Learning Outcomes

The learners will be able to

• understand the postions of the $d$ - & f-blocks elements in the periodic table

• write the electronic configurations of the $d-&$ f-block elements

• describe the general characteritics of the $\mathrm{d}$ - & $\mathrm{f}$-block elements

• explain the physical properties, preparations, structure & uses of $\mathrm{K} _{2} \mathrm{Cr} _{2} \mathrm{O} _{7}$ and $\mathrm{KMnO} _{4}$

• give a comparative account of the lanthanoids and actinoids with respect to their electronic configuration, oxidation states and chemical behaviour

Position in the Periodic Table

$\mathrm{d}$-Block elements which lie between most electropositive s- and most electronegative p-blocks in the long form of the periodic table. They are so called because last electron in them enters in the d-orbital of the ( $n-1$ ) or penultimate shell. They are also called transition elements because their properties lie in between those of s-block and p-block elements and represent a transition (change) from them.

General Electronic Configuration of the transition elements is $(n-1) d^{1-10} n s^{1-2}$ where $n$ is the outermost shell.

The definition excludes $\mathrm{Zn}, \mathrm{Cd}$ and $\mathrm{Hg}$ from transition elements because they do not have incomplete $\mathrm{d}$-subshell in the atomic state or their common oxidation state (viz., $\mathrm{Zn}^{2+}, \mathrm{Cd}^{2+}, \mathrm{Hg}^{2+}$ which are $3 \mathrm{~d}^{10}, 4 \mathrm{~d}^{10}$ and $5 \mathrm{~d}^{10}$ respectively. They do not show properties of transition elements to any appreciable extent except for their ability to form complexes. In order to maintain a rational classification of elements, they are generally studied with the d-block elements and are called non-typical transition elements. Thus, transitions elements are $d$-block elements but all $d$-block elements are not transition elements.

Some Exceptional Electronic Configurations of Transition Elements are as follows:

$\mathrm{Cr}=3 \mathrm{~d}^{5} 4 \mathrm{~s}^{1}, \mathrm{Cu}=3 \mathrm{~d}^{10} 4 \mathrm{~s}^{1}$,

$\mathrm{Nb}=4 d^{4} 5 s^{1}, M o=4 d^{5} 5 s^{1} R u=4 d^{7} 5 s^{1}, R h=4 d^{8} 5 s^{1}, P d=4 d^{10} 5 s^{0}, A g=4 d^{10} 5 s^{1}$.

$P t=5 d^{9} 6 s^{1}, A u=5 d^{10} 6 s^{1}$

The irregularities in the observed configurations of $\mathrm{Cr}, \mathrm{Cu}, \mathrm{Mo}, \mathrm{Pd}, \mathrm{Ag}$ and $\mathrm{Au}$ are explained on the basis of the concept that half filled and completely filled orbitals are relatively more stable than other d-orbital configurations.

Classification

Each horizontal row of the d-block containing 10 elements is called a transition series. There are four transition series:

First transition series or $3 d$ series $(Z=21$ - 30. i.e., $S c$ to $Z n)$ in which $3 d$ subshell is being progressively filled. It is a part of 4 th period.

Second transition series or $4 d$ series $(Z=39-48$, i.e., $Y$ to $C d)$ in which $4 d$ subshell is being progressively filled. It is a part of 5 th period.

Third transition series or $5 \mathrm{~d}$ series ( $\mathrm{Z}=57$, $\mathrm{La}$ and 72 - 80 , i.e., $\mathrm{Hf}$ to $\mathrm{Hg}$ ) in which $5 \mathrm{~d}$ subshell is being progressively filled. It is a part of 6th period.

Fourth transition series or $6 \mathrm{~d}$ series $(Z=89, \mathrm{Ac}$ and 104 - 112, i.e., Rf to $\mathrm{Cn}$ now a complete series containing ten elements in which $6 \mathrm{~d}$ subshell is being progressively filled. It is a part of 7 th period. The outer electronic configuration of $\mathrm{Rf}, \mathrm{Z}=104$ is $6 \mathrm{~d}^{2} 7 \mathrm{~s}^{2}$ and there is a regular increase of one electron in $6 \mathrm{~d}$ orbital as $\mathrm{Z}$ increases. Thus, there are total 40 transition elements at present in the periodic table.

General Characteristics

1. Atomic Radii

$\begin{array}{lllllllllll} \text { 3d series elements } & \mathrm{Sc} & \mathrm{Ti} & \mathrm{V} & \mathrm{Cr} & \mathrm{Mn} & \mathrm{Fe} & \mathrm{Co} & \mathrm{Ni} & \mathrm{Cu} & \mathrm{Zn} \\ \text { At. radius (pm) } & 144 & 132 & 122 & 117 & 117 & 117 & 116 & 115 & 117 & 125 \\ \text { 4d series elements } & \mathrm{Y} & \mathrm{Zr} & \mathrm{Nb} & \mathrm{Mo} & \mathrm{Tc} & \mathrm{Ru} & \mathrm{Rh} & \mathrm{Pd} & \mathrm{Ag} & \mathrm{Cd} \\ \text { At. radius }(\mathrm{pm}) & 162 & 145 & 134 & 129 & 126 & 124 & 125 & 128 & 134 & 141 \\ \text { 5d series elements } & \text { La } & \mathrm{Hf} & \mathrm{Ta} & \mathrm{W} & \mathrm{Re} & \mathrm{Os} & \mathrm{Ir} & \mathrm{Pt} & \mathrm{Au} & \mathrm{Hg} \\ \text { At. radius }(\mathrm{pm}) & 169 & 144 & 134 & 130 & 128 & 126 & 126 & 129 & 134 & 144 \end{array}$

• In a series, the atomic radii first decrease with increase in atomic number upto the middle of the series, then become constant and at the end of the series show a slight increase. This is due to the fact that in the begining, the atomic radius decreases as the nuclear charge increases. But with the increase in the d-electrons, screening effect increases which counter balances the increased nuclear charge due to increase in atomic number. As a result, the atomic radii remain practically same after midway of the series. However, near the end of the series, the electron-electron repulsions between the added electrons in the same d-orbitals are greater than the attractive forces due to increased nuclear charge. This results in the expansion of the electron cloud and thus the atomic radius increases. The atomic radii increase down the group from first transition series to the second transition series but the atomic radii of the second and third transition series are almost the same. The increase in the atomic radii down the group of the second transition series is due to the increase in the number of shells in the atoms but the similarity in the atomic radii of the elements of second and third transition metals is due to lanthanoid (lanthanide) contraction.

2. Metallic Character

All transition elements display typical metallic properties such as high tensile strength, ductility, malleability, high thermal and electrical conductivity and metallic lustre.

All the transition elements are metals having hcp, сср or bcc lattices. Metallic bonding is caused due to the presence of one or two electrons in the outermost energy level (ns) and also unpaired d-electrons. Greater the number of unpaired d-electrons, stronger is the metallic bonding and metals will be hard e.g. $\mathrm{Cr}, \mathrm{Mo}$ and $\mathrm{W}$. Thus, transition elements are more metallic than the representative elements. On the other hand, absence of unpaired electrons results in a weak metallic bonding and the metals will be soft e.g., $\mathrm{Zn}, \mathrm{Cd}$ and $\mathrm{Hg}$.

3. Melting and Boiling Points

• The transition metals have very high m.p. and b.p. The high melting points of these metals are attributed to the involvement of greater number of electron from ( $n-1$ )d in addition to the ns electrons. In any row, the melting points of these metals rise to maximum at $d^{5}$ and fall regularly as the atomic number increases.

• $M n$ and Tc have abnormally low melting points.

• Tungsten has the highest m.p. (3683K) among the d-block elements.

• Due to the absence of unpaired electrons, $\mathrm{Zn}, \mathrm{Cd}$ and $\mathrm{Hg}$ have low m.p.

• $\mathrm{Hg}$ has the lowest m.p. (234K) among the d-block elements & thus it is a liquid at room temperature.

4. Enthalpy of atomization

Transition elements (except $\mathrm{Zn}, \mathrm{Cd}$ and $\mathrm{Hg}$ ) are much harder and less volatile. They exhibit high enthalpies of atomization. The maximum value lies at about middle of each series indicating the interatomic interaction increases with the number of unpaired d-electrons. The metals of $4 \mathrm{~d}$ and $5 \mathrm{~d}$ transition series have greater enthalpies of atomization than the corresponding elements of the first (3d) series. This is due to much more frequent metal-metal bonding in their compounds.

5. Density

Because of large number of valence electrons involving $n s$ and ( $n-1$ ) $d$-orbitals in transition metals, they have strong metallic bonding due to which these metals posses high densities.

6. Ionization Enthalpies

The first ionization enthalpies of $d$-block elements are higher than those of s-block elements and are lesser than those of $p$-block elements. The ionization enthalpies of $3 d$ and $4 d$-series are irregular but increase across the series while those of $5 \mathrm{~d}$-series are much higher than $3 \mathrm{~d}$ and $4 \mathrm{~d}$-elements. This is because of the weak shielding of nucleus by $4 \mathrm{f}$ electrons in $5 d$-transition series which results in greater effective nuclear charge acting on the outer valence electrons.

7. Oxidation States

• All transition elements except the first and the last member in each series show variable oxidation states as the difference of energy in the ( $\mathrm{n}-1) \mathrm{d}$ and $\mathrm{ns}$ orbitals is very little. Hence, electrons from both the energy levels can be used for bond formation.

• The most common oxidation state of $3 \mathrm{~d}$-transition elements is +2 .

• The stable oxidation states shown by 3d-transition series are: Se (III), Ti(IV), V(V), $\mathrm{Cr}(\mathrm{VI}), \mathrm{Mn}(\mathrm{VII}), \mathrm{Fe}(\mathrm{II}, \mathrm{III}), \mathrm{Co}(\mathrm{II}, \mathrm{III}), \mathrm{Ni}(\mathrm{II}), \mathrm{Cu}(\mathrm{II})$ and $\mathrm{Zn}(\mathrm{II})$. The maximum oxidation states correspond in value to the sum of $\mathrm{s}$ and d-electrons upto $\mathrm{Mn}$.

• Mostly ionic bonds are formed in +2 and +3 oxidation states. But higher oxidation states give covalent bonds.

• Higher oxidation states are stabilized by atoms of high electronegativity like 0 or $F$ whereas lower oxidation states (zero or +1 ) are stabilized by ligands which can accept electrons from the metal through $\pi$-bonding (such as $\mathrm{CO}$ ).

• In going down a group, the stability of higher oxidation states increases while that of lower oxidation states decreases.

8. Standard Electrode Potentials $\left(\mathrm{E}^{\circ}\right)$ and Reducing Character

Quantitatively, the stability of transition metal ions in different oxidation states in solution can be determined on the basis of electrode potential data. Lower the electrode potential (i.e., more negative the standard reduction potential) of the electrode, more stable is the oxidation state of the transition metal ion in aqueous solution. Electrode potential values depend upon energy of sublimation of the metal, the ionization enthalpy and the hydration enthalpy. $\mathrm{E}^{\circ}$ values $\left(\mathrm{M}^{2+} / \mathrm{M}\right.$ ) for first row (3d-series) transition metals are given below:

• All the elements of 3d-series are good reducing agents except copper. However they are weaker reducing agents than s-block elements.

$\hspace{1cm}$ The irregular $\mathrm{E}^{\circ}$ values are explained by irregular variation of ionization and sublimation enthalpies.

• More negative $\mathrm{E}^{\circ}$ values than expected for $\mathrm{Mn}$, $\mathrm{Ni}$ and $\mathrm{Zn}$ show greater stability for $\mathrm{Mn}^{2+}$ $\left(\mathrm{d}^{5}\right), \mathrm{Ni}^{2+}\left(\mathrm{d}^{8}\right)$ and $\mathrm{Zn}^{2+}\left(\mathrm{d}^{10}\right)$. The exceptional $\mathrm{E}^{\circ}$ value of $\mathrm{Ni}^{2+}$ is due to its high negative enthalpy of hydration.

  • The comparatively high value of $\mathrm{E}^{\circ}\left(\mathrm{Mn}^{3+} / \mathrm{Mn}^{2+}\right)$ shows that $\mathrm{Mn}^{2+}$ is very stable which is on account of stable $\mathrm{d}^{5}$ configuration of $\mathrm{Mn}^{2+}$.

  • The comparatively low value of $\mathrm{E}^{\circ}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)$ is on account of extra stability of $\mathrm{Fe}^{3+}\left(\mathrm{d}^{5}\right)$. The comparatively low value of $\mathrm{E}^{\circ}\left(\mathrm{V}^{3+} / \mathrm{N}^{2+}\right)$ is on account of the stability of $\mathrm{V}^{2+}$ ion due to its half-filled $\mathrm{t} _{29}{ }^{3}$ configuration.

• Copper is unique in showing positive value of $\mathrm{E}^{\circ}$. This explains why it does not liberate $\mathrm{H} _{2}$ gas from acids. The reason for positive value of $\mathrm{E}^{\circ}$ for $\mathrm{Cu}$ is that the sum of enthalpies of sublimation and ionization is not balanced by hydration enthalpy.

• $\mathrm{E}^{\circ}$ value for the couple $\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}=-0.41 \mathrm{~V}$ and $\mathrm{Mn}^{3+} / \mathrm{Mn}^{2+}=+1.51 \mathrm{~V}$ suggest that $\mathrm{Cr}^{2+}$ is unstable and is oxidised to $\mathrm{Cr}^{3+}$ (which is more stable) and acts as a reducing agent whereas $\mathrm{Mn}^{3+}$ is unstable and is reduced to $\mathrm{Mn}^{2+}$ (which is more stable) and acts as an oxidising agent. It may be noted that both $\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{3+}$ are $\mathrm{d}^{4}$ species.

• $\mathrm{Cu}^{+}$is unstable in water as it may undergo oxidation to $\mathrm{Cu}^{2+}$. The greater stability of $\mathrm{Cu}^{2+}$ (aq) than $\mathrm{Cu}^{+}(\mathrm{aq})$ is due to much more negative enthalpy of hydration of $\mathrm{Cu}^{2+}(\mathrm{aq})$ than $\mathrm{Cu}^{+}(\mathrm{aq})$ which more than compensates for the high second ionization enthalphy of $\mathrm{Cu}$.

9. Magnetic properties

Paramagnetic - Due to presence of the unpaired electrons in(n-1)d orbitals, most of the transition metal ions and their compounds are paramagetic. They are attracted by the magnetic field.

Diamagnetic - They have paired electrons and are repelled by magnetic field.

Magnetic moment $\mu=\sqrt{n(n+2)}$

$\mathrm{n}=$ number of unpaired electrons

Magnetic moment or paramagnetic property increases with increase in the number of unpaired electrons.

Ferromagnetic - Substance which are attracted very strongly are said to be ferromagnetic. Ferromagnetism is an extreme form of paramagnetism.

10. Catalytic Properties

Many transition metals (like Co, Ni, Pt, Fe, Mo) and their compounds are used as catalysts because of the following reasons:

Because of variable oxidation states, they easily absorb and re-emit wide range of energies to provide the necessary activation energy.

Because of variable oxidation states, they easily combine with one of the reactants to form an intermediate which reacts with the second reactant to form the final products.

Because of presence of free valencies on the surface, they can adsorb the reacting molecules, thereby increasing the concentration of the reactants on the surface and hence the rate of reaction. eg. $\mathrm{V} _{2} \mathrm{O} _{5}$ is used in manufacture of $\mathrm{H} _{2} \mathrm{SO} _{4}$ by contact process. Finely divided nickel is used as a catalyst in hydrogenation of oils and fats.

11. Coloured Ions

The transition metals are coloured either due to the $d-d$ transition or charge transfer.

a) d-d transition

This is because of the presence of incompletely filled d-orbitals. When a transition metal compound is formed, the degenerate d-orbitals of the metal split into two sets, one (having three orbitals $d x y$, dyz and $d x z$ called $t _{2 g}$ orbitals) with lower energy and the other (having two orbitals $\mathrm{dx}^{2}-\mathrm{y}^{2}$ and $\mathrm{dz}^{2}$ called eg orbitals) with slightly higher energy in an octahedral field. This is called crystal field splitting. When white light falls on these compounds, some wavelength is absorbed for promotion of electrons from one set of lower energy orbitals to another set of slightly higher energy within the same d-subshell. This is called $d$-d transition. The remainder light is reflected which has a particular colour. $\mathrm{Sc}^{3+}, \mathrm{Ti}^{4+}$ and $\mathrm{V}^{5+}$ have completely empty $\mathrm{d}$-orbitals and hence are colourless. Similarly, $\mathrm{Cu}^{+}, \mathrm{Ag}^{+}, \mathrm{Au}^{+}, \mathrm{Zn}^{2+}$ and $\mathrm{Hg}^{2+}$ have completely filled $\mathrm{d}$-orbitals and there are no vacant $d$-orbitals for promotion of eletrons and hence are colourless.

b) Charge transfer

Charger transfer transition always produces intense colour since the restrictions of selection rules do not apply to transitions between atoms. $\mathrm{MnO} _{4}{ }^{-}$ion has an intense purple colour in solution due to charge transfer transition. In $\mathrm{MnO} _{4}^{-}$, an electron is momentarily transferred from 0 to the metal, thus momentarily changing $0^{2-}$ to $0^{-}$and reducing the oxidation state of the metal from $\mathrm{Mn}(\mathrm{VII})$ to $\mathrm{Mn}(\mathrm{VII})$. Charge transfer transition requires that the energy levels on the two different atoms involved are fairly close.

Colours of $\mathrm{Cr} _{2} \mathrm{O} _{7}{ } _{7}^{2-}, \mathrm{CrO} _{4}{ } _{4}^{2-}, \mathrm{MnO} _{4}^{-}, \mathrm{Cu} _{2} \mathrm{O}$ and Ni-DMG complex are due to charge transfer transitions.

12. Complex Formation

Transition metal ions form a large number of complexes in which the central metal ion is linked to a number of ligands.

• They have high nuclear charge and small size, i.e., charge/size ratio (charge density) is large.

• They have empty d-orbitals to accept the lone pairs of electrons donated by ligands.

• The stability of complexes increase with increase in atomic number of the elements in a series and with decreasing size of its atoms. Moreoever, higher valent cations form more stable complexes.

13. Interstitial compounds

Transition metals form a number of interstitial compounds in which small non-metal atoms such as $\mathrm{H}, \mathrm{C}, \mathrm{B}, \mathrm{N}$ and He occupy the empty spaces (interstitial sites) in their lattices and also form bonds with them.

14. Alloy formation

Due to similarity in atomic sizes, atoms of one transition metal can easily take up positions in the crystal lattice of the other in the molten state and miscible with each other forming homogeneous solid solutions and smooth alloys on cooling.

15. Non-stoichiometric Compounds

Non-stoichiometric compounds are those in which the chemical composition does not correspond to their ideal chemical formulae. The compounds of transition metals with $0, \mathrm{~S}$, Se and Te are generally non-stoichiometric and have indefinite composition. The nonstoichiometry is due to

• Variable valency of transition metals

• presence of defects in their solid state (structure).

e.g. $\mathrm{FeO} _{0.94}, \mathrm{TiH} _{1.7}$

General Properties of First Row Transition Metal Compounds

Oxides and oxometal ions.

• Oxides of metals in low oxidation states +2 and $+3\left(\mathrm{MO}, \mathrm{M} _{3} \mathrm{O} _{4}\right.$ and $\left.\mathrm{M} _{2} \mathrm{O} _{3}\right)$ are generally basic except $\mathrm{Cr} _{2} \mathrm{O} _{3}$ which is amphoteric in character.

• Oxides of metals in higher oxidation states $+5\left(\mathrm{M} _{2} \mathrm{O} _{5}, \mathrm{MO} _{3}, \mathrm{M} _{2} \mathrm{O} _{7}\right)$ are generally acidic in character.

• Oxides of metals in their intermediate oxidation states $+4\left(\mathrm{MO} _{2}\right)$ are generally amphoteric in nature. Besides the oxides, the oxocations, which stabilise $\mathrm{V}(\mathrm{V})$ species is $\mathrm{VO} _{2}{ }^{+}, \mathrm{V}$ (IV) species is $\mathrm{VO}^{2+}$ and $\mathrm{Ti}(\mathrm{IV})$ species is $\mathrm{TiO}^{2+}$.

• As the oxidation number of the metal in the oxide increases, ionic character decreases and acidic character increases. Thus,

$\hspace{0.5cm}\begin{array}{lllll} +2 & +8 / 3 & +3 & +4 & +7 \\ \mathrm{MnO} & \mathrm{Mn}_3 \mathrm{O}_4 & \mathrm{Mn}_2 \mathrm{O}_3 & \mathrm{MnO}_2 & \mathrm{Mn}_2 \mathrm{O}_7 \end{array}$

$\longrightarrow$ Ionic chacracter decreases

$\longrightarrow$ Acidic character increases

Solved Examples

Question 1- Four successive members of the first row transition elements are listed below with their atomic numbers. Which one of them is expected to have the highest third ionisation enthalpy?

1. $\operatorname{Vanadium}(Z=23)$

2. Chromium $(Z=24)$

3. Manganese $(Z=25)$

4. Iron $(Z=26)$

Question 2- The aqueous solution containing which one of the following ions will be colourless?

1. $\mathrm{Sc}^{3+}$

2. $\mathrm{Fe}^{2+}$

3. $\mathrm{Ti}^{3+}$

4. $\mathrm{Mn}^{2+}$

(Atomic number: $\mathrm{Sc}=21, \mathrm{Fe}=26, \mathrm{Ti}=22, \mathrm{Mn}=25$ )

Question 3- Which one of the following elements with the following outer orbital configuration may exhibit the largest number of oxidation states?

1. $3 d^{2} 4 s^{2}$

2. $3 d^{3} 4 s^{2}$

3. $3 d^{5} 4 d^{1}$

4. $3 d^{5} 4 s^{2}$

Question 4- Which one of the following ions has electronic configuraion $[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ ?

1. $\mathrm{Co}^{3+}$

2. $\mathrm{Ni}^{3+}$

3. $\mathrm{Mn}^{3+}$

4. $\mathrm{Fe}^{3+}$

PRACTICE QUESTIONS

Question 1- Transition elements have the maximum tendency to form complexes because

(a) they are metals and all metals form complexes

(b) they contain incompletely filled d-orbitals

(c) their charge/size ratio is quite large

(d) of both (b) and (c)

Question 2- In a transition series, as the atomic number increases, paramagnetism

a) increases gradually

b) decreases gradually

c) first increases to a maximum and then decreases

(d) first decreases to a minimum and then increases

Question 3- Which of the following transition metal ions will have definite value of magnetic moment?

(a) $\mathrm{Sc}^{3+}$

(b) $\mathrm{Ti}^{3+}$

(c) $\mathrm{Cu}^{+}$

(d) $\mathrm{Zn}^{2+}$

Question 4- Cuprous ion is colourless, while cupric ion is coloured because

(a) both have unpaired electrons in the d-orbital

(b) cuprous ion has a complete d-orbital and cupric ion has an incomplete d-orbital

(c) both have half-filled $p$ - and $d$-orbitals

(d) cuprous ion has incomplete d-orbital and cupric ion has a complete d-orbital

Question 5- Amongst $\mathrm{TiF} _{6}{ } _{6}{ }^{2-}, \mathrm{CoF} _{6}{ } _{6}{ }^{3-} \mathrm{Cu} _{2} \mathrm{Cl} _{2}$ and $\mathrm{NiCl} _{4}{ }^{2-}($ At. Nos. $\mathrm{Ti}=22, \mathrm{Co}=27, \mathrm{Cu}=29, \mathrm{Ni}=28)$. The colourless species are

(a) $\mathrm{CoF} _{6}{ } _{6}{ }^{3-}$ and $\mathrm{NiCl} _{4}{ }^{2-}$

(b) $\mathrm{TiF} _{6}{ } _{6}{ }^{2-}$ and $\mathrm{CoF} _{6}{ }^{3-}$

(c) $\mathrm{Cu} _{2} \mathrm{Cl} _{2}$ and $\mathrm{NiCl} _{4}{ } _{4}$

(d) $\mathrm{TiF} _{6}{ } _{6}{ }^{2-}$ and $\mathrm{Cu} _{2} \mathrm{Cl} _{2}$

Question 6- The magnetic moment of a transition metal of $3 \mathrm{~d}$-series is $6.92 \mathrm{~B}$.M. Its electronic configuration would be

(a) $3 d^{5} 4 s^{2}$

(b) $3 d^{5} 4 s^{1}$

(c) $3 d^{6} 4 s^{0}$

(d) $3 d^{5} 4 s^{0}$

Question 7- The first ionization enthalpy of elements of $5 d$-series are higher than those of $3 d$ - and $4 \mathrm{~d}$ series. This is because

(a) the atomic radii of elements of $5 d$-series are smaller than those of $3 d$ - and $4 d$-series

(b) the nuclear charges of elements of $5 d$-series are higher than those of $3 d$ - and $4 d$-series

(c) the valence shell electrons of $5 d$ elements experience greater effective nuclear charge than $3 \mathrm{~d}$ - and $4 \mathrm{~d}$-elements due to poor shielding of $4 \mathrm{f}$ subshell electrons

(d) there is apperciable shielding of 4f-subshell electrons on valence shell electrons of $5 \mathrm{~d}$ elements

Question 8- Which one of the following shows highest magnetic moment?

(a) $\mathrm{V}^{3+}$

(b) $\mathrm{Cr}^{3+}$

(c) $\mathrm{Fe}^{3+}$

(d) $\mathrm{Co}^{3+}$

Question 9- The transition elements are more metallic than the representative elements because they have

(a) the electrons in d-orbitals

(b) electron pairs in d-orbitals

(c) availability of d-orbitals for bonding

(d) unpaired electrons in metallic orbitals

Question 10- Colour in transition metal compounds is attributed to

(a) small sized metal ions

(b) absorption of light in UV region

(c) complete ns subshell

(d) incomplete ( $\mathrm{n}-1$ ) dsubshell

Question 11- Among the following outermost configurations of transition metals which shows the highest oxidation state?

(a) $3 d^{3} 4 s^{2}$

(b) $3 d^{5} 4 s^{1}$

(c) $3 d^{3} 4 s^{2}$

(d) $3 d^{6} 4 s^{2}$

Question 12- In which of the following ions, $d$-d transition is not possible?

(a) $\mathrm{Ti}^{4+}$

(b) $\mathrm{Cr}^{3+}$

(c) $\mathrm{Mn}^{2+}$

(d) $\mathrm{Cu}^{2+}$

$\mathrm{f}$-Block elements

f-block elements (inner transition elements or lanthanoids and actionoids)

The f-block consists of two series, lanthanoids (the fourteen elements following lanthanum) and actinoids (the fourteen elements following actinium).

lanthanoids Actinoids

Both lanthanoids and actinoids are collectively called f-block elements because last electron in them enters into f-orbitals of the antepenultimate (i.e., inner to penultimate) shell partly but incompletely filled in their elementary or ionic states. The name inner transition elements is also given to them because they constitute transition series within transition series ( $\mathrm{d}$-block elements) and the last electron enters into antepenultimate shell (n-2)f.

LANTHANOIDS

The elements with atomic numbers 58 to 71 i.e., cerium to lutetium (which come immediately after lanthanum Z - 57) are called lanthanoids or lanthanides or lanthanones or rare earths. Their general electronic configuration is $[\mathrm{Xe}] 4 \mathrm{f}^{1-14} 5 \mathrm{~d}^{0-1} 6 \mathrm{~s}^{2}$.

General characteristics

1. Atomic and Ionic Radii

The overall decrease in atomic and ionic radii from Lanthanum to Lutetium in a unique feature in the chemistry of the lanthanoids. This regular decrease is known as lanthanoid contraction. It is due to greater effect of the increased nuclear charge than that of the screening effect, which is attributed to the imperfect shielding of one electron by another in the same sub-shell.

Consequences of lanthanoid contraction

a) It results in slight variation in their chemical properties which helps in their separation by ion exchange methods.

b) Each element beyond lanthanum has same atomic radius as that of the element lying above it in the same group (e.g., Zr 145 pm, Hf 144 pm); Nb 134 pm, Ta 134 pm; Mo 129 pm, W 130 pm).

c) The covalent character of hydroxides of lanthanoids increases as the size decreases from $\mathrm{La}^{3+}$ to $\mathrm{Lu}^{3+}$. Hence, the basic strength decreases. Thus, $\mathrm{La}(\mathrm{OH}) _{3}$ is most basic whereas $\mathrm{Lu}(\mathrm{OH}) _{3}$ is least basic. Similarly, the basicity of oxides also decreases in the order from $\mathrm{La}^{3+}$ to $\mathrm{Lu}^{3+}$.

d) Tendency to form stable complexes from $\mathrm{La}^{3+}$ to $\mathrm{Lu}^{3+}$ increases as the size decreases in that order.

2. Oxidation states: Most stable oxidation state of lanthanoids is +3 . Oxidation states +2 and +4 also exist but they revert to +3 e.g., $\mathrm{Sm}^{2+}$, $\mathrm{Eu}^{2+}, \mathrm{Yb}^{2+}$ lose electron to become +3 and hence are good reducing agents, whereas $\mathrm{Ce}^{4+}, \mathrm{Pr}^{4+}, \mathrm{Tb}^{4+}$ in aqueous solution gain electron to become +3 and hence are good oxidizing agents. There is a large gap in energy of $4 \mathrm{f}$ and $5 \mathrm{~d}$ subshells and thus the number of oxidation states is limited.

3. Colour: Most of the trivalent lanthanoid ions are coloured both in the solid state and in aqueous solution. This is due to the partly filled f-orbitals which permit f-transition.

4. Magnetic properties: All lanthanoid ions with the exception of $\mathrm{Lu}^{3+}, \mathrm{Yb}^{2+}$ and $\mathrm{Ce}^{4+}$ are paramagnetic because they contain unpaired electrons in the $4 f$ orbitals. These elements differ from the transition elements in that their magnetic moments do not obey the simple “spin only” formula $\mu _{e f f}=\sqrt{n(n+2)}$ B.M. where $\mathrm{n}$ is equal to the number of unpaired electrons. In transition elements, the orbital contribution of the electron towards magnetic moment is usually quenched by interaction with electric fields of the environment but in case of lanthanoids the 4f-orbitals lie too deep in the atom for such quenching to occur.

Magnetic moments of lanthanoids are calculated by taking into consideration spin as well as orbital contributions and a more complex formula

$$ \mu _{e f f}=\sqrt{4 S(S+1)+L(L+1)} B \cdot M $$

It involves the orbital quantum number $L$ and spin quantum number $S$.

5. Complex formation: Although the lanthanoid ions have a high charge (+3) yet the size of their ions is very large yielding small charge to size ratio, i.e., low charge density. As a consequence, they have poor tendency to form complexes. They form complexes mainly with strong chelating agents such as EDTA, $\beta$-diketones, oxime etc. No complexes with $\beta$-bonding ligands are known.

6. Reducing character: The $\mathrm{E}^{\circ}$ values for $\mathrm{M}^{3+} ; \mathrm{M}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{M}(\mathrm{s})$ lie in the range of -2.2 to $-2.4 \mathrm{~V}$ (exception being $\mathrm{Eu}, \mathrm{E}^{\circ}=-2.0 \mathrm{~V}$ ) indicating thereby that they are highly electropositive, readily lose electrons and thus are good reducing agents.

7. Chemical behaviour:

ACTINOIDS

The elements with atomic numbers 90 to 103, i.e., thorium to lawrencium (which come immediately after actinium, $Z=89$ are called actinoids or actinides or actinones. These elements involve the filling of 5f-orbitals and are also called 5 f-block elements or second inner transition series. Their general electronic configuration is $[R n] 5 f^{1-14} 6 d^{0-1} 7 s^{2}$.

GENERAL CHARACTERSTICS

1. Oxidation states: Exhibition of large number of oxidation states of actinoids is due to the fact that there is a very small energy gap between $5 \mathrm{f}, 6 \mathrm{~d}$ and $7 \mathrm{~s}$ subshells and thus all their electrons can take part in bond formation.

Though $4 f$ and $5 f$ orbitals have similar shapes but $5 f$ is less deeply buried than $4 f$. Hence, $5 f$ electrons can participate in bonding to a far greater extent.

2. Actinoid contraction: There is a regular decrease in ionic radii with increase in atomic number from Th to Lr. This is called actinoid contraction analogous to the lanthanoid contraction. It is caused due to imperfect shielding of one $5 f$ electron by another in the same shell. This results in increase in the effective nuclear charge which causes contraction in size of the electron cloud.

This contraction is greater from element to element in this series due to poor shielding by $5 f$ electrons. This is because $5 f$ orbitals extend in space beyond $6 s$ and $6 p$ orbitals whereas $4 f$ orbitals are buried deep inside the atom.

3. Colour of the ions: Ions of actinoids are generally coloured which is due to f-transitions. It depends upon the number of electrons in $5 f$ orbitals. The cations containing $5 f^{0}$ and $5 f^{7}$ (exactly half filled $f$ subshell) electrons are colourless. The cations containing $5 f^{2-6}$ electrons are coloured both in the crystalline state as well as in aqueous solution. For example, $\mathrm{Ac}^{3+}$ $\left(5 f^{0}\right), \mathrm{Cm}^{3+}\left(5 f^{f}\right), \mathrm{Th}^{4+}\left(5 f^{0}\right)$ are colourless while $\mathrm{U}^{3+}\left(5 f^{3}\right)$ is red, $\mathrm{Np}^{3+}\left(5 f^{4}\right)$ is blue, $\mathrm{Pu}^{3+}\left(5 f^{3}\right)$ is violet, $\mathrm{Am}^{3+}\left(5 f^{6}\right)$ is pink.

4. Complex formation: Actinoids have a greater tendency to form complexes because of higher nuclear charge and smaller size of their atoms. They form complexes even with p-bonding ligands such as alkylphosphines, thioethers etc. besides EDTA, $\beta$-diketones, oxine etc. The degree of complex formation decreases in the order.

$$ \mathrm{M}^{4+}>\mathrm{MO} _{2}^{2+}>\mathrm{M}^{3+}>\mathrm{MO} _{2}^{+} $$

Where $\mathrm{M}$ is element of actinoid series. There is a high concentration of charge on the metal atom in $\mathrm{MO} _{2}{ }^{2+}$ which imparts to it relatively high tendency towards complex formation.

Solved Examples

Question 1- Which of the following Lanthanoid ions is diamagnetic?

(At. No. $\mathrm{Ce}=58, \mathrm{Sm}=62$, $\mathrm{Eu}=63, \mathrm{Yb}=70$ )

1. $\mathrm{Eu}^{2+}$

2. $\mathrm{Yb}^{2+}$

3. $\mathrm{Ce}^{2+}$

4. $\mathrm{Sm}^{2+}$

Question 2- In context with the lanthanoids and actinoids which of the following statements is incorrect?

1. Compared with lanthanoids the lower oxidation states are less important and the higher oxidation states are more important.

2. The $5 f$ electrons of actinoids are more easily removed than the $4 f$ electrons of lanthanoids.

3. Actinoids exhibit greater variation in oxidation states as compared with lanthanoids.

4. The +3 oxidation state in lanthanoid is much less common than that in actinoids.

PRACTICE QUESTIQNS

Question 1- The +3 ion of which one of the following has half filled $4 \mathrm{f}$ subshell?

(a) La

(b) Lu

(c) $\mathrm{Gd}$

(d) $\mathrm{Ac}$

Question 2- Arrange $\mathrm{Ce}^{3+}, \mathrm{La}^{3+}, \mathrm{Pm}^{3+}$ and $\mathrm{Yb}^{3+}$ in increasing order of their ionic radii

(a) $\mathrm{Yb}^{3+}<\mathrm{Pm}^{3+}<\mathrm{Ce}^{3+}<\mathrm{La}^{3+}$

(b) $\mathrm{Ce}^{3+}<\mathrm{Yb}^{3+}<\mathrm{Pm}^{3+}<\mathrm{La}^{3+}$

(c) $\mathrm{Yb}^{3+}<\mathrm{Pm}^{3+}<\mathrm{La}^{3+}<\mathrm{Ce}^{3+}$

(d) $\mathrm{Pm}^{3+}<\mathrm{La}^{3+}<\mathrm{Ce}^{3+}<\mathrm{Yb}^{3+}$

Question 3- Cerium $(Z=58)$ is an important member of the 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 oxidizing agent

(c) The +4 oxidation state of cerium is more stable in solutions

(d) The +3 oxidation state of cerium is more stable than the +4 oxidation state.

Question 4- Lanthanoid contraction is caused due to

(a) the same effective nuclear charge from Ce to Lu

(b) the imperfect sheiding on outer electrons by $4 \mathrm{f}$ electrons from the nuclear charge

(c) the appreciable shielding on outer by $4 f$ electrons from the nuclear charge

(d) the appreciable shielding on outer electrons by 5 d electrons from the nuclear charge.

Question 5- Knowing that the chemistry of lanthanoids $(\mathrm{Ln})$ is dominated by its +3 oxidation state, which of the following statements is incorrect?

(a) The ionic sizes of $\mathrm{Ln}$ (III) decrease in general with increasing atomic number

(b) Ln (III) compounds are generally colourless

(c) Ln (III) hydroxides are mainly basic in character

(d) Because of the large size of the Ln (III) ions, the bonding in ts compounds is predominantly ionic in character

Question 6- Which is not correct statement about the chemistry of $3 \mathrm{~d}$ and $4 \mathrm{f}$ series elements?

(a) $3 d$-elements show more oxidation states than $4 \mathrm{f}$-series elements

(b) The energy difference between $3 \mathrm{~d}$ and $4 \mathrm{~s}$ orbitals is very little

(c) Europium (II) is more stable than Ce (II)

(d) The paramagnetic character in $3 \mathrm{~d}$-series elements increases from scandium to copper

Question 7- The maximum oxidation state exhibited by actinoid ions is

(a) +5

(b) +4

(c) +7

(d) +8

Question 8- In aqueous solution, Eut ${ }^{2+}$ ion acts as

(a) an oxidising agent

(b) a reducing agent

(c) either (a) and (b)

(d) none of these

Question 9- The actinoids showing +7 oxidation state are

(a) $\mathrm{U, Np}$

(b) $\mathrm{Pu}, \mathrm{Am}$

(c) $\mathrm{Np, Pu}$

(d) $\mathrm{Am}, \mathrm{Cm}$

Question 10- Across the lanthanoid series, the basicity of the lanthanoid hydroxides

(a) increases

(b) decreases

(c) first increases and then decreases

(d) first decreases and then increases

Question 11- The reason for the stability of $\mathrm{Gd}^{3+}$ ion is

(a) $4 \mathrm{f}$ sub shell-half filled

(b) $4 \mathrm{f}$ sub shell-completely filled

(c) possesses the general electronic configuration of noble gases

(d) $4 f$ sub shell empty

Question 12- Most common oxidation states shown by cerium are

(a) $+2,+4$

(b) $+3,+4$

(c) $+3,+5$

(d) $+2,+3$