(i) 

(ii)

(i) River water is colloidal solution of clay. Sea water contains a number of electrolytes.
(ii) When river water meets the sea water, the electrolytes present in sea water coagulate the colloidal solution of clay which get deposited with the formation of delta.

(i) Colloidal solutions in which the dispersed phase has very little affinity for the dtspersion medium are termed as lyophobic (solvent hating) colloids.
(ii) Colloidal solutions of metals which have negligible affinity for solvents and sulphur in water are examples of this type.

Emulsion can be separated into two separate layers. The process is called Deemulsification.

(i) The surface of colloidal particle adsorbs one type of ion due to preferential adsorption.
(ii) This layer attracts the oppositely charged ions in the medium and hence at the boundary separating the two electrical double layers are setup. This is called as Helmholtz electrical double layer.

\({ \Lambda }_{ m }={ \Lambda }_{ m }^{ o }-(-A+B{ \Lambda }_{ m }^{ o })\sqrt { C } \)

(i) 
                                             n-Nitro benzoate (Major Product)

(ii) 

Metals having higher oxidation potential will liberate H₂ from H₂SO₄. Hence, the metal M₁ having +xV, oxidation potential will liberate H₂ from H₂SO₄.

Gallium.

(i) 3SiCl₄ + 4 Al ⟶ 4 AICl₃ + 3 Si
(ii) n(CH₃)₂Si(OH)₂

(iii) \(SiC{ l }_{ 4 }+4{ H }_{ 2 }O\longrightarrow \underset { orthosilicic \ acid }{ { H }_{ 4 }Si{ O }_{ 4 } } +4HCl\)
H₄SiO₄ \(\overset { \triangle }{ \longrightarrow } \) SiO₂ + 2H₂O
SiO₂ + Na₂CO₃ ⟶ Na₂SiO₃ + H₂O.

The name alum is given to the double salt of potassium aluminium sulphate.
[K₂SO₄.Al₂(SO₄)₃.24H₂O]

(i) Silicon of is refined by Zone refining method.
(ii) It is based on the principle that melting point of a substance is lowered by the presence of impurities
(iii) Consequently when an impure molten metal is cooled, crystals of the pure metal are solidified and the impurities remain behind the remaining metal.

(i) Graphite rods act a anode during electrolytic reduction of alumina.
(ii) At anode, O₂ gas is produced which react with the carbon of anode (rods) to produce CO₂ gas.
(iii) So these graphite rods are consumed slowly and need to be replaced from time to time.

Arrhenius equation is,
\(k=Ae^\left ({ \frac { -Ea }{ RT } } \right )\)
Here,
A \(\rightarrow\) Frequency factor
E_a \(\rightarrow\) Activation energy of the reaction
R \(\rightarrow\) Gas constant
T \(\rightarrow\) Absolute temperature (in K)

(i) Transition metals have number of unpaired electron. They are involved in metallic bonding. Hence they show high melting point.
(ii) As we move from left to right along the transition metal series melting point first increases reach a maximum value and then decreases as the d-electrons pair up and become less available for bonding.

(i) When excess of liquid ammonia is added to an aqueous solution of copper sulphate gives tetra ammine copper (II) sulphate is formed.
CUso₄ + 4NH₃ \(\rightarrow\) [Cu(NH₃)₄]SO₄
(ii) The co-ordination entity is [Cu(NH₃)₄]²⁺

Ligand	Type of Ligand	Number of donor atoms
NH3	monodentate ligand	1
en	bidentate ligand	2
ox2-	bidentate ligand	2
pyridine	monodentate ligand	1

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)

Number of X type atoms in the unit cell \(=8 \times \frac{1}{8}=1\)
Number of Y type atoms in the unit cell \(=1 \times \frac{1}{1}=1\)
Hence the formula is XY (or) X₁Y₁

(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'.

i) Silica is an acidic flux which is mixed with Copper pyrites ore and is heated in a reverberatory furnace. The ferrous oxide formed from the ore is basic in nature and it combines with silica to form ferrous silicate. It is removed as slag.
\(\mathrm{FeO}_{(s)}+\mathrm{SiO}_{2(s)} \overset { \Delta }{ \longrightarrow } \mathrm{FeSiO}_{3(s)}\\ gangue \quad flux \quad \quad slag\)
ii) Aluminium oxide does not dissolve in water, but it dissolves in molten cryolite. This is an aluminium compound with a lower melting point than aluminium oxide. The use of cryolite reduces some of the energy costs involved in extracting aluminium.
\(\mathrm{Al}_{2} \mathrm{O}_{3} \overset { \Delta }{ \longrightarrow } 2 \mathrm{Al}^{3+}+3 \mathrm{O}^{2-}\)
iii) Van-Arkel method is based on the thermal decomposition of metal compounds which lead to the formation of pure metals. Zirconium can be purified by using this method. Zirconium metal is heated with iodine at a temperature of 550 K to form zirconium tetraiodide. (ZrI₄). The impurities are left behind
\(\mathrm{Zr}_{(\mathrm{s})}+2 \mathrm{I}_{2(\mathrm{~s})} \frac{550 \mathrm{~K}}{\mathrm{D}} \mathrm{ZrI}_{\text {t(vapour })}\)
The volatile (ZrI₄) on further heating around 1800 K, it gives pure zirconium.
\(\mathrm{ZrI}_{4 \text { (vapour) }} \frac{1800 \mathrm{~K}}{\mathrm{D}} \mathrm{Zr}_{(\mathrm{s})}+2 \mathrm{I}_{2} \)    
iv) In the extraction of lead from galena (PbS) by froth flotation process, if ZnS is present as impurity, Sodium cyanide is added to depresses the flotation property of ZnS. It forms a complex. Na₂[Zn(CN)₄] on the surface of zinc sulphide.

(i) Concentration of the ore
(ii) Extraction of crude metal
(iii) Refining of crude metal

(i) Compound A giving violet colour with neutral ferric chloride is phenol (C₆H₅OH).
(ii) (B) and (C) isomers are formed when alkaline solution of (A) is refluxed with CCI₄.

(iii) (A) when treated with C₆H₅N₂CI, forms a red orange dye (D).

Compound	Compound Name	Formula
A	Phenol	C6H5OH
B	o-Hydroxy benzoic acid
C	p-Hydroxy benzoic acid
D	p-Hydroxy azo benzene

(i) Compound (A) with molecular formula C₆H₆O gives violet colour with neutral FeCI₃. 
(ii) (A) reacts with CHCl₃ and NaOH gives two isomers (B) and (C) with molecular formula C₇H₆O₂.

(iii) Compound (A) reacts with ammonia at 473 K in the presence of ZnCl₂ gives compound (D) with molecular formula C₆H₇N.

(iv) Compound (D) undergoes carbylamine test.
\(\underset { (D) }{ { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 } } +CHCl_{ 3 }+3KOH\overset { \triangle }{ \longrightarrow } { C }_{ 6 }{ H }_{ 5 }NC+3KCl+3{ H }_{ 2 }O\)

Compound	Compound Name	Formula
A	Phenol	C6H5OH
B	Orthohydroxy benzaldehyde
C	Para hydroxy benzaldehyde
D	Aniline	C6H5NH2

(i) When alkaline solution of (A) is refluxed with chloroform.isomers (B) and (C) are obtained

(ii) (B) on oxidation gives acid (D),

(iii) The acid (D) is also obtained by treating sodium salt of (A) with CO₂ under pressure followed by hydrolysis.

Compound	Compound Name	Formula
A	Phenol	C6H5OH
B	Ortho hydroxy benzaldehyde
C	Para hydroxy benzaldehyde
D	Ortha hydroxybenzoic acid

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.

Reaction with metals:
(i) Sulphuric acid reacts with metals and gives different product depending on the reactants and reacting condition
(ii) Dilute sulphuric acid reacts with metals like: tin, aluminium, zinc to give corresponding: sulphates.
Zn + H₂SO₄\(\longrightarrow \) ZnSO₄ + H₂ \(\uparrow \)
2AI + 3H₂SO₄ \(\longrightarrow \) Al₂(SO₄)₃+ 3H₂ \(\uparrow \)
(iii) Hot concentrated sulphuric acid reacts with copper and lead to give the respective sulphates as shown below 
Cu + 2H₂SO₄ \(\longrightarrow \) CuSO₄ + 2H₂O + SO₂\(\uparrow \)
Pb + 2H₂SO₄ \(\longrightarrow \) PbSO₄ + 2H₂O + SO₂\(\uparrow \)
(iv) Sulphuric acid doesn't react with noble metals like gold, silver and platinum.

Silicates are classified into various types based on the way in which the tetrahedral units, [SiO₄]^4- are linked together.
(i) Ortho silicates (Neso silicates):
The simplest silicates which contain discrete [SiO₄]^4- tetrahedral units are called ortho silicates or nesosilicates.
Examples: Phenacite - Be₂SiO₄ (Be²⁺ ions are tetrahedrally surrounded by O^2- ions)
(ii) pyro silicate (or) Soro silicates: Silicates:
Which contain [Si₂O₇]^6- ions are called pyro silicates (or) Soro silicates.
Example: Thortveitite - Sc₂Si₂O₇
(iii) Cyclic silicates (or Ring silicates): 
Silicates which contain (SiO₃)₃^2n- ions which are formed by linking three or more tetrahedral SiO₄^4- units cyclically are called cyclic silicates.
Example: Beryl [Be₃Al₂ (SiO₃)₆] (an aluminosilicate with each aluminium is surrounded by 6 oxygen atoms octahedrally)
(iv) Inosilicates: Silicates which contain 'n':
number of silicate units liked by sharing two or more oxygen atoms are called inosilicates.
Example: They are further classified as chain silicates and double chain silicates.
(v) Chain silicates (or pyroxenes):
These silicates contain [(SiO₃)_n]^2n- ions formed: by linking 'n' number of tetrahedral [SiO₄]^4- units linearly. Each silicate unit shares two of its oxygen atoms with other units.
Example: Spodumene - LiAl(SiO₃)₂·
(vi) Double chain silicates (or amphiboles): 
These silicates contains \(\left[ { Si }_{ 4 }{ O }_{ 11 } \right] _{ n }^{ 6n- }\) ions. In these silicates there are two different types of tetrahedra:
(a) Those sharing 3 vertices
(b) those sharing only 2 vertices.
Example:
Asbestos: These are fibrous and non-combustible silicates.
(vii) Sheet or phyllo silicates: 
Silicates which contain \(({ Si }_{ 2 }{ O }_{ 5 })_{ n }^{ 2n- }\) are called sheet or phyllo silicates. In these, Each [SiO₄]^4- tetrahedron unit shares three oxygen atoms with others and thus by forming two dimensional sheets.
Example: Talc, Mica etc.
(viii) Three dimensional silicates (or tectosilicates): 
Silicates in which all the oxygen atoms of [SiO₄]^4- tetrahedra are shared with other tetrahedra to form three dimensional network are called three dimensional or tectosilicates.
Example: Quartz.

(i) Alumina is separated from silica in a bauxite ore through Baeyer's process, in which bauxite ore is concentrated by the method of leaching or chemical separation.
(ii) Chemical method is employed in case where the ore is to be in a very pure form, e.g., aluminium extraction. Bauxite (Al₂O₃), an ore of aluminium, contains SiO₂ and Fe₂O₃ as impurities. When bauxite ore is treated with NaOH, the Al₂O₃ goes into solution as sodium meta aluminate leaving behind the undissolved impurities [Fe₂O₃, SiO₂, Fe(OH)_3' etc.], which are then filtered off.
\({ Al }_{ 2 }{ O }_{ 3 }+{ 2NaOH }\longrightarrow \underset { Sod.meta.aliminate\\ (In \ solution \ form) }{ { 2NaAlO }_{ 2 }+{ H }_{ 2 }O } \)
(iii) The filtrate (containing sodium meta aluminate) on dilution, and stirring gives a precipitate of aluminium hydroxide, which is filtered, and ignited to get pure alumina.
\({ NaAlO }_{ 2 }+2{ H }_{ 2 }O\longrightarrow \underset { Ppt }{ { Al(OH) }_{ 3 } } +NaOH\)
\(2Al\left( OH \right) _{ 3 }\overset { \Delta }{ \longrightarrow } \underset { Pure }{ { Al }_{ 2 }{ O }_{ 3 } } +3{ H }_{ 2 }O\)

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

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) Using a colloid mill, the solid is ground to colloidal dimension.
(ii) The colloid mill consists of two metal plates rotating in opposite direction at very high speed of nearly 7000 revolution I minute.
(iii) The colloidal particles of required colloidal size is obtained by adjusting the distance between two plates.
(iv) By this method, colloidal solutions of ink and graphite are prepared.

(i) Covalent Solids: Silicon carbide, graphite.
(ii) Molecular Solids: Urea, benzene, ammonia, water and argon.
(iii) Ionic Solids: Zinc sulphide, potassium sulphate.
(iv) Metallic solids: Rubidium and tin.

(i) CO being a good reducing agent, reduces several metal oxides into crude metal. Hence it is used in extraction of metals.
(ii) CO forms carboxy-hemoglobin complex with hemoglobin of blood which is about 300 times mores stable than oxygen - hemoglobin complex and thus it stops the supply of oxygen and hence, leads to death of the person.
(iii) Solid CO₂, produce cooling and sublimes directly into vapour state, hence used for refrigeration.

\(k=\frac { 2.303 }{ t } \log\frac { \left( { V }_{ \infty }-{ V }_{ 0 } \right) }{ \left( { V }_{ \infty }-{ V }_{ 0 } \right) } \)
\(k=\frac { 2.303 }{ 20 } \log\frac { 50.4-20.2 }{ 50.4-25.6 } \)
\(=0.1151 \log\frac { 30.2 }{ 24.8 } \)
= 0.1151 log 1.2479
= 0.1151 x 0.0959
= 11.03 x10^-3 min^-1
When t = 40 mts
\(k=\frac { 2.303 }{ 40 } \log\frac { 50.4-20.2 }{ 50.4-25.6 } \)
\(=0.0576\times \log\frac { 30.2 }{ 20.9 } \)
= 9.19 x 10^-3 min^-1
When t = 60 mts
\(k=\frac { 2.303 }{ 60 } \log\frac { 50.4-20.2 }{ 50.4-25.6 } \)
\(=0.03838\times \log\frac { 30.2 }{ 17.6 } \)
= 0.03838 x 0.2343
= 8.99 x 10^-3 min^-1
The constant values of K show that the reaction is of first order.

S.No	Lanthanoids	Actinoids
1.	Differentiating electron enters in 4f orbital	Differentiating electron enters in 5f orbital
2.	Binding energy of 4f orbitals are higher	Binding energy of 5f orbitals are lower
3.	They show less tendency to form complexes	They show greater tendency to form complexes
4.	Most of the lanthanoids are colourless	Most of the actinoids are coloured For Example: U3+ (red) U4+ (green).
5.	They do not form oxo cations	They do form oxo cations such as UO22+, NpO22++ etc.
6.	Besides +3 oxidation states lanthanoids show +2 and +4 oxidation states in few cases	Besides +3 oxidation states actinoids show higher oxidation states such as +4, +5, +6 and +7

(i) In transition series, as we move down from Ti to Zn, the standard reduction potential E⁰_M²⁺/M³ value is approaching towards less negative value and copper has a positive reduction potential, i. e. elemental copper is more stable than Cu²⁺.
(ii) E⁰_M²⁺/M value for manganese and zinc are more negative than regular trend. It is due to extra stability arises due to the half filled d⁵ configuration in Mn²⁺ and completely filled d¹⁰ configuration in Zn²⁺.
(iii) The standard electrode potential for the M³⁺/M²⁺ half cell gives the relative stability between M³⁺ and M²⁺.
(iv) The high reduction potential of Mn³⁺/Mn²⁺ indicates Mn²⁺ is more stable than Mn³⁺.
(v) Mn³⁺ has a 3d⁴ configuration while that of Mn²⁺ is 3d⁵. The extra stability associated with a half filled d sub-shell makes the reduction of Mn³⁺ very feasible \(\left[\mathrm{E}^{\circ}=+1.51 \mathrm{~V}\right]\).

IUPAC defines transition metal as an element whose atom has an incomplete d-sub shell or which can give rise to cations with an incomplete d-sub shell. They occupy the central position of the periodic table, between s and p-block elements.
Examples: Fe, Cu, Ag, Au

(i) Schottky defect arises due to the missing of equal number of cations and anions from the crystal lattice. This effect does not change the stoichiometry of the crystal.
(ii) Ionic solids in which the cation and anion are of almost of similar size show schottky defect.
Example: NaCl.
(iii) Presence of large number of schottky defects in a crystal, lowers its density.
(iv) Presence of Schottky defect in the crystal provides a simple way by which atoms or ions can move within the crystal lattice.

(i) Its small size
(ii) Its high electronegativity
(iii) Its high ionisation energy
(iv) Non-availability of d-orbital in the valence shell.
(v) Rather inert
(vi) High bond energy

(i) Sulphuric acid is used in the manufacture of fertilisers, ammonium sulphate and super phosphates and other chemicals such as HCl, HNO₃.
(ii) It is used as a drying agent and also used in the preparation of pigments, explosives etc.

Generally on descending a group the ionisation energy decreases and hence the metallic character increases.

1. Reduction of oxides of active metals such as sodium, potassium etc., by carbon is thermodynamically not feasible.
2. Such metals are extracted from their ores by using electrochemical methods.
3. In this technique, the metal salts are taken in a fused form or in solution form.
4. The metal ion present can be reduced by some suitable reducing agent or by electrolysis.
5. Gibbs free energy for the electrolysis process is given by
Here,
 \(\Delta\)G^o = -nFE^o
 \(\Delta\)G^o - Standard Gibb's free energy change
n - number of electrons involved,
F - Faraday,
E^o - Standard electrode potential
6. If E^o is positive then the \(\Delta\)G^o is negative and the reduction is spontaneous
7. Hence a redox reaction is planned in such a way that the e.m.f of the net redox reaction is positive.
8. When a more reactive metal is added to less reactive metal salt solution, the more reactive metal will go into the solution.
Example:
\({ Cu }_{ (s) }+2{ Ag }^{ + }_{ (aq) }\longrightarrow { Cu }^{ 2+ }_{ (aq) }+2{ Ag }_{ (s) }\)
\({ Cu }^{ 2+ }_{ (aq) }+{ Zn }_{ (s) }\longrightarrow { Cu }_{ \left( s \right) } +{ Zn }^{ 2+ }_{ (aq) }\)