ΔS is -ve

7MnO₄^- + 5e^- → Mn²⁺ + 4H₂O
5 moles of electrons i.e., 5F charge is required.

Λ = k/W x 10⁻³ mol⁻¹ dm³
\(=\frac{ 5.76 \times 10^{−3} S cm^{−1} \times 10^{−3}}{0.5} = mol^{-1} dm^{3}\)
\(=\frac{ 5.76 \times 10^{−3} S cm^{−1} \times 10^{−3}}{0.5}\) S cm^-1 mol⁻¹ dm³
= 11.52 S cm²mol^-1

Ca(OH)₂ ⇌ Ca²⁺ + 2OH^-
Given that pH = 9
pOH = 14 - 9 = 5
[pOH = - log₁₀ [OH]]
[OH^-] = 10 ^[pOH]
[OH] = 10^-5 M
Ksp = [Ca²⁺] [OH^-]
= 10^-5/2 x (10^-5)²  = 0.5 x 10^-15

(rate constant K is equal to collision frequency A if E_a = 0)
In zero order reactions, increase in the concentration of reactant does not alter the rate.
So statement (i) is wrong.
\(k=A{ e }^{ -\left( \frac { { E }_{ a } }{ RT } \right) }\)
if E_a = 0 so, statement (ii) is correct, and statement (iii) is wrong
k = Ae⁰
k = A
In k = In A - \({ \left( \frac { { E }_{ a } }{ R } \right) }\) \(\left( \frac { 1 }{ T } \right) \)
This equation is of the form of a straight line y = mx+c
A plot of In k Vs \(\left( \frac { 1 }{ T } \right) \) gives a straight line with a negative slope
So statement (iv) and (v) are wrong.

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

a) Zn²⁺ (d¹⁰ ⇒  diamagnetic)
b) Co³⁺ (d⁶ Low spain ⇒  t_2g⁶ e⁰_g ; diamagnetic)
c) Ni²⁺  (d⁸ Low spain ⇒  t_2g⁶ e²_g ; paramagnetic)
d) [Ni(CN)₄]²⁺ (dsp² ; square planar, diamagnetic)

a) Fe²⁺  b) Fe³⁺  c) Fe⁰

(i) The dissociation constant is generally called acidity constant because it measures the relative strength of an acid. The stronger the acid, the large will be its K_a value.
(ii) The strength of carboxylic acid can be expressed in terms of the dissociation constant(K):

(iii) The dissociation constant of an acid can also be expressed in terms of p^Ka value
p^Ka = -log k_a,

Oxidation of alcohols with acidified K₂Cr₂O₂ gives different oxidation products.

Various demulsification techniques are given below:
(i) Distilling of one component.
(ii) Adding an electrolyte to destroy the charge.
(iii) Destroying the emulsifier using chemical methods.
(iv) Using solvent extraction to remove one component.
(v) By freezing one of the components.
(vi) By applying centrifugal force.
(vii) Adding dehydrating agents for water in oil.
(viii) Type.
(viii) Using ultrasonic waves.
(ix) Heating at high pressures.

The galvanic cell is represented as
\({ Zn }_{ (s) }|{ Zn }_{ (aq) }^{ 2+ }||{ Cu }_{ (aq) }^{ 2+ }|{ Cu }_{ (s) }\)
The anode half cell is \({ Zn }_{ (s) }|{ Zn }_{ (aq) }^{ 2+ }\)
The cathode half cell is \({ C }u_{ (Aq) }^{ 2+ }|{ Cu }_{ (s) }\)

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

\({ CH }_{ 3 }-{ CH }_{ 2 }-OH\overset { { H }_{ 2 }{ SO }_{ 4 } }{ \underset { 443k }{ \longrightarrow } } { CH }_{ 2 }={ CH }_{ 2 }+{ H }_{ 2 }O\)
Mechanism
Primary alcohols undergo dehydration by E₂ mechanism

First law:
The mass of the substance (m) liberated at an electrode during electrolysis is directly proportional to the quantity of charge (Q) passed through the cell.
m α Q \(\left[\because \mathrm{I}=\frac{\mathrm{Q}}{\mathrm{t}} \Rightarrow \mathrm{Q}=\mathrm{It}\right]\)
m α It
m = ZIt
Where Z = electro chemical equivalent of the substance
I = current
t = time of passage of current
Second law:
When the same quantity of charge is passed through the solutions of different electrolytes, the amount of substances liberated at the respective electrodes are directly proportional to their electrochemical equivalents.
m α Z
\(\frac{m_{1}}{Z_{1}}=\frac{m_{2}}{Z_{2}}\)
m = mass of the metal deposited
Z = electro chemical equivalent

(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.

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

(i) The rate constant is nearly doubled with a rise in temperature by 10^oC for a chemical reaction.
(ii) The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation.
\(k=A{ e }^{ \frac { -{ E }_{ a } }{ RT } }\)

Commercial method of preparation
(i) Nitric acid prepared in large scales using Ostwald's process. In this method ammonia from Haber's process is mixed about 10 times, of air.
(ii) This mixture is preheated and passed into the catalyst chamber where they come in contact with platinum gauze.
(iii) The temperature rises to about 1275 K and the metallic gauze brings about the rapid catalytic oxidation of ammonia resulting in the formation of NO, which then oxidised to nitrogen dioxide.
\({ 4NH }_{ 3 }+{ 5O }_{ 2 }\longrightarrow 4NO+6{ H }_{ 2 }O+120KJ\)
\(2NO+{ O }_{ 2 }\longrightarrow { 2NO }_{ 2 }\)
(iv) The nitrogen dioxide produced is passed through a series of adsorption towers. It reacts with water to give nitric acid. Nitric acid formed is bleached by blowing air.
\({ 6NO }_{ 2 }+3{ H }_{ 2 }O\longrightarrow { 4HNO }_{ 3 }+2NO+{ H }_{ 2 }O\)

The fundamental equation that gives a simple relation between the wavelength of the X-rays, the interplanar distance in the crystal and the angle of reflection is known as Bragg's equation.
nλ = 2dsinθ
Where n is the order of reflection
λ is the wavelength of X-rays
d is the interplanar distance in the crystal
θ is the angle of reflection 

At high temperatures it forms higher boranes liberating hydrogen.
\(5{ B }_{ 2 }{ H }_{ 6 }\overset { 388K }{ \underset { U-tube }{ \longrightarrow } } 2{ B }_{ 5 }{ H }_{ 11 }+4{ H }_{ 2 }\)
\(2{ B }_{ 2 }{ H }_{ 6 }\overset { 198-373K }{ \longrightarrow } { B }_{ 4 }{ H }_{ 10 }+{ H }_{ 2 }\)
\(5{ B }_{ 2 }{ H }_{ 6 }\overset { 373K }{ \underset { sealed\ tube }{ \longrightarrow } } { B }_{ 10 }{ H }_{ 14 }+8{ H }_{ 2 }\)
\(5{ B }_{ 2 }{ H }_{ 6 }\overset { 473-523K }{ \longrightarrow } 2{ B }_{ 5 }{ H }_{ 9 }+6{ H }_{ 2 }\)
\(10{ B }_{ 2 }{ H }_{ 6 }\overset { 523K }{ \longrightarrow } 2{ B }_{ 5 }{ H }_{ 9 }+2{ B }_{ 5 }{ H }_{ 10 }+11{ H }_{ 2 }\)
\({ B }_{ 2 }{ H }_{ 6 }\overset { Red \ hot }{ \longrightarrow } 2B+3{ H }_{ 2 }\).

(i) This method can be applied to the oxides of the metals (Fe, Pb, Cu) having less electropositive character than hydrogen
\(\mathrm{Ag}_{2} \mathrm{O}_{(s)}+\mathrm{H}_{2(g)} \longrightarrow 2 \mathrm{Ag}_{(s)}+\mathrm{H}_{2} \mathrm{O}_{(l)}\)
(ii) Nickel oxide can be reduced to nickel by using a mixture of hydrogen and carbon monoxide (water gas)
(iii) \({ 2NiO }_{ (s) }+{ CO }_{ (g) }+{ H }_{ 2(g) }\longrightarrow 2{ Ni }_{ (s) }+{ CO }_{ 2(g) }+{ H }_{ 2 }{ O }_{ (l) }\)

Aqueous copper sulphate (blue) is [Cu(H₂O)₄] SO₄.
[Cu(H₂O)₄]SO₄ ⟶ [Cu(H₂O)₄]²⁺ + SO₄^2-
[Cu(H₂O₄)]²⁺ is a liable complex in which H₂O ligand get easily replaced by F^- ions of KF and by Cl^- ions of KCI.
(i) [Cu(H₂O)₄]²⁺_(aq) + 4F^- ⟶ \(\underset { Green \ ppt }{ { \left[ Cu{ F }_{ 4 } \right] }^{ 2- } } +4{ H }_{ 2 }O\)
(ii) [Cu(H₂O)₄]²⁺_(aq) +4Cl^-_(aq) ⟶ \(\underset { Bright \ green\ ppt }{ { \left[ Cu{ F }_{ 4 } \right] }^{ 2- } } +4{ H }_{ 2 }O\)

(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 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.

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) Compound (A) must be acetyl chloride which undergoes Rosenmund's reduction giving acetaldehyde.
\(\underset { (A) }{ { CH }_{ 3 }COCl } \overset { Pd/BaS{ O }_{ 4 } }{ \longrightarrow } \underset { (B) }{ { CH }_{ 3 }CHO } \) 
(B) undergoes iodoform reaction
\(\underset { (B) }{ { CH }_{ 3 }COH } \overset { { I }_{ 2 }/NaOH }{ \longrightarrow } \underset { Iodoform }{ { CHI }_{ 3 } } +HCOONa\) 
(ii) Acetaldehyde (B) undergoes polymerisation giving a cyclic compound.

Compound	Compound name	Formula
A	Acetyl chloride	CH3COCI
B	Acetaldehyde	CH3CHO
C	Paraldehyde

Saytzeff's rule: 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 l.e., the stable alkene.
For example, the dehydration of 3,3 - dimethyl - 2- butanol gives a mixture of alkenes. The secondary carbocation formed in this reaction undergoes rearrangement to form a more stable tertiary carbocation.

Food:
Food stuffs like milk cream, butter, etc are present in colloidal form.
Medicines:
Antibodies such as penicillin and streptomycin are produced in colloidal form for suitable injections. Colloidal gold and colloidal calcium are used as tonics. Milk of magnesia is used for stomach troubles. Silver sol protected by gelatine known as Argyrol is used as eye lotion.
In Industry:
Colloids find many applications in industries
(i) Water purification:
Purification of drinking water is activated by coagulation of suspended impurities in water using alums containing Al³⁺.
(ii) In washing:
The cleansing action of soap is due to the formation of emulsion of soap molecules with dirt and grease.
(iii) Tanning of leather:
Skin and hides are protein containing positively charged particles which are coagulated by adding tannin to give hardened leather for further application. Chromium salts are used for the purpose. Chrome tanning can produce soft and polishable leather.
(iv) Rubber industry:
Latex is the emulsion of natural rubber with negative particles. By heating rubber with sulphur, vulcanized rubbers are produced for tyres, tubes, etc.
(v) Sewage disposal:
Sewage contains dirt, mud and wastes dispersed in water. The passage of electric current deposits the wastes materials which can be used as a manure.
(vi) Cortrell's precipitator:
Carbon dust in air is solidified by Cottrell's precipitator. In it, a high potential difference of about 50,000V is used. The charge on carbon is neutralized and solidified. Thus the air is free from carbon particles.
(vii) The blue colour of the sky in nature is due to Tyndall effect of air particles.
(viii) Formation of delta:
The electrolyte in sea and river water coagulates the solid particles in river water at their intersection. So, the earth becomes a fertile land.
(ix) Analytical application:
Qualitative and quantitative analysis are based on the various properties of colloids.

Given:
E^oFe³⁺/Fe²⁺ = 0.61V
E^oCe³⁺ / Ce²⁺ = 0.85 V (after reversing)
Solution:
E^o = E_R - E_L
= 0.85 - 0.61 = 0.24 V
Cell is,
Fe³⁺ I Fe²⁺ II Ce³⁺ I Ce²⁺
Left                       Right
E^o_cell = 0.24 V
ΔG^o = - nFE^o
ΔG^o = - (1 x 96495 x 0.24)
= -ve value

(HCI strong acid fully ionised)
\(pH=\log\frac { 1 }{ { H }^{ + } } =\log\frac { 1 }{0 .01 } \)
\(=\log\frac { 100 }{ 2 } =1.6990\)

(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) Molar conductivity
Given : C = 0.01 M;
\(\kappa=1.5 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1} \)
\(=1.5 \times 10^{-2} \mathrm{~S} \mathrm{~m}^{-1} \)
\(\lambda_{\text {cation }}^{0}=248.2 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \)
\(\lambda_{\text {anion }}^{0}=51.8 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \)
\(\Lambda_{m}^{0}=\frac{\kappa \times 10^{-3}}{C} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} \)
\(=\frac{1.5 \times 10^{-2} \times 10^{-3}}{0.01} \)
\(=1.5 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\)
ii) \(\alpha=\frac{\Lambda_{m}}{\Lambda_{m}^{0}}\)
\(\Lambda_{\mathrm{m}}^{0}=\lambda_{\text {cation }}^{0}+\lambda_{\text {anion }}^{0} \)
\(=(248.2+51.8) \mathrm{S} \mathrm{cm}^{2} \mathrm{~mol}^{-1} \)
\(=300 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
\(=300 \times 10^{-4} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}  \)
\(\alpha =\frac{1.5 \times 10^{-3}}{300 \times 10^{-4}}=0.05\)
iii) \(\mathrm{K}_{\mathrm{a}} =\frac{\alpha^{2} \mathrm{C}}{1-\alpha} \)
\(\mathrm{K}_{\mathrm{a}} =\frac{(0.05)^{2} \times(0.01)}{1-0.05}=2.6 \times 10^{-5} \)
(or) 
\(\mathrm{K}_{\mathrm{a}} =\alpha^{2} \mathrm{C}  \)
\(=(0.05)^{2} \times(0.01)  \)
\(\mathrm{K}_{\mathrm{a}} =2.5 \times 10^{-5}\)

Given data: Time taken for 25% completion of the reaction = 100 mins
a = 100; x = 25 and a - x = 75; t = 100min
Formula: \(k=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } ;{ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } \)
Solution:
\(k=\frac { 2.303 }{ 100 } \log\frac { 100 }{ 100-25 } \)
\(=\frac { 2.303 }{ 100 } \log\frac { 100 }{ 75 } =\frac { 2.303 }{ 100 } \log\frac { 4 }{ 3 } \)
\(=\frac { 2.303 }{ 100 } \times 0.1249\)
Rate constant(k) = 2.8773 x 10^-3 min^-1
\({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } =\frac { 0.693 }{ 2.8773\times { 10 }^{ -3 }{ ,min }^{ -1 } } \)
\(=\frac { 693 }{ 2.8773 } =240.85\)
Half-life period (t_1/2) = 240.85 minutes

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.

Oxidising property of H₂SO₄: 
Sulphuric acid is an oxidising agent as it produces nascent oxygen as shown below.
\({ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { H }_{ 2 }O+\underset { nascentoxygen }{ { SO }_{ 2 } } +\left( O \right) \)
Sulphuric acid oxidises elements such as carbon, sulphur and phosphorus. It also oxides bromide and iodide to bromine and iodine respectively.
C + 2H₂SO₄ \(\longrightarrow \) 2SO₂ + 2H₂O + CO₂
S + 2H₂SO₄ \(\longrightarrow \) 3SO₂ + 2H₂O
P₄ + 10H₂SO₄ \(\longrightarrow \) 4H₃PO₄ + 10SO₂ + 4H₂O
H₂S + H₂SO₄ \(\longrightarrow \) SO₂ + 2H₂O + S
H₂SO₄ + 2HI \(\longrightarrow \) SO₂ + H₂O + I₂
H₂SO₄ + 2HBr \(\longrightarrow \) 2SO₂ + 2H₂O + Br₂

Density of the element, d = 2.7 x 10³ kg m^-3
Molar mass, M = 2.7 x 10^-2 kg mol^-1
Edge length, a = 405 pm
= 405 x 10^-12 m
= 4.05 x 10^-10 m
Avogadro's number, N_A= 6.022 x 10²³ mol^-1
\(\therefore d=\frac { Z\times M }{ { a }^{ 3 }\times { N }_{ A } } \)
\(\Rightarrow Z=\frac { d\times { a }^{ 3 }{ N }_{ A } }{ M } \)
\(=\frac { 2.7\times { 10 }^{ 3 }kg\quad { m }^{ -3 }{ (4.05\times { 10 }^{ -10 }m) }^{ 3 }\times 6.022\times { 10 }^{ 23 }{ mol }^{ -1 } }{ 2.7\times { 10 }^{ -2 }kg\quad { mol }^{ -1 } } \)
= 4.004 = 4.
This implies that four atoms of the element are present per unit cell. Hence the unit cell is face centred cubic.

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) 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.

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.

a) i) Ellingham diagram for the formation of Al₂O₃ and MgO intersects around 1500^oC. Above this temp Mg lies above the Aluminium line. Hence only above 1500^oC Aluminium might be excepted to reduce magnesia.
ii) Ellingham diagram for the formation of MgO lies below the formation of Al₂O₃. Hence MgO is more stable than Al₂O₃. Hence Magnesium could reduce Alumina.
1. Below 983K, formation of CO line lies below many of the metal oxide formation in Ellingham diagram, hence CO is more effective reducing agent than Carbon.
2. But above this temperature Carbon lies below other metal oxides.
b) Around 1200K Carbon lies below the formation of Fe₂O₃. Hence it is possible to reduce Fe₂O₃ by coke at 1200K.

(i) Benzoic acid is used as an urinary antiseptic.
(ii) Sodium benzoate is used as food preservative.
(iii) Benzoic acid vapours are used to disinfect bronchial tube.

The aqueous solutions of sodium or potassium salts of carboxylic acid on electrolysis gives alkanes at anode. This reaction is called kolbes electrolysis.

Sodium formate solution on electrolysis gives hydrogen

Adsorption is an exothermic process, i.e.,energy factor favours the process. As ΔG = ΔH - TΔS, in adsorption, though ΔS is -ve but ΔH is also -ve and ΔH > TΔS in magnitude so that ΔG is -ve.
Hence, the process is spontaneous.

(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.
\({ N }_{ 2 }+{ 3H }_{ 2 }\overset { Fe }{ \underset { +mo }{ \leftrightharpoons } } { 2NH }_{ 3 }\)

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)

The main functions of the salt bridge are
(i) To complete the electrical circuit by the allowing only ions to flow from one solution to other without mixing the two solutions.
(ii) To maintain electrical neutrality of the solution in the two half cells.

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

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

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.

(i) Acidic buffer solution: a solution containing a weak acid and its salt.
Example: Solution containing acetic acid and sodium acetate
(ii) Basic buffer solution: a solution containing a weak base and its salt.
Example: Solution containing NH₄OH and NH₄Cl.

(i) p^H = -log₁₀ [H₃O⁺]
(ii) The pH of a solution is defined as the negative logarithm of base 10 of the molar concentration of the hydronium ions present in the solution.

(i) Lewis acid: It is a species that accepts an electron pair. Eg: \(\mathrm{Ag}^{+} ; \mathrm{BF}_{3} ; \mathrm{A} / \mathrm{Cl}_{3}\)
(ii) Lewis base: It is a species that donates an electron pair. Eg: \( \mathrm{Cl}^{-} ; \mathrm{NH}_{3} ; \mathrm{H}_{2} \mathrm{O}\)

(i) Environmental conservation
(ii) Use jute bags or cloth bags in place of polythene bags.
(iii) KMnO₄ acts as a strong oxidising agent.

Order of the reaction = \(\frac { 1 }{ 2 } +\frac { 3 }{ 2 } =\frac { 4 }{ 2 } =2\)
The reaction is second order.

Rate of a reaction at any time depends on the concentration of the reactants which keeps on decreasing with time.

(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) Sulphur has great catenation property and forms large molecule having high weight, which results higher melting point.
(ii) While oxygen exists as diatomic molecule of less molecular weight, hence has very low melting point.

The defects which are caused by missing or misplaced atoms or ions in the crystal.

(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) Anhydrous AlCl₃ is an electro-deficient compound while hydrated AICl₃ is not.
(ii) Therefore, anhydrous AICl₃ is more soluble in diethyl either because the oxygen atom of ether donates a pair of electrons to the vacant p-orbital on the Al atom in AlCl₃ forming a co-ordinate bond.
(iii) In case of hydrated AlCl₃, Al is not electron deficient, since, H₂O has already donated a pair of e^- to it.

(i) Zeolites have a three dimensional crystalline structure looks like a honeycomb consisting of a network of interconnected tunnels and cages.
(ii) Water molecules moves freely in and out of these pores hence they are highly porous.

CrO < Cr₂O₃ < CrO₃
∴ Higher the oxidation state, more will be the acidic character.

MA₂B₂
[Pt(NH₃)₂Cl₂]

(i) The stereoisomers of a coordination compound have the same chemical formula and connectivity between the central metal atom and the ligands.
(ii) But they differ in the spatial arrangement of ligands in three dimensional space. They can be further classified as geometrical isomers and optical isomers.

(i) Lighter sulphide ore particles are wetted by pine oil and rise to the surface with the from while gangue particles are wetted by water.
(ii) Therefore, the froth flotation method is selected for the concentration of sulphide ores.

(i) Conversion of ores into oxides.
(ii) Reduction of metal oxides.

Ea = 200 kJ mol^-1=200 \(\times\) 10³J mol^-1
A = 1.6 \(\times\) 10¹³s^-1; T = 600 K; R = 8.314 JK mol^-1
\(k=A{ e }^{ -\left( \frac { Ea }{ RT } \right) }\)
\(k=1.6\times { 10 }^{ 13 }{ s }^{ -1 }{ e }^{ -\left( \frac { 200\times 10^3}{ 8.314 \times 600 } \right) }\)
\(k=1.6\times { 10 }^{ 13 }{ s }^{ -1 }{ e }^{ -\left( 40.09 \right) }\)
\(k=1.6\times { 10 }^{ 13 }\times 3.8\times { 10 }^{ -18 }{ s }^{ -1 }\)
\(k=6.08\times { 10 }^{ -5 }{ s }^{ -1 }\)

An interstitial compound or alloy is a compound that is formed when small atoms like hydrogen, boron, carbon or nitrogen are trapped in the interstitial holes in a metal lattice. They are usually non-stoichiometric compounds. Transition metals form a number of interstitial compounds such as TiC, ZrH_1.92, Mn₄N etc.
Properties of interstitial compound
(i) They are hard and show electrical and thermal conductivity.
(ii) They have high melting points higher than those of pure metals.
(iii) Transition metal hydrides are used as powerful reducing agents 
(iv) Metallic carbides are chemically inert.

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)

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

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) Fluorine is most electronegative atom.
(ii) It has only one unpaired electron.

(i) In heavier post transition metals, the outers electrons (ns) have a tendency to remain inert and show reluctance to take part in the bonding, which is known as inert pair effect.
(ii) This effect is also observed in groups 14, 15 and 16.

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

In p-block elements, the first member of each group differs from the other elements of the corresponding group. The following factors are responsible for this anomalous behaviour
(i) Small size of the first member.
(ii) High ionisation enthalpy and high electronegativity.
(iii) Absence of d-orbitals in their valance shell.

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