Volume of KMnO₄ used is proportional to the amount of H₂O₂ present. If the reaction is of first order, it must obey the equation.
\(k=\frac { 2.303 }{ t } log\frac { \left[ { A } \right] _{ 0 }}{ \left[ A \right] } \)
(or) \(k=\left( \frac { 2.303 }{ t } \right) log\left( \frac { { V }_{ 0 } }{ { V }_{ 1 } } \right) \)
(i) \( \mathrm{V}_{\mathrm{o}}=46.1 \mathrm{~mL} ; \mathrm{t}=10 \mathrm{mins} ; \mathrm{V}_{\mathrm{t}}=29.8 \mathrm{~mL}\)
\(k =\frac{2.303}{10} \log \left(\frac{46.1}{29.8}\right)  \)
\(=0.2303 \log 1.5469  \)
\(=0.2303 \times 0.1894=0.0436 \mathrm{~min}^{-1}\)
\(\mathrm{t}=20 \mathrm{mins} ; \mathrm{V}_{\mathrm{t}}=19.3 ; \mathrm{V}_{\mathrm{o}}=46.1 \mathrm{~mL} \)
\(k =\frac{2.303}{20} \log \left(\frac{46.1}{19.3}\right)  \)
\(=0.11515 \times \log 2.388  \)
\(=0.11515 \times 0.3780  \)
\(=0.0435 \mathrm{~min}^{-1}\)
Since the value of k comes out to be almost constant for the reaction, it is of first order. The mean value of k = 0.04355 min^-1

(i) Order of the reaction =1; \(\mathrm{t}_{1 / 2}=60 \mathrm{~s} ; \mathrm{k}=?\)
\(\mathrm{k}=\frac{0.6932}{\mathrm{t}_{\frac{1}{2}}} \)
\(=\frac{0.6932}{60}  \)
\(k =1.155 \times 10^{-2} \mathrm{~s}^{-1}\)
(ii) \(\left[\mathrm{A}_{0}\right]=100 \% ; \mathrm{t}=180 \mathrm{~s} ;[\mathrm{A}]=? ; \mathrm{k}=1.155 \times 10^{-2} \mathrm{~s}^{-1}\)
For first order reaction
\(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]} \)
\(1.155 \times 10^{-2} =\frac{2.303}{180} \log \left(\frac{100}{[A]}\right)  \)
\(\frac{0.01155 \times 180}{2.303} =\log \left(\frac{100}{[A]}\right)  \)
\(0.9027 =\log 100-\log [\mathrm{A}]  \)
\(\log [\mathrm{A}] =\log 100-0.9027 \)
\(\log [A]=2-0.9027 \)
\(\log [A]=1.0972 \)
[A] = antilog  of (1.0972)
[A] =12.51 %

(i) It does not explain the colour of the complex.
(ii) It considers only the spin only magnetic moments and does not consider the other components of magnetic moments.
(iii) It does not provide a quantitative explanation as to why certain complexes are inner orbital complexes and the others are outer orbital complexes for the same metal. For example, [Fe(CN)₆]^4- is diamagnetic (low spin) whereas [FeF₆]^4- is paramagnetic (high spin).

(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.

1. The crude metal is refined by electrolysis. It is carried out in an electrolytic cell
Anode : Impure metal to be refined with dilute acid.
Cathode : Thin strips of pure metal
Electrolyte : Aqueous solution of the salts of the metal with dilute acid.
2. The metal dissolves from the anode, pass into the solution.
3. At the same amount of metal ions from the solution will be deposited at the cathode.
4. During electrolysis, the less electropositive impurities in the anode, settle down at the bottom and are removed as anode mud.
Example: Electrolytic refining of silver.
Cathode: Pure silver
Anode: lmpure silver rods
Electrolyte: Acidified aqueous solution of silver nitrate
5. When a current is passed through the electrodes the following reactions will take place
(a) Reaction at anode: \({ Ag }_{ (s) }\longrightarrow { Ag }^{ + }_{ (aq) }+{ 1e }^{ - }\)
(b) Reaction at cathode: \({ Ag }^{ + }_{ (aq) }+{ 1e }^{ - }\longrightarrow { Ag }_{ (s) }\)
6. During electrolysis, at anode silver loses electrons and form silver ions and the silver ions migrate towards the cathode and get discharged and deposited on the cathode.
7. Copper, Zinc etc can also be refined by this process.

a) BrF₅

Valence electron of bromine atom 7+ Number of fluorine atom (5) = 12
\(X=\frac{12}{2}=6\) 
Hybridization: sp³d² ;
Geometry: Square Pyramidal
b) BrF₃

Valence electron of bromine atom 7+ Number of fluorine atom (3) = 10 
X = \(\frac{10}{2}=5\)
Hybridization: sp³d²;
Geometry: Triangular bipyramidal (T - shaped)

On heating PCl₅, it decomposes into PCl₃ and chlorine.
\(\mathrm{PCl}_{5(\mathrm{~g})} \rightarrow \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}\)

(i) Zeolites are three-dimensional crystalline solids containing Al, Si and O in their regular three dimensional framework.
(ii) They are hydrated sodium alumino silicates with general formula Na₂O(AI₂O₃).·x(SiO₂)·yH₂O
(x = 2 to 10; y = 2 to 6).
(iii) Zeolites have porous structure in which the monovalent sodium ions and water molecules are loosely held.
(iv) The Si and Al atoms are tetrahedrally coordinated with each other through shared oxygen atoms.
(v) Zeolites are similar to clay minerals but they differ in their crystalline structure.
(vi) Zeolites have a three dimensional crystalline structure looks like a honeycomb consisting of a network of interconnected tunnels and cages.
(vii) Water molecules moves freely in and out of these pores but the zeolite framework remains rigid
(viii) Another special aspect of this structure is that the pore/channel sizes are nearly uniform, allowing the crystal to act as a molecular sieve.

(i) The first transition metal Scandium exhibits only +3 oxidation state, but all other transition elements exhibit variable oxidation states by loosing electrons from (n-1)d orbital and ns orbital as the energy difference between them is very small. At the beginning of the series, +3 oxidation state is stable but towards the end +2 oxidation state becomes stable. The first and last elements show less number of oxidation states and the middle elements with more number of oxidation states
(ii) For example, the first element Sc has only one oxidation state +3; the middle element Mn has six different oxidation states from +2 to +7. The last element Cu shows +1 and +2 oxidation states only.

1. The number of nearest neighbours that surrounding a particle in a crystal is called the coordination number of that particle.
2. The coordination number of atoms in a bcc structure is '8'.