(i) Compound (A) with molecular formula C₃H₆O that undergoes iodoform reaction is CH₃COCH₃ (A) (acetone).
(ii) Two molecules of (A) react with dry HCl to give compound (B).

(iii) Compound (B) reacts with (A) to give compound (C).

Compound	Compound Name	Formula
A	Acetone
B	Mesityl oxide
C	Phorone

(i) Tollens Reagent Test: Tollens reagent is an ammoniacal silver nitrate solution. When an aldehyde is warmed with Tollens reagent a bright silver mirror is produced due to the formation of silver metal. This reaction is also called silver mirror test for aldehydes.
\({ CH }_{ 3 }CHO+2\left[ Ag\left( { NH }_{ 3 } \right) _{ 2 } \right] ^{ + }+3{ OH }^{ - }\longrightarrow { CH }_{ 3 }CO{ O }^{ - }+\underset { Silver\ mirror }{ 4{ NH }_{ 3 }+2Ag+2{ H }_{ 2 }O } \) 
(ii) Fehlings solution Test:
(a) Fehling's solution is prepared by mixing equal volumes of Fehlings solution A containing aqueous copper sulphate and Fehlings solution 'B' containing alkaline solution of sodium potassium tartarate (Rochelle salt).
(b) When aldehyde is warmed with Fehlings solution deep blue colour solution is changed to red precipitate of cuprous oxide.
\(\\ \\ { CH }_{ 3 }CHO+\underset { blue }{ 2{ Cu }^{ 2 } } +5{ OH }^{ - }\longrightarrow { CH }_{ 3 }{ COO }^{ - }+\underset { red }{ { Cu }_{ 2 }O\downarrow } +3{ H }_{ 2 }O\)
(iii) Schiffs' reagent Test: Dilute solution of aldehydes when added to Schiff's reagent (Rosaniline hydrochloride dissolved in water and its red colour decolourised by passing SO₂) yields its red colour. This is known as Schiff's test for aldehydes. Ketones do not give this test. Acetone however gives a positive test but slowly.

(i) CH₃COO^-

(ii) CH₃CONH₂

(iii) C₆H_sCHO

(iv) C₆H₅COO^-

(i) Hormones are classified according to the distance over which they act as, endocrine, paracrine and autocrine hormones.
(ii) Endocrine hormones act on cells distant from the site of their release. Example: insulin and epinephrine are synthesized and released in the bloodstream by specialized ductless endocrine glands.
(iii) Paracrine hormones (alternatively, local mediators) act only on cells close to the cell that released them. For example, interleukin -1 (IL-1) Autocrine hormones act on the same cell that released them. For example, protein growth factor interleukin-2 (IL-2).

(i) Structure:
(a) Linear polymers (long continuous chain) E.g. HDPE, PVC
(b) Branched polymers (one main chain with small chains as branches) E.g. polypropylene, LDPE.
(c) Cross linked polymers (linking of chain polymers) E.g. bakelite, melamine, formaldehyde
(ii) Mode of synthesis:
(a) Addition polymers. Formed by polymerisation of monomers without the elimination of byproduct. E.g. polyethylene, PVC, teflon.
(b) Condensation Polymer formed by the condensation of two or more monomers with the elimination of simple molecules like H₂O, NH₃, etc., E.g. Nylon:6-6, polyester.

(i) Compound (A) with molecular formula C₃H₈O₃ is glycerol.
(ii) Glycerol on heating with P₂O₅ gives an unsaturated compound (B).

(iii) Glycerol with cooled mixture of cone. H₂SO₄ and fuming HNO₃ form compound (C).

(i) Primary amine react with nitrous acid to form alcohols and nitrogen gas
\(\underset { primary \ amine }{ { CH }_{ 3 }NH_{ 2 } } \rightarrow \underset { unstable }{ { [{ CH } }_{ 3 }-N=N-OH] } \rightarrow { CH }_{ 3 }OH+{ N }_{ 2 }\)
Aliphatic diazonium compound is unstable because of absence of resonance stabilisation.
(ii) Secondary amines react with nitrous acid to form N-nitroso amines which are water insoluble yellow oils.
\(\underset { Secondary \ amine }{ { { (CH }_{ 3 } })_{ 2 }NH } +HO-N=O\rightarrow \underset { N-nitroso \ dimethy \ amine-yellow \ oil\\ (insolube \ in \ water) }{ { (CH }_{ 3 })_{ 2 }N-N=O } \)
(iii) Tertiary amine react with nitrous acid to form trialkyl ammonium nitrite salts which are soluble in water
\(\underset { Tertiary \ amine }{ { (CH }_{ 3 })_{ 2 }N } +HONO\rightarrow \underset { trimethyl \ ammonium \ nitrite\\ (salt \ soluble \ in \ water) }{ { { (CH }_{ 3 }) }_{ 3 }{ NH }^{ + }{ NO }_{ 2 }^{ - } } \)

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.

Classification of nitro compounds

(i) Isopropyl alcohol (A).
(ii) (B) on oxidation gives (B)

(B) On further oxidation gives (C) which gives brisk efferescencewith Na₂CO₃.
\({ CH }_{ 3 }-\underset { \overset { || }{ \underset { (B) }{ O } } }{ C } -{ CH }_{ 3 }+{ I }_{ 2 }+KOH\longrightarrow \underset { (C) }{ { CH }I_{ 3 }+{ CH }_{ 3 }COOH } \) 

Helmholtz double layer:
(i) The surface of colloidal particle adsorbs one type of ion due to preferential adsorption.
(ii) This layer attracts the oppositely charged < ions in the medium and hence at the boundary separating the two electrical double layers are setup.
(iii) This is called as Helmholtz electrical double layer.
(iv) As the particies nearby are having similar: charges, they cannot come close and condense.

(i) For a chemical reaction, catalyst is needed in very small quantity.
(ii) There may be some physical changes, but the catalyst remains unchanged in mass and chemical composition in a chemical reaction.
(iii) A catalyst itself cannot initiate a reaction. 
(iv) A solid catalyst will be more effective if it is taken in a finely divided form.
(v) A catalyst are specific in nature.
(vi) In an equilibrium reaction, presence of catalyst reduces the time for attainment of equilibrium and hence it does not affect the position of equilibrium and the value of equilibrium constant.
(vii) A catalyst is highly effective at a particular temperature called as optimum temperature.
(viii) Presence of a catalyst generally does not change the nature of products

According to Ostwald dilution Law,
\({ K }_{ a }=\frac { { \alpha }^{ 2 }C }{ (1-\alpha ) } \)  ...(1)
Substitute a value in the above expression (1)
\({ K }_{ a }=\frac { { \Lambda }_{ m }^{ 2 }C }{ { \Lambda }_{ m }^{ 2 }\left( 1-\frac { { \Lambda }_{ m } }{ { \Lambda }_{ m }^{ o } } \right) } \)
\({ K }_{ a }=\frac { { \Lambda }_{ m }^{ 2 }C }{ { \Lambda }_{ m }^{ 2 }\frac { { \Lambda }_{ m }^{ o }-{ \Lambda }_{ m } }{ { \Lambda }_{ m }^{ o } } } \)
⇒ \({ K }_{ a }=\frac { { \Lambda }_{ m }^{ 2 }C }{ { \Lambda }_{ m }^{ 2 }{ (\Lambda }_{ m }^{ o }-{ \Lambda }_{ m }) } \)

(i) The concentration of hydronium ion in an acidic buffer solution depends on the ratio of the concentration of the weak acid to the concentration of its conjugate base present in the solution i.e.,
\(\left[ { H }_{ 3 }{ O }^{ + } \right] ={ K }_{ a }\frac { [{ acid] }_{ aq } }{ [{ base] }_{ aq } } \)
(ii) The weak acid is dissociated only to a small extent. Moreover, due to common ion effect, the dissociation is further suppressed and hence the equilibrium concentration of the acid is nearly equal to the initial concentration of the unionised acid. Similarly, the concentration of the conjugate base is nearly equal to the initial concentration of the added salt.
\(\left[ { H }_{ 3 }{ O }^{ + } \right] ={ K }_{ a }\frac { [{ acid] } }{ [{ salt] } } \)
(iii) Here [acid] and [salt] represent the initial concentration of the acid and salt, respectively used to prepare the buffer solution
Taking logarithm on both sides of the equation
\(\log\left[ { H }_{ 3 }{ O }^{ + } \right] ={ \log K }_{ a }+\log\frac { [{ acid] } }{ [{ salt] } } \)
reverse the sign on both sides
- \(\log\left[ { H }_{ 3 }{ O }^{ + } \right] ={- \log K }_{ a }-\log\frac { [{ acid] } }{ [{ salt] } } \)
We know that
pH = -log [H₃O⁺] and pK_a = -log K_a
\(\Rightarrow pH={ pK }_{ a }-\log\frac { [{ acid] } }{ [{ salt] } } \)
\(​​\Rightarrow pH={ pK }_{ a }+\log\frac { [{ acid] } }{ [{ salt] } } \)
Similarly for a basic buffer,
pOH = \({ pK }_{ a }+\log\frac { [{ acid] } }{ [{ salt] } } \)

\(pH={ pK }_{ a }+{ \log }\frac { [salt] }{ [acid] } \)
The concentration of the salt and the acid should be in moles/lit.
Number of moles of acetic acid
= \(\frac{weight \ of \ acetic \ acid}{molecular \ weight \ of \ acetic \ acid}=\frac{7.5}{60}\)
Number of moles of sodium acetate
= \(\frac{weight \ of \ sodium \ acetate}{molecular \ weight \ of \ sodium acetate}=\frac{10.25}{82}\)
pK_a = -log K_a
= -Iog 1.8 x 10^-5
\(=log\frac { 1 }{ 1.8\times { 10 }^{ -5 } } \)
= log 1 - log 1.8 - log 10^-5
= 5 - 0.2553 = 4.7447
\(\therefore pH=4.7447+\log\frac { 0.12 }{ 0.08 } \)
pH = 4.7447.

Rate = k [x]² = ka²
[x] = a mol^-1
If the concentration of x is in cross three time, then
(x) = 3a mol L^-1
Rate = R(3a)² = 9 ka²
Hence, the rate of formation will increase by 9 times.

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

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

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) O₂ molecules are held together by weak Vander Waal's force because of small size and high electronegativity of oxygen.
(ii) In contrast, sulphur shows catenation and forms stronger S-S bonds.
(iii) Due to catenation, sulphur forms octa-atomic S₈ molecules having eight membered puckered ring structure.
(iv) Because of its bigger size the force of attraction holding S₈ molecules are much stronger.
(v) Hence sulphur is a solid at room temperature or in other words, that is why there is a large difference between the boiling point (also melting points) of the two elements.

(i) By knowing the density of an unknown metal and the dimension of its unit cell, the atomic mass of the metal can be: determined.
(ii) Let 'a' be the edge length of a unit cell of a crystal, 'd' be the density of the metal, 'm' be the atomic mass of the metal and 'z' be the number of atoms in the unit cell.
(iii) Now,
Density of the unit cell
\(=\frac{Mass\ of\ the\ unit\ cell}{Volume\ of\ the\ unit\ cell}\)
\(d=\frac{Z\times m}{a^3}\) ...(1)
[Since, mass of the unit cell = Number of atoms in the unit cell x Atomic mass]
[Volume of the unit cell = (edge length of the cubic unit cell)³]
(iv) From equation (1), We have
\(m=\frac{d\times a^3}{Z}\) ....(2)
(v) Now, 
Mass of the metal (M) \(=\frac{Atomic\ mass(M)}{Avogadro's\ number(N_A)}\)
M=\(\frac{d\times a^3 \times N_A}{Z}\)
(vi) From equation (3), we can determine the atomic mass of the unknown metal.

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.

Zone refining: 
This method is employed for preparing highly pure metal (such as silicon, tellurium, germanium), which are used as semiconductors. It is based on the principle that melting point of a substance is lowered by the presence of impurities. Consequently, when an impure molten metal is cooled, crystals of the pure metal are solidified, and the impurities remain behind the remaining metal.
The process consists In casting the impure metal in the form of a bar. A circular heater fitted around this bar is slowly moved longitudinally from one end to the other. At the heated zone, the bar melts, and as the heater moves on, pure metal crystallizes, while the impurities pass into the adjacent molten part In this way, the impurities are swept from one end of the bar to the other. By repeating the process, ultra pure metal can be obtained.

Metal carbonyls are classified in two different ways as described below Classification based on the number of metal atoms present.
a. Mononuclear carbonyls
These compounds contain only one metal atom. For example, [Ni(CO)₄] - nickel tetracarbonyl is tetrahedral, [Fe(CO)₅] - Iron pentacarbonyl is trigonal bipyramidal, and [Cr(CO)₆] - Chromium hexacarbonyl is octahedral.
b. Polynuclear carbonyls
Metallic carbonyls containing two or more metal atoms are called polynuclear carbonyls. Polynuclear metal carbonyls may be Homonuclear [Co₂(CO)₈], [Mn₂(CO)₁₀], [Fe₃ (CO)₁₂] or heteronuclear [MnCo(CO)₉], [MnRe(CO)₁₀] etc.

(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.

E = E^O- 2.303 \(\frac { RT }{ 2F } \) log Zn²⁺
E_Zn²⁺/Zn = E_{Zn²⁺/ Zn}- 2.303 \(\frac { RT }{ 2F } \)  log [Zn²⁺]