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

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

Cr ⇒ [Ar]3d⁵4s¹

(i) It is used as a strong oxidizing agent
(ii) It is used for the treatment of various skin infections and fungal infections of the foot.
(iii) It used in water treatment industries to remove iron and hydrogen sulphide from well water.
(iv) It is used as a Bayer's reagent for detecting unsaturation in an organic compound.
(v) It is used in quantitative analysis for the estimation of ferrous salts, oxalates, hydrogen peroxide and iodides.

(i) When potassium dichromate is heated with any chloride salt in the presence of Conc. H₂SO₄, orange red vapours of chromyl chloride (CrO₂CI₂) is evolved.
(ii) This reaction is used to confirm the presence of chloride ion in inorganic qualitative analysis

(iii) The chromyl chloride vapours are dissolved in sodium hydroxide solution and then acidified with acetic acid and treated with lead acetate. A yellow precipitate of lead chromate is obtained

(i) Redox reactions involve transfer of electrons from one reactant to another. Such reactions are always coupled, which means that when one substance is oxidised, another must be reduced.
(ii) The substance which is oxidised is a reducing agent and the one which is reduced is an oxidizing agent

(i) Cobalt (III) ion has greater tendency to form complexes than cobalt (II) ion. Therefore, Co (II) ion being stable in aqueous solution, changes to Co (III) ion in the presence of complexing reagents and get oxidised.
(ii) Ions of transition metals with d¹ configuration tend to lose one electron to acquire d⁰ configuration that is quite stable. Therefore, such ions (with d¹) undergo either oxidation or disproportionation, hence unstable.

(i) The use of 3d electron for formation of I bond increases from Sc to Mn, causing the increase in oxidation state upto +7.
(ii) The reason for Mn having highest oxidation state of +7 is due to the presence of 7 unpaired electrons in its atom.
(iii) As the number of unpaired electrons decrease from Fe to Cu. So there is the decrease in oxidation state.

(i) Atomic size : The atomic size in 3d transition series decrease from Sc to Mn and then Fe, CO, Ni have almost same atomic size while copper has bigger size. It is because number of unpaired electrons in d- orbitals increase in the beginning till Mn. Therefore effective nuclear charge increases hence atomic size decreases then pairing of electrons in d- orbitals takes place, so the atomic size remains the same and finally it increases due to repulsion between paired electrons in d- orbitals which leads to decrease in effective nuclear charge.
(ii) They have high enthalpy of atomisation due to strong metallic bonds and additional covalent bonding due to the presence of unpaired electrons in d- orbitals.

(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.

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.

(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.

(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) 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.

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

Action of heat : When heated, potassium permanganate decomposes to form potassium manganate and manganese dioxide.

(i) An alloy is formed by blending a metal with one or more other elements. The elements may be metals or non-metals or both.
(ii) The bulk metal is named as solvent, and the other elements in smaller portions are called solute.
(iii) The alloys so formed are hard and often have high melting points.
(iv) Example: Ferrous alloys, gold - copper alloy, chrome alloys etc.

(i) When orange solution containing Cr₂O₇^2- ion is treated with an alkali, a yellow solution of Cr₂O₇^2- is obtained.
(ii) Similarly, when H⁺ ions are added to yellow solution, an orange solution of Cr₂O₇^2- is obtained due to interconversion.

(i) Due to lanthanide contraction the size of La³⁺ ions decreases regularly with increase in atomic number.
(ii) According to Fajan's rule decrease in size of Ln³⁺ ions increase the covalent character and decreases the basic character between Ln³⁺& OH^- ion in Ln(OH)³
(iii) Since the order of size Ln³⁺ ions are
La³⁺ > Ce³⁺ ...> Lu³⁺
(iv) Hence La(OH)³ is the strongest base while Lu(OH)³ is the weakest base

This is because the last electron enters into the (n-2) f-subshell.

[Rn]⁸⁸ 5⁺¹⁴ 6d⁰7S²

(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.

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)

(i) The compounds of Cu²⁺ are coloured as it has one free electron its valence shell which absorb I radiation of visible region and get excited to emit its complementary colour.
(ii) Zn has no free electron it has fully filled shells. Due to extra stable orbitals electron can't be excited by radiations of visible light, hence its compounds are colourless.

(i) It oxidises ferrous salts to ferric salts.
 Cr₂O₇^2- + 6Fe²⁺+ 14H⁺ ⟶ 2Cr³⁺ + 6Fe³⁺ + 7H₂O
(ii) It oxidises iodide ions to iodine
Cr₂O₇^2- + 6I^-+ 14H⁺ ⟶ 2Cr³⁺ + 3I₂ + 7H₂O
(iii) It oxidises sulphide ion to sulphur
Cr₂O₇^2- + 3S^2-+ 14H⁺ ⟶ 2Cr³⁺ + 3S + 7H₂O
(iv) It oxidises sulphur dioxide to sulphate ion
Cr₂O₇^2- + 3SO₂+ 2H⁺ ⟶ 2Cr³⁺ + 3SO₄^2- + H₂O
(v) It oxidises stannous salts to stannic salt
 Cr₂O₇^2- + 3Sn²⁺ + 14H⁺ ⟶ 2Cr³⁺ + 3Sn⁴⁺ + 7H₂O
(vi) It oxidises alcohols to acids
2K₂Cr₂O₇ + 8H₂SO₄ + 3CH₃CH₂OH ⟶ 2K₂SO₄ +2Cr₂(SO₄)₃+ 3CH₃COOH + 11H₂O
(vii) It oxidises oxalic acid to CO2
2MnO₄^- + 5(COO)^2- + 6H⁺ ⟶ 2Mn²⁺+ 10CO₂ + 8H₂O
(viii) It oxidises iodide ions to iodine
2MnO4^- + 10I^- + 16H⁺ ⟶ 2Mn²⁺+ 5I₂+ 8H₂O
(ix) It oxidises sulphide ion to sulphur
2MnO₄^- + 5S^2-+ 16H⁺ ⟶ 2Mn²⁺+ 5S + 8H₂O
(x) It oxidises nitrites to nitrates
 2MnO₄^- + 5NO^2- + 6H⁺ ⟶ 2Mn²⁺+ 5NO₃^- + 3H₂O
(xi) It oxidises alcohols to aldehydes.
 2KMnO₄ + 3H₂SO₄ + 5CH₃CH₂OH ⟶ 2K₂SO₄ + 2MnSO₄ + 5CH₃CHO + 8H₂O
(xii) It oxidises sulphite to sulphate
2MnO₄^- + 5SO₃^2- + 6H⁺ ⟶ 2Mn²⁺ + 5SO₄^2- + 3H₂O

(i) Ionization energy of transition element is intermediate between those of sand p block elements.
(ii) As we move from left to right in a transition metal series, the ionization enthalpy increases as expected.
(iii) This is due to increase in nuclear charge corresponding to the filling of d electrons.
(iv) The increase in first ionisation enthalpy with increase in atomic number along a particular series is not regular.
(v) 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 values

The heavier transition elements belong to fourth (4d), fifth (Sd) and sixth (6d) transition series. Their properties are expected to be different form the elements belonging to the first (3d) series due to the following reasons.
(i) Atomic radii: Size of the transition elements 94d and Sd series are larger than those of the corresponding elements of the first transition series though those of 4d and Sd series are very close to each other.
(ii) Ionisation enthalpy of Sd series are higher than the corresponding elements of 3d and 4d series.
(iii) Atomisation enthalpy of 4d and Sd series are higher than the corresponding elements of the first series.
(iv) Melting and boiling points of heavier transition elements are greater than those of the first transition series due to stronger intermetallic bonding.
(v) The elements of the first transition series generally form low or high spin complexes, depending upon the higher of ligand field. However, the heavier transition elements form low spin complexes irrespective of the strength of the ligand filed.