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

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

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

\(k=\frac { 1 }{ t } ln \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
[A₀] = 0.1: [A] = 0.05
\(k= \left[ \frac { 1 }{ t _{1/2}} \right] ln\left[ \frac { { 0.1 } }{ 0.05 }\right] \)
\(k= \left[ \frac { 1 }{ t _{1/2}} \right] ln (2)\)
t_1/2 = In(2)/K

A catalyst provides a new path to the reaction with low activation energy. i.e., it lowers the activation energy.

(a) coordination isomers
(b) no isomerism ( different molecular formula)
(c)⬅NCS; ⬅SCN coordinating atom differs : linkage isomers

Molecular formula: MSO₄Cl.6H₂O
Formation of white precipitate with Barium chloride indicates that SO^2-₄ ions are outside the coordination sphere, and no precipitate with AgNO₃ solution indicates that the Cl^- ions are inside the coordination sphere. Since the coordination number of M is 6.Cl^- and 5 H₂O are ligands, remaining 1 H₂O molecular and SO^2-₄ are in the outer coordination sphere.

(i) Carboxylic acids having an a -hydrogen are halogenated at the a -position on treatment with chlorine or bromine in the presence of small amount of red phsosphorus to form a halo carboxylic acids. This reaction is known as Hell - Volhard - Zelinsky reaction (HVZ reaction)
(Ii) The α - Halogenated acids are convenient starting materials for preparing α - substituted acids.

Acetone is soluble in water due to the intermolecular hydrogen bonding between polar carbonyl group and water molecules. In benzophenone, the phenyl groups are bulkier so C = 0 cannot form hydrogen bonds with water due to steric hindrance.

(i) A substance which, though itself not a catalyst, promotes the activity of a catalyst is called a promotor.
(ii) Eg: In the Haber's process for the synthesis of ammonia, traces of molybdenum increase the activity of finely divided iron which acts as acatalyst.
\({ N }_{ 2 }+{ 3H }_{ 2 }\overset { Fe }{ \underset { +mo }{ \leftrightharpoons } } { 2NH }_{ 3 }\)

TiCl₄ + (C₂H₅)₃AI.

In this technique, metals such as Mg or zinc which is corroded more easily than iron can be used as a sacrificial anode and the iron material acts as a cathode. So iron is protected, but Mg or Zn is corroded.

(i) Anode: Porous graphite
(ii) Cathode: Transition metal oxide such as CoO₂
(iii) Electrolyte: Lithium salt in an organic solvent

During intramolecular dehydration, if there is a possibility to form a carbon-carbon double bond at different locations, the preferred location is the one that gives the more (highly) substituted alkene i.e., the stable alkene.

When ethylene glycol is treated with cone. H₂SO₄ two molecules of water are removed and the product formed is dioxane.

The conjugate base of HF is F^-
For F^-, \({ K }_{ b }=\frac { { K }_{ w } }{ { K }_{ a } } =\frac { { 10 }^{ -14 } }{ { 6.8\times 10 }^{ -14 } } \)
= 1.47 x 10^-11.

Magnesium has to precipitate as Mg(OH)₂ when treated with NH₄OH, but addition of NH₄CI will suppress the ionisation of NH₄OH due to common ion effect. So ionic product of Mg²⁺ and OH^- ions will be less than the solubility product (K_sp) of Mg(OH)₂. Hence will not precipitate.

\(\underset{0.001M}{HCl}\overset{H_{2}O}{\rightleftharpoons }\underset{0.001M}{H_{3}O^{+}}+\underset{0.001M}{Cl^{-}}\)
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

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

(i) Mn (manganese) shows the maximum number of oxidation state.
(ii) Scandium shows only +3 oxidation state

\({ k }=2.303\log { \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } } \)
\(t=\frac { 2.303 }{ { 10 }^{ -3 } } \log { \frac { 10 }{ 2.5 } } \)
t = 2303 x 0.301 x 2 = 1386 s.

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.

(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.

(i) Nitrogen is separated industrially from liquid air by fractional distillation.
(ii) Pure nitrogen gas can be obtained by the thermal decomposition of sodium azide about 575K.
\(2Na{ N }_{ 3 }\longrightarrow 2Na+{ 3N }_{ 2 }\)

No. of atoms of X at the corners = 8 - 1 = 7
(since 1 is missing)
Contribution of X towards unit cell
\(=7\times \frac { 1 }{ 8 } =\frac { 7 }{ 8 } \)
No. of atoms of Y at the face centres = 6
Contribution of Y towards unit cell
= 6 x \(\frac{1}{2}\) = 3
Ratio of X : Y = \(\frac{7}{8}\) : 3 = 7:24
Formula of the compound is X₇ Y₂₄.

(i) The intermolecular forces of attraction that are present in solids are very strong.
(ii) The constituent particles of solids have fixed position.
(iii) Hence, solids have a definite volume.

(i) Carbon
(ii) Lead (metallic character increases on going down the group).

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.
\(Si{ Cl }_{ 4 }+{ C }_{ 2 }{ H }_{ 5 }OH\longrightarrow \underset { Tetraethoxy\ silane }{ Si({ OC }_{ 2 }{ H }_{ 5 })_{ 4 } } +4HCl\).

(i) Hydraulic washing or gravity separation
(ii) Froth flotation
(iii) Electromagnetic separation
(iv ) Chemical method.

Metal oxides to metal : Carbon, carbon monoxide, hydrogen, aluminum and metals like sodium.

Tetrahedral, paramagnetic.

Tetraammine copper (II) sulphate.

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)

The energy of the two e_g orbitals will increase by  \(\frac{3}{5} \Delta_{O}\) and that of the three t_2g will decrease by (2/5) \(\Delta_{O}\)

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.

i) Na₂[Ni(EDTA)] - Sodium 2, 2', 2",2'" - (ethane-1,2 diyldinitrilo tetraacetatonickelate)(II)
ii) [Ag(CN)₂]^- - Dicyanido-kC-argentate (I) ion
iii) [CO(en)₃]₂(SO₄)₃ - Tris (ethane 1, 2 diamine)cobalt (III) sulphate
iv) [CO(ONO)(NH₃)₅]²⁺ - Pentaamminenitrito - KO cobalt (III) ion
v) [Pt(NH₃)₂Cl(NO₂)] - Diammainechloridonitro - kN - platinum (II)

Diborane adds on to alkenes and alkynes in ether solvent at room temperature. This reaction is called hydroboration.
\({ B }_{ 2 }{ H }_{ 6 }+6RCH=CHR\longrightarrow 2B(RCH_2-{ CH }R)_{ 3 }B\)

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

(i) Cannizzaro reaction is a redox reaction which takes place only in aldehydes which do not have α- hydrogen atom.
(ii) Formaldehyde (HCHO) and benzaldehyde (C₆H₅CHO) do not have a-hydrogen atom and so they undergo Cannizzaro reaction.
(iii) But acetaldehyde CH₃CHO contains 3 α-H atoms and it does not undergo Cannizzaro reaction.

"Adsorption is a phenomenon of concentration of substance on the surface of liquid or solid" (or) "The condition in which concentration of a substance in the interfacial layers between two phases is greater than in the bulk of either phase, then the substance is said to be adsorbed at the interface and the phenomenon is known as adsorption".

Oxidation half cell : A metallic zinc strip that dips into an aqueous solution of zinc sulphate taken in a beaker.
Reduction half cell : A copper strip that dips into an aqueous solution of copper sulphate taken in a beaker.

Glycerol occurs in many natural fats and it is also found in long chain fatty acids in the form of glyceryl esters (Triglycerides). The alkaline hydrolysis of these fats gives glycerol and the reaction is known as saponification.

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.
\({ Ca(OH) }_{ 2 }\overset { { H }_{ 2 }O }{ \rightleftharpoons } { Ca }_{ (aq) }^{ 2+ }+{ 2OH^- }_{ (aq) }\)

Given : I = 1.608A;
t = 50 min = 50 \(\times\) 60 = 3000sec ; S = 250 mL;
C = 0.5 M ; \(\eta=100%\)%
i) Q = It     (I = Q / t)
= 1.608 x 3000 = 4824 Columb
No. of Faradays of electricity \(=\frac{4824}{96500}=0.04 \mathrm{~F}\)
ii) Electrolysis of CuSO₄
\(\mathrm{Cu}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}_{(\mathrm{s})}\)
\(\therefore\) 2 F electricity will deposit 1 mole of Cu²⁺ to Cu ​​​​
\(\therefore\) 0.05 F electricity will deposit 
\(x=\frac{1 \times 0.05}{2}=0.025\) mole
Initial No. of moles of Cu²⁺ in 250 ml of solution \(=\frac{0.5}{1000} \times 250 \) mole = 0.125 moles
\(\therefore\) No. of moles of Cu²⁺ after elettrolysis = 0.125 - 0.025 = 0.1 mole
\(\therefore\) concentration of Cu²⁺
\(=\frac{0.1}{250} \times 1000=0.4 \mathrm{M}\)

(i) The electrolytic cell consists of two iron electrodes dipped in molten sodium chloride and they are connected to an external DC power supply via a key as shown in the figure. The electrode which is attached to the negative end of the power supply is called the cathode, and the one which attached to the positive end is called the anode. Once the key is closed, the external DC power supply drives the electrons to the cathode and at the same time pull the electrons from the anode.

Cell reactions:
Na⁺ ions are attracted towards cathode, where they combines with the electrons and reduced to liquid sodium.

Cathode (reduction)
\(N a_{(l)}^{+}+e^{-} \rightarrow N a_{(l)} \quad ; \quad E^{0}=-2.71 V\)
Similarly, Cl^- ions are attracted towards anode where they lose their electrons and oxidised to chlorine gas.
Anode (oxidation)
2CI^-_(l) ⟶ CI_2(g) + 2e^- E⁰ = -1.36V
The overall reaction is
2Na⁺_(l) + 2Cl^-_(l)➝ 2Na_(l) + Cl_2(g) ; E° = - 4.07V
(ii) The negative E° value shows that the above reaction is a non-spontaneous one.
(iii) Hence, we have to supply a voltage greater than 4.07V to cause the electrolysis of molten NaCI.
(iv) In electrolytic cell, oxidation occurs at the anode and reduction occur at the cathode as in a galvanic cell.
(v) But the sign of the electrodes is the reverse i.e., in the electrolytic cell cathode is -ve and anode is +ve.

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

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

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

Direct reaction of nitrogen with hydrogen gives ammonia. This reaction is favored by high pressures and at optimum temperature in presence of iron catalyst.
\(\cfrac { 1 }{ 2 } { N }_{ 2 }+\cfrac { 3 }{ 2 } { H }_{ 2 }\rightleftharpoons { 2NH }_{ 3 }\)\(\Delta H_f { }_{ }=-46.2KJmol^{ -1 }\) 

i) Crystals have long range repeated pattern of arrangement of constituent particles but in the process of crystallisation.
ii) Some deviations from the ideal arrangement may be introduced because the constituent particles may not get sufficient time to arrange themselves in a perfect order. Therefore, crystals are usually not perfect.

(i) When aluminium metal or aluminium hydroxide is treated with hydrochloric acid, aluminium trichloride is formed.
(ii) The reaction mixture is evaporated to obtain hydrated aluminium chloride.
2Al + 6HCl ⟶ 2AICl₃ + 3H₂
Al(OH)₃ + 3HCl ⟶ AlCl₃ + 3H₂O

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

(i) A catalyst is substance which alters the rate of a reaction without itself undergoing any permanent chemical change. They may participate in the reaction, but again regenerated and the end of the reaction. In the presence of a catalyst, the energy of activation is lowered and hence, greater number of molecules can cross the energy barrier and change over to products, thereby increasing the rate of the reaction.
(ii) The reaction between KMnO₄ and H₂SO₄ and oxalic acid is catalysed by MnSO₄ and increases the rate of oxidation of C₂O₄^2- by MnO₄^-.

(i) The step which has the lowest rate value among the other steps of the reaction is called as the rate determining step (or) rate limiting step: (or)
(ii) The overall rate of a reaction is controlled by the slowest step in a reaction called the rate determining step.
Example:
\(2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}\) going by two steps like,
\( \mathrm{A}+\mathrm{B} \stackrel{\mathrm{k}_{1}}{\longrightarrow} \mathrm{C}+\mathrm{Z}-(1) \text { Step }(\text { slow }) \)
\(Z+A \stackrel{k_{2}}{\longrightarrow} D-(2) \text { Step }(\text { fast }) \)
Over all reaction:  \(2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}\)
Here \(A+B \underset{\text { Slow }}{\stackrel{K_{1}}{\longrightarrow}} C+Z\), step is the rate determining step. For the decomposition of hydrogen peroxide catalysed by I^-.
2H₂O_2(aq)\(\rightarrow\) 2H₂O_(I) + O₂(g)
It is experimentally found that the reaction is first order with respect to both H₂,O₂, and I^-, which indicates that I^- is also involved in the reaction. The mechanism involves the following steps.
Step: 1
H₂O_2(aq)+I^-1_(aq) \(\rightarrow\) H₂O_(l)+OI^-1_(aq)
Step: 2
H₂O_2(aq)+OI^-1_(aq)\(\rightarrow\) H₂O + I^-_(aq) + O_(g)
Overall reaction is
2H₂O_2(aq) \(\rightarrow\) 2H₂O_(l) + O_2(g)
These two reactions are elementary reactions. Adding equation (1), and (2) gives the overall reaction. Step 1 is the rate determining step, since it involves both H₂,O₂ and I^-, the overall reaction is bimolecular.

(i) Fe³⁺ - electronic configuration - [Ar] 3d⁵
(ii) It has exactly half-filled stable electronic configuration.
(iii) Fe²⁺ - electronic configuration -[Ar]3d⁶
(iv) It has only partially filled d-orbitals.
Hence Fe³⁺ is more stable than Fe²⁺.

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

(i) 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

a) if both assertion and reason are true and reason is the correct explanation of assertion.

d) both assertion and reason are false

(i) Compound (A) with molecular formula C₃H₆O that undergoes iodoform reaction is CH₃COCH₃ (A) (acetone).
(ii) Two molecules of (A) react with dry HCl to give compound (B).

(iii) Compound (B) reacts with (A) to give compound (C).

Compound	Compound Name	Formula
A	Acetone
B	Mesityl oxide
C	Phorone

Factors affecting adsorption:
The adsorption is well understood by considering I the various factors affecting it. Qualitatively, the extent of surface adsorption depends on
(i) Nature of adsorbent  
(ii) Nature of adsorbate
(iii) Pressure
(iv) Concentration at a given temperature.
(i) 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.
(ii) Nature of adsorbate:
The nature of adsorbate can influence the adsorption. Gases like SO₂, NH₃, HCI 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.
(iii) Effect of temperature:
Chemical adsorption is fast with increase pressure, it cannot alter the amount. In Physisorption, when pressure increases the amount of adsorption increases.
(iv) Effect of pressure:
When pressure increases, adsorption becomes steadily increase in case of chemisorption and increases in case of physisorption.

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:
\(\frac { { 5.55\times 10 }^{ -3 }\times 1000 }{ 1 } =5.55\times { 10 }^{ -3 }\times { 10 }^{ -3 }\)

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

Kohlraush's law:
(i) At infinite dilution, the limiting molar conductivity of an electrolyte is equal to the sum of the limiting molar conductivities of its constituent ions. i.e., the molar conductivity is due to the independent migration of cations in one direction and anions in the opposite direction.
(ii) For a uni - univalent electrolyte such as NaCl, the Kohlraush's law is expressed as
\(\left(\Lambda_{\mathrm{m}}^{0}\right)_{\mathrm{NaCl}}=\left(\lambda_{\mathrm{m}}^{0}\right)_{\mathrm{Na}^{+}}+\left(\lambda_{\mathrm{m}}^{0}\right)_{\mathrm{Cl}}^{-}\)
(iii) In general, according to Kohlraush's law, the molar conductivity at infinite dilution for a electrolyte represented by the formula A_x B_y, is given below.
\(\left(\Lambda_{m}^{0}\right)_{A_{x} B_{y}}=x\left(\lambda_{m}^{0}\right)_{A^{y+}}+y\left(\lambda_{m}^{0}\right)_{B^{x-}}\)
b) Calculation of molar conductance at infinite dilution for weak electrolytes experimentally.
(i) However, the same can be calculated using Kohlraush's Law. For example, the molar conductance of CH₃COOH, can be calculated using the experimentally determined molar conductivities of strong electrolytes HCl, NaCl and CH₃COONa.
\(\Lambda_{\mathrm{CH}_{3} \mathrm{COONa}}^{0}=\lambda_{\mathrm{Na}^{+}}^{0}+\lambda_{\mathrm{CH}_{3} \mathrm{COO}^{-}}^{0} \) ..........(1)
\(\Lambda_{\mathrm{HCl}}^{0}=\lambda_{\mathrm{H}+}^{0}+\lambda_{\mathrm{Cl}^{-}}^{0} \) .........(2)
\(\Lambda_{\mathrm{NaCl}}^{0}=\lambda_{\mathrm{Na}^{+}}^{0}+\lambda_{\mathrm{Cl}^{-}}^{0}\) ...........(3)
(ii) Equation (1) + Equation (2) - Equation (3) gives,
\(\left(\Lambda_{\mathrm{CH}_{3} \mathrm{COONa}}^{0}\right)+\left(\Lambda_{\mathrm{HCl}}^{0}\right)-\left(\Lambda_{\mathrm{NaCl}}^{0}\right)=\lambda_{\mathrm{H}+}^{0}+\lambda_{\mathrm{CH}_{3} \mathrm{COO}-}^{0} \)
\(=\Lambda_{\mathrm{CH}_{2} \mathrm{COOH}}^{0}\)

(i) The differential rate equation will be \(-\frac { d\left[ R \right] }{ dt } \) = k [A] [B]²
(ii) If the concentration of B is increased three times, then
\(-\frac { d\left[ R \right] }{ dt } \) = k [A] [3B]² = 9.k [A] [B]²
Therefore, the rate of reaction will increase 9 times.
(iii) When the concentration of both A and B are doubled.
\(-\frac { d\left[ R \right] }{ dt } \) = k[A] [B]² 
= k [2A] [2B]²
Therefore, the rate of reaction will increase 8 times.

(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.

Oxidising and bleaching action: 
Chlorine is a strong oxidising and bleaching agent because of the nascent oxygen.
\({ H }_{ 2 }O+{ Cl }_{ 2 }\longrightarrow Hcl+\underset { Hypo\ chlorous \ acid }{ HOCl } \)
HOCI \(\longrightarrow \) HCI + (0)
Colouring matter + Nascent oxygen - 7 Colourless oxidation product.
The bleaching of chlorine is permanent. It oxidises ferrous salts to ferric, sulphites to sulphates and hydrogen sulphide to sulphur.
2FeCl₂ + Cl₂ \(\longrightarrow \) 2FeCl₃
Cl₂ + H₂O \(\longrightarrow \)HCI + HOCI
2FeSO₄ + H₂SO₄ + HOCI \(\longrightarrow \)Fe₂ (SO₄)₃ + HCI + H₂O
Overall reaction
2FeSO₄ + H₂SO₄ + Cl₂ \(\longrightarrow \) Fe₂(SO₄)₃ + 2HCI
Cl₂ + H₂O \(\longrightarrow \) HCI + HOCI
Na₂SO₃ + HOCI\(\longrightarrow \) Na₂SO₄ + HCI
Overall reaction
Na₂SO₃ + H₂O+Cl₂ \(\longrightarrow \) Na₂SO₄ + 2HCI
Cl₂ + H₂S\(\longrightarrow \)2HCI + S

(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) Iron is one of the most useful metals and its alloys are used everywhere including bridges, electricity pylons, bicycle chains, cutting tools and rifle barrels.
(ii) Cast iron is used to make pipes, valves and pumps stoves etc.
(iii) Magnets can be made from iron and its alloys and compounds.
(iv) An important alloy of iron is stainless steel, and it is very resistant to corrosion. It is used in architecture, bearings, cutlery, surgical instruments and jewellery.
(v) Nickel steel is used for making cables, automobiles and aeroplane parts.
(vi) Chrome steels are used for manufacturing cutting tools and crushing machines.

(i) It exhibits both geometrical and optical isomerism
(a) Geometrical isomers:

(b) Optical isomers:

(ii) It shows two optical isomers

(iii) Ionisation isomers
Linkage isomers
[Co(NH₃)₅(NO₂)(NO₃)₂], [CO(NH₃)₅ (ONO)](NO₃)₂
(iv) Geometrical isomers