Nitrogen on direct reaction with hydrogen gives ammonia. This reaction is favoured by high pressures and at optimum temperature in presence of iron catalyst.
\(\frac{1}{2} \mathrm{~N}_{2}+\frac{3}{2} \mathrm{H}_{2} \rightleftharpoons \mathrm{NH}_{3} \)
\( \Delta \mathrm{H}_{\mathrm{f}}=-46.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

\(\mathrm{Cr}+\mathrm{nB} \stackrel{150 \mathrm{~K}}{\longrightarrow} \mathrm{CrB}_{\mathrm{n}}\)

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

Kohlrausch deduced the following empirical relationship between the molar conductance (\({ \Lambda }_{ m }\)) and the concentration of the electrolyte (C).
\(​​{ \Lambda }_{ m }={ \Lambda }_{ m }^{ o }-k\sqrt { C } \)

Cathodic protection:
In this technique, unlike galvanising the entire surface of the metal to be protected need not be covered with a protecting metal. Instead, metals such as Mg or zinc which is corroded more easily than iron can be used as a sacrificial anode and the iron material acts as a cathode. So iron is protected, but Mg or Zn is corroded. This known as sacrificial protection. (or) Cathodic protection.

In chemical adsorption, \(\frac { x }{ m } \) increases with rise in temperature and then decreases. The increase illustrates the requirement of activation of the surface for adsorption is due to fact that formation of activated complex requires certain energy. The decrease at high temperature is due to desorption, as the kinetic energy of the adsorbate increases.

∴ Two dipeptides structures are possible from glycine and alanine. They are glycyl alanine and Alanyl glycine.

(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: \( \mathrm{Cl}^{-} ; \mathrm{NH}_{3} ; \mathrm{H}_{2} \mathrm{O}\)

\(\log K=\log A-\frac { { E }_{ a } }{ 2.303R } \left( \frac { 1 }{ T } \right) \)
y = c + mx
\(m=-\frac { { E }_{ a } }{ 2.303R } \)
E_a = -2.303 Rm
E_a = -2.303 x 8.314 x (-4000)
E_a = 76,589J  mol^-1
E_a = 76.589 KJ mol^-1

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

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) sp³d³ hybridisation
(ii) Pentagonal bipyramidal structure.

(i) A basic repeating structural unit of a crystalline solid is called a unit cell.
(ii) A crystal is consisted of large number of unit cells.

i) Silica is an acidic flux which is mixed with Copper pyrites ore and is heated in a reverberatory furnace. The ferrous oxide formed from the ore is basic in nature and it combines with silica to form ferrous silicate. It is removed as slag.
\(\mathrm{FeO}_{(s)}+\mathrm{SiO}_{2(s)} \overset { \Delta }{ \longrightarrow } \mathrm{FeSiO}_{3(s)}\\ gangue \quad flux \quad \quad slag\)
ii) Aluminium oxide does not dissolve in water, but it dissolves in molten cryolite. This is an aluminium compound with a lower melting point than aluminium oxide. The use of cryolite reduces some of the energy costs involved in extracting aluminium.
\(\mathrm{Al}_{2} \mathrm{O}_{3} \overset { \Delta }{ \longrightarrow } 2 \mathrm{Al}^{3+}+3 \mathrm{O}^{2-}\)
iii) Van-Arkel method is based on the thermal decomposition of metal compounds which lead to the formation of pure metals. Zirconium can be purified by using this method. Zirconium metal is heated with iodine at a temperature of 550 K to form zirconium tetraiodide. (ZrI₄). The impurities are left behind
\(\mathrm{Zr}_{(\mathrm{s})}+2 \mathrm{I}_{2(\mathrm{~s})} \frac{550 \mathrm{~K}}{\mathrm{D}} \mathrm{ZrI}_{\text {t(vapour })}\)
The volatile (ZrI₄) on further heating around 1800 K, it gives pure zirconium.
\(\mathrm{ZrI}_{4 \text { (vapour) }} \frac{1800 \mathrm{~K}}{\mathrm{D}} \mathrm{Zr}_{(\mathrm{s})}+2 \mathrm{I}_{2} \)    
iv) In the extraction of lead from galena (PbS) by froth flotation process, if ZnS is present as impurity, Sodium cyanide is added to depresses the flotation property of ZnS. It forms a complex. Na₂[Zn(CN)₄] on the surface of zinc sulphide.

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.

Mn²⁺ + 2e^- ➝ Mn E^o = -1.18V
Mn²⁺ ➝ Mn³⁺ + e^- E^o = -1.51V
3Mn²⁺➝ Mn³⁺ + 2Mn³⁺ E^o_cell = ?
E^o_cell = (E^o_ox )+ (E^o_red) 
= 1.51 - 1.18 and non spontaneous
= -2.69V
Since E°is ve ΔG is +ve and the given forward cell reaction is non- spontaneous

HPO₄²⁻ can have the ability to accept a proton to form H₂PO₄^-
It can also have the ability to donate a proton to form PO₄^-3

[Fe(H₂N-CH₂-CH₂-NH₂)₃]₃(PO₄)₂
[Fe(en)₃]²⁺(\(PO_{4}^{3-}\))

Co(en)₂ Cl₂Br
1. (B) Reacts with silver nitrate to give a white precipitate readily soluble in NH₄OH. This shows (B) has Cl^- "counterion". Hence (B) is [Co(en)₂C/Br] Cl form.
2. (A) Reacts with silver nitrate to give a pale yellow precipitate. This shows (A) has Br^- counterion. hence A is [Co(en)₂Cl₂]Br.
Therefore A: [Co(en)₂Cl₂]Br
              B: [Co(en)₂Cl Br]Cl
A - [Co(en)₂ Cl₂]Br
Oxidation state of Co atom:
1(Co) + 2(en) + 2(Cl^-) = + 1
1(x) + 2(0) + 2(-1) = + 1
x = + 3
\(\therefore\) Co³⁺
Hybridisation and formation of [Co(en)₂,CI₂] Br complex

hybridisation = sp³d²
n = 4, m_s = \(\sqrt{n(n+2)}=\sqrt{4(4+2)}=4.89 BM\)

(i) This oxidation is governed by Popoff's rule.
(ii) It states that during the oxidation of an unsymmetrical ketone, a (C-CO) bond is cleaved in such a way that the keto group stays with the smaller alkyl group.

(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.

By Faraday II law of electrolysis:
\(\frac{\mathrm{m}_{\mathrm{Ni}}}{\mathrm{E}_{\mathrm{Ni}}}=\frac{\mathrm{m}_{\mathrm{cr}}}{\mathrm{E}_{\mathrm{cr}}} \)
\(\mathrm{Ni}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni}_{(\mathrm{s})}  \)
\(\mathrm{Cr}_{(\mathrm{aq})}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Cr}_{(\mathrm{s})}\)

Ni	Cr
\(\mathrm{m}_{\mathrm{Ni}_{\mathrm{i}}} =2.935 \mathrm{~g} \) \(\mathrm{E}_{\mathrm{Ni}} =\frac{58.74}{2} \) \(=29.37 \mathrm{~g} \mathrm{eq}^{-1}\)	\( \mathrm{m}_{\mathrm{cr}} =x \) \(\mathrm{E}_{\mathrm{cr}} =\frac{52}{3} \) \(=17.33 \mathrm{~g} \mathrm{eq}^{-1}\)

\( \frac{2.935}{29.37}=\frac{\mathrm{x}}{17.33} \)
\(x =\frac{2.935 \times 17.33}{29.37} \)
=1.732 g

In this reaction, phenol is first converted into sodium phenoxide which is more reactive than phenol towards electrophilic substitution reaction with CO₂, Treatment of sodium phenoxide with CO₂ at 400 K, 4-7 bar pressure followed by acid hydrolysis gives salicylic acid.

\(\mathrm{Ag}_{2} \mathrm{CrO}_{4(\mathrm{~s})} \rightleftharpoons 2 \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{CrO}_{4{(\mathrm{aq})}}^{2-}\\ \quad s \quad \quad \quad \quad \quad 2s \quad \quad \quad \quad s\)
\(\left[\mathrm{Ag}^{+}\right]=2 \mathrm{~s}+0.01 \)
\(\simeq 0.01 \)
\((\because 2 s<<0.01) \)
\(\left[\mathrm{CrO}_{4}^{2-}\right]=\mathrm{S} \)
\(\mathrm{AgNO}_{3(\mathrm{~s})} \rightleftharpoons \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{NO}_{3_{(\mathrm{aq})}}^{-}\\ 0.01M \quad \quad 0.01M \quad \quad 0.01M \quad\)
\(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{CrO}_{4}^{2-}\right]  \)
\(1 \times 10^{-12}=(0.01)^{2}(\mathrm{~s})  \)
\(\mathrm{S}=\frac{1 \times 10^{-12}}{10^{-4}}=1 \times 10^{-8} \mathrm{M}\)

a) Bakelite - Cross linked polymer
b) Nylon - Linear polymer
c) polythene (HDPE) - Linear polymer
d) polythene (LDPE) - Branched polyner

(i) Cobalt (III) ion has greater tendency to form complexes than cobalt (II) ion. Therefore, Co (II) ion being stable in aqueous solution, changes to Co (III) ion in the presence of complexing reagents and get oxidised.
(ii) Ions of transition metals with d¹ configuration tend to lose one electron to acquire d⁰ configuration that is quite stable. Therefore, such ions (with d¹) undergo either oxidation or disproportionation, hence unstable.

(i) The element of group number 17 and period number 2 is fluorine (A).
(ii) The element of group number 17 and period number 4 is bromine (B).
(iii) (A) and (B) react to give an interhalogen compound (C) bromine pentafluoride
\(\underset { (A) }{ { Br }_{ 3 } } +\underset { (B) }{ { 5F }_{ 2 } } \longrightarrow \underset { (C) }{ { 2BrF }_{ 5 } } \)
(C) undergoes sp³d² hybridisation and has octahedral structure

A	F	Fluorine
B	Br	Bromine
C	BrF5	Bromine pentafluoride

(i) The stability of a coordination complex is a measure of its resistance to the replacement of one ligand by another.
(ii) The stability of a complex refers to the degree of association between two species involved in an equilibrium.
(iii) Let us consider the following complex formation reaction
Cu²⁺ +4NH₃ ⇌ [Cu(NH₃)₄]^{2+
\(\beta =\frac { { \left[ { Cu\left( { NH }_{ 3 } \right) }_{ 4 } \right] }^{ 2+ } }{ \left[ { Cu }^{ 2+ } \right] { { \left[ { NH }_{ 3 } \right] }^{ 4 } } } \)}
(iv) So, as the concentration of [Cu (NH₃)₄]²⁺ increases the value of stability complexes also increases.
(v) Therefore the greater the value of stability constant greater is the stability of the complex.

(i) The rate of a reaction increases with the increase in the concentration of the reactants.
(ii) The effect of concentration is explained on the basis of collision theory of reaction rates.
(iii) According to this theory, the rate of a reaction depends upon the number of collisions between the reacting molecules.
(iv) Higher the concentration, greater is the possibility for collision and hence increase the rate.

Zirconium is a 4d element and Hafnium is a 5d element. There is an unexpected observation in the atomic radius of 5d elements which have nearly the same atomic radius as that of corresponding 4d elements. This is due to lanthanoid contraction. Hence they exhibit similar properties.

Mn³⁺ has large and negative standard electrode potential E⁰ (-1.18 V) than that of Cr²⁺ which has only -0.91 V. If the standard electrode potential of a metal is large and negative, the metal is a powerful reducing agent because it loses electrons easily. Hence Mn³⁺ is strongly oxidizing while Cr²⁺ is strongly reducing.

For cubic closed packed structure
\(r =\frac{a \sqrt{2}}{4}  \)
\(\therefore a =\frac{4 r}{\sqrt{2}}  \)
\(=\frac{4 \times 1.25 \times 10^{-8}}{1.414}=3.53 \times 10^{-8} \mathrm{~cm} \)
= 353 pm

Fusion of urea with B(OH)_3' in an atmosphere of ammonia at 800 - 1200 K gives boron nitride.
B(OH)₃ + NH₃ \(\overset { \Delta }{ \longrightarrow } \) BN+ 3H₂O

(v) CH₂ = CH – CH(Cl)–O–CH₃ : 3 - methoxy,3- choloro propene (mixed ether)
(vi) dibenzyl ether CH₆CH₅ -O-CH₂-C₆H₅ : benzoxy toluene + Simple ether
(vii) vinyl allyl ether : ethoxy  2 - propene
CH₂ = CH –O– CH₂–CH–CH₂(mixed ether)

 

Compound	Name
A	Acetylchloride
B	Acetaldehyde
C	3 - Hydroxy butanal
D	Crotonaldehyde

The intermediate compound formation theory :
A catalyst acts by providing a new path with low energy of activation. In homogeneous catalysed reactions a catalyst may combine with one or more reactant to form an intermediate which reacts with other reactant or decompose to give products and the catalyst is regenerated.
Consider the reactions :
A+B➝AB; C is the catalyst .............(1)
A + C ➝ AC (intermediate) ..........(2)
AC + B ⟶ AB + C ...............(3)
Example 1:
The mechanism of Fridel crafts reaction is given below
\({ C }_{ 6 }{ H }_{ 6 }+{ { CH }_{ 3 }Cl\overset { anhydrous\\ { AlCl }_{ 3 } }{ \longrightarrow } }{ C }_{ 6 }{ H }_{ 5 }{ CH }_{ 3 }+HCl\) 
The action of catalyst is explained as follows
CH₃Cl + AlCl₃ ⟶ [CH₃]⁺ [AlCl₄]^-
It is an intermediate.
\({ C }_{ 6 }{ H }_{ 6 }+\left[ { CH }_{ 3 }^{ + } \right] \left[ { AlCl }_{ 4 } \right] ^{ - }\longrightarrow { C }_{ 6 }{ H }_{ 5 }{ CH }_{ 3 }+{ AlCl }_{ 3 }+Hcl\) 
Example 2:
\({ { 2KClO }_{ 3 }\overset {\\ { MnO }_{ 3 } }{ \longrightarrow } }{ 2KCl }+{ 3O }_{ 2 }\)
Thermal decomposition of KCIO₃ in the presence of MnO₂ proceeds as follows. Steps in the reaction
2KCIO₃ ⟶ 2KCI + 3O₂ Can be given as
2KClO₃ + 6MnO₂   →  6MnO₃ + 2KCl
It is an intermediate
6MnO₃ → 6MnO₂ + 3O₂
Example 3:
Formation of water due to the reaction of H₂ and O₂ in the presence of Cu can be given as
H₂ + 1/2O₂ → H₂O 
2Cu + \(\frac{1}{2}\)O₂   → Cu₂O
It is an intermediate.
Cu₂O + H₂  → H₂O + 2Cu
Advantages:
This theory describes
(a) The specificity of a catalyst and
(b) The increase in the rate of the reaction with increasc inthe concentration of a catalyst
Limitations:
(a) The intermediate compound theory fails to explain the action of catalytic poison and activators (promoters).
(b) This theory is unable to explain the mechanism of heterogeneous catalysed reactions.

S.No	Reagents or Reaction	Primary amine RNH2	Secondary amine R2NH	Tertiary amine R3N
1.	Carbylamine reaction or with CHCl3/KOH	Carbylamine is formed (unpleasant smell)	-	-
2.	Mustard oil reaction or CS2/HgCl2 (Hoffmann's mustard oil test)	Alkyl isothiocyanate is formed (Mustard oil odour)	-	-
3.	HNO2 (or) NaNO2 / HCl	Alcohol is formed +H2	Yellow oily nitrosoamine is formed, insoluble in water. (Liberman's Test)	Forms nitrite in cold, soluble in water.
4.	CH3COCl	N-acetyl derivative is formed	N,N- diacetyl derivative is formed	-
5.	Diethyl oxalate Hoffmann's method	Solid oxamide is formed	Liquid oxamic ester is formed	-
6.	Benzene sulphonyl chloride in presence of excess. KOH (Hinsberg's reaction)	N- alkyl benzene sulphonamide is formed (soluble)	N, N - dialkyl benzene sulphonamide is formed (Insoluble).	-
7.	With RX	1 mol → 2o amine 2 mol → 3o amine 3 mol → Quarternary salt	1 mol → 3o amine 2 mol → Quarternary salt	1 mol → Quarternary salt

Average rate = \(-\frac { \triangle \left( R \right) }{ \triangle t } =-\frac { { \left[ R \right] }_{ 2 }-{ \left[ R \right] }_{ 1 } }{ { t }_{ 2 }-{ t }_{ 1 } } \)
\(=-\frac { 0.02M-0.03M }{ 25min } =\frac { -0.01M }{ 25min } \)
= 4 x 10^-4 M min^-1 and 
= \(-\frac { -0.01m }{ 25\times 60 } \) = 6.66 x 10^-6 Ms^-1

S. No	Double salts	Co-ordination compound
1.	They usually contain two simple salt in equimolar proportions	The simple salts from which they are formed may or may not be in equimolar proportion.
2.	They exists only in the solid state. In aqueous solution they dissociate completely into ions.	They exist in the solid state as well as in aqueous solution. This is because even in solution, the complex ion does not dissociate into ions.
3.	They are ionic compounds and do not contain any co-ordinate bond.	They may or may not be ion but the complex part always contain coordinate bonds
4.	The properties of the double salts are same as those of its constituent compounds.	The properties of the coordination compounds are different for its constituent bonds.
5.	In a double salt, the metal ion show their normal valency.	In a coordinate compound the metal ion satisfies its two types of valence called primary & secondary valenices.
6.	A double salt loses its identity and dissociates into its constitute simple ions in solution.	The complex ion does not lose its identity and never dissociate to give simple ions.
	Example: FeSO4 (NH4)2 SO4.6H2O	Example: K4[Fe(CN)6), K3[Fe(SCN)6)