\(\log K=\log A-\frac { { E }_{ a } }{ 2.303R } \left( \frac { 1 }{ T } \right) \)
y = c + mx
\(m=-\frac { { E }_{ a } }{ 2.303R } \)
E_a = -2.303 Rm
E_a = -2.303 x 8.314 x (-4000)
E_a = 76,589J  mol^-1
E_a = 76.589 KJ mol^-1

The CFSE is defined as the energy of the electronic configuration in the ligand field minus the energy of the electronic configuration in the isotropic field.
CFSE (\(\Delta\)E_o) = {E_Lf}-{E_iso}
={[n_t2g(-0.4) + n_eg(0.6)]\(\Delta\)_o + n_pP} - {n'_pP}
Here, n_t2g is the number of electrons in t_2g orbitals;
n_eg is number of electrons in e_g orbitals;
n_p is number of electron pairs in the ligand field; &
n'_p is the number of electron pairs in the isotropic field (barycenter).
P - pairing energy

Oxidation number of F = -1
Oxidation number of H = +1
Oxidation number of O in HOF =x
(+1) + x + (-1) = 0
x = 0
Oxidation number of O in HOF = 0

For bcc structure \((r)=\frac{\sqrt{3}}{4} a\)
a = 4.3 \(\times\) 10^-8 cm, r = ?
\(=\frac{1.732 \times 4.3 \times 10^{-8}}{4}\)
\(r=1.86 \times 10^{-8} \mathrm{~cm}\)

(i) The extraction of metals from their oxides can be carried out by using different reducing agents.
(ii) Consider the following reaction
\(\frac{2}{\mathrm{y}} \mathrm{M}_{\mathrm{x}} \mathrm{O}_{\mathrm{y}(\mathrm{s})} \rightarrow \frac{2 \mathrm{x}}{\mathrm{y}} \mathrm{M}_{(s)}+\mathrm{O}_{ 2(\mathrm{~g})}\)            (1)
(iii) The above reduction may be carried out with carbon. In this case the reducing agent carbon may be oxidized to either CO or CO₂
\(\mathrm{C}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2(\mathrm{~g})}  \)                  (2)
\(2 \mathrm{C}+\mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{(\mathrm{g})} \)                (3)
(iv) If CO is used as a reducing agent
\(2 \mathrm{CO}+\mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}\)           (4)
(v) A suitable reducing agent is selected based on the thermodynamics considerations.
(vi) We know that for a spontaneous reaction, the change in free energy (\(\triangle\)G) should be negative.
(vii) Therefore, thermodynamically, the reduction of metal oxide with a given reducing agent can occur if the free energy change for the coupled reaction is negative.
(viii) Hence, the reducing agent is selected in such a way that it provides a large negative \(\triangle\)G value for the coupled reaction.

According to Henderson – Hasselbalch equation,
\(pH=pK_{a}+\log\frac{[salt]}{[acid]}\)
\(p{K_{a}}=-\log K_{a}=-\log(1.8\times10^{-5})=4.74\)
[Salt]=\(\frac{\text {Number of moles of sodium acetate}}{\text {Volume of the solution (litre)}}\)
Number of moles of sodium acetate =\(\frac{\text {mass of sodium acetate}}{\text {molar mass of sodium acetate}}\)
\(=\frac{8.2}{82}=0.1\)
\(\therefore [Salt]=\frac{0.1\ mole}{1/2 \ Litre}=0.2M\)
\([acid]=\frac{(\frac{mass \ of \ CH_{3}COOH}{molar \ mass \ of \ CH_{3}COOH})}{\text{Volume of solution in litre}}\)
=\(\frac{(\frac{6}{60})}{\frac{1}{2}}\)=0.2 M
\(\therefore pH=4.74+log\frac{(0.2)}{(0.2)}\)
pH = 4.74 + log1
pH = 4.74 + 0 = 4.74

a) Oxidation at anode :
\(\mathrm{Mg} \rightarrow \mathrm{Mg}^{2+}+2 \mathrm{e}^{-} ; \mathrm{E}_{\mathrm{Ox}}^{0}=2.37 \mathrm{~V}\) ...(1) 
Reduction at cathode:
\(A \mathrm{~g}^{+} +\mathrm{e}^{-} \rightarrow \mathrm{Ag} ; \mathrm{E}_{\text {red }}^{0}=+0.80 \mathrm{~V}  \) ...(2)
\(E_{\text {Cell }}^{0} =\left(\mathrm{E}_{\text {ox }}^{0}\right)_{\text {anode }}+\left(\mathrm{E}_{\text {red }}^{0}\right)_{\text {cathode }} \)
= 2.37 + 0.80 = 3.17V
Overall reaction : (1) + 2 x (2)
\(\mathrm{Mg} \rightarrow M g^{2+}+2 e^{-}  \)
\(2 \mathrm{Ag}^{+}+2 \mathrm{e}^{-} \rightarrow 2 A g \)
__________________
\(M g+2 A g^{+} \rightarrow M g^{2+}+2 A g\)
 b) \( \therefore \Delta G^{0}=-n F E^{0} \)
\(=-2 \times 96500 \times 3.17 \)
\(=-611810=-6.12 \times 10^{5} \mathrm{~J} \)
\(W=6.12 \times 10^{5} \mathrm{~J}\)
c) \(\Delta \mathrm{G}^{0}=-2.303 \mathrm{RT} \log \mathrm{K}_{c} \)
\(\log \mathrm{K}_{\mathrm{c}}=-\frac{\Delta G^{0}}{2.303 R T} \)
\(\log \mathrm{K}_{c}=-\frac{\left(-6.12 \times 10^{5}\right)}{2.303 \times 8.314 \times 298}=107.2 \)
\(K_{c}=A . \log 107.2 \)
\(\mathrm{K}_{c}=\text { Antilog of (107.2) }\)
= 1.58 x 10¹⁰⁷

\(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \rightleftharpoons 3 \mathrm{Ca}^{2+}_{(aq)}+2 \mathrm{PO}_{4_{(aq)}}^{3-}\\\)
(s)                           (3s)             (2s)
\(K_{sp}=[Ca^{2+}]^{3}[PO_{4}^{3-}]^{2}\)
\(K_{sp}=(3s)^{3}(2s)^{2}\)
\(K_{sp}=27s^{3}.4s^{2}\)
\(K_{sp}=108s^{5}\)
(or)
\(K_{s p} =m^{m} \cdot n^{n} \cdot(s)^{m+n} \)
\(K_{\text {sp }} =3^{3} \cdot 2^{2} \cdot(s)^{3+2} \)
\(K_{s p} =27 \times 4 \times(s)^{5} \)
\(=108(s)^{5}=108 s^{5}\)

(i) A polymer containing two or more different kinds of monomer units is called a copolymer.
(ii) For example, SBR rubber(Buna-S) contains styrene and butadiene monomer units.
(iii) Copolymers have properties quite different from the homopolymers.
(iv) Mixture of styrene and 1,3 butadiene to form a copolymer (Buna -S)
Preparation of Buna-S:
It is a co-polymer. It is obtained by the polymerisation of buta -1,3 - diene and styrene in the ratio 3: 1 in the presence of sodium.

(i) Ribonucleic acids are similar to DNA. Cells contain up to eight times high quantity of RNA than DNA. RNA is found in large amount in the cytoplasm and a lesser amount in the nucleus. In the cytoplasm it is mainly found in ribosomes and in the nucleus, it is found in nucleolus.
(ii) RNA molecules are classified according to their structure and function into three major types
i. Ribosomal RNA (rRNA)
ii. Messenger RNA (mRNA)
iii. Transfer RNA (tRNA)
rRNA:
rRNA is mainly found in cytoplasm and in ribosomes, which contain 60% RNA and 40% protein. Ribosomes are the sites at which protein synthesis takes place.
tRNA:
tRNA molecules have lowest molecular weight of all nucleic acids. They consist of 73 - 94 nucleotides in a single chain. The function of tRNA is to carry amino acids to the sites of protein synthesis on ribosomes.
mRNA:
mRNA is present in small quantity and very short lived. They are single stranded, and their synthesis takes place on DNA. The synthesis of mRNA from DNA strand is called transcription. mRNA carries genetic information from DNA to the ribosomes for protein synthesis. This process is known as translation.

(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.

\(Eu (63)-[\mathrm{Xe}] 4 \mathrm{f}^{7} 5 \mathrm{~d}^{0} 6 \mathrm{~s}^{2}, \mathrm{Eu}^{2+}-[\mathrm{Xe}] 4 \mathrm{f}^{7} \)
\(\mathrm{Ce}(58)-[\mathrm{Xe}] 4 \mathrm{f}^{1} 5 \mathrm{~d}^{1} 6 \mathrm{~s}^{2}, \mathrm{Ce}^{2+}-[\mathrm{Xe}] 4 \mathrm{f}^{1} 5 \mathrm{~d}^{1} \)
Eu²⁺ has exactly half filled stable electronic configuration. Hence Europium (II) is more stable than Cerium (II).

For cubic closed packed structure
\(r =\frac{a \sqrt{2}}{4}  \)
\(\therefore a =\frac{4 r}{\sqrt{2}}  \)
\(=\frac{4 \times 1.25 \times 10^{-8}}{1.414}=3.53 \times 10^{-8} \mathrm{~cm} \)
= 353 pm

A hydride of 2^nd period alkali metal (A) is lithium hydride (LiH).
Lithium hydride (A) reacts with diborane (B) to give lithium borohydride (C) which is acts as a reducing agent.
B₂H₆ + 2 LiH \(\xrightarrow[]{ether}\) 2 LiBH₄
[Diborane (B)]   [Lithium hydride (A)]  [Lithium borohydride (C)]
Result:

Compound	Formula	Name
A	LiH	Lithium hydride
B	B2H6	Diborane
C	LiBH4	Lithium borohydride

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) Emulsions are used in food industries. Food stuff like milk, cream, butter, etc., are emulsions.
(ii) Emulsions are very common in pharmaceutical industries. Many medicines are produced in the form of emulsions. Ex: Milk of magnesia is used for stomach troubles. Many lotions and ointments are emulsions.
(iii) Non ionic emulsions are most popular due to their low toxicity.
(iv) Cationic emulsions have anti microbial properties.
(v) In agriculture industry emulsions are used as delivery vehicles for insecticides, fungicides and pesticides.
(vi) In cosmetics, emulsions are the delivery vehicles for many hair and skin conditioning agents.
(v) The blood, protoplasm in plant and animal cells and fats in intestines are emulsions.

(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 %