ΔS is -ve

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

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

MnO^-₄ + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O

(i) For a strong electrolyte, at high concentration, the number of constituent ions of the electrolyte in a given volume is high and hence the attractive force between the oppositely charged ions is also high.
(ii) Moreover the ions also experience a viscous drag due to greater solvation.
(iii) These factors attribute for the low molar conductivity at high concentration.

(i) Neutralisation of chargers of ion will taken place and hence precipitation will take place (ie) Fe³⁺ and S^2- ion changes are neutralized. No new compounds are formed.
(ii) Fe(OH)₃ is a positive Sol
(iii) As₂S₃ is a negative Sol

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

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

(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.

The conversion of ore into metal oxide (oxidation) is called calcination. It is the process in which the ore is subjected to the action of heat at high temperature in the absence of air below its melting point. Example
\({ CaCO }_{ 3 }(limestone)\longrightarrow CaO+{ CO }_{ 2 }\uparrow \)
\({ MgCO }_{ 3 }\left( Magnesite \right) \longrightarrow { MgO }+{ CO }_{ 2 }\uparrow \)

(i) Complex with co-ordination number 8 - Cu₂ [ZrF₈].
(ii) Complex with co-ordination number 7 - (NH₄)₃[ZrF₇].

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.

(i) A - C₆H₅NO₂ - Nitro benzene
(ii) B - C₆H₅NH₂ - Aniline
(iii) C - C₆H₅NC - Phenyl Carbylamine

LDPE is formed by heating ethene at 200° to 300°C under oxygen as a catalyst. The reaction follows free radical mechanism. The peroxides formed from oxygen acts as a free radical initiator.

(i) In the complete cell diagram, the two half cells are separated by a double vertical line (salt bridge) in between.
(ii) The zinc-copper cell can now be written as 

A - CH₃I- Methyl Iodide
B - C₆H₅OH - Phenol
C - C₆H₅ONa - Sodium Phenoxide

Tertiary alcohols do not undergo oxidation reaction under normal conditions, but at elevated temperature under strong oxidizing, agent cleavage of C-C-bond takes place to give a mixture of carboxylic acid.

When two are more solution are mixed, the resulting concentrations are different from the original.
\(\text { Molarity }=\frac{n}{\mathrm{~V}} \text { (or) } \mathrm{n}=\text { Molarity } \times \mathrm{v}  \)
Total Volume of the mixture = 0.15 + 0.1
= 0.25 L
\(\underset{0.1M}{Pb(NO_{3})_{2}}\rightleftharpoons \underset{0.1M}{Pb^{2+}}+2\underset{0.2M}{2NO^{-}_{3}}\)
n_Pb²⁺ \(=0.1\times0.15=0.015 \ mol\)
\([Pb^{2+}]_{mix}= \frac{n}{v} = \frac{0.1\times0.15}{0.25}=0.06M\)
\(\underset{0.2M}{NaCl}\rightleftharpoons \underset{0.2M}{Na^{+}}+\underset{0.2M}{Cl^{-}}\)
\(\mathrm{n}_{\mathrm{Cl^-}}=0.2 \times 0.1=0.02 \mathrm{~mol} \)
\(\left[\mathrm{Cl}^{-}\right]_{\text {mix }}=\frac{0.02}{0.25}=0.08 \mathrm{M} \)
\(\therefore Ionic \ Product =\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^{2} \)
\(=0.06 \times(0.08)^{2} \)
\(IP =3.84 \times 10^{-4}\)
\(\therefore 3.84 \times 10^{-4}>1.2 \times 10^{-5}\)
(or) \(\mathrm{IP}>\mathrm{K}_{\mathrm{sp}}\)
\(\therefore\) PbCl₂ will be precipitated.

(i) Cr₂O₇^2- + 2OH^- ⟶ 2CrO₄^2- + H₂O
(ii) MnO₄^- + 4H⁺ + 3e^- ⟶ MnO₂ + 2H₂O

In Ellingham diagram, the graph of \(CO\rightarrow { CO }_{ 2 }\) conversion remains below \(Fe\rightarrow { Fe }_{ 2 }{ O }_{ 3 }\)  upto 1073 K
(for Fe\(\longrightarrow \) FeO). SO, CO (g) act as reducing agent upto this temperature.
\(3{ Fe }_{ 2 }{ O }_{ 3 }+CO\longrightarrow 2{ Fe }_{ 3 }{ O }_{ 4 }\) (i.e. FeO.Fe₂O₃) + CO₂
\({ Fe }_{ 3 }{ O }_{ 4 }+CO\longrightarrow 3Feo+{ CO }_{ 2 }\)
FeO + CO \(\longrightarrow \) Fe + CO₂
Also, graph of C\(\longrightarrow \) CO is below the graph of
Fe \(\longrightarrow \)Fe₂O₃ after 1123 K. So, carbon acts as reducing agent above this temperature.
Fe₂O₃ + C\(\longrightarrow \) 3 CO + 2 Fe.

(i) Ionic solids have high melting points.
(ii) These solids do not conduct electricity, because the ions are fixed in their lattice positions.
(iii) They are hard so strong external force can change the relative positions of ions.

A hydride of 2^nd period alkali metal (A) is lithium hydride (LiH).
Lithium hydride (A) reacts with diborane (B) to give lithium borohydride (C) which is acts as a reducing agent.
B₂H₆ + 2 LiH \(\xrightarrow[]{ether}\) 2 LiBH₄
[Diborane (B)]   [Lithium hydride (A)]  [Lithium borohydride (C)]
Result:

(i) Hydrolysis: Acid anhydride are slowly hydrolysed, by water to form corresponding carboxylic acids.

(ii) Reaction with alcohol:

Condensation Methods:
Various chemical methods for the formation of colloidal particles.
(i) Oxidation:
Sols of some non-metals are prepared by this method. (a) When hydroiodic acid is treated with iodic acid, 12 sol is obtained.
\({ HIO }_{ 3 }+5HI\longrightarrow { 3H }_{ 2 }O+{ I }_{ 2 }\) (Sol)
(ii) Reduction:
Many organic reagents like phenyl hydrazine, formaldehyde, etc are used for the formation of sols. For example: Gold sol is prepared by reduction of auric chloride using formaldehyde.
\(2{ AuCl }_{ 3 }+3HCHO+{ 3H }_{ 2 }O\longrightarrow 2Au\left( sol \right) +6HCl+3HCOOH\) 
(iii) Hydrolysis: 
Sols of hydroxides of metals like chromium and aluminium can be produced by this method.
For example,
\({ FeCl }_{ 3 }+3H_{ 2 }O\longrightarrow Fe(OH)_{ 3 }+3HCl\) 
(iv) Double decomposition: 
For the preparation of water insoluble sols this method can be used. When hydrogen sulphide gas is passed through a solution of arsenic oxide, a yellow coloured arsenic sulphide is obtained as a colloidal solution.
\({ As }_{ 2 }{ O }_{ 3 }+3{ H }_{ 2 }S\longrightarrow { AS }_{ 2 }{ S }_{ 3 }+{ 3H }_{ 2 }O\) 
(v) Decomposition:
When few drops of an acid is added to a dilute solution of sodium thio sulphate, the insoluble free sulphur produced by decomposition of sodium thiosulphate accumulates into small, clusters which impart various colours blue, yellow and even red to the system. depending on their growth within the size of colloidal dimensions.
\({ S }_{ 2 }{ O }_{ 3 }^{ 2- }+{ 2H }^{ + }\longrightarrow \underset{sol}S+{ H }_{ 2 }O+{ SO }_{ 2 }\) 
By exchange of solvent:
(i) Colloidal solution of few substances like phosphorous or sulphur is obtained by preparing the solutions in alcohol and pouring them into water.
(ii) As they are insoluble in water, they form colloidal solution.
P in alcohol + water ⟶ Psol.

(i) The first transition metal Scandium exhibits only +3 oxidation state, but all other transition elements exhibit variable oxidation states by loosing electrons from (n-1)d orbital and ns orbital as the energy difference between them is very small.
(ii) At the beginning of the series, +3 oxidation state is stable but towards the end +2 oxidation state becomes stable.
(iii) The number of oxidation states increases with the number of electrons available, and it decreases as the number of paired electrons increases.
(iv) Hence, the first and last elements show less number of oxidation states and the middle elements with more number of oxidation states.
(v) For example, the first element Sc has only one oxidation state +3; the middle element Mn has six different oxidation states from +2 to +7. The last element Cu shows +1 and +2 oxidation states only.
(vi) The relative stability of different oxidation - states of 3d metals is correlated with the extra stability of half filled and fully filled electronic configurations. Example: Mn²⁺(3d⁵) is more stable than Mn⁴⁺(3d³).

(i) The element which occupies group number 15 and period number 3 is phosphorus. Therefore A is phosphorus. Phosphorus reacts with chlorine to give PCI₃There fore compound B is phosphorus trichloride and it has a pyramidal shape.
P₄ + 6Cl₂ \(\longrightarrow \) 4PCl₃
PCl₃ further reacts with Cl₂ to give PCl₅
Therefore, the compound C is phosphorus pentachloride and it has a trigonal bipyramidal shape.
PCI + CI\(\longrightarrow \) PCI
Thus, A= Phosphorus
B = Phosphorus trichloride (PCI₃)
C = Phosphorous pentachloride (PCI₅)

(i) The colour develops because of the presence of electrons in the 8 anionic sites.
(ii) These electron absorb energy from the visible region of radiation and get excited. 
(iii) For example when crystals of NaCl are heated in an atmosphere of sodium vapours, the sodium atoms get deposited on the surface of the crystal and the deposited Na atoms.
(iv) During this process, the Na atoms on the surface lose electrons to form Na⁺ ions 
(v) These electrons get excited by absorbing energy from the visible light and impart yellow colour to the crystals.

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.

Volume of KMnO₄ used is proportional to the amount of H₂O₂ present. If the reaction is of first order, it must obey the equation.
\(k=\frac { 2.303 }{ t } log\frac { \left[ { A } \right] _{ 0 }}{ \left[ A \right] } \)
(or) \(k=\left( \frac { 2.303 }{ t } \right) log\left( \frac { { V }_{ 0 } }{ { V }_{ 1 } } \right) \)
(i) \( \mathrm{V}_{\mathrm{o}}=46.1 \mathrm{~mL} ; \mathrm{t}=10 \mathrm{mins} ; \mathrm{V}_{\mathrm{t}}=29.8 \mathrm{~mL}\)
\(k =\frac{2.303}{10} \log \left(\frac{46.1}{29.8}\right)  \)
\(=0.2303 \log 1.5469  \)
\(=0.2303 \times 0.1894=0.0436 \mathrm{~min}^{-1}\)
\(\mathrm{t}=20 \mathrm{mins} ; \mathrm{V}_{\mathrm{t}}=19.3 ; \mathrm{V}_{\mathrm{o}}=46.1 \mathrm{~mL} \)
\(k =\frac{2.303}{20} \log \left(\frac{46.1}{19.3}\right)  \)
\(=0.11515 \times \log 2.388  \)
\(=0.11515 \times 0.3780  \)
\(=0.0435 \mathrm{~min}^{-1}\)
Since the value of k comes out to be almost constant for the reaction, it is of first order. The mean value of k = 0.04355 min^-1