Yb²⁺ - 4f¹⁴ - no unpaired electrons - diamagnetic

As we move from La to Lu, their metallic behaviour because almost similar to that of aluminium.

+7       +2+3          2+              3+       4+
MnO₄^- + FeC₂O₄ ⟶ Mn²⁺ + Fe³⁺ + 2CO₂
5e^- acception       3e^- release
5 moles of FeC₂O₄ \(\equiv\) 3 moles of KMnO₄
1 mole of FeC₂O₄ \(\equiv\) (3/5) moles of KMnO₄
1 mole of FeC₂O₄ \(\equiv\) 0.6 moles of KMnO₄

MnO^-₄ + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O

+5           +6           +7
VO₂⁺ < Cr₂O₇^2- < MnO₄^-
Greater the oxidation state, higher is the oxidising power.

Mn²⁺ ⇒ 3d⁵ contains 5 unpaired electrons
n = 5,
\( \sqrt{n(n+ 2)} \) BM
\(= \sqrt{5(5+ 2)} = \sqrt{35} = 5.92 BM\)

Cr ⇒ [Ar]3d⁵4s¹

Copper exhibits +1 oxidation state most frequently Cu (29) - electronic configuration 3d¹⁰4s¹ copper ready to lose outer most one electron to attain the stable full filled electronic configuration. Hence it exhibits +1 oxidisation state.

(i) The first transition metal Scandium exhibits only +3 oxidation state, but all other transition elements exhibit variable oxidation states by loosing electrons from (n-1)d orbital and ns orbital as the energy difference between them is very small. At the beginning of the series, +3 oxidation state is stable but towards the end +2 oxidation state becomes stable. The first and last elements show less number of oxidation states and the middle elements with more number of oxidation states
(ii) For example, the first element Sc has only one oxidation state +3; the middle element Mn has six different oxidation states from +2 to +7. The last element Cu shows +1 and +2 oxidation states only.

Elemental copper is more stable than Cu²⁺. The electronic configuration of copper is 3d¹⁰ 4s¹ completely filled 3d orbital with stable configuration.
But Cu²⁺ has configuration as 3d⁹. Hence \(\mathrm{E}_{\mathrm{M}^{2+} / \mathrm{M}}^{0}\) value is positive for Cu²⁺.

Cr²⁺ is stronger reducing agent than Fe²⁺. The standard electrode potential (E⁰) of Cr²⁺ is -0.91 V and that of Fe²⁺ is only -0.44 V.
If the standard electrode potential of a metal is large and negative is a powerful reducing agent, because it loses electrons easily.
Hence Cr²⁺ is stronger reducing agent.

Zirconium is a 4d element and Hafnium is a 5d element. There is an unexpected observation in the atomic radius of 5d elements which have nearly the same atomic radius as that of corresponding 4d elements. This is due to lanthanoid contraction. Hence they exhibit similar properties.

\(Eu (63)-[\mathrm{Xe}] 4 \mathrm{f}^{7} 5 \mathrm{~d}^{0} 6 \mathrm{~s}^{2}, \mathrm{Eu}^{2+}-[\mathrm{Xe}] 4 \mathrm{f}^{7} \)
\(\mathrm{Ce}(58)-[\mathrm{Xe}] 4 \mathrm{f}^{1} 5 \mathrm{~d}^{1} 6 \mathrm{~s}^{2}, \mathrm{Ce}^{2+}-[\mathrm{Xe}] 4 \mathrm{f}^{1} 5 \mathrm{~d}^{1} \)
Eu²⁺ has exactly half filled stable electronic configuration. Hence Europium (II) is more stable than Cerium (II).

La(OH)₃ is more basic than Lu(OH)₃. Due to lanthanide contraction, the size of Ln³⁺ ions decreases regularly with increase in atomic number. According to Fajan's rule, decrease in size of Ln³⁺ ions decreases the basic character between Ln³⁺ and OH^- ion in Ln(OH)₃. So La(OH)₃ is more basic than Lu(OH)₃.

(i) In the actinoid series, the elements have poor shielding effect when compared with lanthanide series.
(ii) Hence in the actinoid series, when atomic number increases the effective nuclear charge also increases so actinoid contraction is greater from element to element than the lanthanoid contraction.

As we move from left to right in a transition metal series, the ionization enthalpy increases as expected. This is due to increase in the nuclear charge corresponding to the filling of d electrons. The increase in first ionisation enthalpy with increase in atomic number along a particular series is not regular. The added electron enters (n-1) d orbital and the inner electrons act as a shield and decrease the effect of nuclear charge on valence ns electrons. Therefore, it leads to variation in the ionization energy value.

Mn³⁺ has large and negative standard electrode potential E⁰ (-1.18 V) than that of Cr²⁺ which has only -0.91 V. If the standard electrode potential of a metal is large and negative, the metal is a powerful reducing agent because it loses electrons easily. Hence Mn³⁺ is strongly oxidizing while Cr²⁺ is strongly reducing.

S.No	Lanthanoids	Actinoids
1.	Differentiating electron enters in 4f orbital	Differentiating electron enters in 5f orbital
2.	Binding energy of 4f orbitals are higher	Binding energy of 5f orbitals are lower
3.	They show less tendency to form complexes	They show greater tendency to form complexes
4.	Most of the lanthanoids are colourless	Most of the actinoids are coloured For Example: U3+ (red) U4+ (green).
5.	They do not form oxo cations	They do form oxo cations such as UO22+, NpO22++ etc.
6.	Besides +3 oxidation states lanthanoids show +2 and +4 oxidation states in few cases	Besides +3 oxidation states actinoids show higher oxidation states such as +4, +5, +6 and +7

(i) In transition series, as we move down from Ti to Zn, the standard reduction potential E⁰_M²⁺/M³ value is approaching towards less negative value and copper has a positive reduction potential, i. e. elemental copper is more stable than Cu²⁺.
(ii) E⁰_M²⁺/M value for manganese and zinc are more negative than regular trend. It is due to extra stability arises due to the half filled d⁵ configuration in Mn²⁺ and completely filled d¹⁰ configuration in Zn²⁺.
(iii) The standard electrode potential for the M³⁺/M²⁺ half cell gives the relative stability between M³⁺ and M²⁺.
(iv) The high reduction potential of Mn³⁺/Mn²⁺ indicates Mn²⁺ is more stable than Mn³⁺.
(v) Mn³⁺ has a 3d⁴ configuration while that of Mn²⁺ is 3d⁵. The extra stability associated with a half filled d sub-shell makes the reduction of Mn³⁺ very feasible \(\left[\mathrm{E}^{\circ}=+1.51 \mathrm{~V}\right]\).

(i) Fe³⁺ - electronic configuration - [Ar] 3d⁵
(ii) It has exactly half-filled stable electronic configuration.
(iii) Fe²⁺ - electronic configuration -[Ar]3d⁶
(iv) It has only partially filled d-orbitals.
Hence Fe³⁺ is more stable than Fe²⁺.

(i) The actual position of Lanthanides in the periodic table is at group number 3 and period number 6. However, in the sixth period after lanthanum, the electrons are preferentially filled in inner 4f sub shell and these fourteen elements following lanthanum show similar chemical properties.
(ii) Similarly the fourteen elements following actinium resemble in their physical and chemical properties. Hence they are placed separately bottom of the modern periodic table.

(i) The elements in which the extra electron enters (n-2) f orbitals are called f-block elements. These elements are called as inner transition elements because they form a transition. Series within the transition elements.
(ii) The f-block elements are also called as rare earth elements. They are divided into lanthanoid series (4f block elements) and actinoid series (5f block elements).

IUPAC defines transition metal as an element whose atom has an incomplete d-sub shell or which can give rise to cations with an incomplete d-sub shell. They occupy the central position of the periodic table, between s and p-block elements.
Examples: Fe, Cu, Ag, Au

1. Fe²⁺ the electronic configuration is 3d⁶.
2. Fe³⁺ the electronic configuration is 3d⁵.
3. So it has exactly half filled stable electronic configuration. Hence Fe³⁺ is more stable than Fe²⁺.
4. For manganese Mn²⁺. the electronic configuration is 3d⁵ and that of Mn³⁺ is 3d⁴. Here Mn²⁺ has exactly half-filled stable electronic configuration.

1. The value of the ionisation enthalpies can be used in estimating the relative stability of various transition metal compounds or ions.
2. The relative stability of Pt⁴⁺ & Ni⁴⁺ can be calculated as follows; Ni⁴⁺:I.E₁+1.E₂+ 1.E₃+ IE₄ = 737 + 1753 + 3395 + 5297 = 11,182 kJmol^-1 
Pt⁴⁺:I.E₁+I.E₂+ I.E₃+ IE₄ = 864 + 1791 + 2800 + 4150 = 9,605 kJmol^-1
3. The calculated ionisation enthalpy / energy values of Pt⁴⁺ (9,605 kJmol^-1) is smaller than Ni⁴⁺(11,182 kJmol^-1). 
4. Formation of Pt⁴⁺ requires lesser energy as compared to the formation of Ni⁴⁺.
5. Therefore, Pt⁴⁺ compounds are more stable than Ni⁴⁺ compounds.

(i) Transition metals have number of unpaired electron. They are involved in metallic bonding. Hence they show high melting point.
(ii) As we move from left to right along the transition metal series melting point first increases reach a maximum value and then decreases as the d-electrons pair up and become less available for bonding.

Chromium (24), the electronic configuration is 3d⁵4s¹. It ready to lose its outer most electron (4s¹) to get exactly half-filled stable electronic configuration. The electronic configuration of Zinc is 3d¹⁰4s².ie., It has completely filled stable configuration. From this configuration, the removal of 1e^- from 4s orbital is very difficult & it required more ionisation enthalpy. Due to this reason Zn has higher first Ionisation enthalpy (1.E₁) than that of chromium.

In 3d series as we move from Ti to Zn, the standard reduction potential \(\left(\mathrm{E}_{\mathrm{M}^{2+} / \mathrm{M}}^{0}\right)\) value is approaching towards less negative value and copper has a positive reduction potential. If the standard electrode potential E⁰, of a metal is large and negative, the metal is a powerful reducing agent, because it loses electrons easily. Hence +2 states becomes more and more stable in the first half of the first row transition elements.

Electronic configuration of Ce⁴⁺ = [Xe] 4f⁰5d⁰6s⁰
Electronic configuration of Co²⁺ = [Ar]3d⁷

Electronic configuration of Ti = 3d²4s²
Electronic configuration of Ti³⁺ =3d¹
Hence number of unpaired electron = 1
Spin only magnetic moment \((\mu)=\sqrt{\mathrm{n}(\mathrm{n}+2)}\)
= \(\sqrt{1(1+2)} \)
= \(\sqrt{3}\) 
=1.732 BM
Electronic configuration of \(\mathrm{Mn}=3 \mathrm{~d}^{5} 4 \mathrm{~s}^{2}\)
Electronic configuration of \(\mathrm{Mn}^{2+}=3 \mathrm{~d}^{5}\)
Hence number of unpaired electrons = 5
Spin only magnetic moment 
\((\mu) =\sqrt{5(5+2)}\)
= 5.92 BM

An interstitial compound or alloy is a compound that is formed when small atoms like hydrogen, boron, carbon or nitrogen are trapped in the interstitial holes in a metal lattice. They are usually non-stoichiometric compounds. Transition metals form a number of interstitial compounds such as TiC, ZrH_1.92, Mn₄N etc.
Properties of interstitial compound
(i) They are hard and show electrical and thermal conductivity.
(ii) They have high melting points higher than those of pure metals.
(iii) Transition metal hydrides are used as powerful reducing agents 
(iv) Metallic carbides are chemically inert.

a. 3MnO₄^2- + 4H⁺ ⟶  2MnO₄^- + MnO₂ + 2H₂O
(Manganate ion) (Permanganate ion)  Manganese dioxide
b. C₆H₅CH₃ \(\overset { acidified }{ \underset { KMnO_{ 4 } }{ \longrightarrow } } \) C₆H₅COOH
 Toluene                         Benzoic Acid
c. 2MnO₄^- + 10Fe²⁺+16H⁺ \(\underrightarrow { { 8H }^{ + } } \)  2Mn²⁺+ 10Fe³⁺  + 8H₂O
d. 2KMnO₄ \(\overset { \triangle }{ \underset { Red\ hot }{ \longrightarrow } } \) K₂MnO₄ + MnO₂ + O₂
(Potassium Permanganate)    (Potassium Manganate)
e. Cr₂O₇^2- + 6I^- + 14H⁺ \(\underrightarrow { { (O) }}\) 2Cr³⁺ + 3I₂ + 7H₂O
                 (Iodide ion)                        Iodine
f. Na₂Cr₂O₇ + 2KCl ⟶ K₂Cr₂O₇ + 2NaCl
(Sodium dichromate)    (Potassium dichromate)

Lanthanoid contraction:
As we move across 4f series, the atomic and ionic radii of lanthanoids show gradual decrease with increase in atomic number. This decrease in ionic size is called lanthanoid contraction.
Effects (consequence) of lanthanoid contraction:
1. Basicity difference:
As we move from Ce³⁺ to Lu³⁺, the basic character of Ln³⁺ ions decrease. Due to the decrease in the size of Ln³⁺ ions, the ionic character of Ln -OH bond decreases (covalent character increases) which results in the decrease in the basicity.
2. Similarities among lanthanoid:
In the complete f - series only 10 pm decrease in atomic radii and 20 pm decrease in ionic radii is observed because of this very small change in radii of lanthanoids, their chemical properties are quite similar.
3. The elements of the second and third transition series resemble each other more closely than the elements of the first and second transition series.

Potassium dichromate is prepared from chromate ore (FeCr₂O₄). The ore is concentrated by gravity separation. It is then mixed with excess sodium carbonate and lime and roasted in a reverberatory furnace.

The roasted mass is treated with water to separate soluble sodium chromate from insoluble iron oxide. The yellow solution of Sodium Chromate is treated with concentrated Sulphuric acid which converts Sodium Chromate into Sodium dichromate.

The saturated solution of sodium dichromate is mixed with KCl and filtered, the filtrate is cooled to obtain K₂Cr₂O₇ crystals.
\(\begin{array}{cc} \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{KCl} & \rightarrow \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{NaCl} \\ \begin{array}{c} \text { Sodium dichromate } \\ \text { (orange red) } \end{array} & \begin{array}{c} \text { Potassium dichromate } \\ \text { (orange red) } \end{array} \end{array}\)

The fourteen elements following actinoids is from thorium to lawrencium are called actinides.
Examples: Uranium, Thorium, Neptunium

The oxidation states of 4d metals vary from +3 for Y to +8 for Ru and Os.
The highest oxidation state of 4d elements are found in their compounds with the higher electronegative elements like O, F & Cl.
Example: In RuO₄, OsO₄ & WCl₆
The oxidation state of Ru and Os is +8.
The oxidation state of W is +6.
Generally in going down a group, a stability of higher oxidation state increases while that of lower oxidation state decreases.
4d series (5^th period) - Yttrium to Cadmium (10 elements)

Elements	Oxidation states
Y	+3
Zr	+3, +4
Nb	+2, +3, +4, +5
Mo	+2, +3, +4, +5, +6
Tc	+2, +4, +5, +7
Ru	+2, +3, +4, +5, +6, +7 +8
Rh	+2, +3, +4, +6
Pd	+2, +3, +4
Ag	+1, +2, +3
Cd	+2