As₂S₃ is a negatively charged colloid. It will be most effectively coagulated by the cation with greater valency. i.e., Al³⁺.

Anodic Oxidation: (Reverse the given reaction)
(E^o_ox ) = 0.76 V Cathodic reduction
E^o_cell = (E^o_ox )+ (E^o_red) 
= 0.76 + 0.34 = 1.1 V

x ml of 0.1 M NaOH + x mL of 0.01 M HCI
No. of moles of NaOH = 0.1 x X x 10^-3
= 0.1 X x 10^-3
No. of moles of HCI = 0.01 x X x 10^-3
= 0.01 X x 10^-3
No. of moles of NaOH after mixing
= 0.1 X x 10^-3 - 0.01 X x 10^-3
= 0.09 X x 10^-3
Concentration of NaOH\(= (\frac{0.09x \times 10^{-3}}{2x \times 10^{-3}}) = 0 .045\)
[OH^-] = 0.045
pOH =-log (4.5 x 10^-2)
= 2 -log 4.5
= 2 - 0.65 = 1.35
pH = 14 - 1.35 = 12.65

Hydrolysis of water
At anode:
\(2H_{2}O\rightarrow 4H^{+}+O_{2}+4e^{-}\) ..... (1)
At cathode:
\(2H_{2}O+2e^{-}\rightarrow H_{2}+2OH^{-}\) ....(2) 
Overall reaction 
\(6H_{2}O\rightarrow 4H^{+}+4OH^{-}+2H_{2}+O_{2}\)
(or)
Equation (1) +(2) \(x^2 \Rightarrow 2H_{2}O\rightarrow 2H_{2}+O_{2}\)
\(\therefore\) According to faradays Law of electrolysis, to electrolyse two mole of Water (36g ≃ 36 mL of H₂O), 4F charge is required alternatively, when 36 mL of water is electrolysed, the charge generated = \(4\times 96500\)C.
\(\therefore\) When the whole water which is available on the lake is completely electrolysed the amount of charge generated is equal to \(\frac{4\times96500\quad C}{36 \quad mL}\times9\times10^{12}L\)
\(=\frac{4\times96500\times9\times10^{12}}{36\times10^{-3}}C\)
= \(96500\times10^{15}C\)
\(\therefore\) Given that in 1 second, \(2\times10^{6}\) C is generated therefore, the time required to generate \(96500 \times 10^{15}\) C is = \(\frac{1\quad S}{2\times 10^{6}C}\times 96500 \times10^{15}C\)
=\(48250 \times 10^{9} S\)
\(\therefore\) Number of years = \(\frac{48250 \times 10^{9}}{365 \times 24 \times 60 \times 60}\)   
=\(1.5299 \times 10^{6}\) years
1 year = 365 days
= 365\(\times\)24 hours
= 365\(\times\)24\(\times\)60 min
= 365\(\times\)24\(\times\)60\(\times\)60 sec.

 
Since 1 - methoxy propane has primary alkyl group, it under goes SN2 reaction.

K_sp =1.1 \(\times\)10^-2, [K₂, CrO₄]= 0.1M
\(\underset {s}{Ag_{2}CrO_{4}}\rightleftharpoons \underset{2s}{2Ag^{+}}+\underset{s}{CrO^{2-}_{4}}\)
\(\underset{0.1M}{K_{2}CrO_{4}}\rightleftharpoons \underset{0.2M}{2K^{+}}+\underset{2 \times 0.1M}{CrO^{2-}_{4}}\)
\({\left[\mathrm{Ag}^{+}\right] } =2 \mathrm{~s} ;\left[\mathrm{CrO}_{4}^{2-}\right]=(\mathrm{S}+0.1) \simeq 0.1  \)
\(\mathrm{~K}_{\mathrm{sp}} =\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{CrO}_{4}^{2-}\right]  \)
\(1.1 \times 10^{-12} =(2 \mathrm{~s})^{2}(0.1)  \)
\(1.1 \times 10^{-12} =0.4 \mathrm{~s}^{2}  \)
\(0.4 \mathrm{~s}^{2} =1.1 \times 10^{-12}  \)
\(\mathrm{~s}^{2} =\frac{1.1 \times 10^{-12}}{0.4}=2.75 \times 10^{-12}  \)
\(\mathrm{~s} =\sqrt{2.75 \times 10^{-12}}  \)
\(\mathrm{~s} =1.658 \times 10^{-6} \mathrm{M}\)

(i) Preservatives are chemicals added to food which are capable of inhibiting, retarding or arresting the process of fermentation, acidification or other decomposition of food by the growth of micro organisms (or)
(ii) Chemical substances which are added to food that prevents the spoilage of food materials by destroying the food - spoiling micro-organisms are called food preservatives.
(iii) Eg: sodium benzoate, NaCl, Acetic acid Sodium metabisulphite, potassium meta bisulphite etc.

(i) Redox reactions involve transfer of electrons from one reactant to another. Such reactions are always coupled, which means that when one substance is oxidised, another must be reduced.
(ii) The substance which is oxidised is a reducing agent and the one which is reduced is an oxidizing agent

(i) Diamond is very hard.
(ii) The carbon atoms in diamond are sp³ hybridised and bonded to four neighbouring carbon atoms by a bonds with a C-C bond length of 1.54 Å.
(iii) This results in a tetrahedral arrangement around each carbon atom that extends to the entire lattice.
(iv) Since all four valance electrons of carbon are involved in bonding there is no free electrons for conductivity.
(v) Being the hardest element, it used for sharpening hard tools, cutting glasses, making bores and rock drilling.

(i) Leaching of sulphide ores such as ZnS, PbS etc., can be done by treating them with hot aqueous sulphuric acid
\(2Zn{ S }_{ (s) }+{ 2H }_{ 2 }{ SO }_{ 4(aq) }+{ O }_{ 2(g) }\longrightarrow { 2ZnSO }_{ 4(aq) }+2{ S }_{ (s) }+{ H }_{ 2 }O\)
(ii) In this process the insoluble sulphide is converted into soluble sulphate and elemental sulphur

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

(i) Coordination compounds which possess chirality exhibit optical isomerism similar to organic compounds.
(ii) The pair of two optically active isomers which are mirror images of each other are called enantiomers.
(iii) Their solutions rotate the plane of the plane polarised light either clockwise or anticlockwise and the corresponding isomers are called 'd' (dextrorotatory) and 'I' (levorotatory)  forms respectively.
(iv) The octahedral complexes of type \(\left[\mathrm{M}(\mathrm{xx})_{3}\right]^{\mathrm{n} \pm}, \left[\mathrm{M}(\mathrm{xx})_{2} \mathrm{AB}\right]^{\mathrm{n\pm}}\) and \(\left[\mathrm{M}(\mathrm{xx})_{2} \mathrm{~B}_{2}\right]^{\mathrm{n\pm}}\) exhibit optical isomerism.
Examples:
(i) The optical isomers of \(\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}\) 
(ii) The coordination complex \(\left[\mathrm{CoCl}_{2}(\mathrm{en})_{2}\right]^{+}\) has three isomers, two optically active cis forms and one optically inactive trans form.

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'.

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. 

(i) Benzylchloride to benzylalcohol
\(\underset { Benzyl \ chloride }{ { C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }Cl } + NaOH { \longrightarrow } \underset { Benzyl \ alcohol }{ { C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }OH } +NaCl\)
(ii) benzyalalcohol to benzoic acid
 \(\underset { Benzyl \ alcohol }{\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH} }\stackrel{\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} / \mathrm{H}^{+}}{\longrightarrow}\underset { Benzaldehyde } {\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}}\stackrel{\mathrm{Na}_{2} \mathrm{CrO}_{7} / \mathrm{H}^{+}}{\longrightarrow} \underset { Benzoic \ acid } {\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}}\)

\(\mathrm{HClO}_{4} \rightleftharpoons \mathrm{H}^{+}+\mathrm{ClO}_{4}^{-}\)
Bronsted Acid Proton Conjugate Base
HClO₄ can donate a proton. Therefore HClO₄ is an acid. Its conjugate base is ClO₄^-

Ligand is an atom or a group of atoms which is either negatively charged or has lone pair of electrons and from co-ordinate bond with the central metal or ion, e.g.H₂O.

(i) The compounds of Cu²⁺ are coloured as it has one free electron its valence shell which absorb I radiation of visible region and get excited to emit its complementary colour.
(ii) Zn has no free electron it has fully filled shells. Due to extra stable orbitals electron can't be excited by radiations of visible light, hence its compounds are colourless.

[Ma₂b₂C₂]^\(\pm\)n where a, b, c are monodentate ligands.

[Pt(py₂)(NH₃)₂Cl₂]²⁺. It exhibits both optical and geometrical isomerism. c is isomer exhibit optical isomerism also. While trans isomer exhibits geometrical isomerism only.

(i) \(\mathrm{Mg}\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right) \mathrm{Cl}_{5}\right]^{2-} \rightleftharpoons \mathrm{Mg}^{2+}+\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right) \mathrm{Cl}_{5}\right]^{2-} (2 ions)\)
(ii) \(\begin{aligned} {\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}_3\left[\mathrm{CoF}_6\right]_2 \rightleftharpoons 3\right.} & {\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{2+} } \\ & (5 \text { ions }) \end{aligned}\)\(+2\left[\mathrm{CoF}_6\right]^{3-}\)
(iii) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right]= \text{ No ions}\)
If no of ions increases, molar conductivity increases molar conductivity of the complex also INCREASES.
\(\therefore\) The order of the given compound is
[Cr(NH₃)₃Cl₃]<Mg[Cr(NH₃)₃Cl₃] < [Cr(NH₃)₅Cl₃] [CoF₆]₂
  (No ion)          (2 ions)                    (5 ions)

The structural units of an ionic crystal are cations and anions. They are bound together by strong electrostatic attractive forces. To maximize the attractive force, cations are surrounded by as many anions as possible and vice versa. Hence they are hard and brittle.

Chlorine is prepared by the action of conc. sulphuric acid on chlorides in presence of manganese dioxide
4NaCl + MnO₂ + 4H₂SO₄ \(\longrightarrow \)Cl₂+ MnCl₂ +4NaHSO₄ + 2H₂O

(i) sp³d³ hybridisation
(ii) Pentagonal bipyramidal structure.

(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.

Aniline does not undergo Friedel - Craft's reaction:
Aniline does not undergo Friedel - Craft's reaction (alkylation and acetylation). Aniline is basic in nature and it donates its lone pair of electrons to the lewis acid AlCl₃ to form an adduct which inhibits further electrophilic substitution reaction.
Diazonium salts of aromatic amines are more stable than those of aliphatic amines:
This is due to resonance
Resonance Structure:
The stability of arene diazonium salt is due to the dispersal of the positive charge over the benzene ring.

pK_b of aniline is more than that of methylamine:
pK_b - methylamine -3.35
pK_b - aniline -9.376
In aniline the lone pair of electrons on N - atom is delocalized over the benzene ring. So, the electron density on the N - atom decreases. In methylamine + 1 effect to CH₃ group increases the electron density on the nitrogen atom Hence aniline is a weaker base than methylamine. Due to this, the pK_b value for aniline is more than that of methylamine.
(iv) Gabriel phthalimide synthesis is preferred for synthesising primary amines:
In this method alkyl halides react with pottassium phthalimide to give pure primary amine by nucleophilic substitution. In contrast, Aniline (Aromatic primary amine) can not be prepared by this method because Aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide. Therefore, this method used for the Aliphatic primary. amines only. Aryl halides do not undergo S_N2 mechanism with the ion formed by the phthalimide.
(v) Ethylamine is soluble in water whereas aniline is not:
(a) Ethylamine is soluble in water, as it can form intermolecular H - bonds with water molecules. In aqueous solution, the substituted ammonium cation get stabilized not only by electron releasing (+I) effect of the alkyl group but also by solvation with water molecules. The greater the size of the ion, the lower will be the solvation.
(b) Amiline doesn't form H - bond with water to a very large extent due to the presence of a large hydrophobic -C₆H₅ group.
(vi) Amines are more basic than amides:
This is because, in amides, the carbonyl group is highly electro negative It has a greater power to attract the electrons towards it. It makes the lone pair of electrons on amide nitrogen (-CONH₂) less available to accept a proton.
(vii) Although amino group is o - and p - directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m - nitro aniline:
In strong acid medium, aniline is protonated to form anilinium ion which is m - directing and hence m - nitro aniline is formed.
 

Arrhenius suggested that the rates of most reactions vary with temperature in such a way that the rate constant is directly proportional to \({ e }^{ -\left( \frac { { E }_{ a } }{ RT } \right) }\) and he proposed a relation between the rate constant and temperature.
\(k=A{ e }^{ -\left( \frac { { E }_{ a } }{ RT } \right) }\) ....(1)
Where A the frequency factor,
R the gas constant,
E_a the activation energy of the reaction and,
T the absolute temperature (in K)
(ii) The frequency factor (A) is related to the frequency of collisions (number of collisions per second) between the reactant molecules. The factor A does not vary significantly with temperature and hence it may be taken as a constant.
(iii) E_a is the activation energy of the reaction, which Arrhenius considered as the minimum energy that a molecule must have to posses to react.
(iv) Taking logarithm on both side of the equation (1)
In k = In A + In \({ e }^{ -\left( \frac { { E }_{ a } }{ RT } \right) }\)
In k = In A -\(\left( \frac { { E }_{ a } }{ RT } \right) \)       (∴ In e = 1)
In k = In A - \({ \left( \frac { { E }_{ a } }{ R } \right) }\) \(\left( \frac { 1 }{ T } \right) \)....(2)
y = c = m x
The above equation is of the form of a straight line y = mx+c
(v) A plot of In k Vs \(\left( \frac { 1 }{ T } \right) \) gives a straight line with a negative slope -\(\frac { { E }_{ a } }{ R } \) If the rate constant for a reaction at two different temperatures is known, we can calculate the activation energy as follows.
At temperature T = T¹; the rate constant k = k₁
In k₁ = In A - \(^{ \left( \frac { { E }_{ a } }{ { RT }_{ 1 } } \right) }\)      ....(3)
At temperature T = T₂; the rate constant k = k₂
In k₂ = In A - \(\left( \frac { { E }_{ a } }{ { RT }_{ 2 } } \right) \)    ......(4)
(4) - (3)
In k₂ - In k₁ = - \(\left( \frac { { E }_{ a } }{ { RT }_{ 2 } } \right) \) +  \(^{ \left( \frac { { E }_{ a } }{ { RT }_{ 1 } } \right) }\)
In \(\left( \frac { { k }_{ 2 } }{ { k }_{ 1 } } \right) \) = \(\frac { { E }_{ a } }{ R } \) \(\left( \frac { 1 }{ T_{ 1 } } -\frac { 1 }{ { T }_{ 2 } } \right) \)
2.303 log \(\left( \frac { { k }_{ 2 } }{ { k }_{ 1 } } \right) \)= \(\frac { { E }_{ a } }{ R } \) \(\left( \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right) \)
log \(\left( \frac { { k }_{ 2 } }{ { k }_{ 1 } } \right) \) = \(\frac { { E }_{ a } }{ 2.303R } \) \(\left( \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right) \)
In k₂ - k₁ = - \(\left( \frac { { E }_{ a } }{ R{ T }_{ 2 } } \right) \) + \(\left( \frac { { E }_{ a } }{ R{ T }_{ 1 } } \right) \)
This equation can be used to calculate E_a from rate constants k₁ and k₂ at temperatures T₁ and T₂.

(i) Because of its lightness and noninflammability helium is used to filling balloons for meteorological observations.
(ii) Because of its lightness it is used in inflating aeroplane tyres.
(iii) Helium oxygen mixture is used by deep sea divers in preference to nitrogen oxygen mixtures. This prevents bends when a diver comes to the surface.
(iv) A mixture of oxygen and helium is used in the treatment of asthma.
(v) Liquid helium (b.pt 4.2K) is used as cryogenic agent for carrying out various experiments at low temperatures.
(vi)It is used to produce and sustain powerful super conducting magnets of modern NMR Spectrometers and Magnetic Resonance Imaging system (MRI) for clinical diagnosis.

a) [FeF₆]^4-: Fe atom - outer electronic configuration 3d⁶ 4s²

F^- is weak field ligand
In [FeF₆]^4-the hybridisation takes place is sp³d²
The number of unpaired electrons = 4.
\(\therefore \mu =\sqrt { 4(4+2) } =\sqrt { 24 } \)
The molecule is paramagnetic due to the presence of unpaired electrons.
The geometry of the molecule is octahedral.
b) [Fe(CN)₆]^4-

In [Fe(CN)₆]^4- complex, the CN^- ligand is a powerful ligand, it forces the unpaired electrons in the 3d level to pair up inside.
Hence the species has no unpaired electron after hybridisation So the molecule is diamagnetic.

The geometry of the molecule is octahedral.