Cyclic alcolhol →  sodium cyclic alkoxide- Williamson ether synthesis.

Emulsion-Dispersed phase
Dispersion meduun -liquid

AgCl_(s) ⇌ Ag⁺_(aq) + Cl^-_(aq)
NaCl  → Na⁺ +Cl^-
0.1M      0.1M   0.1M
K_sp = 1.6 × 10^-10
K_sp = [Ag⁺][ Cl^-]
K_sp = (s) (s + 0.1)
0.1 >>> s
∴ s + 0.1 = 0.1
∴ s = 1.6 × 10^-10/ 0.1 = 1.6 × 10^-9

Rate  \(=\frac { d\left[ { N_2O_5 } \right] }{ dt } =\frac { 1 }{ 2 } =\frac { d\left[ { NO}_{ 2 } \right] }{ dt } =\frac { 2d\left[ { O }_{ 2 } \right] }{ dt } \)
Given that \(=\frac { d\left[ { N_2O_5 } \right] }{ dt } \) =6.5 x 10^-2 mol L^-1 s^-1
\(=\frac { d\left[ { NO}_{ 2 } \right] }{ dt }\) =2 x 6.5 x 10^-2 = 1.3 x 10^-1 mol L^-1 s^-1
\(=\frac { d\left[ { O}_{ 2 } \right] }{ dt }\)\(=\frac{6.5 \times 10^{-2}}{2}\) = 3.25 x 10^-2 mol L^-1s^-1

Both are acetylating agents, but acylehloride gives pale fumes of hydrogen chloride also, Hence acid anhydride are preferred. Acid anhydride gives acetic acid as the by product in acylation reaction, lt will also help for further acylation.

1. \(\kappa=\mathrm{C}\left(\frac{1}{\mathrm{~A}}\right)\)
2. \(\operatorname{let} \frac{l}{A}=x\)
i) \(0.01 \mathrm{M} \mathrm{KCl}: \kappa=0.01 x=10^{-2} x\)
ii) \(0.005 \mathrm{M} \mathrm{KCl}: \kappa=0.005 x=5 \times 10^{-3} x\)
iii) \(0.1 \mathrm{M} \mathrm{KCl} \quad: \kappa=0.1 x=10^{-1} x\)
iv) \(0.25 \mathrm{M} \mathrm{KCl}: \kappa=0.25 x=2.5 \times 10^{-1} x\)
v) \(0.5 \mathrm{M} \mathrm{KCl}: \mathrm{K}=0.5 x=5 \times 10^{-1} x\)
\(\therefore 5 \times 10^{-1} x>2.5 \times 10^{-1} x>10^{-1} x>10^{-2} x>5 \times 10^{-3} x\)
\((ie) 0.5 \mathrm{M} \mathrm{KCl}>0.25 \mathrm{M} \mathrm{KCl}>0.1 \mathrm{M} \mathrm{KCl}>0.01 \mathrm{M} \mathrm{KCl}>0.005 \mathrm{M} \mathrm{KCl}\)

\(Al(OH)_{3}\rightleftharpoons Al^{3+}_{(aq)}+3OH^{-}_{(aq)}\)
\(K_{sp}=[Al^{3+}][OH^{-}]^{3}\)
Al(OH)₃ precipitates when ionic product > K_sp
K_s = 1.0 \(\times\) 10^-15m, [Al³⁺] = 1.0 \(\times\) 10^-3m
1.0 \(\times\) 10^-15 = [1.0 \(\times\) 10^-3][OH^-]³
\(\left[\mathrm{OH}^{-}\right]^3=\frac{1.0 \times 10^{-15}}{1.0 \times 10^{-3}}\)
[OH^-]³ = 1.0 \(\times\) 10^-12
(or)
[OH^-]³ = 10^-4M
[H⁺][OH^-] = 10^-4M
[H⁺] = \(\frac{10^{14}}{10^{-4}}\) =10^-10
Here,
pH = 10
ie., At pH = 10, Al(OH)₃ gets precipitated on the addition of NH₄Cl & NH₄OH solution.

(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) The actual position of Lanthanides in the periodic table is at group number 3 and period number 6. However, in the sixth period after lanthanum, the electrons are preferentially filled in inner 4f sub shell and these fourteen elements following lanthanum show similar chemical properties.
(ii) Similarly the fourteen elements following actinium resemble in their physical and chemical properties. Hence they are placed separately bottom of the modern periodic table.

There are seven types of unit cell, Cubic, tetragonal, orthorhombic, hexagonal, monoclinic, triclinic and rhombohedral. They differ in the arrangement of their crystallographic axes and angles.
i) Cubic: a = b = c; α = β = ૪ = 90^o.

ii) Tetragonal: a = b ≠ c; α = β = ૪ = 90°.

iii) Orthorhombic:  a  ≠ b  ≠ c; α = β = ૪ = 90°.

iv) Hexagonal: a = b ≠ c; α = β = 90^o, ૪ = 120^o. 

v) Monoclinic: a ≠ b ≠ c; α = ૪ = 90^o, β ≠ 90^o, 

vi) Triclinic: a ≠ b ≠ c; α  ≠ β ≠ ૪ ≠ 90^o.

vii) Rhombohedral:  a = b = c; α = β = ૪ ≠ 90^o. 

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

Mond process for refining nickel:
The impure nickel is heated in a stream of carbon monoxide at around 350 K. The nickel reacts with the CO to form a highly volatile nickel tetracarbonyl. The solid impurities are left behind. 
\(\mathrm{Ni}_{(\mathrm{s})}+4 \mathrm{CO}_{(\mathrm{g})}\ \rightarrow \ \left[\mathrm{Ni}(\mathrm{CO})_{4}\right]_{(g)}\)
(ii) When the Nickel Carbonyl Complex is heated at 460 K, it decomposes to give pure nickel.
\(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]_{(g)} \stackrel{460 \mathrm{~K}}{\longrightarrow} \mathrm{Ni}_{(s)(\text { pure })}+4 \mathrm{CO}_{(\mathrm{g})}\)

 

Compound	Name
A	Acetylchloride
B	Acetaldehyde
C	3 - Hydroxy butanal
D	Crotonaldehyde

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.

Watson & Crick proposed a 3-dimensional secondary structure of DNA. In this DNA molecule, Monomners of DNA are held together by Phosphodiester linkage. This linkages are occured in 5' & 3' carbon atoms of Pentose sugar.

Bonding in metal carbonyls:
(i) In metal carbonyls, the bond between metal atom and the carbonyl ligand consists of two components.
(ii) The first component is an electron pair donation from the carbon atom of carbonyl ligand into a vacant d-orbital of central metal atom.
(iii) This electron pair donation forms \(M\overset { \sigma \ bond }{ \longleftarrow } \text {CO sigma bond}\)
(iv) This sigma bond formation increases the electron density in metal d orbitals and makes the metal electron rich.
(v) In order to compensate for this increased electron density, a filled metal d-orbital interacts with the empty \({ \pi }^{ \bigstar }\) orbital on the carbonyl ligand and transfers the added electron density back to the ligand.
(vi) This second component is called \(\pi\)-back bonding.
(vii) Thus in metal carbonyls, electron density moves from ligand to metal through sigma bonding and from metal to ligand through pi bonding, this synergic effect accounts for strong M⇽CO bond in metal carbonyls.
(viii) This phenomenon is shown diagrammatically as follows.

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)

Allotropism:
1. Some elements exist in more than one crystalline or molecular forms in the same physical state
(a) In Greek "allos" means ⇒ Another
(b) "trope" means ⇒ Change
2. The different forms of an element are called allotropes.
Allotropy of carbon:
Carbon exists as diamond, graphite, fullerenes, carbon nanotubes and graphene.
Graphite:
1. Graphite is the most stable allotropic form of carbon at normal temperature and pressure.
2. It is soft and conducts electricity.
3. It is composed of flat two dimensional sheets of carbon atoms.
4. Each sheet is a hexagonal.
5. It is "sp²" hybridised.
6. C-C bond length is 1.41 Å
7. Each C-atom forms three σ bonds with three neighbouring carbon atoms using three of its valence electrons and the fourth electron present in the unhybridised p-orbital form a π-bond.
8. The successive C-sheets are held together by weak Vander Waals forces.
9 The distance successive sheet is 3.40 Å.
10. It is used as a lubricant either on its own or as a graphited oil.

Diamond:
1. It is very hard.
2. It is "sp³" hybridised.
3. C-C bond length is 1.54 Å
4. It is used for sharpening hard tools, cutting glasses, making bores and rock drilling.

Fullerenes:
1. These allotropes are discrete molecules such as \(C_{32}, C_{50}, C_{60}, C_{70}, C_{76}\) etc.
2. It has cage like structure
3. The C₆₀ molecules have a "soccer" ball like structure and is called buckminster fullerene or buckyballs.
4. It has a fused ring structure consists of 20 six membered rings and 12 five membered ring.
5. Each carbon atom is "sp²" hybridised.
6. It has three σ bonds and a delocalised π bond giving aromatic character to these molecules.
7. The C-C bond distance is 1.44 Å
8. The C=C bond distance is 1.38 Å.

Carbon nanotubes:
1. Carbon nanotubes, another recently discovered allotropes, have graphite like tubes with fullerene ends.
2. Along the axis, these nanotubes are stronger than steel and conduct electricity.
3. These have many applications in nanoscale electronics, catalysis, polymers and medicine.
Graphene:
 It has a single planar sheet of "sp²" hybridised carbon atoms that are densely packed in a "honeycomb crystal" lattice.

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.

Ag₂S + 4NaCN ⇌ 2NalAg(CN)₂] + Na₂S
                              Sodium Argento cyanide
2Na[Ag(CN)₂] + Zn → Na₂[Zn(CN)_4] + 2Ag↓
Silver is obtained by the reduction of sodium Argento cyanide with zinc. At the same time zinc gets oxidised into its complex. Hence leaching process is a redox reaction.

(i) It is instantaneous.
(ii) It is non-specific.
(iii) No transfer of electrons.
(iv) Multilayer of adsorbate is formula.

\((E^{0}_{ox})_{Fe^{2+}|Fe} = -0.44\)  and 
\((E^{0}_{red})_{Cu^{2+}|Cu} = 0.34\)
These +ve emf values shows that iron will oxidise and copper will get reduced i.e., the vessel will dissolve. Hence it is not possible to store copper sulphate in an iron vessel.

i) Rickets - Vitamin - D (Cholecalciferol - (D₃) Ergocalciferol - (D₂)
ii) Scurvy (bleeding gums) - vitamin - C (Ascorbic acid)

K = 1.54 \(\times\) 10^-3s^-1; t_1/2 = ?
\({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } \)
\(=\frac { 0.693 }{ 1.54\times { 10 }^{ -3 } } \)
= 450 s

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

(i) The Chalcogens belong to group (16).
(ii) The group consists of elements: Oxygen, Sulphur, Selenium, Tellurium and Polonium.
(iii) These are ore forming elements as most of the ores are oxides and sulphides.
(iv) Chalcos meaning 'ore formers'.