Zone refining :
1. Zone refining method is based on the principles of fractional crystallisation.
2. When an impure metal is melted and allowed to solidify, the impurities will prefer to be in the molten region. In this process the impure metal is taken in the form of a rod.
3. One end of the rod is heated using a mobile induction heater which results in melting of the metal on that portion of the rod.
4. When the heater is slowly moved to the other end the pure metal crystallises while the impurities will move on to the adjacent molten zone.
5. As the heater moves further away, the molten zone containing impurities also moves along with it.
6. The process is repeated several times by moving the heater in the same direction again and again to get pure metal.
7. This process is carried out in an inert gas atmosphere to prevent the oxidation of metals.
8. Elements such as germanium (Ge), silicon (Si) and galium (Ga) that are used as semiconductor are refined using this process.

Chlorine reacts with cold dilute alkali to give chloride and hypochlorite, while with hot concentrated alkali chlorides and chlorates are formed.
\(\mathrm{Cl}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HCl}+\underset{\text { Hypochlorous acid }}{\mathrm{HOCl}} \)
\(\mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O} \)
\(\mathrm{HOCl}+\mathrm{NaOH} \rightarrow \mathrm{NaOCl}+\mathrm{H}_{2} \mathrm{O} \)
                   (Sodium hypo chlorite)
Overall Reaction
3Cl₂ + 6NaOH \(\rightarrow\) NaClO₃ + 5NaCl + 3H₂O
                             (Sodium Chlorate)

Most of the elements exhibit, two types of valence namely primary valence and secondary valence and each element tend to satisfy both the valences.
The primary valence is referred the oxidation state of the metal atom.
The secondary valence as the coordination number. For example, according to Werner, the primary and secondary valences of cobalt are 3 and 6 respectively.
The primary valence of a metal ions ae always satisfied by negative ions.
For example in the complex CoCI₃.6NH₃. The primary valence of Co is +3 and is satisfied by 3CI^- ions.
The secondary valence is satisfied by negative ions, neutral molecules, positive ions or the combination of these.
For example, in CoCl₃.6NH₃ complex primary valence of cobalt +3 and it is satisfied by 3 CI^-.
The secondary valence of cobalt is 6 and is satisfied by six neutral ammonia molecules. where as in CoCI₆.NH₃. 
Secondary valence of Co = 5{It is satisfied five neutral molecules and a Cl^- ion}
According to Werner, there are two spheres of attraction around a metal atom/ion in a complex.

The inner /coordination sphere:
The groups present in this sphere are firmly attached to the metal.
The outer sphere / ionisation sphere:
The groups present in this sphere are loosely bound to the central metal ion and hence can be separated into ions upon dissolving the complex in a suitable solvent.
The primary valencies are non-directional. while the secondary valencies are directional.
The geometry of the complex is determined by the special arrangement of the groups which satisfy the secondary valence.

(a) Elementary reaction
Each and Every single step in a reaction mechanism is called an elementary reaction.
Rate = k[A] [B]
(b)

(i) An acid is defined as a substance that has a tendency to donate a proton to another substance and base is a substance that has a tendency to accept a proton form other substance.
(ii) In other words, an acid is a proton donor and a base is a proton acceptor.
(iii) When hydrogen chloride is dissolved in water, it donates a proton to the later. Thus, HCI behaves as an acid and H₂O is base. The proton transfer from the acid to base can be represented as
HCI + H₂O ⇌ H₃O⁺ + Cl^-
(iv) When ammonia is dissolved in water, it accepts a proton from water. In this case, ammonia (NH₃) acts as a base and H₂O is acid. The reaction is represented as
H₂O + NH₃ ⇌ NH₄⁺ + OH^-
(v) Let us consider the reverse reaction following equilibrium.
\(\underset { proton\ donar\\ \quad \quad \ (acid) }{ HCl } +\underset { Proton\ acceptor\\ \quad \quad \quad \quad (base) }{ { H }_{ 2 }O } \leftrightharpoons \underset { Proton\ donar\\ \quad \quad \quad \quad \ (acid) }{ { H }_{ 2 }{ O }^{ + } } +\underset { Proton\ acceptor\\ \quad \quad \quad \quad \ (base) }{ { Cl }^{ - } } \) 
H₃O⁺ donates a proton to Cl^- to form HCI i.e., the products also behave as acid and base.
(vi) In general, Lowry - Bronsted (acid - base) reaction is represented as 
Acid₁ + Base₂ ⇌ Acid₂ + Base₁
(vii) The species that remains after the donation of a proton is a base (Base₁) and is called the conjugate base of the Bronsted acid (Acid₁). In other words, chemical species that differ only by a proton are called conjugate acid - base pairs.

1. Nernst equation is the one which relates the cell potential and the concentration of the species involved in an electrochemical reaction. Let us consider an electrochemical cell for which the overall redox reaction is,
xA + yB ⇌ IC + mD
2. The reaction quotient Q for the above reaction is given below
\(Q=\cfrac { \left[ C \right] ^{ ' }\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \)   .....(1)
3. We have already learnt that,
ΔG = ΔG + RT In Q  .........(2)
4. The Gibbs free energy can be related to the cell emf as follows.
[ஃ equation (1) and (2)]
ΔG = - nFE_cell; ΔG^o = - nFE^o_cell
5. Substitute these values and Q from (1) in the equation (2)
(2) ⇒\(-nF{ E }_{ cell }=-n{ FE }^{ o }_{ cell }+RT\quad In\cfrac { \left[ C \right] '\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \) ....(3) 
6. Divide the whole equation (3) by (-nF)
(4) ⇒\({ E }_{ cell }={ E }_{ cell }-\cfrac { RT }{ nF } In\cfrac { \left[ C \right] ^{ l }\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \) 
(or)
\({ E }_{ cell }={ E }_{ cell }-\cfrac { 2.303RT }{ nF } log\cfrac { \left[ C \right] ^{ l }\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \)  .....(4)
7. The above equation (4) is called the Nernst equation
8. At 25°C (298 K), the above equation (4) becomes,
\({ E }_{ cell }={ E^{ o } }_{ cell }-\cfrac { 2.303\times 8.314\times 298 }{ n(96500) } log\cfrac { \left[ C \right] ^{ l }\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \)       
\({ E }_{ cell }={ E^{ o } }_{ cell }-\cfrac { 0.0591 }{ n } log\cfrac { { \left[ C \right] }^{ l }{ \left[ D \right] }^{ m } }{ { \left[ A \right] }^{ x }\left[ B \right] ^{ y } } \)  ......(5) 
\(\left[\begin{array}{l} \text { Where } \\ \therefore \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\ \mathrm{~T}=298 \mathrm{~K} \\ \mathrm{IF}=96500 \mathrm{C} \mathrm{mol}^{-1} \end{array}\right]\)

The intermediate compound formation theory :
A catalyst acts by providing a new path with low energy of activation. In homogeneous catalysed reactions a catalyst may combine with one or more reactant to form an intermediate which reacts with other reactant or decompose to give products and the catalyst is regenerated.
Consider the reactions :
A+B➝AB; C is the catalyst .............(1)
A + C ➝ AC (intermediate) ..........(2)
AC + B ⟶ AB + C ...............(3)
Example 1:
The mechanism of Fridel crafts reaction is given below
\({ C }_{ 6 }{ H }_{ 6 }+{ { CH }_{ 3 }Cl\overset { anhydrous\\ { AlCl }_{ 3 } }{ \longrightarrow } }{ C }_{ 6 }{ H }_{ 5 }{ CH }_{ 3 }+HCl\) 
The action of catalyst is explained as follows
CH₃Cl + AlCl₃ ⟶ [CH₃]⁺ [AlCl₄]^-
It is an intermediate.
\({ C }_{ 6 }{ H }_{ 6 }+\left[ { CH }_{ 3 }^{ + } \right] \left[ { AlCl }_{ 4 } \right] ^{ - }\longrightarrow { C }_{ 6 }{ H }_{ 5 }{ CH }_{ 3 }+{ AlCl }_{ 3 }+Hcl\) 
Example 2:
\({ { 2KClO }_{ 3 }\overset {\\ { MnO }_{ 3 } }{ \longrightarrow } }{ 2KCl }+{ 3O }_{ 2 }\)
Thermal decomposition of KCIO₃ in the presence of MnO₂ proceeds as follows. Steps in the reaction
2KCIO₃ ⟶ 2KCI + 3O₂ Can be given as
2KClO₃ + 6MnO₂   →  6MnO₃ + 2KCl
It is an intermediate
6MnO₃ → 6MnO₂ + 3O₂
Example 3:
Formation of water due to the reaction of H₂ and O₂ in the presence of Cu can be given as
H₂ + 1/2O₂ → H₂O 
2Cu + \(\frac{1}{2}\)O₂   → Cu₂O
It is an intermediate.
Cu₂O + H₂  → H₂O + 2Cu
Advantages:
This theory describes
(a) The specificity of a catalyst and
(b) The increase in the rate of the reaction with increasc inthe concentration of a catalyst
Limitations:
(a) The intermediate compound theory fails to explain the action of catalytic poison and activators (promoters).
(b) This theory is unable to explain the mechanism of heterogeneous catalysed reactions.

(i) Monomers :Ethylene glycol and terepathalic acid (or) dimethyl terephthalate.
(ii) Catalyst : Zinc acetate and antimony trioxide.
(iii) Temperature : 500 K
(iv) Product : Terylene

(v) Uses : blending with cotton or wool fibres and as glass reinforcing materials in safety helmets