Cyclic alcolhol →  sodium cyclic alkoxide- Williamson ether synthesis.

(CH₃)₂C = C (CH₃)₂

Pyroxylin(nitro cellulose)

dispersion medium-gas
dispersed phase-liquid

α = Λ/Λ^o
= 6/400
K_a = α²C
\(= \frac{6}{400} \times \frac{6}{400} \times \frac{1}{36}\)
= 6.25 x 10^-6

Ca(OH)₂ ⇌ Ca²⁺ + 2OH^-
Given that pH = 9
pOH = 14 - 9 = 5
[pOH = - log₁₀ [OH]]
[OH^-] = 10 ^[pOH]
[OH] = 10^-5 M
Ksp = [Ca²⁺] [OH^-]
= 10^-5/2 x (10^-5)²  = 0.5 x 10^-15

BaSO₄ ⇌ Ba²⁺ + SO₄^2-
K_sp = (s) (s)
K_sp = (s)²
= (2.42 × 10^-3gL^-1)²
\(=\frac{2.42 \times 10^{-3} gL^{-1}}{233 \text{ g mol}^{-1}}\)
= (0.01038 x 10^-3)²
= (1.038 x  x 10^-5)²
= 1.077 x 10^-10
= 1.08 × 10^-10mol²L^-2

In 140days ⇒ initial concentration reduced to (1/2) g
In 280 days ⇒ initial concentration reduced to (1/4) g
In 420 days ⇒ initial concentration reduced to (1/8) g
In 560 days ⇒ initial concentration reduced to (1/16) g

\(3 \sqrt{a} = r_{C_{s+}} + 2r_{C_{f-}} + r_{C_{s+}} \)
\(\left( \frac { \sqrt { 3 } }{ 2 } \right) a = ​​ r_{C_{s+}} + r_{C_{f-}} \)
\(\left( \frac { \sqrt { 3 } }{ 2 } \right) \times 400\)= inter ionic distance

Lattice points are occupied by CO₂ molecules

Three isomers. If we consider any one of the ligands as reference (say Py), the arrangement of other three ligands (NH₃, Br^- and Cl^-) with respect to (Py) gives three geometrical isomers.

(i) 

(ii) 

HCHO + CaCO₃

Powdered charcoal will adsorb more gas because of its larger surface area than a lump of charcoal.

(i) In this method, the colloidal solution is taken in a bag made up of semipermeable membrane.
(ii) It is suspended in a trough of flowing water, the electrolytes diffuse out of the membrane and they are carried away by II water.

The greater the E^o value means greater is the tendency shown by the species to accepts I electrons and undergo reduction. So higher the (E^o) values lesser is the tendency to undergo corrosion.

The emf of the cell depends on the concentration of H₂SO₄ As the cell reaction uses \({ SO }_{ 4 }^{ 2- }\) ions, the concentration H₂SO₄ decreases. When the cell potential falls to about 1.8V; the cell has to be recharged.

This method is useful for the preparation of simple ethers and not suitable for preparing mixed ethers. If a mixture of two different alcohols is used, mixture of different ethers will be formed and they are difficult to separate.

(i) Methanol (CH₃OH) is miscible with water because it is able to form hydrogen bonding with water due whereas iodo-methane (CH₃I)does not contain -OH group.
(ii) Since it is not able to form hydrogen bonding with water it is insoluble in water.

Common ion effect decreases the solubility of the electrolyte.

Buffer is a solution which consists of a mixture of a weak acid and its conjugate base (or) a weak base and its conjugate acid.

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

\(\underset{0.001M}{HCl}\overset{H_{2}O}{\rightleftharpoons }\underset{0.001M}{H_{3}O^{+}}+\underset{0.001M}{Cl^{-}}\)
H₃O⁺ from the auto ionisation of H₂O (10^-7M) is negligible when compared to the H₃O⁺ from 10^-3M HCl.
Hence [H₃O⁺] = 0.001 mol dm^-3
pH = -log₁₀ [H₃O⁺]
= -log₁₀^(0.001)
= -log₁₀(10^-3) = 3

Given that H₃O⁺ = \(2\times10^{-3}M\)
\(K_{w}=[H_{3}O^{+}][OH^{-}]\)
\(\therefore [OH^{-}]=\frac{K_{w}}{[H_{3}O^{+}]}=\frac{1\times10^{-14}}{2\times10^{-3}}=0.5\times10^{-11}M\)
\(2\times10^{-3} >>0.5\times10^{-11}\)
i.e., [H₃O⁺]>>[OH^-], hence the juice is acidic in nature

(i) p^H = -log₁₀ [H₃O⁺]
(ii) The pH of a solution is defined as the negative logarithm of base 10 of the molar concentration of the hydronium ions present in the solution.

(i) Environmental conservation
(ii) Use jute bags or cloth bags in place of polythene bags.
(iii) KMnO₄ acts as a strong oxidising agent.

Transition metal has energetically available d orbitals that can accept electrons from reactant molecule or metal can form bond with reactant molecule using its d electrons. For example, in the catalytic hydrogenation of an alkene, the alkene bonds to an active site by using its π electrons with an empty d orbital of the catalyst.

In \(\frac { { k }_{ 2 } }{ { k }_{ 1 } } =\frac { { E }_{ a } }{ R } \left[ \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right] \)
In 2 = \(\frac { { E }_{ a } }{ R } \) \(\left[ \frac { 10 }{ 300\times 310 } \right] \)
E_a = 9300 R In 2
= 53.4 kJ mol^-1

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

(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.

Element of group 15 have stable half-filled p-orbitals hence large amount of energy is required to remove electrons as compared to group 16, which has an incomplete p subshell.

NaCl is a crystalline solid so possess sharp melting points where as glass is amorphous in nature so does not have sharp melting point.

(i) Zeolites have a three dimensional crystalline structure looks like a honeycomb consisting of a network of interconnected tunnels and cages.
(ii) Water molecules moves freely in and out of these pores hence they are highly porous.

The aqueous solution of carbon dioxide is slightly acidic as it forms carbonic acid.
CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃^-

(i) Gold is usually found in native state
(ii) The leaching process is intended to concentrate the gold metal
\(4Au_{ (s) }+8Na{ CN }_{ (aq) }^{ - }+2{ H }_{ 2 }O_{ (aq) }+{ O }_{ 2(g) }\longrightarrow 4Na[Au\left( CN \right) _{ 2 }]^{ - }_{ (aq) }+4NaO{ H }^{ -1 }_{ (aq) }\)
\(2Na\left[ Au(CN)_{ 2 } \right] +Zn\longrightarrow { Na }_{ 2 }\left[ Zn\left( CN \right) \right] +2Au\downarrow \)
(iii) In this reaction, Zn acts as a reducing agent.

Iron, Lead and Copper

Six

K₃[CoF₆]

(i) First order reaction
(ii) First order reaction
(iii) \(\frac{1}{2}+2=2 \frac{1}{2}\); Pseudo first order reaction

(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.

The oxidation states of 4d metals vary from +3 for Y to +8 for Ru and Os.
The highest oxidation state of 4d elements are found in their compounds with the higher electronegative elements like O, F & Cl.
Example: In RuO₄, OsO₄ & WCl₆
The oxidation state of Ru and Os is +8.
The oxidation state of W is +6.
Generally in going down a group, a stability of higher oxidation state increases while that of lower oxidation state decreases.
4d series (5^th period) - Yttrium to Cadmium (10 elements)

Tetrahedral complexes do not exhibit geometrical isomerism. Because the relative position of donor atoms of ligand the unidentate Iigands (donor atom) attached to the central atom are same with respect to each other.

\({ }_{22} \mathrm{Ti}-{ }_{18}[\mathrm{Ar}] 4 \mathrm{s}^{2} 3 \mathrm{d}^{2} / {}_{21}\mathrm{Sc}-{ }_{18}[\mathrm{Ar}] 4 \mathrm{~s}^{2} 3 \mathrm{~d} \)
\({ }_{22} \mathrm{Ti}^{3+}-{ }_{18}[\mathrm{Ar}] 3 \mathrm{~d}^{1} /{ }_{18} \mathrm{Sc}^{3+}{ }_{18}[\mathrm{Ar}] 3 \mathrm{~d}^{0}\)
(i) In this complex the central metal ion is Ti³⁺, which has d¹ configuration. This single electron occupies one of the t_2g orbitals in the octahedral aqua ligand field. When white light falls on this complex the electron absorbs light and promotes itself to e_g level. The spectral data show the absorption maximum is at 20000 cm^-1 corresponding to the crystal field splitting energy \(\left(\Delta_{o}\right)\) 239.7 kJmol^-1. The transmitted colour associated with this absorption is purple and hence the complex appears purple in colour.
(ii) Thus in \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} \mathrm{d}-\mathrm{d}\) transition takes place.
(iii) But in \(\left[\mathrm{Sc}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} \mathrm{Sc}^{3+}\) has the outer electronic configuration of 3d⁰ where d-d transition is not possible and it is colourless.

\({ Ag }^{ + }2N{ H }_{ 3 }\rightleftharpoons { \left[ { Ag\left( { { NH }_{ 3 } } \right) }_{ 2 } \right] }^{ + }\)
\(k=\frac { { \left[ { Ag\left( { { NH }_{ 3 } } \right) }_{ 2 } \right] }^{ + } }{ \left[ { Ag }^{ + } \right] { \left[ { NH }^{ 3+ } \right] }^{ 2 } } \)
\(=\frac { \left[ { Ag }^{ + } \right] }{ { \left[ { Ag\left( { NH }_{ 3 } \right) }_{ 2 } \right] }^{ + } } =\frac { 1 }{ k{ \left( { NH }_{ 3 } \right) }^{ 2 } } \)
\(=\frac { 1 }{ 1.7\times { 10 }^{ 7 }\times { \left( 0.2 \right) }^{ 2 } } =\frac { { 10 }^{ -7 } }{ 1.7\times 4\times { 10 }^{ -2 } } \)
\(=\frac { { 10 }^{ -5 } }{ 6.8 } =1.47\times { 10 }^{ -5 }\)

(a) B(OH)₃ + NH₃\(\overset { \Delta }{ \longrightarrow } \) BN + 3H₂O
                                (Boron nitride)
(b) Na₂B₄O₇ + H₂SO₄ + 5H₂O \(\longrightarrow \) 4H₃BO₃ + Na₂SO₄
                                                           (Boric acid)
(c) B₂H₆ + 2NaOH + 2H₂O \(\longrightarrow \)2NaBO₂ + 6H₂
                                              (Sodium metaborate)
(d) B₂H₆ + 6CH₃OH  \(\longrightarrow \)2B(OCH₃)₃ + 6H₂O
                                              (Trimethyl borate)
(e) 4BF₃ + 3H₂O \(\longrightarrow \) H₃BO₃ + 3H⁺ + 3[BF₄]^-
                                      (Boric acid)
(f) HCOOH + H₂SO₄ \(\longrightarrow \)CO + H₂SO₄. H₂O
                                        (Carbon monoxide)   
(g) 2SiCl₄ + NH₃ \(\overset { 330K }{ \underset { ether }{ \longrightarrow } } \)Cl₃Si- NH - SiCl₃ + 2HCl
                                        (Chlorosilazane)
(h) SiCl₄ + 4C₂H₅OH\(\longrightarrow \)Si(OC₂H₅)₄ + 4HCI
                                        (Tetraethoxysilane)
(i) 2B + 6NaOH\(\longrightarrow \) 2Na₃BO₃ + 3H₂
(j) H₂B₄O₇ \(\xrightarrow[]{Redhot}\) 2B₂O₃ + H₂O

(a) OF₂
+ 2 + 2(x) = 0 
+2 = -2x 
2 x = -2 ⇒ x = -1
(b) O₂F₂
2(+1) + 2x = 0 
2x = -2 
x = -1
(c) Cl₂O₃
2(x) + 3(-2) = 0 
2x = +6 
x = +3
(d) I₂O₄
2(x) + 4(-2) = 0 
2x = +8 
x = +4

(i) In heavier post transition metals, the outers electrons (ns) have a tendency to remain inert and show reluctance to take part in the bonding, which is known as inert pair effect.
(ii) This effect is also observed in groups 14, 15 and 16.

In this extraction, a basic flux, quick lime (CaO) is used, since the silica gangue present in the ore is acidic in nature. The quick lime combines with it to form calcium silicate (slag).
CaO_(s) + Sio_2(s) ⟶ CaSio_3(s)
Flux       Gangue    Slag

During oxidation of unsymmetric ketones with oxidising agent which brings about the cleavage of C-C bond, the smaller alkyl group goes preferentially with the carbonyl group resulting in the carboxylic acids.
\({ CH }_{ 3 }-{ CH }_{ 2 }-{ CH }_{ 2 }-\underset { \overset { || }{ O } }{ C } -{ CH }_{ 3 }\overset { (O) }{ \underset { Con.HNO_{ 3 } }{ \longrightarrow } } \underset { Propanoic\ acid }{ { CH }_{ 3 }{ CH }_{ 2 }-COOH } +\underset { acetic\ acid }{ { CH }_{ 3 }COOH } \)

(i) Alcohols are weakly basic in nature so easily in presence of strong acids.
(ii) Due to the presence of +ve charge on the oxygen atom, C-O in protonated alcohols became weak hence easily cleaved by halide ions to form alkyl halides.
(iii) Where as phenols are much weaker bases due to delocalisation of the lone pair of electrons on the oxygen atom due to resonance and so are not easily protonated.
(iv) The C-O bond in phenols has some double bond character; hence not easily cleaved by halide ions to form respective alkyl halides.

(i) In certain reactions one of the products formed acts as a catalyst to the reaction.
(ii) Initially the rate of reaction will be very slow but with the increase in time the rate of reaction increases.
(iii) Auto catalysis is observed in the following reactions.
\({ CH }_{ 3 }COO{ C }_{ 2 }{ H }_{ 5 }+{ H }_{ 2 }O\longrightarrow CH_{ 3 }COOH+{ C }_{ 2 }{ H }_{ 5 }OH\) 
Acetic acid acts as the auto catalyst.
(iv) \(2As{ H }_{ 3 }\longrightarrow 2As+{ 3H }_{ 2 }\) 
Arsenic acts as an autocatalyst.

Zinc is oxidised to Zn^{2+ i}ons and the Cu²⁺ ions are reduced to metallic copper. The half reactions are represented as below.
\( { Zn }_{ (s) }\rightarrow { Zn }_{ (aq) }^{ + }+2e^{ - }\) (oxiation)
Loss of election oxidation
\({ Cu }_{ (aq) }^{ 2+ }+{ 2e }^{ - }\rightarrow { Cu }_{ (s) }\) (reduction)
Gain of electron oxidation.

According to Arrhenius, an acid is a substance that dissociates to give hydrogen ions in water. For example, HCl, H₂SO₄ etc., are acids. Their dissociation in aqueous solution is expressed as
\({ HCl }_{ (g) }\overset { { H }_{ 2 }O }{ \rightleftharpoons } { H }_{ (aq) }^{ + }+{ Cl }_{ (aq) }^{ - }\)
Similarly a base is a substance that dissociates to give hydroxyl ions in water. For example, substances like NaOH, Ca(OH)² etc., are bases.
\({ Ca(OH) }_{ 2 }\overset { { H }_{ 2 }O }{ \rightleftharpoons } { Ca }_{ (aq) }^{ 2+ }+{ 2OH^- }_{ (aq) }\)

\({ CH }_{ 3 }-{ CH }_{ 2 }-OH\overset { { H }_{ 2 }{ SO }_{ 4 } }{ \underset { 443k }{ \longrightarrow } } { CH }_{ 2 }={ CH }_{ 2 }+{ H }_{ 2 }O\)
Mechanism
Primary alcohols undergo dehydration by E₂ mechanism

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

The steady decrease in the atomic/ionic radius from La³⁺ to Lu³⁺ is called lanthanoid contraction.

Commercial method of preparation
(i) Nitric acid prepared in large scales using Ostwald's process. In this method ammonia from Haber's process is mixed about 10 times, of air.
(ii) This mixture is preheated and passed into the catalyst chamber where they come in contact with platinum gauze.
(iii) The temperature rises to about 1275 K and the metallic gauze brings about the rapid catalytic oxidation of ammonia resulting in the formation of NO, which then oxidised to nitrogen dioxide.
\({ 4NH }_{ 3 }+{ 5O }_{ 2 }\longrightarrow 4NO+6{ H }_{ 2 }O+120KJ\)
\(2NO+{ O }_{ 2 }\longrightarrow { 2NO }_{ 2 }\)
(iv) The nitrogen dioxide produced is passed through a series of adsorption towers. It reacts with water to give nitric acid. Nitric acid formed is bleached by blowing air.
\({ 6NO }_{ 2 }+3{ H }_{ 2 }O\longrightarrow { 4HNO }_{ 3 }+2NO+{ H }_{ 2 }O\)

(i) Compressibility is the ability of the substance to change its shape by applying external pressure.
(ii) Solids form closed packed structure with negligible intermolecular space and possesses very strong intermolecular forces of attraction.
(iii) Thus, they do not change their shape in the presence of external pressure.

(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.

(i) Metallic oxides such as Cr₂O₃ can be reduced by an aluminothermic process.
(ii) In this process, the metal oxide is mixed with aluminum powder and placed in a fire clay crucible.
(iii)To initiate the reduction process, an ignition mixture (usually magnesium and barium peroxide is used 
\({ BaO }_{ 2 }+Mg\longrightarrow Bao+MgO\)
(iv) During the above reaction a large amount of heat is evolved (temperature up to 2400°C, is generated and the reaction enthalpy is: 852 kJ mol^-1 which facilitates the reduction of Cr₂O₃ by aluminium power.
\({ Cr }_{ 2 }{ O }_{ 3 }+2Al\overset { \Delta }{ \longrightarrow } 2Cr+{ Al }_{ 2 }{ O }_{ 3 }\)

(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.

A Reaction whose rate depends on the reactant concentration raised to the first power is called a first order reaction. For a first order reaction,
A ⟶ product
Rate law can be expressed as
Rate = k[A]^-1
\(\frac { -d\left[ A \right] }{ \left[ A \right] } =k dt\)    ...(1)
Integrate the above equation between the limits of time t = 0 and time equal to t, while the concentration varies from the initial: concentration [A_o] to [A] at the later time.
\(\int _{ { [A }_{ 0 }] }^{ [A] }{ \frac { -d\left[ A \right] }{ \left[ A \right] } } =k\int _{ 0 }^{ t }{ dt } \)
\({ \left( -In\left[ A \right] \right) }_{ \left[ { A }_{ 0 } \right] }^{ \left[ A \right] }=k{ \left( t \right) }_{ 0 }^{ t }\)
\(-In\left[ A \right] -\left( In\left[ { A }_{ 0 } \right] \right) =k\left( t-0 \right) \)
\(-In\left[ A \right] -\left( In\left[ { A }_{ 0 } \right] \right) =kt\)
\(In\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right) =kt\) ...(2)
This equation is in natural logarithm. To convert it into usual logarithm with base 10, we have to multiply the term by 2.303.

\(2.303\quad log\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right) =kt\)
\(k=\frac { 2.303 }{ t } log\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right) \)    ...(3)
Equation (2) can be written in the form
y = mx+c as below
In [A₀]-In[A] = kt
In[A] = In[A₀]-kt
⇒ y = c + mx
If we follow the reaction by measuring the concentration of the reactants at regular time interval 't' a plot of In[A] against 't' yields a straight line with a negative slope. From this, the rate constant is calculated.

(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 \(\left[\mathrm{E}^{\circ}=+1.51 \mathrm{~V}\right]\).

(i) Compound (A) does not reduce Fehling's solution, so it is benzaldehyde.
(ii) Benzaldehyde (A) reacts with acetone in the presence of NaOH to give on a, p unsaturated compound (B).
\(\\ \\ \\ \underset { (A) }{ { C }_{ 6 }{ H }_{ 5 }CHO } +{ CH }_{ 3 }-{ COCH }_{ 3 }\overset { NaoH }{ \longrightarrow } \underset { (B) }{ { C }_{ 6 }{ H }_{ 5 } } -\underset { Acetone }{ CH=CH-{ COCH }_{ 3 } } \) 
(iii) (A) reacts with dimethyl aniline in the presence of conc. H₂SO₄ to give compound (C).

Compound	Compound Name	Formula
A	Benzaldehyde	C6H5CHO
B	Benzal acetone	C6H5CH = CH-COCH3
C	Triphenyl methane dye

Ethers are a class of organic compound in which an oxygen atom is connected to two alkyl/aryl groups Ethers can be considered as the derivatives of hydrocarbon in which one hydrogen atom is replaced by an alkoxy (-OR) or an aryloxy (-OAr ) group. The general formula of aliphatic ether is C_nH_2n+2O.
Classification:

(i) The presence of electric field increases the speed of removal of electrolytes from colloidal solution.
(ii) The colloidal solution containing an electrolyte as impurity is placed between two dialysing membranes enclosed into two compartments filled with water.
(iii) When current is passed, the impurities pass into water compartment and get removed periodically.
(iv) This process is faster then dialysis, as the rate of diffusion of electrolytes is increased by the application of electricity.

Given:
Specific conductance = K = 5.55 x 10^-3 mho cm^-1
Concentration C = 1M
Volume = 1 litre.
Formula: Equivalent conductance = K x V
= \(\frac { k\times 1000 }{ C } \)
Solution:
\(\frac { { 5.55\times 10 }^{ -3 }\times 1000 }{ 1 } =5.55\times { 10 }^{ -3 }\times { 10 }^{ -3 }\)

A relation between pH and pOH can be established using their following definitions
pH = -log₁₀ [H₃O⁺] ....(1)
pOH = -log₁₀ [OH^-] ...(2)
Adding equation (1) and (2)
pH + pOH = -log₁₀ [H₃O⁺] -log₁₀[OH^-]
= -(log₁₀ [H₃O⁺] +log₁₀[OH^-])
pH + pOH = -log₁₀ [H₃O⁺] [OH^-]
[∵ log a + log b = log ab]
We know that [H₃O⁺][OH^-] = K_w
⇒ pH + pOH = -log₁₀ K_w
⇒ pH + pOR = pK_w
[∵ pK_w = -log₁₀^K_w ]
at 25°C, the ionic product of water,
pK_w= 1 x 10^-14
pK_w = -log₁₀ 10^-14= 14log₁₀ 10
= 14
∴ (2) ⇒ ∴ At 25°C, pH + pOH = 14.

Factors affecting adsorption
Qualitatively, the extent of surface adsorption depends on
(i) Nature of adsorbent
(ii) Nature of adsorbate
(iii) Pressure
(iv) Concentration at a given temperature.
1. Surface area of adsorbent:
As the adsorption is a surface phenomenon it depends on the surface area of adsorbent. i.e., higher the surface area, higher is the amount adsorbed.
2. Nature of adsorbate:
The nature of adsorbate can influence the adsorption. Gases like SO₂, NH₃, HCl and CO₂ are easily liquefiable as have greater vander waal's force of attraction. On the other hand, permanent gases like H₂, N₂ and O₂ cannot be liquefied easily. These permanent gases are having low critical temperature and adsorbed slowly, while gases with high critical temperature are adsorbed readily.
3. Effect of temperature:
When temperature is raised chemisorption first increases and then decreases. whereas physisorption decreases with increases in temperature.
4. Effect of Pressure:
Chemical adsorption is fast with increase in pressure, it cannot alter the amount of adsorption. In physisorption, the extend of adsorption increases with increase in pressure.

i) Molar conductivity
Given : C = 0.01 M;
\(\kappa=1.5 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1} \)
\(=1.5 \times 10^{-2} \mathrm{~S} \mathrm{~m}^{-1} \)
\(\lambda_{\text {cation }}^{0}=248.2 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \)
\(\lambda_{\text {anion }}^{0}=51.8 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \)
\(\Lambda_{m}^{0}=\frac{\kappa \times 10^{-3}}{C} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} \)
\(=\frac{1.5 \times 10^{-2} \times 10^{-3}}{0.01} \)
\(=1.5 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\)
ii) \(\alpha=\frac{\Lambda_{m}}{\Lambda_{m}^{0}}\)
\(\Lambda_{\mathrm{m}}^{0}=\lambda_{\text {cation }}^{0}+\lambda_{\text {anion }}^{0} \)
\(=(248.2+51.8) \mathrm{S} \mathrm{cm}^{2} \mathrm{~mol}^{-1} \)
\(=300 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
\(=300 \times 10^{-4} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}  \)
\(\alpha =\frac{1.5 \times 10^{-3}}{300 \times 10^{-4}}=0.05\)
iii) \(\mathrm{K}_{\mathrm{a}} =\frac{\alpha^{2} \mathrm{C}}{1-\alpha} \)
\(\mathrm{K}_{\mathrm{a}} =\frac{(0.05)^{2} \times(0.01)}{1-0.05}=2.6 \times 10^{-5} \)
(or) 
\(\mathrm{K}_{\mathrm{a}} =\alpha^{2} \mathrm{C}  \)
\(=(0.05)^{2} \times(0.01)  \)
\(\mathrm{K}_{\mathrm{a}} =2.5 \times 10^{-5}\)

1. Daniel cell is a galvanic cell. This is a voltaic cell also.
(a) The separation of half reaction is the basis for the construction of Daniel cell. It consists of two half cells.
(i) Oxidation half cell: A metallic zinc strip  that dips into an aqueous solution of zinc sulphate taken in a beaker, as shown in Figure
(ii) Reduction half cell: A copper strip that dips into an aqueous solution of copper sulphate taken in a beaker, as shown in Figure

(iii) Joining the half cells:
(a) The zinc and copper strips are externally connected using a wire through a switch (k) and a load (example: volt meter). The electrolytic solution present in the cathodic and anodic compartment are connected using an inverted U tube containing a agar-agar gel mixed with an inert electrolyte such as KCI, Na₂SO₄ etc.,
(b) The ions of inert electrolyte do not react with other ions present in the half I cells and they are not either oxidised (or) reduced at the electrodes. The solution in the salt bridge cannot get poured out, but through which the ions can move into (or) out of the half cells.
(c) When the switch (k) closes the circuit, the electrons flows from zinc strip to copper strip. This is due to the following redox reactions which are taking place at the respective electrodes.
(iv) Anodic oxidation:
(i) zinc strip acts as the anode.
(ii) Here,oxidation occurs.
The electrode at which the oxidation occur is called the anode. In Daniel cell, the oxidation take place at zinc electrode, i.e., zinc is oxidised to Zn²⁺ ions and the electrons.
The Zn²⁺ ions enters the solution and the electrons enter the zinc metal, then flow through the external wire and then enter the copper strip.
Electrons are liberated at zinc electrode and hence it is negative (-ve).
\(Zn_{ (s) }\longrightarrow { { Zn }^{ 2+ }_{ (aq) }+{ 2e }^{ - } } \)  (loss of electron-oxidation)  
(v) Cathodic reduction:
As discussed earlier. the electrons flow through the circuit from zinc to copper, where the Cu²⁺ ions in the solution accept the electrons, get reduced to copper and the same get deposited on the electrode Here, the electrons are consumed and hence it is positive (+ve).
\({ Cu }_{ (aq) }^{ 2+ }+{ 2e }^{ - }\longrightarrow { { Cu }_{ (s) } } \)(gain of electron - reduction)
b) When a Zinc metal strip is placed in a copper sulphate solution, the blue colour of the solution fades and the copper is deposited on the zinc strip as red - brown crust due to the following spontaneous chemical reaction.
\(\mathrm{Zn}_{(\mathrm{s})}+\mathrm{CuSO}_{4(\mathrm{aq})} \rightarrow \mathrm{ZnSO}_{4(\mathrm{aq})}+\mathrm{Cu}_{(\mathrm{s})}\)
The energy produced in the above reaction is lost to the surroundings as heat.
In the above redox reaction, Zinc is oxidised to Zn²⁺ ions and the Cu²⁺ ions are reduced to metallic copper. The half reactions are represented as below.
\(\mathrm{Zn}_{(\mathrm{s})} \rightarrow \mathrm{Zn}^{2+}{ }_{(\mathrm{aq})}+2 \mathrm{e}^{-} \text {(oxidation) }  \)
\(\mathrm{Cu}^{2+}{ }_{(\mathrm{aq})}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}_{(\mathrm{s})} \text { (reduction) }\)
If we perform the above two half reactions separately in an apparatus as shown in figure, some of the energy produced in the reaction will be converted into electrical energy.

(i) It oxidises ferrous salts to ferric salts.
 Cr₂O₇^2- + 6Fe²⁺+ 14H⁺ ⟶ 2Cr³⁺ + 6Fe³⁺ + 7H₂O
(ii) It oxidises iodide ions to iodine
Cr₂O₇^2- + 6I^-+ 14H⁺ ⟶ 2Cr³⁺ + 3I₂ + 7H₂O
(iii) It oxidises sulphide ion to sulphur
Cr₂O₇^2- + 3S^2-+ 14H⁺ ⟶ 2Cr³⁺ + 3S + 7H₂O
(iv) It oxidises sulphur dioxide to sulphate ion
Cr₂O₇^2- + 3SO₂+ 2H⁺ ⟶ 2Cr³⁺ + 3SO₄^2- + H₂O
(v) It oxidises stannous salts to stannic salt
 Cr₂O₇^2- + 3Sn²⁺ + 14H⁺ ⟶ 2Cr³⁺ + 3Sn⁴⁺ + 7H₂O
(vi) It oxidises alcohols to acids
2K₂Cr₂O₇ + 8H₂SO₄ + 3CH₃CH₂OH ⟶ 2K₂SO₄ +2Cr₂(SO₄)₃+ 3CH₃COOH + 11H₂O
(vii) It oxidises oxalic acid to CO2
2MnO₄^- + 5(COO)^2- + 6H⁺ ⟶ 2Mn²⁺+ 10CO₂ + 8H₂O
(viii) It oxidises iodide ions to iodine
2MnO4^- + 10I^- + 16H⁺ ⟶ 2Mn²⁺+ 5I₂+ 8H₂O
(ix) It oxidises sulphide ion to sulphur
2MnO₄^- + 5S^2-+ 16H⁺ ⟶ 2Mn²⁺+ 5S + 8H₂O
(x) It oxidises nitrites to nitrates
 2MnO₄^- + 5NO^2- + 6H⁺ ⟶ 2Mn²⁺+ 5NO₃^- + 3H₂O
(xi) It oxidises alcohols to aldehydes.
 2KMnO₄ + 3H₂SO₄ + 5CH₃CH₂OH ⟶ 2K₂SO₄ + 2MnSO₄ + 5CH₃CHO + 8H₂O
(xii) It oxidises sulphite to sulphate
2MnO₄^- + 5SO₃^2- + 6H⁺ ⟶ 2Mn²⁺ + 5SO₄^2- + 3H₂O

(i) The magnitude of order of a reaction may be zero or fractional or integral values. Order is never fractional for elementary reaction.
(ii) It should be determined only by experiments.
(iii) Simple reactions possess low values of order like n = 0, 1, 2 reactions with order greater than or equal to 3.0 are called complex reactions.
(iv) Some reactions show fractional order depending on rate.
(v) Higher order reactions may be experimentally converted into simpler order (pseudo) reactions by using excess concentrations of one or more reactants.

Manufacture of sulphuric acid by contact process:
The contact process involves the following steps.
(i) Initially sulphur dioxide is produced by burning sulphur or iron pyrites in oxygen/ air.
\(S+{ O }_{ 2 }\longrightarrow { SO }_{ 2 }\)
\({ 4FeS }_{ 2 }+{ 11O }_{ 2 }\longrightarrow { 2Fe }_{ 2 }{ { O }_{ 3 } }+8{ SO }_{ 2 }\)
(ii) Sulphur dioxide formed is oxidised to sulphur trioxide by air in the presence of a catalyst such as V₂O₅ or platinised asbestos.
(iii) The sulphur trioxide is absorbed in concentrated sulphuric acid and produces oleum (H₂S₂O₇). The oleum is converted into sulphuric acid by diluting it with water. 
\(\mathrm{SO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7} \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} 2 \mathrm{H}_{2} \mathrm{SO}_{4}\)
(iv) To maximise the yield the plant is operated at 2 bar pressure and 720 K. The sulphuric acid obtained in this process is over 96 % pure.

(i) By knowing the density of an unknown metal and the dimension of its unit cell, the atomic mass of the metal can be: determined.
(ii) Let 'a' be the edge length of a unit cell of a crystal, 'd' be the density of the metal, 'm' be the atomic mass of the metal and 'z' be the number of atoms in the unit cell.
(iii) Now,
Density of the unit cell
\(=\frac{Mass\ of\ the\ unit\ cell}{Volume\ of\ the\ unit\ cell}\)
\(d=\frac{Z\times m}{a^3}\) ...(1)
[Since, mass of the unit cell = Number of atoms in the unit cell x Atomic mass]
[Volume of the unit cell = (edge length of the cubic unit cell)³]
(iv) From equation (1), We have
\(m=\frac{d\times a^3}{Z}\) ....(2)
(v) Now, 
Mass of the metal (M) \(=\frac{Atomic\ mass(M)}{Avogadro's\ number(N_A)}\)
M=\(\frac{d\times a^3 \times N_A}{Z}\)
(vi) From equation (3), we can determine the atomic mass of the unknown metal.

(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) Many heat exchangers/sinks and our day to day cooking vessels are made of aluminum.
(ii) It is used as wraps (aluminum foils) and is used in packing materials for food items.
(iii) Aluminum alloys with copper, manganese, magnesium and silicon are light weight and strong and they are used in design of aeroplanes and other forms of transport.
(iv) Aluminum shows high resistance, to corrosion, so it is used in the design of chemical reactors, medical equipments, refrigeration units and gas pipelines.
(v) Aluminium is a good electrical conductor and cheap, hence used in electrical overhead electric cables with steel core for strength.

(i) [NiCI₄]^2-
IUPAC name - Tetrachloridonickelate (II) ion
(ii) Ni²⁺ = 3d⁸,4s⁰

Cl^- being a weak field ligand cannot pair up the unpaired electron. So, it is sp³ hybridised, and it has tetrahedral geometry.