(i) They are hard and show electrical land 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

i) Cr² + 2e^- ⟶ Cr
ii) Mn²⁺ + 2e^- ⟶ Mn
iii) Fe²⁺ + 2e^- ⟶ Fe
iv) CO²⁺ + 2e^- ⟶ Co

(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) The ionisation enthalpy values can be used to predict the thermodynamic stability of their compounds.
(ii) Let us compare the ionisation energy required to form Ni²⁺and Pt²⁺ ions.
(iii) For Nickel, IE₁ + IE₂ = (737 + 1753) 2490 kJmol^-1
(iv) For Platinum, IE₁ + IE₂ = (864 + 1791) = 2655 kJmol^-1
(v) Since, the energy required to form Ni²⁺ is less than that of Pt²⁺, Ni(II) compounds are thermodynamically more stable than Pt(II) compounds.

(i) The highest oxidation state in oxides and fluorides is due to small size and high electro negativity of F and O.
(ii) Many Cu⁺ compounds are unstable in aqueous solution and undergo disproportionation.
2Cu⁺ ⟶ Cu²⁺+ Cu
This suggest that in aqueous solution Cu+_(aq), converts into Cu²⁺_(aq) which is due to much more negative Δ_hyd H^- of Cu²⁺_(aq) than Cu⁺, which compensates more for the second ionisation enthalpy of Cu.

(i) (a) Vanadium
(b) Iron
(c) Silver
(ii) H₂S ⟶ 2H⁺+ S^2-
[MnO₄^- + 8H⁺ + 5e^- ⟶ Mn²⁺ + 4H₂O] x 2
[S2^- ⟶ S + 2e^-] x 5
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2MnO₄^- + 5S^2- + 16H⁺⟶ 2Mn²⁺ + 5S + 8H₂O
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(i) The colour of the transition elements is due to the d-d transition.
(ii) Since the oxidation state and polarising power of Mn in Mn₂O₇ is higher, it is acidic in nature
(iii) μ = \(\sqrt{n(n+2)}=\sqrt{4(4+2)}\)
= 4.90 BM

(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) Mn³⁺/Mn²⁺ has large positive E^o value. Hence, Mn³⁺ can be easily reduced to Mn²⁺ because Mn²⁺ has half-filled electronic, configuration, so it is stable and Mn³⁺ is least stable. It is a good oxidising agent.

(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) Cr₂O₇^2- + 2OH^- ⟶ 2CrO₄^2- + H₂O
(ii) MnO₄^- + 4H⁺ + 3e^- ⟶ MnO₂ + 2H₂O

(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.

The steady decrease in the atomic/ionic radius from La³⁺ to Lu³⁺ is called lanthanoid contraction.