Structure of amines:
(i) Like, ammonia, nitrogen atom of amines is trivalent and carries a lone pair of electron and sp³ hybridised, out of the four sp³ hybridised orbitals of nitrogen, three sp³ orbitals overlap with orbitals of hydrogen (or) alkyl groups of carbon, the fourth sp³ orbital contains a lone pair of electron.
(ii) Hence, amines possess pyramidal geometry. Due to presence of lone pair of electron C-N-H (or) C-N-C bond angle is less than the normal tetrahedral bond angle 109.5°.
For example, the C-N-C bond angle of trimethylamine is 108° which is lower than tetrahedral angle and higher than the H-N-H bond angle of 107°.
(iii) This increase is due to the repulsion between the bulky methyl groups.

(i) The nitrogen in amines possess an unshared pair of electrons (Lone Pair).
(ii) The lone pair of electrons is available for the formation of a new bond with a proton or lewis acid.
(iii) Thus amines are basic in nature and they react with acids to form salts.
(iv) The greater is the number of electron releasing alkyl. groups, the greater the availability of nitrogen's lone pair and stronger the base.
NH₃ < CH₃ - NH₂ < CH₃ - NH - CH₃

Benzaldehyde condenses with tertiary aromatic amines like N, N - dimethyl aniline in the presence of strong acids to from triphenyl methane dye or malachite green dye.

When alkylcyanides are reduced using SnCl₂ HCl, imines are formed, which on hydrolysis gives corresponding aldehyde.
\({ CH }_{ 3 }-C\equiv N\overset { { SnCl }_{ 2 }/HCl }{ \longrightarrow } { CH }_{ 3 }-CH={ NH }\overset { { H }_{ 3 }{ O }^{ + } }{ \longrightarrow } { CH }_{ 3 }-CHO\)

(i) Antioxidants are substances which retard the oxidative deteriorations of food.
(ii) Food containing fats and oils is easily oxidised and turn rancid.
(iii) To prevent the oxidation of the fats and oils, chemical BHT(butylhydroxy toluene), BHA(Butylated hydroxy anisole) are added as food additives.
(iv) They are generally called antioxidants. These materials readily undergo oxidation by reacting with free radicals generated by the oxidation of oils, thereby stop the chain reaction of oxidation of food.

(i) Carbohydrates are polyhydroxy aldehydes or polyhydroxy ketones.
(ii) They are naturally occurring organic substances.
(iii) Carbohydrates are formed in the plants by photosynthesis from water in sunlight.
(iv) They are classified into two types
(a) Sugars Eg: Glucose, Fructose.
(b) Non-sugars Eg: Starch, Cellulose.

Glucose reacts with HCN to form cyanohydrin.

When ethers are heated to dryness,
(i) They form peroxide by the action of air or oxygen.
(ii) Ether oxygen is capable of forming a co-ordinate covalent bond with electron deficient species.

(ii) These peroxides are unstable and decomposes violently with explosion on heating. Hence, ether should not be heated to dryness.

(i) Phenol are only sparingly soluble in water. Actually they form only negligible hydrogen bonding with water since the size of the phenyl group is large and it almost masks the polar character of -OH group. So, phenol has a smaller dipole moment.

(ii) But methanol is soluble in soluble in water due to the formation of hydrogen bonding with water and it results in higher dipole moment. In henol, there is +ve charge on oxygen, therefore it does not undergo protonation easily.

(i) A negative ion causes the precipitation of positively charged sol and vice versa.
(ii) When the valency of ion is high, the precipitation power is increased. For example, the precipitation power of some cations and anions varies in the following order.
\({ Al }_{ 3 }+>Ba^{ 2+ }+>Na^{ + },Similarly[Fe(CN)_{ 6 }]^{ 3- }>SO_{ 4 }^{ 2- }>{ Cl }^{ - }\)
(iii) The precipitation power of electrolyte is determined by finding the minimum concentration (millimoles/lit) required to cause precipitation of a sol in 2 hours.
(iv) This value is called flocculation value. The smaller the flocculation value greater will be precipitation

(i) When it is placed on the right-hand side of the zinc electrode, the hydrogen electrode reaction is,
2H⁺+ 2e^- ➝ H²
The electrons flow to the SHE and it acts as the cathode.
(ii) When the SHE is placed on the left hand side, the electrode reaction is
H₂ ➝ 2H⁺+ 2e^- 
The electrons flow to the copper electrode and the hydrogen electrode acts as the anode.

(i) Oxidation occurs at anode:

(ii) Reduction occurs at cathode:

(iii) Overall reaction:
\({ Zn }_{ (s) }+{ HgO }_{ (s) }\rightarrow { ZnO }_{ (s) }+Hg(l)\)
(iv) Cell emf: about 1.35V.

Specific conductance k = \(C(\frac{l}{a})\)
k = \(\frac{1}{R}(\frac{l}{a})\)
That is k α \(\frac{1}{R}\)

Anisole undergoes Friedel Craft's reaction in presence of anhydrous AlCl₃ as a catalyst.

NH₄CI is a salt of a strong acid HCI and weak base NH₄OH.
\({ HCl }_{ (aq) }+{ NH }_{ 4 }OH_{(aq)}\rightleftharpoons { { NH }_{ 4 }Cl }_{ (aq) }+{ H }_{ 2 }O(I)\)
\({ NH }_{ 4 }^{ + }\) is a strong conjugate acid of the weak base NH₄OH and it has a tendency to react with OH^- from water to produce unionised NH₄OH shown below.
\({ NH }_{ 4 }^{ + }+{ H }_{ 2 }O(1)\rightleftharpoons { NH }_{ 4 }{ { { OH }_{ (aq) }+ }H }_{ (aq) }^{ + }\)
There is no such tendency shown by Cl^- and therefore [H⁺] > [OH^-]; the solution is acidic and the pH is less than 7.
As discussed in the salt hydrolysis of strong base and weak acid. In this case also, we can establish a relationship between the K_h and K_b as 
K_h·K_b = K_w
Let us calculate the K_h value in terms of degree of hydrolysis (h) and the concentration of salt
K_b = h²C and \([{ H }^{ + }]=\sqrt { { K }_{ h }.C } \)
= \({ [H }^{ + }]=\sqrt { \frac { { K }_{ w } }{ { K }_{ b } } .C } \)
pH = - log [H⁺]
\(={ \left( \frac { { K }_{ w }.C }{ { K }_{ b } } \right) }^{ \frac { 1 }{ 2 } }\)
= \(-\frac { 1 }{ 2 } \log { K }_{ w }-\frac { 1 }{ 2 } \log C+\frac { 1 }{ 2 } \log{ K }_{ b }\)
\(pH=7-\frac { 1 }{ 2 } p{ K }_{ b }-\frac { 1 }{ 2 } \log C\)

Lr Z = 103, is the last element of actinoid series. Its electronic configuration is [Rn]⁸⁶ 5f¹⁴6d¹7S² the possible oxidation state shown by it is +3.

Potassium permanganate is prepared from pyrolusite (MnO₂) ore. The preparation involves the following steps
(i) Conversion of MnO₂ to potassium manganate: Powdered ore is fused with KOH in the presence of air or oxidising agents like KNO₃ or KCIO₃. A green coloured potassium manganate is formed.

(ii) Oxidation of potassium manganate to potassium permanganate: Potassium manganate thus obtained can be oxidised in two ways, either by chemical oxidation or electrolytic oxidation.

It is know that,
\(t=\frac { 2.303 }{ k } \log { \frac { { \left[ R \right] }_{ 0 } }{ \left[ R \right] } } \)
= \(\frac { 2.303 }{ 60{ s }^{ -1 } } \log\frac { 1 }{ \frac { 1 }{ 16 } } \)
= \(\frac { 2.303 }{ 60{ s }^{ -1 } } \log16\)
= 4.6 x 10^-2 s (approximately)
Hence the required time is 4.6 x 10^-2 s

Formula:
\(\log { \frac { { k }_{ 2 } }{ { k }_{ 1 } } } =\frac { { E }_{ a }\left( { T }_{ 2 }-{ T }_{ 1 } \right) }{ 2.303R{ T }_{ 1 }{ T }_{ 2 } } \)
Given:
Energy of activation: E = 150 kJ = 150000 J
Temperatures: T₁ = 673 K; T₂ = 773 K
Rate constant: k₁ = 2.3 x 10^-3
Gas constant: R = 8.314 J k^-1 mol^-1
Solution:
\(\log { \frac { { k }_{ 2 } }{ { k }_{ 1 } } } =\frac { { k }_{ 2 } }{ 2.3\times { 10 }^{ -3 } } =\frac { 150000(773-673) }{ 2.303\times 8.314\times 673\times 773 } \)
\(=\frac { 15000000 }{ 2.303\times 8.314\times 673\times 773 } \)
\(\log { \frac { { k }_{ 2 } }{ 2.3\times { 10 }^{ -3 } } } \) = 1.505
\(\frac { { k }_{ 2 } }{ 2.3\times { 10 }^{ -3 } } \) Antilog 1.5905 = 32
∴ k₂ = 2.3 x 10^-3 x 32 = 7.36 x 10^-2
k₂ = 7.36 x 10^-2

(i) The highest oxidation state in oxides and fluorides is due to small size and high electro negativity of F and O.
(ii) Many Cu⁺ compounds are unstable in aqueous solution and undergo disproportionation.
2Cu⁺ ⟶ Cu²⁺+ Cu
This suggest that in aqueous solution Cu+_(aq), converts into Cu²⁺_(aq) which is due to much more negative Δ_hyd H^- of Cu²⁺_(aq) than Cu⁺, which compensates more for the second ionisation enthalpy of Cu.

(l) Cl₂+ H₂O\(\longrightarrow \) HCI + HOCI
(ii) XeF₆ + 2H₂O \(\longrightarrow \) XeO₃ + 6HF
(Iii) XeF₆ + 2H₂O\(\longrightarrow \) XeO₂F₂+ 4HF

Structure of ammonia
(i) Ammonia molecule is pyramidal in shape N-H bond distance is 1.016Å and H-H bond, distance is 1.645Å with a bond angle 107°.

(ii) The structure of ammonia may be regarded as a tetrahedral with one lone pair of electrons in one tetrahedral position hence it has a pyramidal shape.

crystal lattice and unit cell:

Simple Cubic Lattice
100,101,111

\({ d }_{ hkl }=\frac { a }{ \sqrt { { h }^{ 2 }+{ k }^{ 2 }+{ l }^{ 2 } } } \)
\({ d }_{ 100 }=\frac { 1 }{ \sqrt { { 1 }^{ 2 }+0^{ 2 }+0^{ 2 } } } =1\)
\({ d }_{ 101 }=\frac { 1 }{ \sqrt { { 1 }^{ 2 }+0^{ 2 }+0^{ 2 } } } =\frac { 1 }{ \sqrt { 2 } } \)
\({ d }_{ 111 }=\frac { 1 }{ \sqrt { { 1 }^{ 2 }+1^{ 2 }+1^{ 2 } } } =\frac { 1 }{ \sqrt { 3 } } \)
\(\\ { d }_{ 100 }:{ d }_{ 101 }:{ d }_{ 111 }=1:\frac { 1 }{ \sqrt { 2 } } :\frac { 1 }{ \sqrt { 3 } } (or)\)
=1:0.707:0.577

\(M=\frac { d\times { a }^{ 3 }\times NA }{ g } \)
d = Density of the material
a = Length of the edge of the cell.
N_A = Avogadro number
Z = No. of atoms
\(M=\frac { 10.5{ gcm }^{ -3 }{ (4.07\times { 10 }^{ -6 }cm) }^{ 3 }\times \left( 6.023\times { 10 }^{ 23 }{ mol }^{ -1 } \right) }{ 4 } \)
Atomic mass of silver M = 107.08 g mol^-1.

(i) Equimolar mixture of hydrogen and carbon monoxide - water gas and the mixture of carbon monoxide and nitrogen - producer gas are important industrial fuels.
(ii) Carbon monoxide is a good reducing agent and can reduce many metal oxides to metals.
(iii) Carbon monoxide is an important ligand and forms carbonyl compound with transition metals.

(i) Boron trifluoride is obtained by the treatment of calcium fluoride with boron trioxide in presence of conc. sulphuric acid.
B₂O₃ + 3CaF₂ + 3H₂SO₄ \(\overset { \triangle }{ \longrightarrow } \) 2BH₃ + 3CaSO₄ + 3H₂O
(ii) It can also be obtained by treating boron trioxide with carbon and fluorine.
B₂O₃ + 3C + 3F₂ ⟶ 2BF₃ + 3CO
(iii) In the laboratory pure BF₃ is prepared by the thermal decomposition of benzene diazonium tetrafluoro borate.
PhN₂BF₄ \(\overset { \triangle }{ \longrightarrow } \) BF₃ + PhF + N₂

(i) Copper is used for making coins and ornaments along with gold and other metals.
(ii) Copper and its alloys are used for making wires, water pipes and other electrical parts.

(i) In this method, the ore is treated with aqueous alkali to form a soluble complex.
(ii) Bauxite, an important ore of aluminum is heated with a solution of sodium hydroxide or sodium carbonate in the temperature range 470 - 520 K at 35 atm to form soluble sodium meta-aluminate leaving behind the impurities, iron oxide and titanium oxide.
\({ Al }_{ 2 }{ O }_{ 3(s) }+2NaO{ H }_{ (aq) }+3{ H }_{ 2 }{ O }_{ (l) }\longrightarrow 2Na[Al({ OH })_{ 4 }]_{ (aq) }\)
(iii) The hot solution is decanted, cooled, and diluted. This solution is neutralised by passing CO₂ gas, to the form hydrated Al₂O₃ precipitate
\(2Na\left[ Al\left( OH \right) _{ 4 } \right] _{ (aq) }+{ CO }_{ 2(g) }\longrightarrow { Al }_{ 2 }{ { O }_{ 3 }.x{ H }_{ 2 }O_{ (s) }+2NaHCO_{ 3(aq) } }\)
(iii) The precipitate is filtered off and heated around 1670 K to get pure alumina Al₂O₃

(i) A red blood corpuscle (RBC) is composed of heme group, which is Fe²⁺- Porphyrin complex. it plays an important role in carrying oxygen from lungs to tissues and carbon dioxide from tissues to lungs.
(ii) Chlorophyll, a green pigment present in green plants and algae, is a coordination complex containing Mg²⁺ as central metal ion surrounded by a modified Porphyrin ligand called corrin ring. It plays an important role in photosynthesis, by which plants converts CO₂ and water into carbohydrates and oxygen.
(iii) Vitamin B₁₂ (cyanocobalamin) is the only vitamin consist of metal ion. it is a coordination complex in which the central, metal ion is Co⁺ surrounded by Porphyrin like ligand.
(iv) Many enzymes are known to be metal complexes, they regulate biological processes. For example, Carboxypeptidase is a protease enzyme that hydrolytic enzyme important in digestion contains a zinc ion coordinated to the protein.