phenol

CH₃ CH (OH) CH₂CH₃
2R-OH+ 2Na ⟶ 2RONa + H₂ ↑
2 moles of alcohol gives 1 mole of H₂ which occupies 22.4 L at 273 K and 1 atm number of moles of alcohol
= 2 moles of R -OH/22.4L of H₂ x 560 ml
= 0.05 moles
No of moles = mass/molar mass = m/M
= 3.7/0.05 = 74 g mol^-1
General formula for (R-OH) Cn H_{2n + 1}OH
n(12) + (2n + 1) (1) + 16 + 1 = 74
14n = 74 - 18
14n = 56
n = 56/14 = 4
The  "2" alcohol which contains 4 carbon is CH₃ - CH (OH) CH₂ - CH₃

Emulsion-Dispersed phase
Dispersion meduun -liquid

(Λ^o_HoAC) = [(Λ^o)_HCl + Λ^o)_NaOAC ] - (Λ^o)_NaCl
= (426.2 + 91) - (126.5)
= 390.7

pH = -log₁₀ [H⁺]
∴  [H⁺] = 10^-pH
Let the volume be x ml.
V₁M₁ +V₂M₂ +V₃M₃ = VM
∴   x ml of 10^-1 M+ x ml of 10^-2 M + x ml of 10^-3 M
= 3 x ml of [H⁺]
\(∴  [H^+] = \frac{ x[0.1 + 0.01 + 0.001]}{3x}\)
\(= \frac{ [0.1 + 0.01 + 0.001]}{3}\)
\(= \frac{ [0.111]}{3}\)
= 0.037
= 3.7 x 10^-2

t_1/2 α \(\frac{1}{[A_{0}]^{n-1}}\)...(1)
If [A₀ = 2[A₀]; then t_1/2 = 2t_1/2
2t_1/2 α \(\frac{1}{[2A_{0}]^{n-1}}\)...(2)
(2)/(1) = \(2= \frac{1}{[2A_{0}]^{n-1}} \times \frac{1}{[A_{0}]^{n-1}}\)
\(2= \frac {[2A_{0}]^{n-1}} {[A_{0}]^{n-1}}\)
\(2= \frac {1} {2}^{n-1}\)
2 = (2^-1)^n-1
2¹ = (2^-n+1)
n = 0

\(k=\frac { 2.303 }{ t } \log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
[A₀] = 100: [A] = 25
\(6.909=\frac { 2.303 }{ t } \log\frac { \left[ {100 } \right] }{ \left[ 25\right] } \)
\(t =\frac { 2.303 }{ 6.909 } \log(4)\)
\(t =\frac { 1 }{ 3 } \log(2^2)\)
\(= \left( \frac { 2 }{ 3 } \right) \log2\)

Lattice points are occupied by CO₂ molecules

Option (a) and (b) -geometrical isomerism is possible
Option (c) - ionization isomerism is possible
Option (d) - no possibility to show either constitutional isomerism or stereo isomerism

In the complex [M(en)₂(Ox)]Cl For the central metal ion M3+
The primary valence is = +3
The secondary valence = 6
sum of primary valence and secondary valence = 3 + 6 = 9

(i) 

(ii) 

It is used in the manufacture of polymeric resin called bakelite, a thermo setting plastic.

Finely divided form of a catalyst has greater surface area when compared to their crystalline form. Greater the surface area, greater the adsorption and higher the catalytical activity.

There are two types of emulsions.
(i) Oil in gwater (O/W)
(ii) Water in oli (W/O)

(i) Leclanche cell
(ii) Mercury button cell

Conductivity, molar conductance and equivalent conductance increases with dilution whereas Conductance (C) decreases.

FeCl₃ when dissolved in water gives strong acid HCl and a weak base Fe(OH)₃ A strong acid ionises completely so concentration of H⁺ ion becomes much greater than OH^- ion. So, aqueous solution of FeCl₃ is acidic.

With increase in temperature the ionic product of water will increase since the concentration of H₃O⁺ and OH^- ions will increase.

Molar conductance = Λ_{m =} \( \frac{k\ (Sm^{-1})\times10^{-3}}{M} mol^{-1}m^{3} \)
\(= \frac{(12.04 \times 10^{-2} Sm^{-1}) \times 10^{-3} (mol^{-1}m^{3})} {0.025}\)
= 481.6 x 10^-5 Sm²mol^-1

By addition of suitable electrolytes precipitated particles can be brought into colloidal state. The process is termed as peptisation and the electrolyte added is called peptising or dispersing agent.
\(\underset { Precipitate }{ AgCl } \overset { HCl }{ \longrightarrow } \underset { Colloid }{ AgCl } \)

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

[H₃O⁺] = 4 \(\times\) 10^-5M
pH = - log₁₀[H₃O⁺]
pH=-log₁₀[4 \(\times\) 10^-5]
pH = -log₁₀[4] - log₁₀[10^-5]        log₁₀ 10 = 1
pH = -log 4 + 5log₁₀10
= 5 - log 4
= 5 - 0.6021
=4.3979
Since pH is less than 7, the solution is acidic.

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

Rate = K[X]ⁿ[y]^m
4 x 10^-3 mol L^-1s^-1= 2 x 10^-2s^-1(0.2mol L^-1)ⁿ(0.2mol L^-1)^m
\(\frac { 4\times { 10 }^{ -3 }mol\quad { L }^{ -1 }{ s }^{ -1 } }{ 2\times { 10 }^{ -2 }{ s }^{ -1 } } =\) (0.2)^n+m(mol L^-1)^n+m
0.2(mol L^-1) = (0.2)^n+m(mol L^-1)^n+m
Comparing the powers on both sides
The overall order of the reaction n + m = 1

A mixture of TiCl₄ and trialkyl aluminium is used for polymerization.

(i) I Order reaction
(ii) S^-1
(iii) \({ t }_{ \frac { 1 }{ 2 } }\) = 0.693/k

Rate decreases with time as the reaction proceeds and the average rate cannot be used to predict the rate of the reaction at any instant. The rate of the reaction, at a particular instant during the reaction is called the instantaneous rate. So instantaneous rate in prepared over average rate.

1 Mole of hexagonal packed structure contains 1 mole of octahedral voids and two moles of tetrahedral voids. Therefore, 0.5 moles of hexagonal packed structure contains 0.5 moles of octahedral voids and 1 mole of tetrahedral voids.
No. of tetrahedral voids = 6.022 x 10²³
No. of octahedral voids\(\frac{1}{2}\) x no. of tetrahedral voids
No. of octahedral voids = \(\frac{1}{2}\) x 6.022 x 10²³
= 3.011 x 10²³
Total No. of voids = (6.022+3.011) x 10²³
= 9.033 x 10²³.

(i) H₂O undergoes extensive H-bonding due to high electronegativity of O-atom and hence exist as an associated molecule.
(ii) Therefore H₂O is a liquid. On the other hand H₂O does not undergo H-bonding and hence exists as a discrete molecules and as a gas.

Ammonia is prepared in the laboratory by heating an ammonium salt with a base.
\({ 2NH }_{ 4 }^{ + }+{ OH }^{ - }\longrightarrow { 2NH }_{ 3 }+{ H }_{ 2 }O\)
\({ 2NH }_{ 4 }Cl+CaO\longrightarrow { CaCl }_{ 2 }+{ 2NH }_{ 3 }+{ H }_{ 2 }O\)

A unit cell that contains only one lattice point is called a primitive unit cell, which is made up from the lattice points at each of the corners.

(i) Borax - Na₂[B₄O₅.(OH)₄].8H₂O
(ii) Kernite - Na₂[B₄O₅(OH)₄]·2H₂O

Tetraethoxy silane is Si(OC₂H₅)₄·The chloride ion in silicon tetrachloride can be substituted by nucleophile such as OH, OR, etc .. using suitable reagents. For example, it forms silicic esters with alcohols.
\(Si{ Cl }_{ 4 }+{ C }_{ 2 }{ H }_{ 5 }OH\longrightarrow \underset { Tetraethoxy\ silane }{ Si({ OC }_{ 2 }{ H }_{ 5 })_{ 4 } } +4HCl\).

5f^1-14 6d^0-1 7s²

Metallurgy is the process of extraction of metal in the purest form from its ores.

(i) The complex ion of the coordination compound containing the central metal atom/ion and the ligands attached to it is collectively called coordination sphere and are usually enclosed in square brackets with the net charge.
(ii) The other ionisable ions, are written outside the bracket are called counter ions. For example, the coordination compound K₄[Fe(CN)₆] contains the complex ion [Fe(CN)₆]^4- and is referred as the coordination sphere.
(iii) The other associated ion K⁺ is called the counter ion.

(i) Coordination entity is an ion or a neutral molecule, composed of a central atom, usually a metal and the array of other atoms or groups of atoms (ligands) that are attached to it. In the formula, the coordination entity is enclosed in square brackets.
(ii) For example, in potassium ferrocyanide, K₄[Fe(CN)₆], the coordination entity is [Fe(CN)₆]^4-.

(i) The end product of both the processes is oxide of metal.
(ii) Volatile impurities are removed from the ore and surface area for the further reaction increases.

An interstitial compound or alloy is a compound that is formed when small atoms like hydrogen, boron, carbon or nitrogen are trapped in the interstitial holes in a metal lattice. They are usually non-stoichiometric compounds. Transition metals form a number of interstitial compounds such as TiC, ZrH_1.92, Mn₄N etc.
Properties of interstitial compound
(i) They are hard and show electrical and thermal conductivity.
(ii) They have high melting points higher than those of pure metals.
(iii) Transition metal hydrides are used as powerful reducing agents 
(iv) Metallic carbides are chemically inert.

(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.

Medicinal uses
(i) Ca-EDTA chelate, is used in the treatment of lead and radioactive poisoning.
(ii) That is for removing lead and radioactive metal ions from the body.
(iii) Cis-platin is used as an antitumor drug in cancer treatment.
Biological Importance compounds:-
(i) Fe²⁺ - porphyrin complex: It plays an important role in carrying oxygen from lungs to tissues and tissue carbon dioxide from to lungs.
(ii) Chlorophyll - useful in photosynthesis in plants.

a) potassiumhexacyanidoferrate(II) - Potassiumhexacyanidoferrate(II) - K₄[Fe(CN)₆]
b) Pentacarbonyliron(0) - [Fe(CO)₅]
c) Pentaamminenitrito −kNcobalt(III)ion - [Co(NH₃)₅(NO₂)]²⁺
d) Hexaamminecobalt(III) Sulphate  - [CO(NH₃)₆](SO₄)₃
e) Sodiumtetrafluoridodihydroxidochromate(III)  - Na₂[CrF₄(OH)₂]

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

(i) Icosogens → B, Al, Ga, In, Tl
(ii) Tetragens → C, Si, Ge, Sn, Pb
(iii) Pnictogens → N, P, As, Sb, Bi
(iv) Chalcogens → O, S, Se, Te, Po

In p-block elements, the first member of each group differs from the other elements of the corresponding group. The following factors are responsible for this anomalous behaviour
(i) Small size of the first member.
(ii) High ionisation enthalpy and high electronegativity.
(iii) Absence of d-orbitals in their valance shell.

This method is commonly used to concentrate sulphide ores such as galena (PbS), zinc blende (ZnS) etc.
In this method, the metallic ore particles which are preferentially wetted by oil can be separated from gangue.

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

(i) Tollens Reagent Test: Tollens reagent is an ammoniacal silver nitrate solution. When an aldehyde is warmed with Tollens reagent a bright silver mirror is produced due to the formation of silver metal. This reaction is also called silver mirror test for aldehydes.
\({ CH }_{ 3 }CHO+2\left[ Ag\left( { NH }_{ 3 } \right) _{ 2 } \right] ^{ + }+3{ OH }^{ - }\longrightarrow { CH }_{ 3 }CO{ O }^{ - }+\underset { Silver\ mirror }{ 4{ NH }_{ 3 }+2Ag+2{ H }_{ 2 }O } \) 
(ii) Fehlings solution Test:
(a) Fehling's solution is prepared by mixing equal volumes of Fehlings solution A containing aqueous copper sulphate and Fehlings solution 'B' containing alkaline solution of sodium potassium tartarate (Rochelle salt).
(b) When aldehyde is warmed with Fehlings solution deep blue colour solution is changed to red precipitate of cuprous oxide.
\(\\ \\ { CH }_{ 3 }CHO+\underset { blue }{ 2{ Cu }^{ 2 } } +5{ OH }^{ - }\longrightarrow { CH }_{ 3 }{ COO }^{ - }+\underset { red }{ { Cu }_{ 2 }O\downarrow } +3{ H }_{ 2 }O\)
(iii) Schiffs' reagent Test: Dilute solution of aldehydes when added to Schiff's reagent (Rosaniline hydrochloride dissolved in water and its red colour decolourised by passing SO₂) yields its red colour. This is known as Schiff's test for aldehydes. Ketones do not give this test. Acetone however gives a positive test but slowly.

Inter molecular dehydration of alcohol: We have already learnt that when ethanol is treated with con.HSO₂ 4 at 443K, elimination takes place to form ethene. If the same reaction is carried out at 413K, substitution competes over elimination to form ethers.

(i) For a chemical reaction, catalyst is needed in very small quantity.
(ii) There may be some physical changes, but the catalyst remains unchanged in mass and chemical composition in a chemical reaction.
(iii) A catalyst itself cannot initiate a reaction. 
(iv) A solid catalyst will be more effective if it is taken in a finely divided form.
(v) A catalyst are specific in nature.
(vi) In an equilibrium reaction, presence of catalyst reduces the time for attainment of equilibrium and hence it does not affect the position of equilibrium and the value of equilibrium constant.
(vii) A catalyst is highly effective at a particular temperature called as optimum temperature.
(viii) Presence of a catalyst generally does not change the nature of products

Given: Initial concentration of the solution = 1 M
The concentration remaining after electrolysis = 0.667 M
Solution:
∴ The amount deposited = 1 - 0.667 M
= 0.333 M
1F = Faraday = 3 x 0.333 M
= 0.999M
= 1M
∴1 Faraday current is used.

(i) The concentration of hydronium ion in an acidic buffer solution depends on the ratio of the concentration of the weak acid to the concentration of its conjugate base present in the solution i.e.,
\(\left[ { H }_{ 3 }{ O }^{ + } \right] ={ K }_{ a }\frac { [{ acid] }_{ aq } }{ [{ base] }_{ aq } } \)
(ii) The weak acid is dissociated only to a small extent. Moreover, due to common ion effect, the dissociation is further suppressed and hence the equilibrium concentration of the acid is nearly equal to the initial concentration of the unionised acid. Similarly, the concentration of the conjugate base is nearly equal to the initial concentration of the added salt.
\(\left[ { H }_{ 3 }{ O }^{ + } \right] ={ K }_{ a }\frac { [{ acid] } }{ [{ salt] } } \)
(iii) Here [acid] and [salt] represent the initial concentration of the acid and salt, respectively used to prepare the buffer solution
Taking logarithm on both sides of the equation
\(\log\left[ { H }_{ 3 }{ O }^{ + } \right] ={ \log K }_{ a }+\log\frac { [{ acid] } }{ [{ salt] } } \)
reverse the sign on both sides
- \(\log\left[ { H }_{ 3 }{ O }^{ + } \right] ={- \log K }_{ a }-\log\frac { [{ acid] } }{ [{ salt] } } \)
We know that
pH = -log [H₃O⁺] and pK_a = -log K_a
\(\Rightarrow pH={ pK }_{ a }-\log\frac { [{ acid] } }{ [{ salt] } } \)
\(​​\Rightarrow pH={ pK }_{ a }+\log\frac { [{ acid] } }{ [{ salt] } } \)
Similarly for a basic buffer,
pOH = \({ pK }_{ a }+\log\frac { [{ acid] } }{ [{ salt] } } \)

(i) It is generally expected a steady decrease in atomic radius along a period as the nuclear charge increases and the extra electrons are added to the same sub shell.
(ii) But for the 3d transition elements, the expected decrease in atomic radius is observed from Sc to V, thereafter up to Cu the atomic radius nearly remains the same.
(iii) As we move from Sc to Zn in 3d series the extra electrons are added to the 3d orbitals, the added 3d electrons only partially shield the increased nuclear charge and hence the effective nuclear charge increases slightly.
(iv) However, the extra electrons added to the 3d sub shell strongly repel the 4s electrons and these two forces are operated in opposite direction and as they tend to balance each other, it leads to constancy in atomic radii.
(v) At the end of the series, d - orbitals of Zinc contain 10 electrons in which the repulsive interaction between the electrons is more than the effective nuclear charge and hence, the orbitals slightly expand and atomic radius slightly increases.

The initial rate of the reaction is
Rate = k [A] [B] 2
= (2.0 x 10^-6 mol^-2 L² S^-1) (0.1  mol L^-1)²
= 8.0 x 10^-9 mol^-2 L² s^-1
When [A] is reduced from 0.1 mol L^-1 to 0.06 mol^-1, the concentration of A reacted = (0.1 - 0.06) mol L^-1 = 0.04 mol L^-1.
∴ The concentration of B reaction 
= \(\frac { 1 }{ 2 } \) x 0.04 mol L^-1
= 0.02 mol L^-1
Then, concentration of B available [B] = (0.2 - 0.02) mol L^-1
= 0.18 mol L^-1.
After [A] is reduced to 0.06 mol L^-1, the rate of the reaction is given by,
Rate = k [A] [B] 2
= (2.0 x 10^-6 mol^-2 L² s^-1) (0.06 mol L^-1) (0.18 mol L^-1)²
= 3.89 mol L^-1 s^-1.

(i) O₂ molecules are held together by weak Vander Waal's force because of small size and high electronegativity of oxygen.
(ii) In contrast, sulphur shows catenation and forms stronger S-S bonds.
(iii) Due to catenation, sulphur forms octa-atomic S₈ molecules having eight membered puckered ring structure.
(iv) Because of its bigger size the force of attraction holding S₈ molecules are much stronger.
(v) Hence sulphur is a solid at room temperature or in other words, that is why there is a large difference between the boiling point (also melting points) of the two elements.

Molecular solids: 
In molecular solids, the constituents are neutral molecules. They are held together by weak Vander Waals forces. Generally molecular solids are soft and they do not conduct electricity. These molecular solids are further classified into three types.
(i) Non-polar molecular solids:
(a) In non-polar molecular solids constituent molecules are held together by weak dispersion forces or London forces.
(b) They have low melting points and are usually in liquids or gaseous state at room temperature.
Ex: Naphthalene, anthracene etc.,
(ii) Polar molecular solids:
(a) The constituents are molecules formed by polar covalent bonds. 
(b) They are held together by relatively strong dipole-dipole interactions.
(c) They have higher melting points than the nonpolar molecular solids. 
Ex: Solid CO₂, solid NH₃ etc.

(iii) Hydrogen bonded molecular solids:
(a) The constituents are held together by hydrogen bonds.
(b) They are generally soft solids under room temperature.
(c) Examples: solid ice (H₂O), glucose, urea etc.

(i) Carbon monoxide can be prepared by the reaction of carbon with limited amount of oxygen.
2C + O₂ ⟶ 2CO
(ii) (a) On industrial scale carbon monoxide is produced by the reaction of carbon with air.
(b) The carbon monoxide formed will contain nitrogen gas also and the mixture of nitrogen and carbon monoxide is called producer gas.
(c) \(2C+{ O }_{ 2 }/{ N }_{ 2 }(air)\longrightarrow \underset { Producers \ Gas }{ 2CO } +{ N }_{ 2 }\)
(d) The producer gas is then passed through a solution of copper(I) chloride under pressure which results in the formation of CuCI(CO).2H₂O.
(e) At reduced pressures this solution releases the pure carbon monoxide.
(iii) Pure carbon monoxide is prepared by warming methanoic acid with concentrated sulphuric acid which acts as a dehydrating agent.
HCOOH + H₂SO₄ ⟶ CO + H₂O + H₂SO₄

(i) The process in which no external reducing. When ore is heated in air, a part of the ore gets oxidised which combines with remaining sulphide to give metal.
Eg: Reduction copper glance (Cu₂S)
\({ 2Cu }_{ 2 }S+{ 3O }_{ 2 }\longrightarrow { 2Cu }_{ 2 }O+2{ SO }_{ 2 }\uparrow \)
\({ 2Cu }_{ 2 }O+{ Cu }_{ 2 }S\longrightarrow 6Cu+{ SO }_{ 2 }\uparrow \)
(ii) Electrolytic reduction is used for highly electro positive metals. In this method, fused metal ore is electrolysed and pure metal is deposited at cathode.
Eg: Al is obtained by electrolysis of Al₂O_{3
\({ 2Al }_{ 2 }{ O }_{ 3 }+3C\longrightarrow \underset { Cathode\quad Anode }{ 4Al+{ 3CO }_{ 2 } } \)}
(iii) \({ Cr }_{ 2 }{ O }_{ 3 }+2Al\longrightarrow {Al }_{ 2 }{ O }_{ 3 }+2Cr\)
\(3 \mathrm{Mn} _3 O_{4}+8 \mathrm{Al} \longrightarrow 4 \mathrm{Al}_{2} \mathrm{O}_{3}+9 \mathrm{Mn}\)

Valance bond theory helps us to visualize the bonding in complexes. However, it has limitations as mentioned above. Hence Crystal Field Theory to explain some of the properties, like colour, magnetic behavior, etc., This theory I was originally used to explain the nature of bonding in ionic crystals. Later on, it is used to explain the properties of transition metals and their complexes. The salient features of this theory are as follows.
(i) Crystal Field Theory (CFT) assumes that the bond between the ligand and the central metal atom is purely ionic. i.e. the bond is formed due to the electrostatic attraction between the electron rich ligand and the electron deficient metal.
(ii) In the coordination compounds, the central metal atom/ion and the ligands are considered as point charges (in case of I charged metal ions or ligands) or electric dipoles (in case of neutral metal atoms or ligands).
(iii) According to crystal field theory, the complex formation is considered as the following series of hypothetical steps.
Step 1: In an isolated gaseous state, all the five d orbitals of the central metal ion are degenerate. Initially, the ligands form a spherical field of negative charge around the metal. In this filed, the energies of all the five d orbitals will increase due to the repulsion between the electrons of the metal and the ligand.
Step 2: The ligands are approaching the metal atom in actual bond directions. To illustrate this let us consider an octahedral field, in which the I central metal ion is located at the origin and the six ligands are coming from the +x, -x, +y, -y, +z  and -z directions as shown below.

As shown in the figure, the orbitals lying along the axes dx²-y² and dz² orbitals will experience strong repulsion and raise in energy to a greater extent than the orbitals with lobes directed between the axes (d_xy, d_yz, and d_zx). Thus the degenerate d orbitals now split into two sets and the process is called crystal field splitting.
Step 3: Up to this point the complex formation would not be favored. However, when the ligands approach further, there will be an attraction between the negatively charged electron and the positively charged metal ion, that results in a net decrease in energy. This decrease in energy is the driving force for the complex formation.
Crystal field splitting in octahedral complexes: During crystal field splitting in octahedral field, in order to maintain the average energy of the orbitals (barycentre) constant, the energy of the orbitals d_x2-_y2 and d _z2 (represented as e_g orbitals) will increase by 3/5 \({ \triangle }_{ o }\) while that of the other three orbitals d_{xy '} d_yz and d_zx (represented as t_2g orbitals) decrease by 2/5  \({ \triangle }_{ o }\) , Here, \({ \triangle }_{ o }\)   represents the crystal field splitting energy in the octahedral field.

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.

The product obtained is ethanol

(a) 2-methyl-heptan-3-ol (2°)
(b) 2,2-dimethyl-propan-1-01-(1°)
(c) 2-phenyl-2-propanol (3°)
(d) 2,2-dimethyl-S-bromo-3-pentanol (2°)
(e) l-Butene-ol (2°)
(f) Phenyl methanol (1°)
(g) 4-phenyl-1-butanol (1°)
(h) 3-ethyl-3-pentanol (3°)

Physisorption is an exothermic process
\(\underset { (Adsorbent) }{ Solid } +\underset { (Adsorbate) }{ Gas } \rightleftharpoons \underset { (Gas\ adsorbed \ on\ solid) }{ Gas/solid } +Heat\) 
According to Le chatelier's principle, if we increase the temperature, equilibrium will shift in the backward direction, i.e., gas is released from the surface on which it is adsorbed.

Other metals such as aluminium, copper and silver also undergo corrosion, but at a slower rate than iron. Let us consider the reduction of aluminium,
\({ Al }_{ (s) }\rightarrow { Al }_{ (aq) }^{ 3+ }+{ 3e }^{ - }\)
Al³⁺, which reacts with oxygen in air to form a protective coating of Al₂O₃. This coating act as a protective film for the inner surface. So, further corrosion is prevented.

Buffer index β, as a quantitative measure of the buffer capacity. It is defined as the number of gram equivalents of acid or base added to 1litre of the buffer solution to change its pH by unity.
\(\beta =\frac { dB }{ d(pH) } \)
Here,
dB = number of gram equivalents of acid / base added to one litre of buffer solution
d(pH) = The change in the pH after the addition of acid / base.

(i) Chloro benzene:
When Chlorobenzene is hydrolysed with 6-8% NaOH at 300 bar and 633K in a closed vessel, sodium phenoxide is formed which on treatment with dilute HCl gives phenol.

(ii) isopropyl benzene:
A mixture of benzene and propene is heated at 523K in a closed vessel in presence of H₃PO₄ catalyst gives cumene (isopropylbenzene). On passing air to a mixture of cumene and 5% aqueous sodium carbonate solution, cumene hydro peroxide is formed by oxidation. It is treated with dilute acid to get phenol and acetone. Acetone is also an important byproduct in this reaction.

First law:
The mass of the substance (m) liberated at an electrode during electrolysis is directly proportional to the quantity of charge (Q) passed through the cell.
m α Q \(\left[\because \mathrm{I}=\frac{\mathrm{Q}}{\mathrm{t}} \Rightarrow \mathrm{Q}=\mathrm{It}\right]\)
m α It
m = ZIt
Where Z = electro chemical equivalent of the substance
I = current
t = time of passage of current
Second law:
When the same quantity of charge is passed through the solutions of different electrolytes, the amount of substances liberated at the respective electrodes are directly proportional to their electrochemical equivalents.
m α Z
\(\frac{m_{1}}{Z_{1}}=\frac{m_{2}}{Z_{2}}\)
m = mass of the metal deposited
Z = electro chemical equivalent

(i) Let us consider the reactions between a strong acid, HCI, and a weak base, NH₄OH, to produce a salt, NH₄CI, and water.
HCI_(aq) + NH₄OH_(aq) ⇌ NH₄CI_(aq) + H₂O _(I)
NH₄CI_(aq) ⟶ NH₄⁺ +CI^-_(aq)
(ii) NH₄⁺ 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
NH₄⁺ _(aq) + H₂O_(I) ⇌ NH₄OH_(aq) + H⁺_(aq)
(iii) There is no such tendency shown by Cl^- and therefore [H⁺] > [OH^-]; the solution is acidic and the pH is less than 7.
(iv) The K_h and K_b are related by
\(\mathrm{K}_{\mathrm{h}} \cdot \mathrm{K}_{\mathrm{b}}=\mathrm{K}_{\mathrm{w}}\)
(Or)
\(\mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}}}\)
Degree of hydrolysis (h)
\(\underset{(1-h)}{\mathrm{NH}_{4}^{+}(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(j)} \rightleftharpoons \underset{h} {\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})}} +\mathrm{H}^{+} \underset{h}{(\mathrm{aq})}\\ \)
\(\mathrm{K}_{\mathrm{h}} =\frac{\left[\mathrm{NH}_{4} \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{NH}_{4}^{+}\right]} \\ \)
\(=\frac{\mathrm{hc} \times \mathrm{h}}{(1-\mathrm{h}) \mathrm{c}} \)
\(K_{h}=\frac{h^{2} c}{(1-h)} \)
\(\text{If } \mathrm{h}<<1 ; \mathrm{K}_{\mathrm{h}} \simeq \mathrm{h}^{2} \mathrm{c} \)
\(h^{2}=\frac{K_{h}}{c}\)
\(h=\sqrt{\frac{K_{h}}{c}} \ (or) \ h=\sqrt{\frac{K_{w}}{K_{b} \cdot C}} \quad\left(\because K_{h}=\frac{K_{w}}{K_{b}}\right)\)
Also; \(\left[\mathrm{H}^{+}\right]=\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{C}} \) (or) \(\left[\mathrm{H}^{+}\right]=\sqrt{\frac{\mathrm{K}_{\mathrm{w}} \cdot \mathrm{C}}{\mathrm{K}_{\mathrm{b}}}}\)
pH = -log [H⁺]

Let [A₀] = 100;
When t = t_99.9%; [A] = (100-99.9) = 0.1
\(k=\frac { 2.303 }{ t } \log\left( \frac { [{ A }_{ 0 }] }{ [A] } \right) \)
\({ t }_{ 99.9\% }=\frac { 2.303 }{ K } \log\left( \frac { 100 }{ 0.1 } \right) \)
\({ t }_{ 99.9\% }=\frac { 2.303 }{ K } \log1000\)
\({ t }_{ 99.9\% }=\frac { 2.303 }{ K } (3)\)
\({ t }_{ 99.9\% }=\frac { 6.909 }{ K } \)
\({ t }_{ 99.9\% }=10\times \frac { 0.69 }{ K } \)
\({ t }_{ 99.9\% }={ 10 } t_{ 1/2 }\)

(i) The use of 3d electron for formation of I bond increases from Sc to Mn, causing the increase in oxidation state upto +7.
(ii) The reason for Mn having highest oxidation state of +7 is due to the presence of 7 unpaired electrons in its atom.
(iii) As the number of unpaired electrons decrease from Fe to Cu. So there is the decrease in oxidation state.

(i) When the concentration of the reactant is increased by 'n' times, the rate of reaction is also increased by n times. That is, if the concentration of the reactant is doubled, the rate is doubled.
(ii) The unit of rate constant of a first order reaction is sec^-1 or time^-1.
\({ k }_{ 1 }=\frac { rate }{ (a-x) } =\frac { { mol.lit }^{ -1 }{ sec }^{ -1 } }{ { mol.lit }^{ -1 } } \)
(iii) The time required to complete a definite fraction of reaction is independent of the initial concentration, of the reactant if t_1/u is the time of one 'u' th fraction of reaction to take place then from equation.
\({ k }_{ 1 }=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } \)
\(x=\frac { a }{ u } and\quad { t }_{ \frac { 1 }{ u } }=\frac { 2.303 }{ { k }_{ 1 } } \log { \frac { a }{ a-\frac { a }{ u } } } ;\)
\({ t }_{ \frac { 1 }{ u } }=\frac { 2.303 }{ { k }_{ 1 } } \log { \frac { u }{ (u-1) } } \)
since k₁ = rate constant, t_1/u is independent of initial concentration 'a'.

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.

(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) CaC₂
2 + 2x = 0
2x = -2
x = -1
(ii) H₂CO₃
2 + x + (-6) = 0
x = +4
(iii) HCN
1 + x + (-3) = 0
x = +3 - 1 = +2
(iv) CO
x + (-2) = 0
x= +2

(i) into metal using carbon as a reducing agent. In this method the oxide ore of the metal is mixed with coal (coke) and heated strongly in a furnace (usually in a blast furnace).
(ii) This process can be applied to the metals which do not form carbides with carbon at the reduction temperature
\({ ZnO }_{ (s) }+{ C }_{ (s) }\longrightarrow { Zn }_{ (s) }+{ { CO }_{ (g) }\uparrow }\)

(i) The stability of a coordination complex is a measure of its resistance to the replacement of one ligand by another.
(ii) The stability of a complex refers to the degree of association between two species involved in an equilibrium.
(iii) Let us consider the following complex formation reaction
Cu²⁺ +4NH₃ ⇌ [Cu(NH₃)₄]^{2+
\(\beta =\frac { { \left[ { Cu\left( { NH }_{ 3 } \right) }_{ 4 } \right] }^{ 2+ } }{ \left[ { Cu }^{ 2+ } \right] { { \left[ { NH }_{ 3 } \right] }^{ 4 } } } \)}
(iv) So, as the concentration of [Cu (NH₃)₄]²⁺ increases the value of stability complexes also increases.
(v) Therefore the greater the value of stability constant greater is the stability of the complex.

(i) Reaction Rate = \(k{ \left[ A \right] }^{ 2 }{ \left[ B \right] }{ \left[ L \right] }^{ 3/2 }\) ...(1)
When [L] = [4L]
Rate = \(k{ \left[ A \right] }^{ 2 }{ \left[ B \right] }{ \left[ 4L \right] }^{ 3/2 }\)
The Reaction Rate = \(8(k{ \left[ A \right] }^{ 2 }{ \left[ B \right] }{ \left[ L \right] }^{ 3/2 })\) ...(2)
Comparing (1) and (2) rate is increased by 8 times
(ii)  [A] = [2A] and [B] = [2B]
Reaction Rate = \(k{ \left[ 2A \right] }^{ 2 }\left[ 2B \right] { \left[ L \right] }^{ 3/2 }\)
Reaction Rate = \(\\ 8(k{ \left[ A \right] }^{ 2 }{ \left[ B \right] }{ \left[ L \right] }^{ 3/2 })\)  ...(3)
Comparing (1) and (3) rate is increased by 8 times
(iii) \(\left[ A \right] =\left[ \frac { A }{ 2 } \right] \)
Reaction Rate = \(k{ \left[ \frac { A }{ 2 } \right] }^{ 2 }{ \left[ B \right] }{ \left[ L \right] }^{ 3/2 }\)
Reaction Rate = \(\frac { 1 }{ 4 } \left( k\left[ { A }^{ 2 } \right] { \left[ B \right] }{ \left[ L \right] }^{ 3/2 } \right) \) ...(4)
Comparing (1) and (4); rate is reduced to (1/4) times.
(iv) \(\left[ A \right] =\left[ \frac { 1 }{ 3 }A \right] and\left[ L \right] =\left[ 4L \right] \)
Rate = \(k{ \left[ \frac { 1 }{ 3 }A \right] }^{ 2 }{ \left[ B \right] }{ \left[ 4L \right] }^{ 3/2 }\)
Rate = \(\left( \frac { 8 }{ 9 } \right) \left( k{ \left[ A \right] }^{ 2 }{ \left[ B \right] }{ \left[ L \right] }^{ 3/2 } \right)\) ...(5)
Comparing (1) and (5); rate is reduced to \(\frac { 8 }{ 9 } \) times.

(i) The elements in which the extra electron enters (n-2) f orbitals are called f-block elements. These elements are called as inner transition elements because they form a transition. Series within the transition elements.
(ii) The f-block elements are also called as rare earth elements. They are divided into lanthanoid series (4f block elements) and actinoid series (5f block elements).

(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

Each halogen combines with other halogens to form a series of compounds are called interhalogen compounds.
Example: AB type: BrF
AB₃ type: ICI₃