(i) Zinc (Z = 30, 1822822p6 3823p6 3d10 482)does not have partially filled d-subshell in its elementary state or in its commonly occurring oxidation state (Zn²⁺: 3d10). Therefore, it is not regarded as a transition element.
(ii) Transition elements or the d-block elements from a large number of coordination complexes. The transition metal ions bind to a number of anions or neutral molecules in these complexes. The common examples are [Ni(NHJJ²⁺, [Co(NHJJ^3-, [Fe(CN)J³⁺, [Fe(CN)J^4-, [Cu(NH3)4]²⁺ etc. The high tendency of transition metal ions to form complexes is due to
(a) small size of the atoms and ions of transition metals
(b) high nuclear charge
(c) availability of vacant d-orbitals of suitable energy to accept lone pairs of electrons donated by other groups (called ligands).
(iii) Mn²⁺ has 3d5 electronic configuration. It is stable because of the half-filled configuration of d-subshell. Therefore, Mn has very high third ionization enthalpy for the change from d5 to d4 and it is responsible for much more positive E_o value for Mn³⁺/Mn²⁺couple in comparison to Cr³⁺/Cr²⁺couple.

La(OH)3 is more basic than Lu(OH)3 As the size of the lanthanoid ions decreases from La3+ to Lu3+, the covalent character of the hydroxides increases (Fajan's rules). Hence, the basic strength decreases from La(OH)3 to Lu(OH)3.

(i) Transition metals form interstitial compounds. Transition metals have a unique character to form interstitial compounds with small non-metallic elements such as hydrogen, boron, carbon and nitrogen. The small atoms of these non-metallic elements (H, B, C, N, etc.) fit into the vacant spaces of the lattices of the transition metal atoms. As a result of the filling up of the interstitial spaces, the transition metals become rigid and hard. These interstitial compounds have similar chemical properties as the parent metals but have different physical properties, particularly, density, hardness, and conductivity. For example, steel and cast iron are hard because of the formation of interstitial compounds with carbon. These interstitial compounds have a variable composition and cannot be expressed by a simple formula. Therefore, these are called nonstoichiometric compounds.
(ii) There is a gradual increase in density from Ti to Cu because as we move along a transition series from left to right, the atomic radii decrease due to increasing in effective nuclear charge. Therefore, the atomic volume decreases but at the same time atomic mass increases. Therefore, density (mass/volume)increases.
(iii) Lanthanoids show limited the number of oxidation states, such as + 2, + 3 and + 4 (+ 3 is the principal oxidation state). This is because of the large energy gap between 5d and 4f subshells. On the other hand, actinoids also show principal oxidation state of + 3 but show a number of other oxidation states also. For example, uranium (Z = 92) exhibits oxidation states of + 3, + 4, + 5, + 6 and + 7 and neptunium (Z = 93) shows oxidation states of + 3, + 4, + 5, + 6 and + 7. This is because of the small energy difference between 5f, 6d and 7s orbitals.

(i)Transition metals form interstitial compounds. Transition metals have a unique character to form interstitial compounds with small non-metallic elements such as hydrogen, boron, carbon and nitrogen. The small atoms of these non-metallic elements (H, B, C, N, etc.) fit into the vacant spaces of the lattices of the transition metal atoms. As a result of the filling up of the interstitial spaces, the transition metals become rigid and hard. These interstitial compounds have similar chemical properties as the parent metals but have different physical properties, particularly, density, hardness, and conductivity. For example, steel, and cast iron are hard because of the formation of interstitial compounds with carbon. These interstitial compounds have a variable composition and cannot be expressed by a simple formula. Therefore, these are called nonstoichiometric compounds.
(ii) The metallic radii of the third (5d) series of transition metals are virtually same as those of corresponding group members of(4d) series due to the phenomenon of lanthanoid contraction. The steady decrease in atomic and ionic sizes of lanthanoid elements with increasing atomic number is known as lanthanoid contraction.
(iii) In Actinoids Sf, 6d and 7s orbitals have comparable energies and electrons from these orbitals can take part to show higher oxidation states.

All the lanthanoids have similar outer electronic configuration and show +3 oxidation state in their compounds. Therefore, all the lanthanoids have similar chemical properties. The different lanthanoids differ mainly in the number of4f-electrons which are buried deep in the atoms and hence, do not influence the properties. Moreover, due to lanthanoid contraction, there is the very small difference in the size of all the trivalent lanthanoid ions. Thus, the size of their ions is also almost identical which results in similar chemical properties.

The electronic configuration of Hg(I), i.e., Hg⁺ is [Xe] 4f¹⁴5d¹⁰ 6s¹ and thus has one electron in the valence 6s-orbital. If this were so, all Hg (I) compounds should be paramagnetic but actually they are diamagnetic. This behaviour can be explained if we assume that the singly filled 6s-orbitals of the two Hg⁺ ions overlap to form a Hg-Hg covalent bond. Thus, Hg+ions exist as dimeric species, i.e., Hg_2²⁺. In contrast, the electronic configuration of Cu(l) ion, i.e., Cu⁺ is [Ar] 3d¹⁰.Therefore, it has no unpaired b electrons to form dimeric species, i.e., Cu₂²⁺ and hence it always exists as Cu⁺ ion.

As we proceed along a transition series, the nuclear charge increases which tends to decrease the size but the addition of electrons in the d-subshell increases the screening effect which tends to counterbalance the effect of the increased nuclear charge

In Cu, all the d-electrons are paired (3d¹⁰ 4s¹). In Cr, all the d-electrons are unpaired (3d⁵ 4s¹). Hence, d-d electron re~ulsions in Cu are much greater than those in Cr. Therefore, Cu atom is larger in size than Cr. In Cu²⁺ (3d⁹), d-d electron repulsions decrease due to presence of one unpaired d-electron. Moreover, the electrons are attracted by 29 protons of the nucleus whereas in Cr²⁺, three unpaired electrons are still present but they are attracted by ony 24 protons of the nucleus. Thus, Cu2+is smaller in size than Cr²⁺.

\(2Na_2CrO_4 + H_2SO_4\longrightarrow Na_2SO_4 + Na_2Cr_2O_7 + H_20\)
\(Na_2Cr_2O_7 + 2KCI\longrightarrow K2Cr_2O_7 + 2NaCI\)
or
\(MnO_2+2KOH+{1\over 2}O_2\xrightarrow{heat}K_2MnO_4+H_2O\)
\(mNo_4^{2-}\xrightarrow{electrolysis}MnO_4^{-}+e^-\)

The decrease in atomic and ionic size with increase in atomic number in lanthanoids is called lanthanoid contraction. Lanthanoids are used for production of alloy steels for plates and pipes. Mixed oxides of lanthanoid metals are used as catalyst in petroleum cracking.

\((i) 2MnO_2 + 4KOH + O_2\xrightarrow{\Delta}2K2MnO_4 + 2H_2O\)
\((ii)\underset{Green}{MnO_4^{2-}}\xrightarrow{electrolysis}\underset{Purple}{MnO_4^-}+e^-\)

It is because hydration energy and lattice energy of Cu²⁺ is more than that of Cu⁺.

Copper exhibits +1 oxidation state frequently due to stable electronic configuration.

Preparation of KMnO₄ It is prepared by the fusion of pyrolusite ore (MnO₂) with an alkali metal hydroxide and an oxidising agent like KNO₃. Dark green K₂MnO₄ is obtained which on disproportionation in neutral or acidic solution gives potassium permanganate.
\(\underset{Pyrolusite}{2MnO_4}+4KOH+O_2⟶\underset{Potassium\ manahanate\\ ( Green\ mass)}{2KMnO_4}+2H_2O\) 
The green mass can be extracted with water and then oxidized either electrolytically or by passing chlorine/ozone into the solution.
Electrolytic oxidation
\(K_{2} M n O_{4} \leftrightarrow 2 K^{+}+M n O_{4}^{2-}\) 
\(H_{2} O \leftrightarrow H^{+}+O H^{-}\) 
At anode, manganate ions are oxidized to permanganate ions.
\(M n O_{4}^{2-} \leftrightarrow M n O_{4}^{-}+e^{-}\) 
Green       Purple
Oxidation by chlorine
\(2 \mathrm{~K}_{2} \mathrm{MnO}_{4}+\mathrm{Cl}_{2} \rightarrow 2 \mathrm{KMnO}_{4}+2 \mathrm{KCl}\) 
\(2 \mathrm{MnO}_{4}^{2-}+\mathrm{Cl}_{2} \rightarrow 2 \mathrm{MnO}_{4}^{-}+2 \mathrm{Cl}^{-}\) 
Oxidation by ozone
\(2 \mathrm{~K}_{2} \mathrm{MnO}_{4}+\mathrm{O}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{KMnO}_{4}+2 \mathrm{KOH}+\mathrm{O}_{2}\) 
\(2 \mathrm{MnO}_{4}^{2-}+\mathrm{O}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{MnO}_{4}^{2-}+2 \mathrm{OH}^{-}+\mathrm{O}_{2}\) 
(i) Acidified KMnO₄ solution oxidizes Fe (II) ions to Fe (III) ions i.e., ferrous ions to ferric ions.
\(\mathrm{MnO}_{4}^{-}+8 H^{+}+5 e^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}\) 
\(\left.F e^{2+} \rightarrow F e^{3+}+e^{-}\right] \times 5\) 
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\(\mathrm{MnO}_{4}^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_{2} \mathrm{O}\) 
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(ii) Acidified potassium permanganate oxidizes SO₂ to sulphuric acid.
\(\left.\mathrm{MnO}_{4}^{-}+6 \mathrm{H}^{+}+5 e^{-} \rightarrow \mathrm{Mn}^{2+}+3 \mathrm{H}_{2} \mathrm{O}\right] \times 2\) 
\(\left.2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightarrow 4 H^{+}+2 S O_{4}^{2-}+2 e^{-}\right] \times 5\) 
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\(2 \mathrm{MnO}_{4}^{-}+10 \mathrm{SO}_{2}+5 \mathrm{O}_{2}+4 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{SO}_{4}^{2-}+8 \mathrm{H}^{+}\) 
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(iii) Acidified potassium permanganate oxidizes oxalic acid to carbon dioxide.
\(\left.\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+} 5 e^{-} \rightarrow \mathrm{Mn}^{2-}+4 \mathrm{H}_{2} \mathrm{O}\right] \times 2\) 
\(\left.\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow 2 \mathrm{CO}_{2}+2 e^{-}\right] \times 5\) 
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\(2 \mathrm{MnO}_{4}^{-}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}\) 
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