Hydroboration - Anti Markownikoff product
CH₃ – CH₂ –CH₂  – CH₂ –  CH₂ – OH

ΔS is -ve

As₂S₃ is a negatively charged colloid. It will be most effectively coagulated by the cation with greater valency. i.e., Al³⁺.

Z is tie strongest oxidising agent (High SRP)
X is the strongest reducing agent (Low SRP)

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)

Rate = k[A]ⁿ
Rate = \(\frac{-\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\)
unit of rate = \(\frac{mol L^{-1}}{s}\)=mol L^-1/s^-1
unit of rate constant 
\(=\frac{ (mol{ L }^{ -1 }{ S }^{ -1 }) }{ ({ mol }{ L }^{ -1 })^n } \)
= mol^1-nL^n-1s^-1
in the case
rate = k [Acetone]^3/2
n = 3/2
= mol^1-(3/2)L^(3/2)-1s^-1
(mol^-(1/2) L^(1/2) s^-1).

For a first order reaction
t_1/2 = \(\frac { 0.693 }{ { k }}\)
t_1/2 does not depend on the initial concentration and it remains constant (whatever may be the initial concentration)
t_1/2 = 2.5 hrs

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

Three isomers. If we consider any one of the ligands as reference (say Py), the arrangement of other three ligands (NH₃, Br^- and Cl^-) with respect to (Py) gives three geometrical isomers.

Esters containing at least one ∝- hydrogen atom undergo self condensation in the presence of a strong base such as sodium ethoxide to form β-keto ester.

HCHO + CaCO₃

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

Its is because oxidation half reaction and reduction half reaction cannot take place alone. It can be measured only by using a reference electrode.

Conductivity, molar conductance and equivalent conductance increases with dilution whereas Conductance (C) decreases.

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

According to Lewis concept of Acid and bases, any species capable of accepting an electron pair is an acid. BF₃ is electron deficient so accepts a pair of electron, Hence termed as acid

When the product of molar concentration of the constituent ions i.e., ionic product, exceeds the solubility product then the compound gets precipitated. Ionic product > K_sp, precipitation will occur and the solution is super saturated.

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

(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^-

Average rate = \(-\frac { 1 }{ 2 } \frac { \triangle \left[ A \right] }{ \triangle L } \)
\(=-\frac { 1 }{ 2 } \frac { { \left[ A \right] }_{ 2 }-{ \left[ A \right] }_{ 1 } }{ { t }_{ 2 }-{ t }_{ 1 } } \)
\(=-\frac { 1 }{ 2 } \left( \frac { 0.4-0.5 }{ 10 } \right) \)
\(=-\frac { 1 }{ 2 } \left( \frac { -0.1 }{ 10 } \right) \)
= 0.005 mol L^-1 min^-1
= 5 x 10^-3 M min^-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) The given statement is true.
(ii) The relative acidic strength of HF, HCI, HBr and HI depends upon their bond dissociation enthalpies.
(iii) Since, the bond dissociation enthalpy of H-X bond from H-F to H-I as the size of atom increases from F to l. Therefore, the acidic strength increases in the opposite order as:
H - F < H - CI < H - Br < H - I

Yellow phosphorus reacts with alkali on boiling in an inert atmosphere liberating phosphine. Phosphorus acts as a reducing agent.
\({ P }_{ 4 }+NaOH+{ H }_{ 2 }O\longrightarrow \underset { Sodiumhypophosphite }{ { 3NaH }_{ 2 }{ PO }_{ 2 } } +\underset { Phosphine }{ { PH }_{ 3 } } \uparrow \)

Total number of atoms per unit cell in a fcc structure = 4 atoms.

(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) 3SiCl₄ + 4 Al ⟶ 4 AICl₃ + 3 Si
(ii) n(CH₃)₂Si(OH)₂

(iii) \(SiC{ l }_{ 4 }+4{ H }_{ 2 }O\longrightarrow \underset { orthosilicic \ acid }{ { H }_{ 4 }Si{ O }_{ 4 } } +4HCl\)
H₄SiO₄ \(\overset { \triangle }{ \longrightarrow } \) SiO₂ + 2H₂O
SiO₂ + Na₂CO₃ ⟶ Na₂SiO₃ + H₂O.

The name alum is given to the double salt of potassium aluminium sulphate.
[K₂SO₄.Al₂(SO₄)₃.24H₂O]

(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

+3

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

It is essential to design an eco friendly metallurgical process that would minimize waste, maximize energy efficiency. Such advances in metallurgy is vital for the economic and technical progress in the current era.

Tetraammine copper (II) sulphate.

K₃[CoF₆]

In Gd⁺³ there are 64 electrons. Hence electronic configuration will be [Xe]4f⁷ 5d¹ 6s². Hence no electrons are there in outer d - orbital. Due to this it is colourless.

The fourteen elements following actinoids is from thorium to lawrencium are called actinides.
Examples: Uranium, Thorium, Neptunium

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

(a) 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

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.

In this extraction, a basic flux, quick lime (CaO) is used, since the silica gangue present in the ore is acidic in nature. The quick lime combines with it to form calcium silicate (slag).
CaO_(s) + Sio_2(s) ⟶ CaSio_3(s)
Flux       Gangue    Slag

(i) Compound A is identified as benzaldehyde C₆H₅CHO from its molecular formula. C₆H₅CHO undergoes Cannizzaro reaction and it does not reduce Fehling's solution.
\({ C }_{ 6 }{ H }_{ 5 }CHO+{ C }_{ 6 }{ H }_{ 5 }CHO\overset { NaOH }{ \longrightarrow } { C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }OH+{ C }_{ 6 }{ H }_{ 5 }COONa\) 
(ii) Benzaldehyde reacts with aniline to form Schiff's base C₆H₅CH = NC₆H₅ and it is (B)
\({ C }_{ 6 }{ H }_{ 5 }-\overset { \underset { | }{ H } }{ C } =\underset { Aniline }{ O+{ C }_{ 6 }H_{ 5 }{ NH }_{ 2 } } \longrightarrow \underset { Schiff's \ base(B) }{ { C }_{ 6 }{ H }_{ 5 }CH={ NC }_{ 6 }{ H }_{ 5 }+{ H }_{ 2 }O } \)
(iii) Benzaldehyde reacts with chlorine in the presence of catalyst to give m-chlorobenzaldehyde and it is (C).

Compound	Compound Name	Formula
A	Benzaldehyde	C6H5CHO
B	Schiff's base	C6H5CH = NC6H5
C	m-cholorobenzaldehyde

(i) Compound (A) with molecular formula C₆H₆O gives violet colour with neutral FeCI₃. 
(ii) (A) reacts with CHCl₃ and NaOH gives two isomers (B) and (C) with molecular formula C₇H₆O₂.

(iii) Compound (A) reacts with ammonia at 473 K in the presence of ZnCl₂ gives compound (D) with molecular formula C₆H₇N.

(iv) Compound (D) undergoes carbylamine test.
\(\underset { (D) }{ { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 } } +CHCl_{ 3 }+3KOH\overset { \triangle }{ \longrightarrow } { C }_{ 6 }{ H }_{ 5 }NC+3KCl+3{ H }_{ 2 }O\)

Compound	Compound Name	Formula
A	Phenol	C6H5OH
B	Orthohydroxy benzaldehyde
C	Para hydroxy benzaldehyde
D	Aniline	C6H5NH2

(i) Suppose the reactant of a reaction is present in one solvent and the other reactant is present in an another solvent.
(ii) The reaction between them is very slow, if the solvents are immisible.
(iii) As the solvents from separate phases the reactants have to migrate across the boundary to react.
(iv) But migration of reactants across the boundary is not easy. For such situations a third solvent is added which is miscible with both.
(v) So, the phase boundary is eliminated reactants freely mix and react fast.
(vi) But for large scale production of any product, use of a third solvent is not convenient as it may be expensive.
(vii) For such problems phase transfer catalysis provides a simple solution, which avoides the use of solvents.
(viii) It directs the use a phase transfer catalyst (a phase transfer reagent) to facilitate transport of a reactant in one solvent to the other solvent where the second reactant is present.
(ix) As the reactants are now brought together they rapidly react and form the product.

Given: K = 0.335;
Eo = (Ag+, Ag) = 0.7991 V
n = 1, F = 96495 C
Formula:
\({ E }^{ o }=\frac { -2.303RT }{ nF } \log K\)
Solution:
\(=\frac { -0.591 }{ 1 } \log 0.335\)
Standard emf of the cell
= - 0.0280 V
Eocell = EoF - EoL; EoR = ?
∴ EoR = Eocell - EoL
= -0.0280 + 0.7991 = 0.771 V
Eo(Fe3+,Fe2+) = 0.771 V.

The degree of ionisation and the dissociation constant of the weak acid are related by the equation.
\({ K }_{ a }=\frac { { \alpha }^{ 2 }{ C } }{ 1-\alpha } \cong { \alpha }^{ 2 }C\)
α = 5% (or) 0.05
C = 0.02M
K_a = (0.05)2 x 0.02 = 0.00005
K_a = 5 x 10^-5

1. Daniel cell is a galvanic cell. This is a voltaic cell also.
(a) The separation of half reaction is the basis for the construction of Daniel cell. It consists of two half cells.
(i) Oxidation half cell: A metallic zinc strip  that dips into an aqueous solution of zinc sulphate taken in a beaker, as shown in Figure
(ii) Reduction half cell: A copper strip that dips into an aqueous solution of copper sulphate taken in a beaker, as shown in Figure

(iii) Joining the half cells:
(a) The zinc and copper strips are externally connected using a wire through a switch (k) and a load (example: volt meter). The electrolytic solution present in the cathodic and anodic compartment are connected using an inverted U tube containing a agar-agar gel mixed with an inert electrolyte such as KCI, Na₂SO₄ etc.,
(b) The ions of inert electrolyte do not react with other ions present in the half I cells and they are not either oxidised (or) reduced at the electrodes. The solution in the salt bridge cannot get poured out, but through which the ions can move into (or) out of the half cells.
(c) When the switch (k) closes the circuit, the electrons flows from zinc strip to copper strip. This is due to the following redox reactions which are taking place at the respective electrodes.
(iv) Anodic oxidation:
(i) zinc strip acts as the anode.
(ii) Here,oxidation occurs.
The electrode at which the oxidation occur is called the anode. In Daniel cell, the oxidation take place at zinc electrode, i.e., zinc is oxidised to Zn²⁺ ions and the electrons.
The Zn²⁺ ions enters the solution and the electrons enter the zinc metal, then flow through the external wire and then enter the copper strip.
Electrons are liberated at zinc electrode and hence it is negative (-ve).
\(Zn_{ (s) }\longrightarrow { { Zn }^{ 2+ }_{ (aq) }+{ 2e }^{ - } } \)  (loss of electron-oxidation)  
(v) Cathodic reduction:
As discussed earlier. the electrons flow through the circuit from zinc to copper, where the Cu²⁺ ions in the solution accept the electrons, get reduced to copper and the same get deposited on the electrode Here, the electrons are consumed and hence it is positive (+ve).
\({ Cu }_{ (aq) }^{ 2+ }+{ 2e }^{ - }\longrightarrow { { Cu }_{ (s) } } \)(gain of electron - reduction)
b) When a Zinc metal strip is placed in a copper sulphate solution, the blue colour of the solution fades and the copper is deposited on the zinc strip as red - brown crust due to the following spontaneous chemical reaction.
\(\mathrm{Zn}_{(\mathrm{s})}+\mathrm{CuSO}_{4(\mathrm{aq})} \rightarrow \mathrm{ZnSO}_{4(\mathrm{aq})}+\mathrm{Cu}_{(\mathrm{s})}\)
The energy produced in the above reaction is lost to the surroundings as heat.
In the above redox reaction, Zinc is oxidised to Zn²⁺ ions and the Cu²⁺ ions are reduced to metallic copper. The half reactions are represented as below.
\(\mathrm{Zn}_{(\mathrm{s})} \rightarrow \mathrm{Zn}^{2+}{ }_{(\mathrm{aq})}+2 \mathrm{e}^{-} \text {(oxidation) }  \)
\(\mathrm{Cu}^{2+}{ }_{(\mathrm{aq})}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}_{(\mathrm{s})} \text { (reduction) }\)
If we perform the above two half reactions separately in an apparatus as shown in figure, some of the energy produced in the reaction will be converted into electrical energy.

(i) Ostwald's dilution law relates the dissociation constant of the weak acid (K_a) with its degree of dissociation (α) and the concentration (c).
where  \(\alpha=\frac{\text { Number of moles dissociated }}{\text { Total number of moles }}\)
(ii) The dissociation of acetic acid can be represented as
\(\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{H}^{+}+\mathrm{CH}_{3} \mathrm{COO}^{-}\)
The dissociation constant of acetic acid is,
\({ K }_{ a }=\frac { \left[ { H }^{ + } \right] \left[ { CH }_{ 3 }COO^{ - } \right] }{ \left[ { CH }_{ 3 }COOH \right] } \)  ........(1) 

Substituting the equilibrium concentration in equation (1)
\({ K }_{ a }=\cfrac { \left( \alpha C \right) \left( \alpha C \right) }{ \left( 1-\alpha \right) C } \) 
\({ K }_{ a }=\cfrac { { \alpha }^{ 2 }C }{ 1-\alpha } \) .......(2) 
(iii) We know that weak acid dissociates only to a very small extent compared to one, a is so small and hence in the denominator (1 - α) ⋍1. The above expression (2) now becomes,
k_a =a²C \(\Rightarrow { \alpha }^{ 2 }=\cfrac { { k }_{ a } }{ C } \) ; \(\alpha =\sqrt { \cfrac { { K }_{ a } }{ C } } \) 
(iv) When dilution increases, the degree of dissociation of weak electrolyte also increases. This is called Ostwald's dilution law
Also \(;\left[\mathrm{H}^{+}\right]=\alpha \mathrm{C}\) and \(\left[\mathrm{H}^{+}\right]=\left(\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{C}}}\right) \mathrm{C}=\sqrt{\frac{\mathrm{K}_{\mathrm{a}} \mathrm{C}^{2}}{\mathrm{C}}}=\sqrt{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{C}}\)
Similarly for a weak base
\(\begin{aligned} & \mathrm{K}_{\mathrm{b}}=\alpha^2 \mathrm{C} ; \quad \therefore \alpha=\sqrt{\frac{\mathrm{k}_{\mathrm{b}}}{\mathrm{C}}}, \\ \end{aligned}\)
\(\begin{aligned} & {\left[\mathrm{OH}^{-}\right] \alpha \mathrm{C}=\sqrt{\frac{\mathrm{K}_{\mathrm{b}}}{\mathrm{C}}} \times \mathrm{C}=\sqrt{\frac{\mathrm{K}_{\mathrm{b}} \mathrm{C}^2}{\mathrm{C}}}=\sqrt{\mathrm{K}_{\mathrm{b}} \mathrm{C}}} \end{aligned}\)

Given data: Rate = 5.2 x 10^-6 mol lit^-1 sec^-1
Initial concentration [A] = 2.6 x 10^-3
Formula: Rate = k[A]¹
Solution: Rate = 5.2 x 10^-6 mol lit^-1 sec^-1 = k x 2.6 x 10^-3 lit^-1
= \(\frac { 5.26\times { 10 }^{ -6 }mol{ \ lit }^{ -1 }{ sec }^{ -1 } }{ 2.6\times { 10 }^{ -3 }mol \ { lit }^{ -1 } } \)= 2 x 10^-3
∴ k = 2 x 10^-3 s^-1

(i) It is generally expected a steady decrease in atomic radius along a period as the nuclear charge increases and the extra electrons are added to the same sub shell.
(ii) But for the 3d transition elements, the expected decrease in atomic radius is observed from Sc to V, thereafter up to Cu the atomic radius nearly remains the same.
(iii) As we move from Sc to Zn in 3d series the extra electrons are added to the 3d orbitals, the added 3d electrons only partially shield the increased nuclear charge and hence the effective nuclear charge increases slightly.
(iv) However, the extra electrons added to the 3d sub shell strongly repel the 4s electrons and these two forces are operated in opposite direction and as they tend to balance each other, it leads to constancy in atomic radii.
(v) At the end of the series, d - orbitals of Zinc contain 10 electrons in which the repulsive interaction between the electrons is more than the effective nuclear charge and hence, the orbitals slightly expand and atomic radius slightly increases.

(I) Since the byproduct of roasting to sulphide ore is SO₂ It turns lime water milky.
Therefore, gas 'B' must be SO₂
(ii) As the gas 'B' is obtained when amorphous solid 'A' burns in air therefore, amorphous solid 'A' must be sulphur S₈
\(\underset { (A) }{ { S }_{ g } } +{ 8O }_{ 2 }\overset { \Delta }{ \longrightarrow } \underset { (B) }{ { 8SO }_{ 2 } } \)
(iii) Gas (B) reduces acidified aqueous KMnO₄ solution and reduces Fe³⁺ to Fe²⁺ salts as shown below:
\(\underset { (yellow) }{ { 2MnO }_{ 4 } } ^{ - }+\underset { (b) }{ { SO }_{ 2 } } +2{ H }_{ 2 }O\longrightarrow { 2Fe }^{ 2+ }+\underset { (Green) }{ { SO }_{ 4 }^{ 2- } } +{ 4H }^{ + }\)
(iv) Thus, solid 'A' is S₈ and gas 'B' is SO₂

(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) The process in which no external reducing. When ore is heated in air, a part of the ore gets oxidised which combines with remaining sulphide to give metal.
Eg: Reduction copper glance (Cu₂S)
\({ 2Cu }_{ 2 }S+{ 3O }_{ 2 }\longrightarrow { 2Cu }_{ 2 }O+2{ SO }_{ 2 }\uparrow \)
\({ 2Cu }_{ 2 }O+{ Cu }_{ 2 }S\longrightarrow 6Cu+{ SO }_{ 2 }\uparrow \)
(ii) Electrolytic reduction is used for highly electro positive metals. In this method, fused metal ore is electrolysed and pure metal is deposited at cathode.
Eg: Al is obtained by electrolysis of Al₂O_{3
\({ 2Al }_{ 2 }{ O }_{ 3 }+3C\longrightarrow \underset { Cathode\quad Anode }{ 4Al+{ 3CO }_{ 2 } } \)}
(iii) \({ Cr }_{ 2 }{ O }_{ 3 }+2Al\longrightarrow {Al }_{ 2 }{ O }_{ 3 }+2Cr\)
\(3 \mathrm{Mn} _3 O_{4}+8 \mathrm{Al} \longrightarrow 4 \mathrm{Al}_{2} \mathrm{O}_{3}+9 \mathrm{Mn}\)

(i) Every metal atom has two types of valencies Primary valency or ionisable valency Secondary valency or non ionisable valency
(ii) The primary valency corresponds to the oxidation state of the metal ion. It is always satisfied by negative ions.
(iii) Secondary valency corresponds to the coordination number of the metal ion or atom. It is satisfied by either negative ions or neutral molecules.
(iv) The molecules or ions that satisfy secondary valencies are called ligands.
(v) The ligands which satisfy secondary valencies must project in definite directions in space. So the secondary valencies are directional in nature whereas the primary valencies are non - directional in nature.
(vi) The ligands have unshared pair of electrons. These unshared pair of electrons are donated to central metal ion or atom in a compound. Such compounds are called coordination compounds.
Werner's representation
Eg: [Co(NH)₆]Cl₃
Cl: primary valency (dotted lines)
NH₃: secondary valency (solid lines).
Defects of Werner's theory
Werner's theory describes the structures of many co-ordination compounds successfully. However, it does not explain the magnetic and spectral properties.

(i) Collision theory is based on the kinetic theory of gases. According to this theory, a chemical reaction occurs as a result of collisions between the reacting molecules.
(ii) Let us understand this theory by considering the following reaction.
A₂(g) + B₂(g) ⟶ 2AB(g)
(iii) If we consider that, the reaction between A₂ and B₂ molecules proceeds through collisions between them, then the rate would be proportional to the number of collisions per second.
(iv) Rate ∝ number of molecules colliding per litre per second (collision rate).
(v) The number of collisions is directly proportional to the concentration of both A₂ and B₂.
Collison rate ∝ [A₂][B₂]
Collision rate = Z [A₂][B₂]
(vi) Where, Z is a constant
(vii) A fraction of effective collisions (f) is given by the following expression
\(f={ e }^{ \frac { { -E }_{ a } }{ RT } }\)
(xiv) This fraction of collisions is further reduced due to orientation factor i.e., even if the reactant collides with sufficient energy, they will not react unless the orientation of the reactant molecules is suitable for the formation of the transition state.
(viii) The diagram illustrates the importance of proper alignment of molecules which leads to reaction.

(ix) The fraction of effective collisions (f) having proper orientation is given by the steric factor p.
⇒ Rate = p x f x collision rate
\(\Rightarrow Rate=p\times { e }^{ \frac { -Ea }{ RT } }\times Z\left[ { A }_{ 2 } \right] \left[ { B }_{ 2 } \right] \quad ...(1)\)
As per the rate law,
Rate = \(k=\left[ { A }_{ 2 } \right] \left[ { B }_{ 2 } \right] \quad ...(2)\)
Where k is the rate constant
On comparing equation (1) and (2), the rate constant k is
\(k=pZ{ e }^{ \frac { -Ea }{ RT } }\)

Chlorine reacts with cold dilute alkali to give chloride and hypochlorite, while with hot concentrated alkali chlorides and chlorates are formed.
\(\mathrm{Cl}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HCl}+\underset{\text { Hypochlorous acid }}{\mathrm{HOCl}} \)
\(\mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O} \)
\(\mathrm{HOCl}+\mathrm{NaOH} \rightarrow \mathrm{NaOCl}+\mathrm{H}_{2} \mathrm{O} \)
                   (Sodium hypo chlorite)
Overall Reaction
3Cl₂ + 6NaOH \(\rightarrow\) NaClO₃ + 5NaCl + 3H₂O
                             (Sodium Chlorate)

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.

Zone refining :
1. Zone refining method is based on the principles of fractional crystallisation.
2. When an impure metal is melted and allowed to solidify, the impurities will prefer to be in the molten region. In this process the impure metal is taken in the form of a rod.
3. One end of the rod is heated using a mobile induction heater which results in melting of the metal on that portion of the rod.
4. When the heater is slowly moved to the other end the pure metal crystallises while the impurities will move on to the adjacent molten zone.
5. As the heater moves further away, the molten zone containing impurities also moves along with it.
6. The process is repeated several times by moving the heater in the same direction again and again to get pure metal.
7. This process is carried out in an inert gas atmosphere to prevent the oxidation of metals.
8. Elements such as germanium (Ge), silicon (Si) and galium (Ga) that are used as semiconductor are refined using this process.

(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,

When ethers are heated to dryness,
(i) They form peroxide by the action of air or oxygen.
(ii) Ether oxygen is capable of forming a co-ordinate covalent bond with electron deficient species.

(ii) These peroxides are unstable and decomposes violently with explosion on heating. Hence, ether should not be heated to dryness.

Brownian movement enables us,
(i) to calculate Avogadro number.
(ii) to confirm kinetic theory which considers the ceaseless rapid movement of molecules that increases with increase in temperature.
(iii) to understand the stability of colloids: As the particles are in continuous rapid movement they do not come close and hence not get condensed.
(iv) That is Brownian movement does not allow the particles to be acted on by force of gravity.

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

(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.
\(\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{CH}_{3}-\mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{H}_{3} \mathrm{O}_{(\mathrm{aq})}^{+} \)
\(\mathrm{CH}_{3} \mathrm{COONa}_{(\mathrm{s})} \stackrel{\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}}{\longrightarrow} \mathrm{CH}_{3}-\mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{Ha}_{(\mathrm{aq})}^{+}\)

(i) The half cell reactions are :
\(2Br^{-} \rightarrow Br_{2}+2e^{-}\)     \(E^{0}_{ox}=-1.09V\) ...(1)
\(2Fe^{3+}+2e^{-}\rightarrow2Fe^{2+}\)  \(E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{0}=+0.771V\) ..(2)
(ii) Adding (1) of (2) :
\(2Fe^{3+}+2Br^{-}\rightarrow 2Fe^{2+}+Br_{2}\)   \(E^{0}_{cell}=?\) ...(3)
\(E^{0}_{cell}=E^{0}_{ox}+E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{0}\)
= (-1.09 + 0.771)V
= -0.319V
(iii) E⁰_cell is – ve; \(\Delta G\) is +ve and the cell reaction is non spontaneous.
(iv) Hence Fe³⁺ cannot oxidises Br^- to Br₂.

Formula:
\(\log { \frac { { k }_{ 2 } }{ { k }_{ 1 } } } =\frac { { E }_{ a }\left( { T }_{ 2 }-{ T }_{ 1 } \right) }{ 2.303R{ T }_{ 1 }{ T }_{ 2 } } \)
Given:
Energy of activation: E = 150 kJ = 150000 J
Temperatures: T₁ = 673 K; T₂ = 773 K
Rate constant: k₁ = 2.3 x 10^-3
Gas constant: R = 8.314 J k^-1 mol^-1
Solution:
\(\log { \frac { { k }_{ 2 } }{ { k }_{ 1 } } } =\frac { { k }_{ 2 } }{ 2.3\times { 10 }^{ -3 } } =\frac { 150000(773-673) }{ 2.303\times 8.314\times 673\times 773 } \)
\(=\frac { 15000000 }{ 2.303\times 8.314\times 673\times 773 } \)
\(\log { \frac { { k }_{ 2 } }{ 2.3\times { 10 }^{ -3 } } } \) = 1.505
\(\frac { { k }_{ 2 } }{ 2.3\times { 10 }^{ -3 } } \) Antilog 1.5905 = 32
∴ k₂ = 2.3 x 10^-3 x 32 = 7.36 x 10^-2
k₂ = 7.36 x 10^-2

(i) The use of 3d electron for formation of I bond increases from Sc to Mn, causing the increase in oxidation state upto +7.
(ii) The reason for Mn having highest oxidation state of +7 is due to the presence of 7 unpaired electrons in its atom.
(iii) As the number of unpaired electrons decrease from Fe to Cu. So there is the decrease in oxidation state.

(i) Ammonia reacts with chlorine and chlorides to give ammonium chloride as a final product.
(ii) The reactions are different under different conditions as given below.
(iii) With excess ammonia
\({ 2NH }_{ 3 }+3{ Cl }_{ 2 }\longrightarrow { N }_{ 2 }+6HCl\)
\(6HCl+6{ NH }_{ 3 }\longrightarrow { 6NH }_{ 4 }Cl\)
(iv) With excess of chlorine ammonia reacts to give nitrogen trichloride, an explosive substance.
\({ 2NH }_{ 3 }+{ 6Cl }_{ 2 }\longrightarrow { 2NCl }_{ 3 }+6HCl\)
\({ 2NH }_{ 3 }(g)+{ HCl }{ (g) }\longrightarrow NH_4Cl(s)\)

(i) H-bonding is the intermolecular force present in ice.
(ii) Covalent bonds are present in network solids.

(i) Diamond is very hard.
(ii) The carbon atoms in diamond are sp³ hybridised and bonded to four neighbouring carbon atoms by a bonds with a C-C bond length of 1.54 Å.
(iii) This results in a tetrahedral arrangement around each carbon atom that extends to the entire lattice.
(iv) Since all four valance electrons of carbon are involved in bonding there is no free electrons for conductivity.
(v) Being the hardest element, it used for sharpening hard tools, cutting glasses, making bores and rock drilling.

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) 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.

Average rate of reaction:
The average rate is defined as the ratio of change in the final concentration of reactants and the initial concentration of reactants over the entire time period of reaction.
Average rate=\(\frac{-[Final \space concentration \space of \space reactants - Initial \space concentration \space of \space reactants}{(Change \space in \space time)}\)
R \(=\frac{-\left(\left[A_{2}\right]-\left[A_{1}\right]\right)}{\left(t_{2}-t_{1}\right)}=-\left(\frac{\Delta[A]}{\Delta t}\right)\)
[A]₁ = Concentration of reactant A₁ at time t₁
[A]₂ = Concentration of reactant A₂ at time t₂
Instantaneous rate of reaction:
The rate of reaction at any particular instant during the course of reaction is called as instantaneous rate.
Instantaneous rate \(=(\text { Average rate })_{\Delta t \rightarrow 0}\)
Rate of the reaction \(=\left(\frac{-\Delta \mathrm{A}}{\Delta \mathrm{t}}\right)\)

(i) In transition series, as we move down from Ti to Zn, the standard reduction potential E⁰_M²⁺/M³ value is approaching towards less negative value and copper has a positive reduction potential, i. e. elemental copper is more stable than Cu²⁺.
(ii) E⁰_M²⁺/M value for manganese and zinc are more negative than regular trend. It is due to extra stability arises due to the half filled d⁵ configuration in Mn²⁺ and completely filled d¹⁰ configuration in Zn²⁺.
(iii) The standard electrode potential for the M³⁺/M²⁺ half cell gives the relative stability between M³⁺ and M²⁺.
(iv) The high reduction potential of Mn³⁺/Mn²⁺ indicates Mn²⁺ is more stable than Mn³⁺.
(v) Mn³⁺ has a 3d⁴ configuration while that of Mn²⁺ is 3d⁵. The extra stability associated with a half filled d sub-shell makes the reduction of Mn³⁺ very feasible \(\left[\mathrm{E}^{\circ}=+1.51 \mathrm{~V}\right]\).

These two are ionisation isomers. [Co(NH₃)₅Cl]SO₄ gives white precipitate with BaCl₂ solution, but not with AgNO₃ solution. [Co(NH₃)₅SO₄]Cl gives curdy white precipitate with AgNO₃ solution but not with BaCl₂ solution.

\(\text {(a) }\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}:{ }_{24} \mathrm{Cr} \Rightarrow{ }_{18}[\mathrm{Ar}] 4 \mathrm{~s}^{2} 3 \mathrm{~d}^{4} \)
\({ }_{21} \mathrm{Cr}^{3+} \Rightarrow{ }_{18}[\mathrm{Ar}] 3 \mathrm{~d}^{3}\)

(i) d²sp³ hybridisation (octahedral)
(ii) It has three unpaired electrons (n = 3)
(iii) So it is paramagnetic
(iv) Magnetic moment \(\left(\mu_{\mathrm{s}}\right)=\sqrt{\mathrm{n}(\mathrm{n}+2)} \mathrm{BM}\)
\(=\sqrt{3(3+2)}=\sqrt{15}=3.87 \mathrm{BM}\)
\((b) \ \left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-:}{ }_{28} \mathrm{Ni} \Rightarrow[\mathrm{Ar}] 4 \mathrm{~s}^{2} 3 \mathrm{~d}^{8} ;{ }_{26} \mathrm{Ni}^{2+} \Rightarrow{ }_{18}[\mathrm{Ar}] 3 \mathrm{~d}^{8}\)

(i) dsp² hybridisation
(ii) Geometry - square planar
(iii) No unpaired electrons- It is Diamagnetic
(iv) Magnetic moment (μ_s) = 0.

(i) Helium is used to provide inert atmosphere in electric-arc welding of metals.
(ii) Helium has lowest boiling point hence used in cryogenics.
(iii) It is much less denser than air and hence used for filling air balloons.