d and f Block Elements | Chapter 8 | Chemistry | Class 12 | CBSE |

d and f Block Elements | Chapter 8 | Chemistry | Class Xll | CBSE |

D – Block Elements 

d and f Block Elements

Transition Elements : A transition element may be defined as the element whose atom in ground state or ion in one of the common oxidation states , has incomplete d-subshell (partly filled) , i.e. having electrons between 1 and 9 . 

D- block elements are called transition elements since they have unpaired electrons in their d orbital.

Note : All transition elements are d block elements but all d block elements are not transition elements.

Zinc , Cadmium and Mercury are not transition elements because they do not partly filled d orbital in their  ground state or stable oxidation state . 

General Electronic Configuration of d block element : [Noble gas ] (n-1)d 1-10 ns 1-2 


Some elements have exceptional electronic configuration since there is very less energy difference between ns and (n-1)d orbital . 

d and f Block Elements

Atomic Size

Question : Why atomic size decreases upto middle (upto Cr)  but after that it remains nearly same ?

Answer : Atomic Radii decreases upto middle due to increase in nuclear charge . After midway , the electrons enter the last but penultimate shell , added d electron shields the outermost shell . Hence with increase in d electron , shielding effect increases . This counterbalance increased nuclear charge due to increase in atomic number . As a result , atomic radii remain practically same after chromium . 

Question : Why at the end of period , there is slight increase in atomic number ?

Answer : At the end of series , increased electron – electron repulsion between added electrons in same orbital are greater than the attractive force due to increased nuclear charge . This results in expansion of electron cloud and thus atomic radii increases . 

Question : Why the atomic radii of  second and third transition series are almost same ?

Answer : The atomic radii of second and third transition series are nearly same due to lanthanoid contraction . 

Metallic Character 

  • Greater the number of unpaired electron , stronger is the metallic bonding .
  • Strength of metallic bonding is inversely proportional to stable electronic configuration . 

Question : Why do transition metals are much harder than alkali metals ?

Answer : Transition metals have more number of unpaired electrons in their valence shells . As a result , they are able to form strong metallic bonding and hence they are much harder than alkali metals.

Enthalpy of Atomisation

The energy which is required to dissociate one mole of substance into gaseous metal atoms .

Question : Why enthalpies of atomisation of transition metals is quite high ?

Answer : Transition metals have high enthalpies of atomisation due to strong metallic bonding and additional covalent bonding . Metallic bonding is due to presence of unpaired electron while covalent bonding is due to d-d overlapping .

Question : What is periodicity of enthalpies of atomisation across a period  for transition metals ?

 Answer : Upto middle enthalpies of atomisation increases because  strength of metallic bonding increases due to increase in number of unpaired electron. After midway , electrons starts pairing resulting in dec in strength of metallic bonding thereby decreasing enthalpy of atomisation. 

You May Also Read The Solid State, Solutions, Electrochemistry, Chemical Kinetics, Surface Chemistry, General Principles and Processes of Isolation of Elements, The P Block Elements Group 15 Part 1, The P Block Elements Group 16 Part 2, The P Block Element Group 17 and 18 Part 3, Coordination Compounds, Haloalkanes and Haloarenes, Alcohols, Phenols and Ethers, Aldehydes, Ketones & Carboxylic Acids, Amines, Biomolecules, Polymers, Chemistry in Everyday Life for acquiring more knowledge about Chemistry and its chapters.

Magnetic Properties 

Those substances which are attracted by the applied magnetic field are paramagnetic whereas those which are repelled by the magnetic field are called diamagnetic . Substances which are very strongly attracted by the applied field are called ferromagnetic .

The paramagnetic behaviour is measured in terms of magnetic moment (μ )  which is calculated from spin only formula   μ = √ n (n + 2) B.M 

where n is number of unpaired electron and B.M. stand for Bohr Magneton.

Question : Why transition metals and their compound are usually paramagnetic ?

Answer : Paramagnetism arises due to presence of unpaired electrons . When transition metal ions have unpaired electrons in d orbital . They exhibit paramagnetic behaviour . 

Question : Why paramagnetic nature increases upto middle but after that it starts to decrease ?

Answer : Upto middle the number of unpaired electron increases ; hence paramagnetic nature increases but after that electron starts pairing which results in decrease in paramagnetic nature .

No of unpaired electron  Magnetic moment (B.M)   [ μ = √ n (n + 2)] 
n=1 μ = √3 = 1.73BM
n=2 μ = √8 = 2.83 BM
n=3 μ = √15 = 3.87 BM
n=4 μ = √24 = 4.91 BM
n=5 μ = √35 = 5.92 BM 

Melting And Boiling Point 

Melting and boiling point ∝ strength of metallic bonding 

Transition metals have very high melting and boiling points . The melting point of transition metals rise upto middle and then fall as atomic number decreases since metallic bonds between the atom of these elements are responsible for high melting and boiling point . 

Question : Why Cr , Mo and W has maximum melting points in their respective series ?

Answer : These elements have highest melting points in their respective series due to presence of maximum unpaired of electron which results in strong metallic bonding . 

Note : Tungsten has the highest melting point among the d block elements.

Ionisation Enthalpies

The first ionisation enthalpies of d block elements lie between s-block and p-block elements . They are higher than those of s block elements and are lesser than those of p-block elements .

Question : Why the ionisation enthalpies gradually increases with increase in atomic number along a given transition series ?

Answer : The increasing ionisation enthalpies are due to increased nuclear charge with increase in atomic number which reduces the size of atom making the removal of outer electron difficult . 

Question : Why first ionization enthalpy of Zn , Cd and Hg are very high ?

Answer : Zn , Cd and Hg have stable fully filled electronic configuration . Thus for removing electron high amount of energy is required . 

Question : Why Cr has low value of first ionization enthalpy ?

Answer : Cr has low value of first ionization enthalpy because loss of one electron gives stable configuration(3d 5) . 

Question : Why there is a fall in second ionisation enthalpy from Cr to Mn and from Cu to Zn ?

Answer : This is because after the removal of first electron , Cr and Cu acquire a stable configuration (d5 and d10 ) and the removal of second electron is very difficult . 

Question : Why third ionization enthalpy of Mn is very high ?

Answer : This is because the 3rd electron has to be removed from the stable half – filled 3d orbital .  

Question : Why the first ionisation enthalpies of 5d elements are higher as compared to those of 3d and 4d elements ?

Answer : This is because the weak shielding of nucleus by 4f electrons in 5d elements results in greater effective nuclear charge acting on the outer valence electrons. 

Standard Electrode Potential (Eº)

In solution , the stability of compound depends upon electrode potentials rather than ionisation enthalpies . 

Electrode potential value depend upon following factors : 

  • Enthalpy of sublimation    Δ sub H = +ve 
  • Ionisation Enthalpy          Δi H = +ve 
  • Hydration Enthalpy        Δhyd H = – ve 

Δ T H = Δ sub H + Δi H + Δhyd

Trends in M2+ / M Standard Electrode Potential 

Question : Why there is no regular trend in E º value for 3d series elements ?

Answer : There is no regular trend in Eº (M2+ / M) values because their ionisation enthalpy and sublimation enthalpy do no show any regular trend . 

Question : Why Eº value for copper is positive ?

Answer : This is because the sum of enthalpies of sublimation and ionization is not balanced by its hydration enthalpy . 

Question : Why Eº for Mn , Ni and Zn are more negative than expected form general trend ?

Answer : This is due to the greater stability of half filled d- subshell (d5) in Mn 2+ and completely filled d -subshell (d10) in Zn 2+ . The exceptional behaviour of Ni towards Eº value from the regular trend is due to its high negative value of hydration . 

Trends in M3+ / M2+ Standard Electrode Potential

Question : Why Zinc has highest value of Eº (M3+ / M2) ?

Answer : The highest value of Zn is on account of very high stability of Zn 2+ ion with d10 configuration . It is difficult to remove an electron from it to change it into +3 state . 


Question : Why is copper (I) not known to exist in aqueous solution?

Answer : Copper (I) ions are unstable in aqueous solution and undergo disproportionation . The stability of  Cu2+ rather  than Cu + is due to much more negative Δhyd H of Cu2+ than Cu+ , which compensates more for the second enthalpy of Cu .  2Cu + → Cu 2+ + Cu.

Question : Cr2+ is a strong reducing agent whereas Mn3+ with the same (d4) configuration is an oxidising agent. Give reasons. 

Answer : Cr2+ is a strong reducing agent because it can loose an electron to form Cr3+ which has stable 3d3 configuration (as it has half – filled t2g) . On the other hand , Mn3+ can accept an electron to form Mn2+ resulting in half filled (d5) configuration which has extra stability . Thus it behaves as an oxidising agent . 

Question : Out of Cr2+ and Fe2+ , which one is more reducing in nature and why?

Answer : Cr2+ is a strong reducing agent than Fe2+ because both Cr3+ and Fe3+ have stable half filled t2g (3d3) and half filled d orbital (3d5) respectively , Cr releases its electron easily as compared to Fe since removing an electron from a paired orbital is rather difficult . 

Oxidation States

Question : Why do transition element show variable oxidation state?

Answer : ns and (n-1)d electrons of transition metal have very less difference in energies and hence both can participate in bonding , which results in variable oxidation state . When ns electrons take part in bonding they exhibit lower oxidation state whereas when  (n-1)d electrons along with ns electrons participate in bonding , they exhibit variable oxidation state . 

Question : Name the oxo metal anion of one of the transition metals in which the metal exhibits the oxidation state equal to the group number?

Answer : In MnO4 ion , the oxidation state of Mn is +7 . It is equal to its group number . 

In CrO4 2- ion , the oxidation state of Cr is +6 . It is equal to its group number 6 .

Mn show all the oxidation state from +2 to +7 

Question : Why Sc , Ti , Cu , Zn or extreme end elements show fewer oxidation state ?

Answer : The lesser number of oxidation state on the extreme ends are either due to too few electrons to loose or share (Sc , Ti) or too many d electrons so that fewer orbitals are available to share electrons with others (Cu , Zn ) .

Note : The highest oxidation state shown by any transition metal is +8 (by osmium or in a very few compounds by ruthenium)

Question : Why is highest oxidation state exhibited in oxo-anions of transition metal?

Answer : This is because oxygen has small size , high electronegativity and is also a strong oxidising agent . Oxygen also has tendency to form π bond with transition metal . 

Question : Why the higher oxidation states are usually exhibited by the members in the middle of a series of transition elements?

Answer : This is due to presence of large number of unpaired electrons in d orbitals in the middle of the series and involvment of all ns and (n-1)d electrons in the bonding . e.g, Mn has oxidation states from +2 to +7 . 

Question: The highest fluoride of Mn is MnF4 whereas the highest oxide is Mn2O7 . Explain?

Answer : This is because oxygen can form multiple bond with transition metal whereas Fluorine can form only single bond with transition metal . Thus , oxygen stabilizes the highest oxidation state even more than fluorine . 

Question: How is variability in oxidation state of transition metal different from non-transition metal?

Answer : The oxidation state of transition metal differ by unity while that of non-transition metal differ by two 

Note: The most common oxidation state of first row transition metal is +2 except in case of Scandium (which has +3) 

Catalytic Properties

Many of the transition metals and their compounds are used as catalysts 

  • Vanadium as V2O5 in manufacture of H2SO4 by contact process
  • Cobalt as Cobalt thorium in synthesis of gasoline (petrol).
  • Nickel in finely divided state in the hydrogenation reactions .
  • Platinum in oxidation of ammonia to nitric acid by Ostwald’s process
  • Iron in synthesis of ammonia by Haber’s process  

Question : Why mostly transition metal and their compounds are used as catalysts?

Answer : The transition metals and their compounds are used as catalyst because of following reasons :

  • Transition metal have unpaired electron in their d orbital and hence possess the capacity to absorb and re-emit wide range of energies . This makes the required energy of activation available . 
  • They provide a suitable large surface area with free valencies on which the reactants are adsorbed . 
  • They have variable oxidation states.

Coloured Ions

  • Most of the transition metal compounds are coloured both in the solid state and in aqueous solution in contrast to the compounds of s- and p- block elements . 
  • Colour is due to the presence of incomplete d – subshell . Further , when ligands approach the transition metal ions , their d orbital do no remain degenerate . They split into two sets , one consisting of lower energy called t2g and other consisting of higher energy orbital called eg . This is called crystal field splitting . 
  • Thus , electrons can jump from lower energy d orbital to higher one . The required amount of energy is obtained by adsoption of light of particular wavelength in the region of visible light . Radiations of light corresponding to such small amount of energy are available within visible region of light . 

Formation of Interstitial Compound

Transition metals have a unique ability to form interstitial compound by accomodating small size non metal atoms such as H , B , C , N etc  in the voids in their crystal lattice. They are represented by formulae like TiC , TiH2 etc .The bonds present in them are neither typically covalent nor ionic . Some of their important characteristics are as follows :

  • They are very hard and rigid .
  • They show conductivity like that of pure metal .
  • They acquire chemical inertness. 
  • They have high melting points which are higher than those of pure metal . 

Alloy Formation 

  • Alloy are homogenous solid solution of two or more metals obtained by melting the components and then cooling the melt . 
  • They are formed by those metals whose atomic radii differ by not more than 15%.
  • Transition metal form a large number of alloy because they have similar atomic size and atom of one metal can easily take up position in crystal lattice of other . 
  • Alloys are generally harder , have high melting and boiling point and more resistant to corrosion than individual metals . 
  • Ex : Brass ( Copper + Zinc ) 

Complex Formation 

Transition metal ions form a large number of complex compound due to following reasons :

  • Comparatively smaller size of their metal ions.
  • They have high ionic charges .
  • Availability of vacant d orbital so that these orbital can accept lone pairs of electron donated by their ligands. 

Oxides of Transition metals 

  • The metals of first transition series form oxides with oxygen at elevated temperature . 
  • All the metals except Scandium form the oxides with the formula MO which are ionic in nature . As oxidation number of the metal increases , ionic character decreases , e.g., Mn2O7 is a covalent green oil.
  • In general , oxides in the lower oxidation states of metals are basic and in higher oxidation state they are acidic whereas in the intermediate oxidation state , the oxides are amphoteric .
  • Thus , Mn2Odissolves in water to give the acid HMnO4

Salts Containing Oxoanions of Transition Metals

Potassium Dichromate (K2Cr2O7)


It is an acidic compound in which Cr is present in +6 oxidation state . 

i) Preparation of Sodium Chromate:

4FeCr2O4 + 8Na2CO3 + 7O2 → 8Na2CrO4 (water soluble ) + 2Fe2O3 (water insoluble) + 8CO2

ii) Conversion of Sodium Chromate into Sodium dichromate:

2Na2CrO4 + 2H+ → Na2Cr2O7 + 2Na+ + H2O

iii) Conversion of Sodium dichromate into potassium dichromate:

Na2Cr2O7 + 2KCl → K2Cr2O7 (less soluble) + 2NaCl (More soluble)


  • Colour and Melting point : It forms orange crystals which melts at 669 K . 
  • Solubility : It is moderately soluble in cold water but freely soluble in hot water . 
  • Action of Heat : When heated , it decomposes with evolution of oxygen .

4 K2Cr2O7 → 4K2CrO4 + 2Cr2O3 + 3O2                                    

  • Action of Alkalies : When an alkali is added to an orange red solution of dichromate , a yellow solution results due to formation of chromate. 

Cr2O72- + 2OH → 2CrO42- + H2O

2CrO42- + 2H+ → Cr2O72- + H2O

  • Oxidising Properties : 

Question : Explain how acidified K2Cr2O7 oxidises alcohol and is used as test for drunken drivers ?

Answer : When drunken driver is asked to breathe into acidified K2Cr2O7 ;solution is taken in the test tube. If orange colour of solution changes to green colour , then driver is drunk . 

K2Cr2O7  + 4H2SO4 → K2SO4 + Cr2(SO4)3 + 4H2O + 3O

Cr2O72 + 14H+ + 6e →2Cr3+ + 7H2O

Structure of Chromate and Dichromate Ion

The structure of chromate ion is tetrahedral and dichromate ion consists of consists of two tetrahedral sharing one corner with Cr—O—Cr bond angle of 126º . 

Potassium Permanganate (KMnO4)

Potassium permanganate is also an acidic compound in which Mn is present in +7 oxidation state . It is a dark purple coloured crystalline solid . 


On large scale , it is prepared from the mineral pyrolusite , MnO2 which is fused with alkali metal hydroxide in presence of oxygen .

3MnO2 + 4KOH + O2 →2K2MnO4 + 2H2O

3MnO42- + 4H+ → 2MnO4 + MnO2 + 2H2O

In laboratory it is prepared by oxidation of manganese (II) ion salt by peroxodisulphite. 

2Mn2+ + 5S2O82- (peroxodisulphite) + 8 H2O → 2MnO4(permanganate) + 10SO42- + 16H+


  • Colour : It exists as deep purple black prism with a greenish lusture which become dull in air due to superficial reduction . 
  • Solubility : It is moderately soluble in water at room temperature and it is more soluble in hot water . 
  • Action of Heat : When heated to 513K , it readily decomposes giving oxygen .

    2KMnO4 →K2MnO4 (potassium manganate) + MnO2 + O2

  • Oxidising Property : It acts as oxidising agent in all the three medium i.e . acidic , alkaline and neutral 

It oxidises sodium thiosulphite to sodium sulphate

8KMnO4 + 3Na2S2O3 + H2O → 3K2SO4 + 8MnO2 + 3Na2SO4 + 2KOH 

Structure of Manganate and Permanganate Ion 

F-Block Elements 

Elements in which last electron ( also called differentiating electron ) enters the ante-penultimate energy level are called F-Block elements . These elements have also been called inner transition elements. 

The general electronic configuration is (n-2)f1-14 (n-1)d0-1 ns2 

Classification of F-Block Elements 

Depending upon whether last electron enters in 4f orbital or 5f orbital , the f block elements have been divided into two categories : Lanthanoids and Actinoids 


The elements in which last electron enters one of the 4f orbital are called 4f- block elements or first inner transition elements . They are also called Lanthanides or lanthanons or lanthanoids because they come immediately after lanthanum . 

  • General configuration of lanthanoid is 4f 1-14 5d 0-1 6s2 
  • Most common oxidation state of lanthanoid is +3 . However , Ce shows +4 , Eu and Yb shows +2 oxidation state because they acquire stable configuration . 
  • The regular decrease (contraction) in the atomic and ionic radii of lanthanoids with increasing atomic number is known as lanthanoid contraction
  • The radii of members of 5d-series are similar to those of corresponding members of 4d series due to lanthanoid contraction . 
  • Consequences of Lanthanoid contraction : difficulty in seperation of elements due to similar chemical properties and decrease in basic strength from La to Lu due to decrease in size. 
  • These are silvery white metals .
  • These are highly electropositive and reactive metals . 
  • Paramagnetism is shown by positive ions of lanthanoids except La3+ and Lu3+


The elements in which last electron enters one of the 5f orbital are called 5f-block elements or second inner transition elements . These 14 elements are also called actinides or actinoids because they come immediately after actinum . 

  • General configuration of actinoids is 5f 1-14 6d0-1 7s
  • There is a greater range of oxidation state in actinoids . It is due to fact that 5f , 6d and 7s are of comparable energies.
  • The steady decrease in atomic and ionic size along actinoid series is called actinoid contraction and it is due to poor shielding effect of 5f electrons . 
  • They show less tendency to form complexes but greater than lanthanoids.
  • They tarnish rapidly in air forming coating of oxide and thus are not attacked by alkalies.
  • They are strongly paramagnetic and have low ionisation energies

Do share the post if you liked it. For more updates, keep logging on BrainyLads

One Comment

Add a Comment

Your email address will not be published. Required fields are marked *

error: Content is protected !!