(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