Adsorption leads to decrease in randomnes (entropy).i.e, ΔS < 0 for the adsorption to occur, ΔG slhould be -ve. We know that ΔG = ΔH - TΔS if ΔS is -ve, TΔS is +ve.It means that ΔG will become negative only when ΔH is -ve and ΔH > TΔS.

In general, specific conductance of an electrolyte decreases with dilution.So,0.002N solution has least specific conductance.

BF₃ → electron deficient →  Lewis acid
PF₃ → electron rich →  Lewis base
CO → having lone pair of electron →  Lewis base
F → unshared pair of electron →  Lewis base

Basic buffer is the solution which has weak base and its salt.
NH₄OH + HCI → NH₄CI +H₂O + NH₄OH
200 ml   100 ml    salt       100 ml  weak base

If a is the length of the side, then the length of the leading diagonal passing through the body centered atom is \(\sqrt{3a} \)
Required distance =  \(\left( \cfrac { \sqrt { 3 } }{ 2 } \right) a\)

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

Complexes given in other options (a), (c) and (d) have symmetry elements and hence they are optically inactive.

The reagent used is NaOH + CaO (sodalime). The reaction is a decarboxylation reaction.

Addition reaction with NaHSO₃.
In the addition of sodium bisulphite to give bisulphite compound - OSO₂Na anion is the nucleophile.

The continuous rapid, zig-zag, chaotic, random I and ceaseless movement executed by a colloidal particle in the dispersion medium is called Brownian movement of colloidal particles.

The two types of catalysis are
(i) Homogeneous
(ii) Heterogeneous catalysis.

At the anode oxidation occurs
Li_(s) ⟶ \({ Li }_{ (aq) }^{ + }\) + e^-
At the cathode reduction occurs
Li⁺+ CoO_2(s) + e^- ⟶ Li CoO_2(s)
Overall reactions
Li_(s)+ CoO₂ ⟶ LiCoO_2(s)

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

 Tertiary butyl alcohol does not give iodoform reaction. All other compounds contains α H atoms and they undergo iodoform reaction.

According to Lewis concept of Acid and bases, any species capable of accepting an electron pair is an acid. BF₃ is electron deficient so accepts a pair of electron, Hence termed as acid

(i) Arrhenius theory does not explain the behaviour of acids and bases in non aqueous solvents such as acetone, Tetrahydrofuran etc ..
(ii) This theory does not account for the basicity of the substances like ammonia (NH₃) which do not possess hydroxyl group.

l = 1.5 cm = 1.5 × 10^-2m 
A = 4.5 cm² = 4.5 × (10^-4)m²
R = 15Ω
\(\kappa = \frac{1}{R}(\frac{l}{A})\)
\(\kappa = \frac{1}{15\Omega}\times\frac{1.5\times10^{-2}m}{4.5\times10^{-4}m^2}\)
= 2.22 ohm^-1 m^-1
= 2.22 Sm^-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

On dilution the concentration decreases. Conductivity decreases with decrease in concentration (or dilution) as the number of ions per unit volume that carry the current in a solution decrease on dilution.

[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.

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

The rate of a reaction is affected by the following factors.
(i) Nature and state of the reactant
(ii) Concentration of the reactant
(iii) Surface area of the reactant
(iv) Temperature of the reaction
(v) Presence of a catalyst

Average rate = \(-\frac { 1 }{ 2 } \frac { \triangle \left[ A \right] }{ \triangle L } \)
\(=-\frac { 1 }{ 2 } \frac { { \left[ A \right] }_{ 2 }-{ \left[ A \right] }_{ 1 } }{ { t }_{ 2 }-{ t }_{ 1 } } \)
\(=-\frac { 1 }{ 2 } \left( \frac { 0.4-0.5 }{ 10 } \right) \)
\(=-\frac { 1 }{ 2 } \left( \frac { -0.1 }{ 10 } \right) \)
= 0.005 mol L^-1 min^-1
= 5 x 10^-3 M min^-1

(i) Bleaching action of Cl₂ is due to its oxidising property.
(ii) In aqueous solution, Cl₂ liberates nascent oxygen which brings about the oxidation of coloured substances present in vegetables and organic matter to colourless substances.
\(\mathrm{Cl}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{HCl}+\mathrm{[O}]\)
Coloured substance +[O] → colorless substance.

Yellow phosphorus reacts with alkali on boiling in an inert atmosphere liberating phosphine. Phosphorus acts as a reducing agent.
\({ P }_{ 4 }+NaOH+{ H }_{ 2 }O\longrightarrow \underset { Sodiumhypophosphite }{ { 3NaH }_{ 2 }{ PO }_{ 2 } } +\underset { Phosphine }{ { PH }_{ 3 } } \uparrow \)

Frenkel defect occurs in a Silver halides due to small size of Ag⁺ ions which occupy interstitial position. Whereas in alkali metal halides, size of the cation and anion are almost similar so alkali metal ions do not fit into the interstitial sites.

The empty spaces present between the metal atom or the ions when they are packed within the crystal are called voids.

In a transition metal ion, all the five d orbitals are degenerate but when it is involved in a complex formation, the degeneracy is split. This is called crystal field splitting.

(i) Silicon tetrachloride is used in the production of semiconducting silicon.
(ii) It is used as a starting material in the synthesis of silica gel, silicic esters, a binder for ceramic materials.

Cr(24): [Ar] 3d⁵4s¹

Cu, Ag and Au which have been used in making of coins in ancient times are called coinage metals

(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.

(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.

Sodium cyanide and sodium carbonate.

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

(i) 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.

In [CuCI₄]^-2 Cu²⁺, is reduced to Cu⁺ by I^-. Hence Cupric Iodide in converted to cuprous Iodide so [CuI₄]^-2 does not exist. In [CuCI₄]^-2 Cl^- cannot effect this change and so exists.

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

(i) 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) 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.

Catenation is an ability of an element to form chain of atoms.
The conditions for catenation.
(a) The valency of element is greater than or equal to two.
(b) Element should have an ability to bond with itself
(c) The self bond must be as strong as Its bond with other elements
(d) Kinetic inertness of catenated compound towards other molecules.
(e) Carbon possesses all the above properties and forms a wide range of compounds with itself and with other elements such as H, O, N, S and halogens.

In this method, the metal is treated with a suitable reagent which can form a volatile compound with the metal.
Then the volatile compound is decomposed to give the pure metal.

Ellingham diagram for the formation of ZnO and CO₂ intersects around 1200 K. Above this temperature ZnO lies above Carbon which indicates Carbon is the better reducing agent CO is more effective reducing agent only below 983 K.

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

(i) Perkin's Reactions:
When an aromatic aldehyde is heated with an aliphatic acid anhydride in the presence of the sodium salt of the acid corresponding to the anhydride, condensation takes place and an α, β unsaturated acid is obtained
This reaction is known as Perkin's reaction.

(ii) Knoevengal Reaction:
(a) Benzaldehyde condenses with malonic acid in the presence of pyridine forming cinnamic acid
(b) Catalyst - Pyridine
(c) Carbanion formed from malonic acid.

Glycerol can give rise to a variety of oxidation products depending on the nature of the oxidising agent used for oxidation.
(i) Oxidation of glycerol with dil. HNO₃ gives glyceric acid and tartronic acid.
(ii) Oxidation of glycerol with Cone. HNO₃ gives mainly glyceric acid.
(iii) Oxidation of glycerol with bismuth nitrate gives as meso oxalic acid.
(iv) Oxidation of glycerol with Br/H₂O (or) NaOBr (or) Fenton reagent (FeSO₄ + H₂O₂) gives a mixture of glyceraldehyde I and dihydroxy acetone (This mixture is named as glycerose).

(i) Nano materials such a metallic nano particles, metal oxides, etc., are used as catalyst in many chemical transformation. Nanocatalysts carry the advantages of both homogeneous and heterogeneous catalyses.
(ii) Like homogeneous catalysts, the nanocatalysts give 100% selective transformations and excellent yield and show extremely high activity.
(iii) Like the heterogeneous catalysts, nanocatalysts can be recovered and recycled.

(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.

(i) To resist changes in its pH on the addition of an acid (or) a base, the buffer solution should contain both acidic as well as basic components so as to neutralize the effect of added acid (or) base and at the same time, these components should not consume each other.
(ii) Let us explain the buffer action in a solution containing CH₃COOH and CH₃COONa.
(iii) The dissociation of the buffer components occurs as below.
\(\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{CH}_{3}-\mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{H}_{3} \mathrm{O}_{(\mathrm{aq})}^{+} \)
\(\mathrm{CH}_{3} \mathrm{COONa}_{(\mathrm{s})} \stackrel{\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}}{\longrightarrow} \mathrm{CH}_{3}-\mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{Ha}_{(\mathrm{aq})}^{+}\)

In this reaction, phenol is first converted into sodium phenoxide which is more reactive than phenol towards electrophilic substitution reaction with CO₂, Treatment of sodium phenoxide with CO₂ at 400 K, 4-7 bar pressure followed by acid hydrolysis gives salicylic acid.

(a) If we apply DC current through the conductivity cell, it will lead to the electrolysis of the solution taken in the cell.
(b) So, AC current is used for this measurement to prevent electrolysis.

\(\mathrm{M} =\frac{\mathrm{V}_{1} \mathrm{M}_{1}-\mathrm{V}_{2} \mathrm{M}_{2}}{\mathrm{~V}_{1}+\mathrm{V}_{2}}  \)
\(=\frac{(50 \times 0.05)-(50 \times 0.025)}{50+50}  \)
\(\text { Molarity }=\frac{\text { Number of millimoles }}{\mathrm{V}_{\mathrm{m} l}}\)
\(=\frac{2.5-1.25}{100}=\frac{1.25}{100}=0.0125 \mathrm{M}  \)
\(\mathrm{pH} =-\log _{10}\left[\mathrm{H}^{+}\right]  \)
\(=-\log _{10} 0.0125  \)
Normality = Molarity \(\times\) basicity
\(=-\log _{10}\left(1.25 \times 10^{-2}\right)  \)
\(=-\left[\log _{10} 1.25-2 \log _{10} 10\right]  \)
\(=2-\log _{10} 1.25=2-0.0969 \)
pH = 1.9031

(i) HCl cannot be used for making the medium acidic since it reacts with KMnO₄ as follows.
2MnO₄^- + 10 Cl^- + 16H⁺ ⟶ 2Mn²⁺ + 5Cl₂+ 8H₂O
(ii) HNO₃ also cannot be used since it is good oxidising agent and reacts with reducing agents in the reaction.
(iii) However, H₂SO₄ is found to be most suitable since it does not react with potassium permanganate.

\({ t }_{ \frac { 3 }{ 4 } }=\frac { 2.303 }{ K } \log { \frac { { \left[ R \right] }_{ 0 } }{ \frac { 1 }{ 4 } { \left[ R \right] }_{ 6 } } } \)
\({ t }_{ \frac { 3 }{ 4 } }=\frac { 2.303 }{ K } \log 4\)
\(=\frac { 2.303\times 0.6021 }{ K } =\frac { 1.386 }{ K } \)
\(=2\times \frac { 0.693 }{ K } \)
\(=2{ t }_{ \frac { 1 }{ 2 } }\)

(i) All halogens react with hydrogen to form volatile covalent hydrides of formula HX.
(ii) These hydrides are called hydracids.
(iii) Hydracids are the reducing agents.
(iv) Except HF, all hydrogen halides are gases. HF is a liquid because of intermolecular hydrogen bonding.
H-F...H-F....H-F....H-F
(v) The acidic character of HX are in the following order.
HF < HCI < HBr < HI.

(i) A face centred cubic lattice has 4 atoms per I unit cell.
(ii) The lattice has 8 atoms in the corners (N_c) and 6 atoms on the faces (N_f) (one on each face)
Contribution by atoms on the corners to unit cell =\(\frac{N_c}{8}=\frac{8}{8}=1\)
Contribution by atoms on the faces to unit cell =\(\frac{N_f}{2}=\frac{6}{2}=3\)
∴ The number of atoms per unit cell in fcc
\(=\frac { { N }_{ c } }{ 8 } +\frac { { N }_{ f } }{ 2 } =\frac { 8 }{ 8 } +\frac { 6 }{ 2 } =1+3=4\)

(i) Carbon nanotubes, have graphite like tubes with fullerene ends.
(ii)  Along the axis, these nanotubes are stronger than steel and conduct electricity.
(iii) These have many applications in nanoscale electronics, catalysis, polymers and medicine.

(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) 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.

As we move from left to right in a transition metal series, the ionization enthalpy increases as expected. This is due to increase in the nuclear charge corresponding to the filling of d electrons. The increase in first ionisation enthalpy with increase in atomic number along a particular series is not regular. The added electron enters (n-1) d orbital and the inner electrons act as a shield and decrease the effect of nuclear charge on valence ns electrons. Therefore, it leads to variation in the ionization energy value.

\(\text {(a) }\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}:{ }_{24} \mathrm{Cr} \Rightarrow{ }_{18}[\mathrm{Ar}] 4 \mathrm{~s}^{2} 3 \mathrm{~d}^{4} \)
\({ }_{21} \mathrm{Cr}^{3+} \Rightarrow{ }_{18}[\mathrm{Ar}] 3 \mathrm{~d}^{3}\)

(i) d²sp³ hybridisation (octahedral)
(ii) It has three unpaired electrons (n = 3)
(iii) So it is paramagnetic
(iv) Magnetic moment \(\left(\mu_{\mathrm{s}}\right)=\sqrt{\mathrm{n}(\mathrm{n}+2)} \mathrm{BM}\)
\(=\sqrt{3(3+2)}=\sqrt{15}=3.87 \mathrm{BM}\)
\((b) \ \left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-:}{ }_{28} \mathrm{Ni} \Rightarrow[\mathrm{Ar}] 4 \mathrm{~s}^{2} 3 \mathrm{~d}^{8} ;{ }_{26} \mathrm{Ni}^{2+} \Rightarrow{ }_{18}[\mathrm{Ar}] 3 \mathrm{~d}^{8}\)

(i) dsp² hybridisation
(ii) Geometry - square planar
(iii) No unpaired electrons- It is Diamagnetic
(iv) Magnetic moment (μ_s) = 0.

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

\(\underset { Acetaldehyde }{ { CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -H+4(H) } \overset { { NH }_{ 2 }{ NH }_{ 2 } }{ \underset { { C }_{ 2 }{ H }_{ 5 }ONa }{ \longrightarrow } } \underset { Ethane }{ { CH }_{ 3 }-{ CH }_{ 3 }+{ H }_{ 2 }O+{ N }_{ 2 } } \)
(i) Clemmensen reduction: Aldehydes and Ketones when heated with zinc amalgam and concentrated hydrochloric acid gives hydrocarbons. 
Example:
\(\underset { Acetald4ehyde }{ { CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -{ H }^{ + }4(H) } \overset { Zn^{ - }Hg }{ \underset { Conc.HCl }{ \longrightarrow } } \underset {Ethane}{{ CH }_{ 3 }-{ CH }_{ 3 }\pm { H }_{ 2 }O}\) 
\(\underset { Acetone }{ { CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -{ CH }_{ 3 }+4\left( H \right) } \overset { { Zn }^{ - }Hg }{ \underset { Conc.HCl }{ \longrightarrow } } { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 3 }+{ H }_{ 2 }O\)
(ii) Wolff-Kishner Reduction: Aldehydes and Ketones when heated with hydrazine (NH₂NH₂) and sodium ethoxide, hydrocarbons are formed Hydrazine acts as a reducing agent and sodium ethoxide as a catalyst.
Example :
\(\underset { Acetaldehyde }{ { CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -H+4(H) } \overset { { NH }_{ 2 }{ NH }_{ 2 } }{ \underset { { C }_{ 2 }{ H }_{ 5 }ONa }{ \longrightarrow } } \underset { Ethane }{ { CH }_{ 3 }-{ CH }_{ 3 }+{ H }_{ 2 }O+{ N }_{ 2 } } \) 
\(\underset { Acetone }{ { CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -{ CH }_{ 3 }+4(H) } \overset { { NH }_{ 2 }{ NH }_{ 2 } }{ \underset { { C }_{ 2 }{ H }_{ 5 }ONa }{ \longrightarrow } } { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 3 }+{ H }_{ 2 }O+{ N }_{ 2 }\)
Aldehyde (or) ketones is first converted to its hydrazone which on heating with strong base gives hydrocarbons.
(iii) Reduction to pinacols: Ketones, on reduction with magnesium amalgam and water, are reduced to symmetrical diols known as pinacol.

(i) Methyl alcohol and Ethyl alcohol

(ii) Benzyl alcohol and phenol

(iii) Ethyl alcohol and benzyl alcohol

Freundlich adsorption isotherm:
According to Freundlinch
\(\frac { x }{ m } =kp^{ \frac { 1 }{ n } }\) 
where x is the amount of adsorbate or adsorbed on 'm' gm of adsorbent at a pressure of p. K and n are constants Value is always less than unity.
This equation is applicable for adsorption of gases on solid surfaces. The same equation becomes  \(\frac { x }{ m } =Kc^{ \frac { 1 }{ n } }\) when used for adsorption in solutions with c as concentration.
These equation quantitively predict the effect of pressure(or concentration) on the adsorption of gases(or adsorbates) at constant temperature.
Taking log on both sides of equation \(\frac { x }{ m } ={ kp }^{ \frac { 1 }{ n } }\) 
\(log\frac { x }{ m } =logK+\frac { 1 }{ n } logP\) 

Hence the intercept represents the value of log k and the slope \(\frac { b }{ q } \) gives \(\frac { 1 }{ n } \)
This equation explains the increase of \(\frac { x }{ m } \) with increase in pressure. But experimental values show the deviation at low pressure.

According to Ostwald dilution Law,
\({ K }_{ a }=\frac { { \alpha }^{ 2 }C }{ (1-\alpha ) } \)  ...(1)
Substitute a value in the above expression (1)
\({ K }_{ a }=\frac { { \Lambda }_{ m }^{ 2 }C }{ { \Lambda }_{ m }^{ 2 }\left( 1-\frac { { \Lambda }_{ m } }{ { \Lambda }_{ m }^{ o } } \right) } \)
\({ K }_{ a }=\frac { { \Lambda }_{ m }^{ 2 }C }{ { \Lambda }_{ m }^{ 2 }\frac { { \Lambda }_{ m }^{ o }-{ \Lambda }_{ m } }{ { \Lambda }_{ m }^{ o } } } \)
⇒ \({ K }_{ a }=\frac { { \Lambda }_{ m }^{ 2 }C }{ { \Lambda }_{ m }^{ 2 }{ (\Lambda }_{ m }^{ o }-{ \Lambda }_{ m }) } \)

The degree of ionisation and the dissociation constant of the weak acid are related by the equation.
\({ K }_{ a }=\frac { { \alpha }^{ 2 }{ C } }{ 1-\alpha } \cong { \alpha }^{ 2 }C\)
α = 5% (or) 0.05
C = 0.02M
K_a = (0.05)2 x 0.02 = 0.00005
K_a = 5 x 10^-5

Rate = k [x]² = ka²
[x] = a mol^-1
If the concentration of x is in cross three time, then
(x) = 3a mol L^-1
Rate = R(3a)² = 9 ka²
Hence, the rate of formation will increase by 9 times.

(i) Ionization energy of transition element is intermediate between those of sand p block elements.
(ii) As we move from left to right in a transition metal series, the ionization enthalpy increases as expected.
(iii) This is due to increase in nuclear charge corresponding to the filling of d electrons.
(iv) The increase in first ionisation enthalpy with increase in atomic number along a particular series is not regular.
(v) The added electron enters (n-1)d orbital and the inner electrons act as a shield and decrease the effect of nuclear charge on valence ns electrons. Therefore, it leads to variation in the ionization energy values

Oxidising property : 
Sulphur dioxide, oxidises hydrogen sulphide to sulphur and magnesium to magnesium oxide.
\({ 2H }_{ 2 }S+{ SO }_{ 2 }\longrightarrow 3S+{ 2H }_{ 2 }O\)
\(2Mg+{ SO }_{ 2 }\longrightarrow 2MgO+S\)
Reducing property : 
As it can readily be oxidised, it acts as a reducing agent. It reduces chlorine into hydrochloric acid.
\({ SO }_{ 2 }+2{ H }_{ 2 }O+{ { Cl }_{ 2 }\longrightarrow { H }_{ 2 }{ SO }_{ 4 }+2HCl }\)
It also reduces potassium permanganate and dichromate to Mn²⁺ and Cr³⁺ respectively.
\({ 2KMnO }_{ 4 }+5{ SO }_{ 2 }+2{ H }_{ 2 }O\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+{ 2MnSO }_{ 4 }+2{ H }_{ 2 }{ SO }_{ 4 }\)
\({ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }+{ 3SO }_{ 2 }+{ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+{ Cr }_{ 2 }\left( SO_{ 4 } \right) _{ 3 }+{ H }_{ 2 }O\)

(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) 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.
\({ NaAlO }_{ 2 }+2{ H }_{ 2 }O\longrightarrow \underset { Ppt }{ { Al(OH) }_{ 3 } } +NaOH\)
\(2Al\left( OH \right) _{ 3 }\overset { \Delta }{ \longrightarrow } \underset { Pure }{ { Al }_{ 2 }{ O }_{ 3 } } +3{ H }_{ 2 }O\)

The stability of a complex or co-ordination compound refers to the extent up to which it exists in a solution as co-ordination sphere.
(i) Change on the central metal ion: Greater the charge on the central metal ion, greater the stability of complex.
(ii) Nature of the metal ion: Group 3 and 6 and inner transition elements form stable complexes when donor atoms of the ligands are N, O and F. The elements after group 6 of the transition metals form stable complex when the donor atoms of the ligands are the heavier members of N, O and F family.
(iii) Basic nature of the ligands: Greater the basic strength, greater is the stability of the complex.
(iv) Presence of chelate rings: Its presence increases the stability of the complex. it is called chelate effect. It is maximum for the 5 and 6 membered rings.
(v) Effect of multidentate cyclic ligand: If the ligands are IT multidentate and cyclic without any steric effect the stability of the complex get increased.

d) both assertion and reason are false