a) i) Ellingham diagram for the formation of Al₂O₃ and MgO intersects around 1500^oC. Above this temp Mg lies above the Aluminium line. Hence only above 1500^oC Aluminium might be excepted to reduce magnesia.
ii) Ellingham diagram for the formation of MgO lies below the formation of Al₂O₃. Hence MgO is more stable than Al₂O₃. Hence Magnesium could reduce Alumina.
1. Below 983K, formation of CO line lies below many of the metal oxide formation in Ellingham diagram, hence CO is more effective reducing agent than Carbon.
2. But above this temperature Carbon lies below other metal oxides.
b) Around 1200K Carbon lies below the formation of Fe₂O₃. Hence it is possible to reduce Fe₂O₃ by coke at 1200K.

(i) Ellingham diagram is constructed based only on thermodynamic considerations. It gives information about the thermodynamic feasibility of a reaction. It does not tell anything about the rate of the reaction. More over, it does not give any idea about the possibility of other reactions that might be taking place.
(ii) The interpretation of \(\triangle\)G is based on the assumption that the reactants are in equilibrium with the product which is not always true.

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.

In \(\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}\) Cobalt is in +3 oxidation state
\(\mathrm{Co}=[\mathrm{Ar}] 3 \mathrm{~d}^{7} 4 \mathrm{~s}^{2}  \)
\(\mathrm{Co}^{3+}=3 \mathrm{~d}^{6} 4 \mathrm{~s}^{\circ}\)

It is diamagnetic; n = 0
d²sp³ hybridisation; μ_s = 0

In bcc unit cell, ΔABC
AC² = AB² + BC²
\(AC=\sqrt { { AB }^{ 2 }+{ BC }^{ 2 } } \)
\(\\ AC=\sqrt { { a }^{ 2 }+{ a }^{ 2 } } =\sqrt { { 2a }^{ 2 } } =\sqrt { 2 } a\)

In ΔACG
AG² = AC² + CG²
\(AG=\sqrt { { AC }^{ 2 }+{ CG }^{ 2 } } \)
\(AG=\sqrt { { \left( \sqrt { 2a } \right) }^{ 2 }+{ a }^{ 2 } } \)
\(AG=\sqrt { { 2a }^{ 2 }+{ a }^{ 2 } } =\sqrt { { 3a }^{ 2 } } \)
\(AG=\sqrt { 3a } \)
\(\sqrt { 3 } a=4r\)
\(r=\frac { \sqrt { 3 } }{ 4 } a\)
∴ Volume of the sphere with radius 'r' \(=\frac { 4 }{ 3 } { \pi r }^{ 3 }\)
\(=\frac{4}{3}\pi { \left( \frac { \sqrt { 3 } }{ 4 } a \right) }^{ 3 }\)\(=\frac { \sqrt { 3 } }{ 16 } \pi { a }^{ 3 }\)
Number of spheres belong to a unit cell in BCC arrangement is equal to two and hence the total volume of all spheres.
(i) Packing fraction = \(=\frac{Total \quad volume \quad occupied \quad by \quad spheres \quad in \quad a \quad unit \quad cell}{volume \quad of \quad the \quad unit \quad cell}\times100\)
\(\therefore\)Volume of all spheres \(=2\times \left( \frac { \sqrt { 3 } \pi { a }^{ 3 } }{ 16 } \right) =\frac { \sqrt { 3 } \pi { a }^{ 3 } }{ 8 } \)
Packing fraction \(=\frac { \left( \frac { \sqrt { 3 } \pi { a }^{ 3 } }{ 8 } \right) }{ ({ a }^{ 3 }) } \times 100\)
\(=\frac { \sqrt { 3 } \pi }{ 8 } \times 100\)
\(\\ =\sqrt { 3 } \pi \times 12.5\)
= 1.732 x 3.14 x 12.5
= 68%

(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.

Kohlraush's law:
(i) At infinite dilution, the limiting molar conductivity of an electrolyte is equal to the sum of the limiting molar conductivities of its constituent ions. i.e., the molar conductivity is due to the independent migration of cations in one direction and anions in the opposite direction.
(ii) For a uni - univalent electrolyte such as NaCl, the Kohlraush's law is expressed as
\(\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{NaCl}}=\left(\lambda_{\mathrm{m}}^{0}\right)_{\mathrm{Na}^{+}}+\left(\lambda_{\mathrm{m}}^{0}\right)_{\mathrm{Cl}}^{-}\)
(iii) In general, according to Kohlraush's law, the molar conductivity at infinite dilution for a electrolyte represented by the formula A_x B_y, is given below.
\(\left(\Lambda_{m}^{0}\right)_{A_{x} B_{y}}=x\left(\lambda_{m}^{0}\right)_{A^{y+}}+y\left(\lambda_{m}^{0}\right)_{B^{x-}}\)
b) Calculation of molar conductance at infinite dilution for weak electrolytes experimentally.
(i) However, the same can be calculated using Kohlraush's Law. For example, the molar conductance of CH₃COOH, can be calculated using the experimentally determined molar conductivities of strong electrolytes HCl, NaCl and CH₃COONa.
\(\Lambda_{\mathrm{CH}_{3} \mathrm{COONa}}^{0}=\lambda_{\mathrm{Na}^{+}}^{0}+\lambda_{\mathrm{CH}_{3} \mathrm{COO}^{-}}^{0} \) ..........(1)
\(\Lambda_{\mathrm{HCl}}^{0}=\lambda_{\mathrm{H}+}^{0}+\lambda_{\mathrm{Cl}^{-}}^{0} \) .........(2)
\(\Lambda_{\mathrm{NaCl}}^{0}=\lambda_{\mathrm{Na}^{+}}^{0}+\lambda_{\mathrm{Cl}^{-}}^{0}\) ...........(3)
(ii) Equation (1) + Equation (2) - Equation (3) gives,
\(\left(\Lambda_{\mathrm{CH}_{3} \mathrm{COONa}}^{0}\right)+\left(\Lambda_{\mathrm{HCl}}^{0}\right)-\left(\Lambda_{\mathrm{NaCl}}^{0}\right)=\lambda_{\mathrm{H}+}^{0}+\lambda_{\mathrm{CH}_{3} \mathrm{COO}-}^{0} \)
\(=\Lambda_{\mathrm{CH}_{2} \mathrm{COOH}}^{0}\)

1. Langmuir explained the action of catalyst in heterogeneous catalysed reactions based on adsorption. The reactant molecules are adsorbed on the catalyst surfaces, so this can also be called as contact catalysis.
2. According to this theory, various steps involved in a heterogeneous catalyse reaction are given as follows:
(i) Reactant molecules diffuse from bulk to the catalyst surface.
(ii) The reactant molecules are adsorbed on the surface of the catalyst.
(iii) The adsorbed reactant molecules are activated and form activated complex which is decomposed to form the products.
(iv) The product molecules are desorbed.
(v) The product diffuse away from the surface of the catalyst.

Mechanism:
This mechanism undergoes Saytzeff rule.
During intramolecular dehydration, if there is a possibility to form a carbon - carbon double bond at different locations, the preferred location is the one that gives the more (highly) substituted alkene i.e., the stable alkene.

1. The materials that are readily decomposed by microorganisms in the environment are called biodegradable.
2. Natural polymers degrade on their own after certain period of time but the synthetic polymers do not.
3. It leads to serious environmental pollution. One of the solution to this problem is to produce biodegradable polymers which can be broken down by soil micro organism.
Examples:
(i) Polyhydroxy butyrate (PHB)
(ii) Polyhydroxy butyrate-co-A- hydroxyl valerate (PHBV)
(iii) Polyglycolic acid (PGA), Polylactic acid (PLA)
(iv) Poly ( E caprolactone) (PCL)
(v) Biodegradable polymers are used in medical field such as surgical sutures, plasma substitute etc...
4. these polymers are decomposed by enzyme action and are either metabolized or excreted from the body.

(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

Preparation:
As discussed earlier diborane can be prepared by the action of metal hydride with boron. This method is used for the industrial production.
Diborane can also be obtained in small quantities by the reaction of iodine with sodium borohydride in diglyme.
2NaBH₄ + I₂ ⟶ B₂H₆ + 2NaI + H₂
On heating magnesium boride with HCl a mixture of volatile boranes are obtained.
2Mg₃B₂ + 12HCl ⟶ 6MgCl₂ + B4H₁₀ + H₂
B₄H₁₀ + H₂ ⟶ 2B₂H₆

(i) KCIO₃ \(\overset { \Delta }{ \underset { { MnO }_{ 2 } }{ \longrightarrow } } \) 2KCI + 3O₂
(ii) 2 ZnS + 3O₂\(\longrightarrow \) 2 ZnO + 2 SO₂
(iii) AI_zO₃(s) + 6 NaOH_(aq) + 3 H₂O_(l) \(\longrightarrow \) 2 Na₃[AI(OH)₆]_(aq)
(iv) 2NaOH + SO₂\(\longrightarrow \)Na₂SO₃ + H₂O
(v) 2KCl + H₂SO₄ \(\longrightarrow \) 2 HCl + K₂SO₄

(a) NaCl + MnO₂ + 4H₂SO₄ \(\longrightarrow \) MnCl₂+ 4NaHSO₄ + 2H₂O + Cl₂
(b) 6XeF₄ + 12H₂O \(\longrightarrow \) 4Xe + 2XeO₃ + 24HF + 3O₂

(i) PCl₅ + 2Ag \(\longrightarrow \) 2 AgCl + PCl₃
(ii) AgCI_(s)+ 2NH_3(aq) \(\longrightarrow \)[Ag(NH₃)₂]CI_(aq)
(iii) PCl₃ + 3H₂O \(\longrightarrow \) H₃PO₃ + 3HCI
(iv) PCl₅ + H₂O \(\longrightarrow \) POCl₃ + 2HCI
(v) NaH₂PO₂ + HCI \(\longrightarrow \) H₃PO₂ + NaCl

(i) It is generally expected a steady decrease in atomic radius along a period as the nuclear charge increases and the extra electrons are added to the same sub shell.
(ii) But for the 3d transition elements, the expected decrease in atomic radius is observed from Sc to V, thereafter up to Cu the atomic radius nearly remains the same.
(iii) As we move from Sc to Zn in 3d series the extra electrons are added to the 3d orbitals, the added 3d electrons only partially shield the increased nuclear charge and hence the effective nuclear charge increases slightly.
(iv) However, the extra electrons added to the 3d sub shell strongly repel the 4s electrons and these two forces are operated in opposite direction and as they tend to balance each other, it leads to constancy in atomic radii.
(v) At the end of the series, d - orbitals of Zinc contain 10 electrons in which the repulsive interaction between the electrons is more than the effective nuclear charge and hence, the orbitals slightly expand and atomic radius slightly increases.

(i) It exhibits both geometrical and optical isomerism
(a) Geometrical isomers:

(b) Optical isomers:

(ii) It shows two optical isomers

(iii) Ionisation isomers
Linkage isomers
[Co(NH₃)₅(NO₂)(NO₃)₂], [CO(NH₃)₅ (ONO)](NO₃)₂
(iv) Geometrical isomers

(i) The defects in ionic solids is by adding impurity ions.
(ii) If the impurity ions are in different valance state from that of host, vacancies are created in the crystal lattice of the host.
(iii) For example, addition of CdCl₂ to silver chloride yields solid solutions where the divalent cation Cd²⁺ occupies the position of Ag⁺.
(iv) This will disturb the electrical neutrality of the crystal.
(v) In order to maintain the same, proportional number of Ag⁺ ions leaves the lattice.
(vi) This produces a cation vacancy in the lattice, such kind of crystal defects are called impurity defects.

Given data: Time taken for completion of 30% of the reaction = 12 mins.
a = 100; x = 30; a - x = 70 and t =12 min
Formula: \(k=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } \)
Solution:
For a first order reaction,
\(\frac { 2.303 }{ 12 } \log\frac { 100 }{ 70 } \)
\(=\frac { 2.303 }{ 12 } \times 0.1549\) = 0.02972 min^-1
k = 2.97 x 10^-2 min^-1
If t = 65.33 minutes, x = ?
\(k=\frac { 2.303 }{ 65.33 } \log\frac { 100 }{ 100-x } \)
\(0.02972=\frac { 2.303 }{ 65.33 } \log\frac { 100 }{ 100-x } \)
\(\log\frac { 100 }{ 100-x } \)= \(\frac { 0.02972\times 65.33 }{ 2.303 } =0.8430\)
\(\frac { 100 }{ 100-x }\)  =Antilog of 0.8430
100 = 6.966 (100 - x)
100 = 696.6 - 6.966x
-6.966x = 100-696.6 = 596.6
\(x=\frac { 596.6 }{ 6.966 } =85.62%\)
The reaction completed in 65.33 minutes

pH = - log [H₃O⁺]
(i) pH - log [10⁴]
pH = log 1 - log 10⁴
pH =  - 4
(ii) pH = - log [10^-7]
pH = log 1 - log 10^-7.
pH = 7
(iii) pH = - log [H₃O⁺]
pH = - log [6.8 x 10^-3]
pH = log 1 - log 6.8 - log 10^-3
= 3 - 0.8325
pH = 2.17 (or) 2.2
(iv) pH = - log [3.2 x 10^-5]
pH = log 1 - log 3.2 - log 10^-5
= 5 - 0.5051
pH = 4.49 (or) 4.5
(v) pH = - log [0.035]
pH = log 1 - log 0.035
= 2 - 0.5441
pH = 1.46 (or) 1.5
(vi) pH = - log [0.25]
pH = log 1 - log 0.25
= 1 - 0.3979
pH = 0.602 (or) 0.60
(vii) pH = -log [H₃O⁺]
pH = log 1 - log 5.4 - log 10^-9
= 7 - 0.7324
pH = 8.267 (or) 8.3
(viii) pH = - log [7.1 x 10^-7]
pH = log 1 - log 7.1 - log 10^-7
= 7 - 0.8513
pH = 6.2.

 Sn⁴⁺ + 2e^- ⟶ Sn²⁺ E⁰ = 0.15V
Given: n = 2 electrons 
F = 96495 coulombs
Formula: ΔG = - nFE^O
Solutlon: ∴ ΔG = - 2 x 96495 x 0.15
= 28.948
Free energy = -28.948 kJ.