ΔS is -ve

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.

E_cell = E^o_cell - \(\frac{0.0591}{2} \log [\frac{Zn^{2+}}{Cu^{2+}}]\)
E_l = E^o_cell - \(\frac{0.0591}{2} \log [\frac{10^{-2}}{1}]\)
E_l = E^o_cell + 0.0591 ...(1)
E₂ = E^o_cell - \(\frac{0.0591}{2} \log [\frac{1}{10^{-2}}]\)
E₂ = E^o_cell - 0.0591 ...(2)
E₁ > E₂
Zn_(s)| ⟶ Zn²⁺_(aq) + 2e^-
Cu²⁺_(aq) + 2e^- ⟶ Cu_(s)
Zn_(s) + Cu²⁺_(aq)  ⟶ Zn²⁺_(aq) + Cu_(s)

pH = -log₁₀ [H⁺]
\(\therefore\) [H⁺] = 10^-pH
10⁰ = 1
[H⁺] = 1 M
The solution is strongly acidic

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

\(k=\frac { 2.303 }{ t } log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
2.303 x 10^-2 hour^-1 = \(\frac { 2.303 }{ 1806 min } log\frac { \left[ { 0.25 }_{ } \right] }{ \left[ A \right] } \)
\(=\left(\frac{2.303 \times 10^{-2} hour^{-1 }\times 1806 min}{2.303}\right) = \log \left(\frac{0.25}{A}\right)  \)
\(=\left(\frac{ 1806 \times 10 ^{-2}}{60}\right) = \log \left(\frac{0.25}{A}\right)  \)
\(= 0.301 = \log \left(\frac{0.25}{A}\right)  \)
\(=\log2 = \log \left(\frac{0.25}{A}\right)  \)
\(2 = \log \left(\frac{0.25}{A}\right)  \)
\([A] = \log \left(\frac{0.25}{2}\right)  = 0.125 M\)

let edge length = a
\(\sqrt{2a} = 4r\)
\(a = \frac{4 \times 300}{\sqrt {2}}\)
a = 600 x 1.414
a = 848.4 pm

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

(i) Decarboxylation: When anhydrous: sodium salt of carboxylic acids are heated with soda lime, carboxyl group is removed with the· formation of hydrocarbon containing one carbon atom less.
\(\underset { Sodium \ acetate }{ { CH }_{ 3 }COONa } \overset { NaOH/CaO }{ \longrightarrow } \underset { Methane }{ { CH }_{ 4 } } +{ Na }_{ 2 }{ CO }_{ 3 }\) 
(ii) Heating benzoic acid with soda lime gives benzene.
\({ C }_{ 6 }{ { H }_{ 5 }COOH }\underset { Sodalime }{ \overset { NaOH }{ \underset { CaO\Delta }{ \longrightarrow } } } { C }_{ 6 }{ H }_{ 6 }+{ CO }_{ 2 }\) 

DIBAL -H is Diisobutyl aluminium hydride. Its is reducing agent.

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 separation of sol particles from electrolyte by filteration through an ultrafilter is called ultrafiltration.

\(\alpha =\frac { { \Lambda }_{ m } }{ { \Lambda }_{ m }^{ o } } \)

(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.

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.

Ethanol

With increase in temperature the ionic product of water will increase since the concentration of H₃O⁺ and OH^- ions will increase.

Given that
κ = 1.3 Sm^-1 (for 0.1M KCl solution)
R = 190 Ω
\(\kappa = \frac{1}{R}(\frac{l}{A})\)
κ . R =\((\frac{l}{A})\) =  (1.3 Sm^-1) (190Ω)
= 247 m^-1
\(\kappa_{(NaCl)} = \frac{1}{R_{(NaCl)}} (\frac{l}{A})\)
\(= \frac{1}{6.3 K\Omega}(247 m^{-1})\)  (6.3KΩ = 6.3 x 10³Ω)
= 39.2 x 10^-3 Sm^-1
\(\Lambda_m = \frac{\kappa \times 10^{-3} mol^{-1} m^3}{M}\)
\(=\frac{39.2 \times 10 ^{-3}(Sm^{-1})10^-3 (mol^{-1} m^3)}{0.003}\)
\(\Lambda_m\) = 13.04 \(\times\) 10^-3  Sm² mol^-1

a) \(Rate=\frac { -1 }{ 2 } \frac { d[NO] }{ dt } =\frac { -d[{ O }_{ 2 }] }{ dt } =\frac { 1 }{ 2 } \frac { d\left[ { NO }_{ 2 } \right] }{ dt } \)
b) \(\frac { -d\left[ { O }_{ 2 } \right] }{ dt } =\frac { 1 }{ 2 } \frac { d\left[ { NO }_{ 2 } \right] }{ dt } \)
\(\frac { d[{ NO }_{ 2 }] }{ dt } =2 \times\left( \frac { -d\left[ { O }_{ 2 } \right] }{ dt } \right) =2\times 0.2 \ { mol \ L }^{ -1 }{ s }^{ -1 }\)
= 0.4 mol L^-1s^-1

For n^th reaction
\(K\alpha \frac { 1 }{ { C }^{ n-1 } } \)

Minus sign i.e.  \(\frac { -dA }{ dt } \) or \(\frac { -dB }{ dt } \) indicates decreases in the concentration of reactants whereas + sign indicates in the concentration of products with time i.e. \(\frac { +dC }{ dt } \)

(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) HCl < HBr < HI < HF
(ii) Fluorine being the most electronegative element cannot have positive oxidation state. Other halogens have d-orbitals, therefore can expand their octet.

No. of Ca²⁺ ions per unit cell
= 8 x \(\frac{1}{8}+6\times \frac{1}{2}\)
= 1 + 3 = 4
Hence No. of F^- ions per unit cell = 2 x 4 = 8

When carbon monoxide is treated with chlorine in presence of light or charcoal, it forms a poisonous gas carbonyl chloride, which is also known as phosgene.
CO + Cl₂ ⟶ COCl₂

(i) Boron trifluoride is used for preparing HBF₄, a catalyst in organic chemistry
(ii) It is also used as a fluorinating reagent.

(i) Due to lanthanide contraction the size of La³⁺ ions decreases regularly with increase in atomic number.
(ii) According to Fajan's rule decrease in size of Ln³⁺ ions increase the covalent character and decreases the basic character between Ln³⁺& OH^- ion in Ln(OH)³
(iii) Since the order of size Ln³⁺ ions are
La³⁺ > Ce³⁺ ...> Lu³⁺
(iv) Hence La(OH)³ is the strongest base while Lu(OH)³ is the weakest base

Cr₂O₇^2- + 14H⁺ + 6Fe²⁺ ⟶ 2Cr³⁺ + 6Fe³⁺ + 7H₂O

(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.

The pair of two optically active isomers which are mirror images of each other are called enantiomers.

Magnetite (Fe₃O₄), haematite (Fe₂O₃) are the ores which can be separated by the magnetic separation method. In these ores, one component is magnetic in nature.

(i) CFSE is higher when the complex contains strong field ligand as per spectro chemical series.
(ii) Thus, crystal field splitting energy increases in the order 
[CrCI₆]^3- <   [Cr(NH₃)₆]³⁺ <  [Cr(CN)₆]^3-, Because the order of field strength is Cl^- < NH³ < CN^-

(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.

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

a) potassiumhexacyanidoferrate(II) - Potassiumhexacyanidoferrate(II) - K₄[Fe(CN)₆]
b) Pentacarbonyliron(0) - [Fe(CO)₅]
c) Pentaamminenitrito −kNcobalt(III)ion - [Co(NH₃)₅(NO₂)]²⁺
d) Hexaamminecobalt(III) Sulphate  - [CO(NH₃)₆](SO₄)₃
e) Sodiumtetrafluoridodihydroxidochromate(III)  - Na₂[CrF₄(OH)₂]

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

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

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

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.

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

Two possible isomers of C₃H₆O are
\(\underset { \overset { (A) }{ Acetone } }{ { CH }_{ 3 }CO{ CH }_{ 3 } } \quad \underset { \overset { (B) }{ Propanal } }{ { CH }_{ 3 }{ CH }_{ 2 }CHO } \) 
\(\\ \underset { \quad \quad (A)\\ Proponal }{ { CH }_{ 3 }CO{ CH }_{ 3 } } \overset { { I }_{ 2 }/NaOH\Delta }{ \underset { Iodo\ form\ \\ reaction }{ \longrightarrow } } \underset { Sodium\\ acetate }{ { CH }_{ 3 }COONa } +\underset { Iodoform\\ Yellow\ ppt }{ { CHI }_{ 3 } } \) 
\(\underset { \quad \quad \quad (B)\\ Propanal }{ { CH }_{ 3 }{ CH }_{ 2 }CHO } \overset { { I }_{ 2 }/NaOH }{ \underset { \Delta }{ \longrightarrow } } No\quad ppt\quad of\quad { CHI }_{ 3 }\) 
Acetone answers Iodo form test whereas proponal does not.

1-Butanol, being primary alcohol gets oxidised by dilute KMnO₄. The brown precipitate is due to the formation of manganese dioxide.

(i) The peptide glycyl L-glutamyl L-lyrosin is hydrolysed by an enzyme called pepsin.
(ii) The enzyme diastase hydrolyses starch into maltose 
\(2\left( { C }_{ 6 }{ H }_{ 10 }{ O }_{ 5 } \right) +{ nH }_{ 2 }O\longrightarrow { nC }_{ 12 }{ H }_{ 22 }{ O }_{ 11 }\) 
(iii) The yeast contains the enzyme zymase which converts glucose into ethanol.
\({ C }_{ 6 }{ H }_{ 12 }{ O }_{ 6 }\longrightarrow { 2C }_{ 2 }{ H }_{ 5 }OH+{ 2CO }_{ 2 }\)

(i) Anode: Zinc container
(ii) Cathode: Graphite rod in contact with MnO₂
(iii) Electrolyte: Ammonium chloride and zinc chloride in water
(iv) Oxidation at anode:
\( { Zn }_{ (s) }\rightarrow { Zn }_{ (aq) }^{ 2+ }+2e^{ - }\)
(v) Reduction at cathode:
\({ 2NH }_{ 4(aq) }^{ + }+{ 2e }^{ - }\rightarrow { 2NH }_{ 3(aq) }+{ H }_{ 2(g) }\)

(i) Glycerol is prepared in a large scale by the hydrolysis of oils or fats either by using alkali or by super heated steam.
(ii) During this process, soap is formed along with the byproduct glycerol and this process is called saponification.

Uses of glycerol:
(i) Glycerol is used as a sweetening agent in confectionery and beverages.
(ii) It is used in the manufacture of cosmetics and transparent soaps.
(iii) It is used in making printing inks and stamp pad ink and lubricant for watches and clocks.
(iv) It is used in the manufacture of explosive like dynamite and cordite by mixing it with chaina clay.

NH₄CI is a salt of a strong acid HCI and weak base NH₄OH.
\({ HCl }_{ (aq) }+{ NH }_{ 4 }OH_{(aq)}\rightleftharpoons { { NH }_{ 4 }Cl }_{ (aq) }+{ H }_{ 2 }O(I)\)
\({ NH }_{ 4 }^{ + }\) is a strong conjugate acid of the weak base NH₄OH and it has a tendency to react with OH^- from water to produce unionised NH₄OH shown below.
\({ NH }_{ 4 }^{ + }+{ H }_{ 2 }O(1)\rightleftharpoons { NH }_{ 4 }{ { { OH }_{ (aq) }+ }H }_{ (aq) }^{ + }\)
There is no such tendency shown by Cl^- and therefore [H⁺] > [OH^-]; the solution is acidic and the pH is less than 7.
As discussed in the salt hydrolysis of strong base and weak acid. In this case also, we can establish a relationship between the K_h and K_b as 
K_h·K_b = K_w
Let us calculate the K_h value in terms of degree of hydrolysis (h) and the concentration of salt
K_b = h²C and \([{ H }^{ + }]=\sqrt { { K }_{ h }.C } \)
= \({ [H }^{ + }]=\sqrt { \frac { { K }_{ w } }{ { K }_{ b } } .C } \)
pH = - log [H⁺]
\(={ \left( \frac { { K }_{ w }.C }{ { K }_{ b } } \right) }^{ \frac { 1 }{ 2 } }\)
= \(-\frac { 1 }{ 2 } \log { K }_{ w }-\frac { 1 }{ 2 } \log C+\frac { 1 }{ 2 } \log{ K }_{ b }\)
\(pH=7-\frac { 1 }{ 2 } p{ K }_{ b }-\frac { 1 }{ 2 } \log C\)

(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.

(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.

Considering Ba(OH)₂ to be a strong base:
\(\text { Normality } =\text { Molarity } \times \text { Acidity }  \)
\(=1.5 \times 10^{-3} \times 2\)
\({\left[\mathrm{OH}^{-}\right] } =3 \times 10^{-3}  \)
\(\mathrm{pOH} =-\log _{10}[\mathrm{OH}^-]  \)
\(=-\log _{10}\left(3 \times 10^{-3}\right)  \)
\(=-\left[\log _{10} 3+3 \log 10\right]  \)
= 3-log 3 
= 3-0.4771 
= 2.5229 
\(\mathrm{pH} =14-\mathrm{pOH} \)
= 14-2.5229 
pH = 11.4771 = 11.48

\(\text { Normality }=\text { Molarity } \times \text { Basicity }  \)
\(=0.04 \times 1  \)
\({\left[\mathrm{H}_{3} \mathrm{O}\right]^{+}=0.04=4 \times 10^{-2} } \)
\(\mathrm{pH} =-\log _{10}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]  \)
\(=-\log \left[4 \times 10^{-2}\right]  \)               log₁₀ 10 =1
\(=-\left[\log _{10} 4+\log _{10} 10^{-2}\right]  \)
\(=-\left[\log _{10} 4-2 \log _{10} 10\right]=2-\log _{10} 4  \)
= 2 - 0.6021 
= 1.3979 \(\simeq\) 1.40

Potassium permanganate is a strong oxidising agent, its oxidising action differs in different reaction medium.
In neutral medium:
In neutral medium, it is reduced to MnO₂
MnO₄^- + 2H₂O + 3e^- ⟶ MnO₂ + 4OH^-
(i) It oxidises H₂S to sulphur
2MnO₄^- + 3H₂S ⟶ 2MnO₂ + 3S + 2OH^- + 2H₂O
(ii) It oxidises thiosulphate into sulphate
8MnO₄^- + 3S₂O₃^-2 + H₂O ⟶ 6SO₄^-2 + 8MnO₂ + 2OH^-

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

(i) Bond angel of HiX (H-O-H=104.5^o) is larger than that of H₂S(H-S-H=92^o)
(ii) Since oxygen is more electronegative than sulphur, therefore bond pair electrons of O-H bond will be closer to oxygen which will be little away from the sulphur atom.
(iii) As a result bond pair-bond pair repulsions between bond pairs of two O-H bonds would be stronger than in the S-H bonds.

Number of tetrahedral voids formed = 2 x No. of atoms of element Y
No. of atoms of element Y in the ccp unit cell = 4
No. of tetrahedral voids by atoms of Y = 2 x 4 = 8
∴ No. of tetrahedral voids occupied by atoms of 
X = \(\frac{2}{3}\times 8=\frac{16}{3}\)
Ratio of the no. of atoms of X and Y is = \(\frac{16}{3}:4\)
= 16:12
= 4:3
Hence, formula would be X₄Y₃.

(i) Diborane is used as a high energy fuel for propellant.
(ii) It is used as a reducing agent in organic chemistry.
(iii) It is used in welding torches.

If we perform a qualitative analysis to identify the constituent ions present in both the compounds, Mohr's salt answers the presence of Fe²⁺, NH₄^+, and SO₄^2- ions, whereas the potassium ferrithiocyanate will not answer Fe³⁺ and SCN ions. We can infer that the double salts loose their identity and dissociate into their constituent simple ions in solutions, whereas the complex ion in coordination compound, does not loose its identity and never dissociate to give simple ions.

\(k=\frac { 2.303 }{ t } \log\frac { \left( { V }_{ \infty }-{ V }_{ 0 } \right) }{ \left( { V }_{ \infty }-{ V }_{ 0 } \right) } \)
\(k=\frac { 2.303 }{ 20 } \log\frac { 50.4-20.2 }{ 50.4-25.6 } \)
\(=0.1151 \log\frac { 30.2 }{ 24.8 } \)
= 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) 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.

(i) The elements in which the extra electron enters (n-2) f orbitals are called f-block elements. These elements are called as inner transition elements because they form a transition. Series within the transition elements.
(ii) The f-block elements are also called as rare earth elements. They are divided into lanthanoid series (4f block elements) and actinoid series (5f block elements).

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

(i) From the molecular formula compound (A) is identified as Benzaldehyde I C₆H₅CHO and it reduces Tollen's reagent.
\(\underset { Tollen's\ regant\ silver\ mirror }{ { C }_{ 6 }{ H }_{ 5 }CHO+{ Ag }_{ 2 }O\longrightarrow 2Ag+{ C }_{ 6 }{ H }_{ 5 }COOH } \) 
(ii) Benzaldehyde on treatment with alkali undergoes Cannizzaro reaction to give benzoic acid (B) and benzyl alcohol (C).
\(\underset { (A) }{ { C }_{ 6 }{ H }_{ 5 }CHO+{ C }_{ 6 }{ H }_{ 5 }CHO } \overset { NaOH }{ \longrightarrow } \underset { (B) }{ { C }_{ 6 }{ H }_{ 5 }COOH } +\underset { (C) }{ { C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }OH } \) 
(iii) Benzoic acid on treatment with sodalime gives benzene.
\(\underset { (B) }{ { C }_{ 6 }{ H }_{ 5 }COOH } \overset { NaOH }{ \underset { CaO }{ \longrightarrow } } \underset { Benzene }{ { C }_{ 6 }{ H }_{ 5 }+{ CO }_{ 2 } } \)
(iv) Benzyl alcohol acts as an antiseptic.

(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.
\({ CH }_{ 3 }-\underset { \overset { | }{ \underset { (C) }{ Cl } } }{ C } H-{ CH }_{ 3 }\overset { alc.KOH }{ \longrightarrow } \underset { (D) }{ { CH }_{ 3 }CH={ CH }_{ 2 }+HCl } \) 

(i) Effective and efficient conversion is the special characteristic of enzyme catalysed reactions. An enzyme may transform a million molecules of reactant in a minute 
For Eg: \({ 2H }_{ 2 }{ O }_{ 2 }\longrightarrow { 2H }_{ 2 }O+{ O }_{ 2 }\)
For this reaction, the activation energy is 18k cal/mole without a catalyst  With colloidal platinum as a, catalyst the activation energy is 11.7kcal /mole. But with the enzyme catalyst the activation energy of this reaction is less than 2kcal/ mole.
(ii) Enzyme catalysis is highly specific in nature.
(iii) Enzyme catalysed reaction has maximum rate at optimum temperature
(iv) The rate of enzyme catalysed reactions varies with the pH of the system. The rate is maximum at a pH called optimum pH.
(v) Enzymes can be inhibited i.e. poisoned activity of an enzyme is decreased and destroyed by a poison. The physiological action of drugs is related to their inhibiting action.
(vi) Catalytic activity of enzymes is increased by coenzymes or activators.

 Sn⁴⁺ + 2e^- ⟶ Sn²⁺ E⁰ = 0.15V
Given: n = 2 electrons 
F = 96495 coulombs
Formula: ΔG = - nFE^O
Solutlon: ∴ ΔG = - 2 x 96495 x 0.15
= 28.948
Free energy = -28.948 kJ.

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

Active centres:
The surface of a catalyst is not smooth. It bears steps, cracks and corners. Hence the atoms on such locations of the surface are co-ordinatively unsaturated. So, they have much residual force of attraction. Such sites are called active centres. So, the surface carries high surface free energy. The presence of such active centres increases the rate of reaction (activity) by adsorbing and activating the reactants.
The adsorption theory explains the following:
(i) Increase in the activity of a catalyst by increasing the surface area. Increase in the surface area of metals and metal oxides by reducing the particle size increases the rate of the reaction.
(ii) The action of catalytic poison occurs when the poison blocks the active centres of the catalyst.
(iii) A promoter or activator increases the number of active centres on the surfaces
Selectivity:
A Catalyst can catalyse a particular type of reaction. Hence they are said to the specific (selectivity) in nature. Enzyme catalysis is highly specific in nature.
\(\mathrm{NH}_{2} \mathrm{CONH}_{2}+\mathrm{H}_{2} \mathrm{O} \stackrel{\text { Unease }}{\longrightarrow} 2 \mathrm{NH}_{3}+\mathrm{CO}_{2}\)
The enzyme urease which catalyses their reaction of Urea does not catalyse the reaction of methyl Urea.
\(\mathrm{NH}_{2} \mathrm{CONH} \mathrm{CH_3}+\mathrm{H}_{2} \mathrm{O} \stackrel{\text { Urease }}{\longrightarrow} \text { No reaction }\)
Intermediate compound formation theory explains the specificity of a catalyst.

(i) An acid is defined as a substance that has a tendency to donate a proton to another substance and base is a substance that has a tendency to accept a proton form other substance.
(ii) In other words, an acid is a proton donor and a base is a proton acceptor.
(iii) When hydrogen chloride is dissolved in water, it donates a proton to the later. Thus, HCI behaves as an acid and H₂O is base. The proton transfer from the acid to base can be represented as
HCI + H₂O ⇌ H₃O⁺ + Cl^-
(iv) When ammonia is dissolved in water, it accepts a proton from water. In this case, ammonia (NH₃) acts as a base and H₂O is acid. The reaction is represented as
H₂O + NH₃ ⇌ NH₄⁺ + OH^-
(v) Let us consider the reverse reaction following equilibrium.
\(\underset { proton\ donar\\ \quad \quad \ (acid) }{ HCl } +\underset { Proton\ acceptor\\ \quad \quad \quad \quad (base) }{ { H }_{ 2 }O } \leftrightharpoons \underset { Proton\ donar\\ \quad \quad \quad \quad \ (acid) }{ { H }_{ 2 }{ O }^{ + } } +\underset { Proton\ acceptor\\ \quad \quad \quad \quad \ (base) }{ { Cl }^{ - } } \) 
H₃O⁺ donates a proton to Cl^- to form HCI i.e., the products also behave as acid and base.
(vi) In general, Lowry - Bronsted (acid - base) reaction is represented as 
Acid₁ + Base₂ ⇌ Acid₂ + Base₁
(vii) The species that remains after the donation of a proton is a base (Base₁) and is called the conjugate base of the Bronsted acid (Acid₁). In other words, chemical species that differ only by a proton are called conjugate acid - base pairs.

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

(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).

Reaction with metals:
(i) Sulphuric acid reacts with metals and gives different product depending on the reactants and reacting condition
(ii) Dilute sulphuric acid reacts with metals like: tin, aluminium, zinc to give corresponding: sulphates.
Zn + H₂SO₄\(\longrightarrow \) ZnSO₄ + H₂ \(\uparrow \)
2AI + 3H₂SO₄ \(\longrightarrow \) Al₂(SO₄)₃+ 3H₂ \(\uparrow \)
(iii) Hot concentrated sulphuric acid reacts with copper and lead to give the respective sulphates as shown below 
Cu + 2H₂SO₄ \(\longrightarrow \) CuSO₄ + 2H₂O + SO₂\(\uparrow \)
Pb + 2H₂SO₄ \(\longrightarrow \) PbSO₄ + 2H₂O + SO₂\(\uparrow \)
(iv) Sulphuric acid doesn't react with noble metals like gold, silver and platinum.

(i) Solids have definite volume and shape 
(ii) Solids are rigid and incompressible
(iii) Solids have strong cohesive forces.
(iv) Their constituents have fixed positions and can only oscillate about their mean positions.

(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) Van-Arkel method is based on the thermal decomposition of metal compounds which lead to the formation of pure metals.
(ii) Titanium and zirconium can be purified, using this method.
(iii) For example, the impure titanium metal is heated in an evacuated vessel with iodine at a temperature of 550 K to form the volatile titanium tetra-iodide(Til₄)
(iv) The impurities are left behind, as they do not react with iodine
\({ Ti }_{ (s) }+{ 2I }_{ 2(s) }\longrightarrow { Til }_{ 4 }(vapour)\)
(v) The volatile titanium tetraiodide vapour is passed over a tungsten filament at a temperature around 1800 K.
(vi) The titanium tetraiodide is decomposed and pure titanium is deposited on the filament
(vii) The iodine is reused.
\({ Til }_{ 4 }(vapour)\longrightarrow { { Ti }_{ (s) } }+{ 2I }_{ 2(s) }\)

The postulates of valence bond theory
(i) The central metal atom/ion makes available a number of vacant orbitals equal to its coordination number.
(ii) These vacant orbitals form covalent bonds with the ligand orbitals.
(iii) A covalent bond is formed by the overlap of a vacant metal orbital and filled ligand orbitals. This complete overlap leads to the formation of a metal ligand, σ (sigma) bond.
(iv) A strong covalent bond is formed only when the orbitals overlap to the maximum extent.
(v) This maximum overlapping is possible only when the metal vacant orbitals undergo a process called 'hybridisation'.
(vi) A hybridised orbital has a better directional characteristic than an unhybridized one. The following table gives the coordination number, orbital hybridisation, and geometry of the complexes.

Magnetic moment 
The paramagnetic moment is given by the following spin-only formula.
\({ \mu }_{ s }=\sqrt { n(n+2) } \) BM
BM = Bohr magneton
\({ \mu }_{ s }\) = spin -only magnetic moment
n = number of unpaired electrons.

Allotropism:
1. Some elements exist in more than one crystalline or molecular forms in the same physical state
(a) In Greek "allos" means ⇒ Another
(b) "trope" means ⇒ Change
2. The different forms of an element are called allotropes.
Allotropy of carbon:
Carbon exists as diamond, graphite, fullerenes, carbon nanotubes and graphene.
Graphite:
1. Graphite is the most stable allotropic form of carbon at normal temperature and pressure.
2. It is soft and conducts electricity.
3. It is composed of flat two dimensional sheets of carbon atoms.
4. Each sheet is a hexagonal.
5. It is "sp²" hybridised.
6. C-C bond length is 1.41 Å
7. Each C-atom forms three σ bonds with three neighbouring carbon atoms using three of its valence electrons and the fourth electron present in the unhybridised p-orbital form a π-bond.
8. The successive C-sheets are held together by weak Vander Waals forces.
9 The distance successive sheet is 3.40 Å.
10. It is used as a lubricant either on its own or as a graphited oil.

Diamond:
1. It is very hard.
2. It is "sp³" hybridised.
3. C-C bond length is 1.54 Å
4. It is used for sharpening hard tools, cutting glasses, making bores and rock drilling.

Fullerenes:
1. These allotropes are discrete molecules such as \(C_{32}, C_{50}, C_{60}, C_{70}, C_{76}\) etc.
2. It has cage like structure
3. The C₆₀ molecules have a "soccer" ball like structure and is called buckminster fullerene or buckyballs.
4. It has a fused ring structure consists of 20 six membered rings and 12 five membered ring.
5. Each carbon atom is "sp²" hybridised.
6. It has three σ bonds and a delocalised π bond giving aromatic character to these molecules.
7. The C-C bond distance is 1.44 Å
8. The C=C bond distance is 1.38 Å.

Carbon nanotubes:
1. Carbon nanotubes, another recently discovered allotropes, have graphite like tubes with fullerene ends.
2. Along the axis, these nanotubes are stronger than steel and conduct electricity.
3. These have many applications in nanoscale electronics, catalysis, polymers and medicine.
Graphene:
 It has a single planar sheet of "sp²" hybridised carbon atoms that are densely packed in a "honeycomb crystal" lattice.

d) Both assertion and reason are false