coagulating power ∝ 1/coagulation value

(E_cell)_A = -1.82 + 1.5 = -0.32V
(E_cell)_B = -1.5 + 1.595 = + 0.095V
(E_cell)_C = -1.595 + 1.0652 = -0.529V
The species undergoing disproportionation is HBrO

E^o_Zn²⁺[Zn] = 0.76V and E^o_Fe²⁺[Fe] = -0.44V
Zinc has higher negative electrode potential than iron, iron cannot be coated on zinc

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

\(\mathrm{Ag}_{2} \mathrm{C_2O_4} \rightleftharpoons 2 \mathrm{Ag}_{}^{+}+\mathrm{C_2O}_{4{}}^{2-}\)
\(\left[\mathrm{Ag}^{+}\right]=2 .24 \) ×10^-4mol L^-1
\(\mathrm{C_2O}_{4{}}^{2-} = \frac {2.24 \times 10 ^{-4}}{2}\) mol L^-1
= 1.12 ×10^-4mol L^-1
K_sp = [Ag]² [C₂O₄^2-]
= (2.24 ×10^-4mol L^-1) (1.12 ×10^-4mol L^-1)
= 5.619 × 10^-12mol³L^-3

Given:
At low pressure the reaction follows first order therefore,
Rate α [reactant]¹
Rate α (surface area)
At high pressure due to the complete coverage of surface area, the reaction follows zero order.
Rate α [reactant]⁰
Therefore the rate is independent of surface area.

sc ⇒ 2r = a
⇒ r = a/2
bcc ⇒ 4r = \( \sqrt{3a} \) ⇒ r = \(\frac { \sqrt{3a}} {4}\)
fcc ⇒ 4r ⇒  \( \sqrt{2a} \) ⇒ r = \(\frac { \sqrt{2a}} {4} = \frac { a} {2\sqrt{2a}}\)
\(\left( \frac { a }{ 2 } :\frac { \sqrt { 3 } }{ 4 } a:\frac { a }{ 2\sqrt { 2 } } \right) \)

\(k=\frac { 2.303 }{ t } log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
\(k=\frac { 1 }{ t } ln \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
\(e^{kt}=\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] }\)
[A] = [A₀] e^kt
In this case
k = x min^-1 and [A₀]  = 0.01 M = 1 x 10^-2M
t = 1 hour = 60 min
[A] = (1 x 10^-2)e^-60x

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

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) Iodoform test
(ii) Tollens test
(iii) Bendicts test.

(i) Ethanal (CH₃CHO) is more reactive towards nucleophilic addition reaction than propanone (\({ CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -{ CH }_{ 3 }\)) because of steric and electronic reasons.
(ii) Greater the number of alkyl groups attached, less will be the reactivity of carbonyl compound towards nucleophilic addition.
(iii) Soethanal is more reactive than propanone.

(i) Temperature
(ii) Pressure
(iii) Nature of the gas and
(iv) Nature of the adsorbent

Emulsion can be separated into two separate layers. The process is called Deemulsification.

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)

IF = 96500 C
It is defined as the quantity of electricity which deposits one gram equivalent of the substance or it is the charge carried by one mole of electrons

When ethylene reacts with alkaline KMnO₄ solution (Baeyer's reagent), it forms glycol (diol).

Alcohols react with carboxylic acids in the presence of an acid to give esters.

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.

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

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

The oxidation potential of M₁ is more +ve than the oxidation potential of Fe which indicates that it will prevent iron from rusting.

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.

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

(i) According to Hume-Rothery rule to, form a substitute alloy the difference between the atomic radii of solvent and solute is less than 15%.
(ii) Both the solvent and solute must have the same crystal structure and valence and their electro negativity difference must be close to zero.

(i) When orange solution containing Cr₂O₇^2- ion is treated with an alkali, a yellow solution of Cr₂O₇^2- is obtained.
(ii) Similarly, when H⁺ ions are added to yellow solution, an orange solution of Cr₂O₇^2- is obtained due to interconversion.

(i) Catalyst lowers the activation energy and changes the path of the reaction and thus increases the rate of the reaction. A catalyst does not alter Gibbs energy of a reaction.
(ii) Higher the temperature faster the reaction. This is because increasing the temperature increases the Kinetic energy, and hence more molecules would have energy greater than the activation energy resulting in increased rate and completion of the reaction. Hence oxalic acid should be heated in redox titration against KMnO₄.

The order of the reaction is zero.

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) Fluorine due to its small size high electronegativity and low F-F bond energy is more reactive than CIF₃
(ii) While CI-F bond in CIF₃ is weaker than CI-CI bond in Cl₂ therefore, CIF₃ is more reactive than Cl₂.
(iii) So these graphite rods are consumed slowly and need to be replaced from time - to - time

(i) The electron gain enthalpy for oxygen is less negative because of its small size due to which the electron repulsions in the relatively small 2p-subshell are comparatively large.
(ii) Hence the Incoming electrons are not accepted with the same ease as in case of sulphur as it has relatively large size.

There are seven primitive crystal systems; cubic, tetragonal, orthorhombic, hexagonal, monoclinic, triclinic and rhombohedral.

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

(i) Boric acid reacts with calcium fluoride in presence of conc. sulphuric acid and gives boron trifluoride.
3CaF₂ + 3H₂SO₄ + 2B(OH)₃ ⟶ 3CaSO₄ + 2BF₃ + 6H₂O
(ii) Borax when heated with soda ash it gives borax
Na₂CO₃ + 4B(OH)₃ ⟶ Na₂B₄O₇ +.CO₂ + 6H₂O

(i) The standard free energy for the formation of CuFeS₂ (copper pyrites) is greater than those of CS₂ and H₂S.
(ii) Hence, CuFeS₂ cannot be reduced carbon or hydrogen.
(iii) However, the free energy of copper oxide less than that of CO₂,
(iv) That is the reason, extraction of copper easier from its oxide through reduction.

(i) Iron is used in the manufacture of wrought iron and steel.
(ii) Cast iron is used for casting stoves, railway sleepers, toys, etc.
(iii) Stainless steel is used for cycles, automobiles, utensils, pens, etc.
(iv) Chrome steel is used for cutting tools and crushing machines

MA₂B₂
[Pt(NH₃)₂Cl₂]

(i) The central atom or ion is the one that occupies the central position in a coordination entity and binds other atoms or groups of atoms (ligands) to itself, through a coordinate covalent bond.
(ii) For example, in K₄[Fe(CN)₆], the central metal ion is Fe²⁺. In the coordination entity [Fe(CN)₆]^4-, the Fe²⁺ accepts an electron pair from each ligand, CN^- and thereby forming six coordinate covalent bonds with them.
(iii) It is referred to as a Lewis acid.

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)

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)

(i) The general electronic configuration of 15 group elements is ns²np³.
(ii) Nitrogen \(\Rightarrow 2 \mathrm{~s}^{2} 2 \mathrm{p}^{3}\) (Valence shell)
(iii) Phosphorus \(\Rightarrow 3 \mathrm{~s}^{2} 3 \mathrm{p}^{3}\)
(iv) Arsenic \(\Rightarrow 4 \mathrm{~s}^{2} 4 \mathrm{p}^{3}\)
(v) Antimony \(\Rightarrow 5 s^{2} 5 p^{3}\)
(vi) Bismuth \(\Rightarrow 6 \mathrm{~s}^{2} 6 \mathrm{p}^{3}\)

(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

(i) Fluorine is more electro negative than other halogens.
(ii) Because except fluorine all the other halogens have positive oxidation state.
(iii) It has high electron affinity character
Example:
\(\mathrm{Cl}_{2}\mathrm{O} \Rightarrow+1 \text { oxidation state }(\mathrm{Cl}) \)
\(\mathrm{OF}_{2} \Rightarrow-1 \text { oxidation state }(\mathrm{F})\)

The reaction of CO with hydrogen at a pressure of less than 50 atm using metal catalysts at 500 - 700 K yields saturated and unsaturated hydrocarbons.
\(nCO+(2n+1){ H_2 }\longrightarrow C_{ n }{ H }_{ (2n+2) }+{ nH }_{ 2 }O\)
\(nCO+2n{ H }_{ 2 }\longrightarrow { C }_{ n }{ H }_{ 2n }+{ nH }_{ 2 }O\)

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

(i) When reduced with hydrogen in the presence of 'poisoned' palladium catalyst, they form aldehydes. This reaction is called Rosenmund reduction.
\(\underset { acetyl\ chloride }{ { CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -Cl } +H-H\overset { Pd }{ \underset { Ba{ SO }_{ 4 } }{ \longrightarrow } } \underset { acetaldehyde }{ CH_{ 3 }-\underset { \overset { || }{ O } }{ C } -H+HCl } \) 
(ii) BaSO₄ is used as a catalytic poi on to stop the reduction at the stage of aldehyde.
(iii) Otherwise, the aldehyde formed will be further reduced to primary alcohol.

On heating phenol with phthalic anhydride in presence of con.H₂SO₄ phenolphthalein is obtained.

(i) Using a colloid mill, the solid is ground to colloidal dimension.
(ii) The colloid mill consists of two metal plates rotating in opposite direction at very high speed of nearly 7000 revolution I minute.
(iii) The colloidal particles of required colloidal size is obtained by adjusting the distance between two plates.
(iv) By this method, colloidal solutions of ink and graphite are prepared.

Solubility can be calculated from the molar solubility i.e., the maximum number of moles of solute that can be dissolved in one litre of the solution.
For a solute X_mY_n,
\({ X }_{ m }{ Y }_{ n(s) }\rightleftharpoons m{ X }_{ (aq) }^{ n+ }+{ nY }_{ (aq) }^{ m- }\)
From the above stoichiometrically balanced equation we have come to know that 1 mole of X_m Y_n(s) dissociated to furnish 'm' moles of Xⁿ⁺ and 'n' moles of Y^m- if 's' is molar solubility of X_mY_n then
[Xⁿ⁺] = ms and [Y^m-] = ns
ஃK_sp = [Xⁿ⁺]^m [Y^m-]ⁿ
K_sp = (ms)^m (ns)ⁿ
K_sp = (m)^m (n)ⁿ (s)^m+n

Action of conc H₂SO₄ : On treating with cold cone H₂SO₄, it decomposes to form manganese heptoxide, which subsequently decomposes explosively. 

But with hot Cone H₂SO₄, potassium permanganate give MnSO₄

Let the concentration of the reactant be [A] = a
Rate of reaction R = k [A]²
R=ka²
(i) R = k (2a)²
= 4 ka²
= 4R
∴ The rate of the reaction would increase by 4 times.
(ii) If the concentration of the reactant is reduced to half [A] = \(\frac { 1 }{ 2 } \) a than the rate of reaction would be
\(R=k\left( \frac { 1 }{ 2 } a \right) 2\)
\(=\frac { 1 }{ 4 } k{ a }^{ 2 }\)
\(=\frac { 1 }{ 4 } kR\)
∴ The rate of the reaction would reduced by \({ \frac { 1 }{ 4 } }^{ th }\)

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

Simple Cubic Lattice
100,101,111

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

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

(i) Leaching of sulphide ores such as ZnS, PbS etc., can be done by treating them with hot aqueous sulphuric acid
\(2Zn{ S }_{ (s) }+{ 2H }_{ 2 }{ SO }_{ 4(aq) }+{ O }_{ 2(g) }\longrightarrow { 2ZnSO }_{ 4(aq) }+2{ S }_{ (s) }+{ H }_{ 2 }O\)
(ii) In this process the insoluble sulphide is converted into soluble sulphate and elemental sulphur

If the monomer is represented by X. Then
\(2 \mathrm{M} \rightarrow(\mathrm{M})_{2}\)
Since the reaction is of second order, the rate of reaction will be given by,
\(\text { Rate }=\mathrm{k}[\mathrm{M}]^{n}  \)
\(7.5 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}=\mathrm{k}\left(0.05 \mathrm{~mol} \mathrm{} \mathrm{~L}^{-1}\right)^{2} \)
\(\mathrm{k} =\frac{Rate}{[M]^{n}}  \)
\(\mathrm{k} =\frac{7.5 \times 10^{-3}}{(0.05)^{2}}  \)
\(=3 \mathrm{~mol}^{-1} \mathrm{~L} \mathrm{~s}^{-1}\)

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) The actual position of Lanthanides in the periodic table is at group number 3 and period number 6. However, in the sixth period after lanthanum, the electrons are preferentially filled in inner 4f sub shell and these fourteen elements following lanthanum show similar chemical properties.
(ii) Similarly the fourteen elements following actinium resemble in their physical and chemical properties. Hence they are placed separately bottom of the modern periodic table.

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

(i) An organic compound (A) is identified as C₆H₅CHO benzaldehyde. Benzaldehyde reduces Tollen'sreagent and also undergoes.
Cannizaro reaction:
\(\\ \underset { (A) }{ { C }_{ 6 }{ H }_{ 5 }CHO } +{ Ag }_{ 2 }O\longrightarrow 2Ag+{ C }_{ 6 }{ H }_{ 5 }COOH\) 
Cannizaro reaction:
\({ C }_{ 6 }{ H }_{ 5 }CHO+{ C }_{ 6 }{ H }_{ 5 }CHO\overset { NaOH }{ \longrightarrow } { C }_{ 6 }{ H }_{ 5 }{ OH }OH+\underset { (B) }{ { C }_{ 6 }{ H }_{ 5 }COOH } \) 
(ii) Benzaldehyde on oxidation gives Benzoic acid C₆H₅COOH and it is (B).
\({ C }_{ 6 }{ H }_{ 5 }CHO\overset { (O) }{ \longrightarrow } \underset { (B) }{ { C }_{ 6 }{ H }_{ 5 }COOH } \) 
(iii) Calcium salt of benzoic acid (calcium benzoate) on dry distillation gives benzophenone C₆H_sCOC₆H_s as (C).

(i) Compound (A) is carbonyl compound, it is acetone
(ii) (A) on reduction with LiAlH₄ gives (B) it gives blue colour in Victor Meyer'stest.
\({ CH }_{ 3 }-\underset { \overset { || }{ \underset { (A) }{ O } } }{ C } -{ CH }_{ 3 }\overset { { LiAIH }_{ 4 } }{ \underset { \left[ H \right] }{ \longrightarrow } } { CH }_{ 3 }-{ CH }_{ 3 }-\underset { \overset { | }{ \underset { (B) }{ OH } } }{ CH } -{ CH }_{ 3 }\) 
(iii) (B) reacts with SOCl₂to give (C).

(iv) (C) on treatment with alcoholic KOH, forms (D) by elimination reaction.
\({ CH }_{ 3 }-\underset { \overset { | }{ \underset { (C) }{ Cl } } }{ C } H-{ CH }_{ 3 }\overset { alc.KOH }{ \longrightarrow } \underset { (D) }{ { CH }_{ 3 }CH={ CH }_{ 2 }+HCl } \) 

Helmholtz double layer:
(i) The surface of colloidal particle adsorbs one type of ion due to preferential adsorption.
(ii) This layer attracts the oppositely charged < ions in the medium and hence at the boundary separating the two electrical double layers are setup.
(iii) This is called as Helmholtz electrical double layer.
(iv) As the particies nearby are having similar: charges, they cannot come close and condense.

Given:
E = - 0.79 V
n = 2 [Zn⁺] = 0.095 M
Formula: \(E={ E }^{ o }-\frac { 0.0591 }{ n } \log{ Zn }^{ 2+ }\)
Solution:
\(E={ { E }_{ { Zn }^{ 2+ }/Zn }^{ 0 } }=E+\frac { 0.0591 }{ n } \log{ Zn }^{ 2+ }\)
\(E={ { E }_{ { Zn }^{ 2+ }/Zn }^{ 0 } }=-0.76+\frac { 0.0591 }{ n } \log{ 0.095 }\)
= - 0.79 + 0.02889 = - 0.76 V
E^o = - 0.76 V.

pH = -log [H⁺]
∴ [H⁺] = antilog [-pH]
= antilog [-7.41]
∴ [H⁺] = 3.9 x 10^-8 M.

(i) Enzymes are complex protein molecules with three dimensional structures. They catalyse the chemical reaction in living organism. They are often present in colloidal state and extremely specific in catalytic action. Each enzyme produced in a particular living cell can catalyse a particular reaction in the cell.
Some common examples for enzyme catalysis:
(ii) The peptide glycyl L-glutamyl L-tyrosin is hydrolysed by an enzyme called pepsin.
(iii) The enzyme diastase hydrolyses starch into maltose
\(2\left(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{5}\right)_{\mathrm{n}}+\mathrm{nH}_{2} \mathrm{O} \rightarrow \mathrm{nC}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\)
(iv) The yeast contains the enzyme zymase which converts glucose into ethanol.
\(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \rightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+2 \mathrm{CO}_{2}\)
(v) The enzyme micoderma aceti oxidises alcohol into acetic acid.
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{O}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{H}_{2} \mathrm{O}\)
(vi) The enzyme urease present in soya beens hydrolyses the urea.
\(\mathrm{NH}_{2}-\mathrm{CO}-\mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{NH}_{3}+\mathrm{CO}_{2}\)
Mechanism of enzyme catalysed reaction
(vii) The following mechanism is proposed for the enzyme catalysis
\(​​​​​​​\mathrm{E}+\mathrm{S} \rightleftharpoons \mathrm{ES} \rightarrow \mathrm{P}+\mathrm{E}\).
(viii) Where E is the enzyme, S the substrate (reactant), ES represents activated complex and P the products.

 

(i) The dissociation of a weak acid (CH₃COOH) is suppressed in the presence of a salt (CH₃COONa) containing an ion common to the weak electrolyte. It is called the common ion effect.
(ii) Consider the dissociation of a weak acid, acetic acid whose ionisation is incomplete.
CH₃COOH_(aq) ⇌ H⁺_(aq) + CH₃COO^-_(aq)
(iii) If the salt sodium acetate with common ion CH₃COO^- is added to the above equilibrium, it dissociates completely increasing.
CH₃COONa_(aq)⟶Na⁺_(aq) + CH₃COO^-_(aq)
(iv) Hence, the overall concentration of CH₃COO^- is increased, and the acid dissociation equilibrium is disturbed.
(v) So, in order to maintain the equilibrium, the excess CH₃COO^- ions combines with H⁺ ions to produce much more unionized CH₃COOH i.e, the equilibrium will shift towards the left. In other words, the dissociation of CH₃COOH is suppressed.

The reaction is 2NH_3(g) \(\overset { Pt }{ \longrightarrow } \) N_2(g) + 3H_2(g)
Here k = 2.5 x 10^-4 mol L^-1 s^-1
The order of the reaction is zero i.e.,
Rate = k[Reactant]^o
Rate = 2.5 x 10^-4 x 1 = 2.5 x 10^-4 mol L^-1 s^-1
\(\therefore \frac { d }{ dt } \left[ { H }_{ 2 } \right] =\frac { 1 }{ 3 } \frac { d }{ dt } \left[ { H }_{ 2 } \right] \)
The rate of formation of N₂ = 2.5 x 10^-4 mol L^-1 s^-1
\(\therefore 2.5\times { 10 }^{ -4 }=\frac { 1 }{ 3 } \frac { d }{ dt } \left[ { H }_{ 2 } \right] \)
\(\therefore \frac { d }{ dt } \left[ { H }_{ 2 } \right] \) = 7.5 x 10^-4
Therefore, rate of formation of H₂ = 7.5 x 10^-4 mol L^-1 s^-1

The heavier transition elements belong to fourth (4d), fifth (Sd) and sixth (6d) transition series. Their properties are expected to be different form the elements belonging to the first (3d) series due to the following reasons.
(i) Atomic radii: Size of the transition elements 94d and Sd series are larger than those of the corresponding elements of the first transition series though those of 4d and Sd series are very close to each other.
(ii) Ionisation enthalpy of Sd series are higher than the corresponding elements of 3d and 4d series.
(iii) Atomisation enthalpy of 4d and Sd series are higher than the corresponding elements of the first series.
(iv) Melting and boiling points of heavier transition elements are greater than those of the first transition series due to stronger intermetallic bonding.
(v) The elements of the first transition series generally form low or high spin complexes, depending upon the higher of ligand field. However, the heavier transition elements form low spin complexes irrespective of the strength of the ligand filed.

(a) Rhombic Sulphur (α - Sulphur):
(a) It is yellow in colour.
(b) Its melting point is 385.8K and specific gravity is 2.06
(c) It is stable form of sulphur at room temperature.
(d) It is formed on evaporating the solution of sulphur in CS₂.
(e) It in insoluble in water, readily soluble in CS₂ and dissolves to some extent in benzene, alcohol and ether.
(b) Monoclinic sulphur \(\left( \beta -sulphur \right) \):
(a) Its melting point is 393K and specific gravity is 1.98
(b) It is prepared by melting rhombic sulphur in a dish and cooling, till crust is formed. Two holes are made in crust and remaining liquid is powered out. On removing crust, colourless needle - shaped crystals of β - sulphur is formed.
(c) Monoclinic sulphur is stable above 369K and below 369K α - sulphur is stable.
(d) At 369K both forms are stable and this temperature is called transition temperature.
(e) Both rhombic and monoclinic sulphur have S₈ molecules, these are packed to give different crystal structure S₈ form is puckered and has crown shape.
Several other modifications containing 6-20 sulphur atoms per ring are synthesised 

(f) In Cyclo-S₆ the ng adopts chair form.

(g) At elevated temperatures (～1000K), S₂ is dominant species and is, paramagnetic like O₂

(i) Similarities between ionic and metallic crystals:
(a) Ionic and metallic crystals have electrostatic forces of attraction.
(b) Both the crystals exhibit high melting point.
(c) The bonds in metallic and ionic crystals are non-directional.
(ii) Differences between ionic and metallic crystals:

Silicates are classified into various types based on the way in which the tetrahedral units, [SiO₄]^4- are linked together.
(i) Ortho silicates (Neso silicates):
The simplest silicates which contain discrete [SiO₄]^4- tetrahedral units are called ortho silicates or nesosilicates.
Examples: Phenacite - Be₂SiO₄ (Be²⁺ ions are tetrahedrally surrounded by O^2- ions)
(ii) pyro silicate (or) Soro silicates: Silicates:
Which contain [Si₂O₇]^6- ions are called pyro silicates (or) Soro silicates.
Example: Thortveitite - Sc₂Si₂O₇
(iii) Cyclic silicates (or Ring silicates): 
Silicates which contain (SiO₃)₃^2n- ions which are formed by linking three or more tetrahedral SiO₄^4- units cyclically are called cyclic silicates.
Example: Beryl [Be₃Al₂ (SiO₃)₆] (an aluminosilicate with each aluminium is surrounded by 6 oxygen atoms octahedrally)
(iv) Inosilicates: Silicates which contain 'n':
number of silicate units liked by sharing two or more oxygen atoms are called inosilicates.
Example: They are further classified as chain silicates and double chain silicates.
(v) Chain silicates (or pyroxenes):
These silicates contain [(SiO₃)_n]^2n- ions formed: by linking 'n' number of tetrahedral [SiO₄]^4- units linearly. Each silicate unit shares two of its oxygen atoms with other units.
Example: Spodumene - LiAl(SiO₃)₂·
(vi) Double chain silicates (or amphiboles): 
These silicates contains \(\left[ { Si }_{ 4 }{ O }_{ 11 } \right] _{ n }^{ 6n- }\) ions. In these silicates there are two different types of tetrahedra:
(a) Those sharing 3 vertices
(b) those sharing only 2 vertices.
Example:
Asbestos: These are fibrous and non-combustible silicates.
(vii) Sheet or phyllo silicates: 
Silicates which contain \(({ Si }_{ 2 }{ O }_{ 5 })_{ n }^{ 2n- }\) are called sheet or phyllo silicates. In these, Each [SiO₄]^4- tetrahedron unit shares three oxygen atoms with others and thus by forming two dimensional sheets.
Example: Talc, Mica etc.
(viii) Three dimensional silicates (or tectosilicates): 
Silicates in which all the oxygen atoms of [SiO₄]^4- tetrahedra are shared with other tetrahedra to form three dimensional network are called three dimensional or tectosilicates.
Example: Quartz.

(i) The impure nickel is heated in a stream of carbon monoxide at around 350 K.
(ii) The nickel reacts with the CO to form a highly volatile nickel tetracarbonyl.
(iii) The solid impurities are left behind 
\({ Ni }_{ (s) }+4{ CO }_{ (g) }\longrightarrow { Ni(CO) }_{ 4(g) }\)
(iv) On heating the nickel tetracarbonyl around 460 K, the complex decomposes to give pure metal.
\({ Ni(CO) }_{ 4(g) }\longrightarrow { Ni }_{ (s) }+{ 4CO }_{ (g) }\)

a) [FeF₆]^4-: Fe atom - outer electronic configuration 3d⁶ 4s²

F^- is weak field ligand
In [FeF₆]^4-the hybridisation takes place is sp³d²
The number of unpaired electrons = 4.
\(\therefore \mu =\sqrt { 4(4+2) } =\sqrt { 24 } \)
The molecule is paramagnetic due to the presence of unpaired electrons.
The geometry of the molecule is octahedral.
b) [Fe(CN)₆]^4-

In [Fe(CN)₆]^4- complex, the CN^- ligand is a powerful ligand, it forces the unpaired electrons in the 3d level to pair up inside.
Hence the species has no unpaired electron after hybridisation So the molecule is diamagnetic.

The geometry of the molecule is octahedral.

Most of the elements exhibit, two types of valence namely primary valence and secondary valence and each element tend to satisfy both the valences.
The primary valence is referred the oxidation state of the metal atom.
The secondary valence as the coordination number. For example, according to Werner, the primary and secondary valences of cobalt are 3 and 6 respectively.
The primary valence of a metal ions ae always satisfied by negative ions.
For example in the complex CoCI₃.6NH₃. The primary valence of Co is +3 and is satisfied by 3CI^- ions.
The secondary valence is satisfied by negative ions, neutral molecules, positive ions or the combination of these.
For example, in CoCl₃.6NH₃ complex primary valence of cobalt +3 and it is satisfied by 3 CI^-.
The secondary valence of cobalt is 6 and is satisfied by six neutral ammonia molecules. where as in CoCI₆.NH₃. 
Secondary valence of Co = 5{It is satisfied five neutral molecules and a Cl^- ion}
According to Werner, there are two spheres of attraction around a metal atom/ion in a complex.

The inner /coordination sphere:
The groups present in this sphere are firmly attached to the metal.
The outer sphere / ionisation sphere:
The groups present in this sphere are loosely bound to the central metal ion and hence can be separated into ions upon dissolving the complex in a suitable solvent.
The primary valencies are non-directional. while the secondary valencies are directional.
The geometry of the complex is determined by the special arrangement of the groups which satisfy the secondary valence.

B) if both assertion and reason are true but reason is not the correct explanation of assertion.

d) Both assertion and reason are false