3+ /M2+ ). EƟ of M2+ /M is more negative the M2+ ion will be stable,if M3+/M2+ is more
negative, M3+ ion will be more stable than M2+ .
Q5. The EƟ M2+/M value for Cu is +ve (+0.34v) why ?
ΔhydH < ΔaH & 1 E . Here M is stable than M2+because 29Cu – [ Ar] 3d104S1 and Cu2+ - [ Ar] 3d9
Q6 . The EƟ M3+/M2+ of Zn is very high ( +ve) .Explain on the basis of electronic configuration .
ΔhydH < Δa H & 1E . Reason – Due to stable electronic configuration of Zn2+. High +ve EƟ value
indicates that M3+ is unstable Zn 2+ →[ Ar ] 3d10 Zn3+ → [Ar] 3d9.
Q7. EƟ of Fe3+/ Fe2+ is very low (–ve). Explain
Fe3+ is stable. Due to extra stable electronic configuration Fe2+ →[Ar] 3d6 Fe3+ →[Ar]3d5
Q8. Why is the EƟ value for Mn3+/Mn2+ much more +ve than that for Fe3+/Fe2+. Explain
Because Mn3+ is less stable than Mn2+ and Fe3+ is stable than Fe2+. Therefore in case of Mn3+/Mn2+
more positive EƟ value and less EƟ value for Fe3+ /Fe2+.
Mn2+ →[Ar]3d5, Mn3+ →[Ar]3d4, Fe2+→[Ar]3d6 , Fe3+ →[Ar]3d5
Q9. Why is Cr2+ reducing and Mn3+ oxidising when both have d4 configuration?
Cr2+ is reducing as its configuration changes from d4 to d3, the latter having a half-filled t2g level. But
the change from Mn2+ to Mn3+ results in the half-filled (d5) configuration which has extra stability.
Q10. Name a transition element which does not exhibit variable oxidation states
scandium (Z=21)
Q11. Which of the 3d series of the transition metals exhibits the largest number of oxidation states and Why?
Manganese (Z=25), Because it has maximum number of unpaired electrons.
Q12. How would you account for the increasing oxidising power in the series VO2+<Cr2O72- < MnO4-?
This is due to the increasing stability of the lower species to which they are reduced.
Q13. How would you account for the irregular variation of ionization enthalpies (first and second) in
the first series of the transition elements?
Irregular variation of ionisation enthalpies is mainly attributed to varying degree of stability of different
3d-Configurations,(eg:-d0,d5,d10 are exceptionally stable)
Q14. For the first row transition metals the EƟ values are :
Metals
V
Cr
Mn
Fe
Co
Ni
Cu
EƟ(M2+/M)
-1.18 -0.91 -1.18 -0.44 -0.28 -0.25 +0.34
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Explain the irregularity in the above values.
Due to irregular variation of ionisation enthalpies (∆iH1+∆iH2) and sublimation enthalpies
which are relatively much less than for Mn and V.
Q15.Why is the EƟ value for the Mn3+/Mn2+ couple much more positive than that of Cr3+/Cr2+ or
Fe3+/Fe2+? Explain.
It is due to much larger third ionization energy of Mn(d5 →d4). ie, the +3 state of Mn is of little
importance.
Q16. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Because of small size and high electro negativity of oxygen or fluorine can oxidise the metal to
its highest oxidation state.
Q17. Which is a stronger reducing agent Cr2+ or Fe2+ and Why?
Cr2+ is stronger reducing agent than Fe2+.
Reason: d4→d3 occurs in the case of Cr2+ to Cr3+
But d6 →d5 occurs in case of Fe2+ to Fe3+.
In a medium (like water) d3 is more stable as compared to d5.
• Magnetic properties:- Most of the transition elements show paramagnetism due to the presence
of unpaired electrons in them. Greater the number of unpaired electrons, greater is the paramagnetic
behaviour. The maximum paramagnetism is noted in the d5 system. Paramagnetism is expressed in
terms of magnetic moment(measured in Bohr Magneton, BM). B = √𝑛(𝑛 + 2)
where 'n' is the number of unpaired electrons and 'B' is the magnetic moment in Bohr Magneton.
Q18. Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25.
n=5  = √5(5 + 2) =√35 = 5.92 BM
Q19. Calculate the 'spin only' magnetic moment of M 2+(aq) ion (Z=27)
7 2
2+
7
therefore n=3  =√3(3 + 2) = √15 = 3.9 BM
27Co-[Ao]3d 4s ; Co -[Ar]3d
Q20. Calculate the magnetic moment of Fe2+
Fe2+ → [Ar]3d6 therefore n=4 = √4(4 + 2)
= √24 = 4.9 BM
• Formation of coloured ions:- The transition metal ions have unpaired electrons which on
absorbing visible light can jump from one d orbital to another. ie, d-d transition takes place.
Thus, when light falls, certain visible wavelengths are absorbed, the transmitted /reflected light
appears coloured and give coloured compounds. The ions having no d-d transition are
colourless. Eg:- Zn compounds are colourless.
• Formation of complex compounds:- The transition metals form a large number of complex
compounds. This is due to the comparatively smaller sizes of the metal ions, their high ionic
charges and the availability of d orbitals for bond formation. Eg:- [Fe(CN)6]3- and [Fe(CN)6]4• Catalytic properties:- Many transition metals and their compounds show catalytic properties,
the most common being them are Fe, pt, Ni, V2O5 etc. This property may be due to their variable
valency which enable them to form unstable intermediate compounds.
• Formation of interstitial compounds:- Transition metals can take up atoms of small size
(H,B,C,N etc ) in vacant spaces in their lattices . They form a large number of interstitial
compounds.
Eg:- steel is an interstitial compound of Fe and C.
Characteristics
➢ They have high melting points ,higher than those of pure metals.
➢ They are very hard ,some borides approach diamond in hardness.
➢ They retain metallic conductivity.
➢ They are chemically inert.
• Alloy Formation :- Alloys may be homogeneous solid solutions in which the atoms of one
metal are distributed randomly among the atoms of the other .Due to their almost equal
atomic size , they can mutually substitute one another in the crystal lattice to form alloys.
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Alloys of transition metals with non transition metals such as brass (Cu&Zn) and bronze
(Cu&Sn ) are examples .
Q21. What is meant by ` disproportionation ' of an oxidation state ? Give an example
When a particular oxidation states, becomes less stable relative to other oxidation states, one lower ,
one higher, it is said to undergo disproportionation
+6
+7
+4
Eg:- 3MnO42- + 4H+→2MnO4- + MnO2+ 2H2O
Q22 .Explain why Cu+ ion is not stable in aqueous solutions ?
Because Cu+ ion in aqueous solution undergoes disproportionation 2Cu+(aq) →Cu2+(aq)+ Cu(s)
Some Important compounds of Transition Elements
Oxides and oxoanions of Metals :- Oxides are generally formed by the reaction of metals with oxygen
at high temperatures. All the metals except Sc form MO oxides which are ionic. As the oxidation
number of a metal increases, ionic character decreases.
• Potassium dichromate , K2 Cr2 O7
It is prepared from chromite ore (Fe Cr2 O4)
➢ Fusion of Chromite ore with Na2CO3
4FeCr2O4+ 8Na2CO3+7O2 → 8Na2 CrO4+2Fe2O3+8CO2
➢ Sodium chromate is filtered and acidified with H2SO4 to form orange sodium dichromate
2Na2CrO4+2H+→Na2 Cr2O7+2Na++H2O
➢ Na2Cr2O7 treating with KCl give K2Cr2O7
Na2 Cr2O7+2KCl→K2 Cr2O7+2NaCl
The chromates and dichromates are inter convertible in aqueous solution depending upon P H of the
solution.
The oxidation state of chromium in chromate and dichromate is same .
2CrO42-+ 2H+ →Cr2O72-+H2O
Cr2O72-+2OH- →2CrO42-+H2O
Structure :- The chromate ion is tetrahedral where as the dichromate ion consists of two tetrahedra
sharing one corner with Cr - O -Cr bond angle of 1260
Properties :- It acts as a strong oxidising agent in acidic medium.Its oxidising action can be
represented as Cr2O72-+14H++6e-→2Cr3++7H2O
Acidified K2Cr2O7 will oxidise iodides to iodine ,sulphides to sulphur ,tin (II) to tin (IV) and iron (II)
salts to iron (III) .The half reactions are 6 I - →3 I2+6e- ; 3H2S→6H++3S+6e;
3Sn2+→3Sn4++6e- ; 6Fe2+ →6Fe3++6eThe full ionic equation may be obtained by adding the half - reaction for K2 Cr2O7 to the half - reaction
for the reducing agent .
Eg:- Cr2O72- +14H++6Fe2+ → 2Cr3++6Fe3++7H2O
Uses :➢ It is used in leather industry.
➢ It is also used in chrome plating.
➢ It is used in volumetric analysis.
➢ It is an oxidising agent.
• Potassium permanganate, KMnO4.
It is prepared by fusion of MnO2 with an alkali and oxidising agent like KNO 3. This produces
K2MnO4 and which disproportionates in a neutral or acidic solution to give permanganate.
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2MnO2 + 4KOH+ O2 →2K2MnO4+2H2O
3MnO42-+4H+ →
2MnO4-+MnO2+2H2O
Commercially it is prepared from fusion of MnO2 followed by the electrolytic oxidation of
Manganate(Vl).
MnO2 fused with KOH, oxidised with air MnO42- MnO42- Electrolytic oxidation in alkaline solution MnO4or KNO
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Manganate ion
Permanganate ion
In lab, a Manganese (II) ion salt is oxidised by peroxo disulphate to permanganate ion
2Mn2++5S2O82-+8H2O →2MnO-4+10SO42-+16H+
KMnO4 forms dark purple crystals and that are isostructural with those of KClO4
Structure
(Paramagnetic)
(Diamagnetic)
Properties
Action of heat:- On heating KMnO4 changes or decomposes to potassium manganate and oxygen gas.
2KMnO4→K2MnO4 + MnO2+ O2
Oxidising properties:- Acidified Permanganate solution oxidises oxalates to CO2, iron (II) to
iron(lll), nitrites to nitrates and iodides to free iodine. A few important oxidising reactions of KMnO4
are given below
1. In acid solutions
a. iodide to iodine
10I-+ 2MnO4-+16H+→2Mn2++8H2O+5I2
b.Ferrous to Ferric
5Fe2++MnO4-+8H+→Mn2++4H2O+5Fe3+
c. Oxalate to CO2
5C2O42- +2MnO4-+16H+→2Mn2++8H2O+10CO2
d.H2S to S
5S2-+2MnO4-+16H+→2Mn2++8H2O+5S
e.Sulphite to Sulphate
5SO32-+2MnO4-+6H+→2Mn2++3H2O+5SO42f.Nitrite to nitrate
5NO2-+2MnO4-+6H+→2Mn2++5NO3-+3H2O
2. In neutral or faintly alkaline solutions.
a. iodide to iodate
2MnO4-+H2O + I- → 2MnO2+ 2OH-+ IO3b. Thiosulphate to sulphate
8MnO4-+3S2O32- +H2O→8MnO2+6SO42-+2OHc. Manganous to MnO2
2MnO4-+3Mn2++2H2O→5MnO2+4H+
Uses
➢ It is used in qualitative and quantitative analysis.
➢ Alkaline KMnO4 is Known as Baeyer’s reagent which is used in organic chemistry .
➢ It is used as an oxidising agent in laboratory and industry
➢ It is also used as disinfectant .
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f – Block elements (The inner Transition Elements )
The inner transition elements or f block elements may be defined as elements whose atoms or
simple ions contain partially filled f-orbitals. Their general outer electronic configuration is (n-2)
f0-14(n-1) d0-2 ns2. In these elements electrons are added to the antipenultimate f orbital. These are
divided into 2 series.
i. The Lanthanide series or Lanthanoids
ii.
The actinide series or Actinoids
i. The Lanthanoids :- Lanthanoids were first known as the 'rare earths' elements , Lanthanum
and the next fourteen elements are called lanthanoids .They are characterised by the filling up of 4f
energy level.
General characteristics
1. Electronic configuration →[Xe](n-2)f0-14(n-1)d0-1ns2
2. Atomic and Ionic sizes:- From left to right its size decreases. Both are vary to a lesser extent as
observed for other transition and non-transition elements. This is due to lanthanoid contraction.
3.Ionization Enthalpies:- Successive ionization enthalpies are much higher as expected. These
behaviour can be seen for those lanthanoids having half-filled or completely filled electronic
configuration.
4. Colour: Many of the lanthanoid ions are coloured which may be linked to the electrons in 4f
orbitals. the colour phenomenon may be called as due to f-f transition just like d-d transition.
5. Oxidation states:- Principal oxidation state is +3. +2 and +4 are also known.
Chemical reactions
• Ln burns inO2 Ln2O3
Ln with c, 2773K LnC2
• Ln reacts with acids H2
Ln heated with N LnN
with halogens
• Ln
LnX3
Ln heated with S Ln2S3
• Ln with H2O Ln(OH)3+H2
Uses
➢ Lanthanum is used to produce Misch metal(alloy). This alloy is added to steel to improve
its strength and workability.
➢ La2O3 is used in making Crooke's lenses which protect eyes from UV radiations.
➢ CeO2 is used to polish glass.
ii.The actinoids:- The second inner transition series, the actinoids include the fourteen elements
from Th to Lr after Ac. The elements in actinoid after Uranium are called transuranium elements.
General characteristics
1. Electronic configuration →[Rn](n-2)f0-14(n-1)d0-2 ns2
2.Ionic size:-The gradual but slow decrease is observed in the ionic sizes of actinoids due to
actinoid contraction. It is similar to lanthanoid contraction. Actinoid contraction is caused due to
the poor shielding power of 5f electrons. With the increase in atomic number nuclear charge
increases, but the electrons are added in the same orbital(5f). This way the inward pull experienced
by the 5f electrons increases. Consequently, the steady decrease in the size occurs in the actinoid
series.
3. Oxidation states:- General oxidation state +3. Elements Pa,U, Np, Pu and Am show oxidation
states from +3 to +7.
Uses
➢ Thorium is used for making incandascent gas mandles.
➢ U and Th are used as fuels in atomic reactors.
Some applications of d- and f- block elements.
1. When Ti is alloyed with Al and Sn, an alloy of very high strength is obtained which is used in
the manufacture of gas - turbine engines. Ti is also used in aircraft industry. Zr finds use in
nuclear reactors as moderator.
2.V2O5 is used as a catalyst to convert SO2 into SO3 in the contact process in the manufacture of
H2SO4.
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3. Use of iron in steel industry and copper in electrical appliances is well established.
4. Cu is extensively used in making alloys with Zn and Al.
5. Among actinoids perhaps the most important metal is Uranium (as U-235) to be used in nuclear
power plants for power generation.
Comparison between Lanthanoids and Actinoids
Similarities
• Both exhibit +3 oxidation state predominantly.
• In both, f orbitals are being progressively filled.
• Both are electropositive and very reactive.
• Both exhibit magnetic and spectral properties.
• Like lanthanoid contraction, actinides show actinoid contraction due to the poor shielding
effect of 5f electrons.
Differences
Lanthanoids
Actinoids
Besides +3 oxidation state, they show +2 and Besides the common oxidation state +3, they
+4.
show +4, +5 +6 and +7.
The tendency to form complex is less.
Greater tendency to form complexes.
Except Pm, all elements are non radioactive.
All actinides are radioactive.
Lanthanide compounds are less basic.
Actinide compounds are more basic.
Lanthanoid contraction
The steady decrease in atomic and ionic size of lanthanide elements(La to Lu) with increasing
atomic number is called lanthanoid contraction. This is due to the fact that for every proton
added in the nucleus, the extra electron goes to fill 4f orbitals. As a result, the whole of 4f
electron shell contracts on passing across the lanthanides. The sum of the successive reductions
give the total lanthanoid contraction.
Consequence of lanthanoid contraction
• Similarity of second and third transition series:- This makes it difficult to separate the
elements in the pure state.
• Separation of lanthanides:- Since the change in ionic radii in the lanthanides is very small,
their chemical properties are similar. This make the separation of elements in the pure state is
difficult.
• Variation in basic strength of hydroxides:- Due to lanthanoid contraction, covalent character
between lanthanide ion and OH- ions increases from La3+ to Lu3+. Therefore, the basic strength
of the hydroxides decreases with increase in atomic number. Thus La(OH)3 is most basic while
Lu(OH)3 is the least basic.
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