(i) An organic compound (A) is identified as C₆H₅CHO benzaldehyde. Benzaldehyde reduces Tollen'sreagent and also undergoes. Cannizaro reaction.
\(\underset { (A) }{ C_{ 6 }{ H }_{ 5 }CHO } +{ Ag }_{ 2 }O\longrightarrow 2Ag+\underset { (B) }{ { C }_{ 6 }{ H }_{ 5 }COOH } \) 
(ii) Benzaldehyde reacts with acetic anhydride in the presence of sodium acetate gives cinnamic acid and it is (B) an unsaturated acid
\(\underset { (A) }{ { C }_{ 6 }{ H }_{ 5 }CHO } +\left( { CH }_{ 3 }CO \right) _{ 2 }O\overset { { CH }_{ 3 }COONa }{ \longrightarrow } \underset { (B) }{ { C }_{ 6 }{ H }_{ 5 }CH } { =CHCOOH }+{ CH }_{ 3 }COOH\) 
(iii) Compound A reacts with acetone to form compound (C) benzal acetone.
\(\underset {(A)}{{ C }_{ 6 }{ H }_{ 5 }CHO}+{ CH }_{ 3 }COC{ H }_{ 3 }\overset { NaOH }{ \longrightarrow } { C }_{ 6 }{ H }_{ 5 }CH=\underset { (C) }{ CH } -CO{ CH }_{ 3 }\) 

Compound	Compound Name	Formula
A	Benzaldehyde	C6H5CHO
B	Cinnamic acid	C6H5CH = CHCOOH
C	Benzal acetone	C6H5CH - CH-COCH3

(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

(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) Formic acid (HCOOH) is unique because it contains both an aldehyde group and carboxyl group also.

(ii) Hence it can act as a reducing agent. It reduces Fehling's solution Tollen's reagent and decolourises pink coloured KMnO₄ solution.
(iii) Whereas in acetic acid, there is no aldehyde group and it cannot act as reducing agent.
(iv) Formic acid reduces ammoniacal silver nitrate solution (Tollen's reagent) to metallic silver.
HCOOH + Ag₂O⟶H₂O + CO₂ + 2Ag↓ (metallic silver)
(v) Formic acid reduces Fehling's solution. It reduces blue coloured cupric ions to red coloured cuprous ions.
\(\mathrm{HCOO}^{-}+2 \mathrm{Cu}^{2+}+5 \mathrm{OH}^{-} \longrightarrow \mathrm{CO}_{3}^{2-}+\mathrm{Cu}_{2} \mathrm{O}+3 \mathrm{H}_{2} \mathrm{O}\\ \quad \quad \quad \quad (blue) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (red)\)

This reaction is catalysed by base. The carbanion generated is nucleophilic in nature. Hence it can bring about nucleophilic attack on carbonyl group
Step 1: The carbanion is formed as the a-hydrogen atom is removed as a proton by the base

Step 2: The carbanion attacks the carbonyl carbon of another unionised aldehyde molecule

Step 3: The alkoxide ion formed is protonated by water to give 'aldol'.

(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).

Importance of carbohydrates:
(i) Carbohydrates, widely distributed in plants and animals, acts mainly as energy sources and structural polymers.
(ii) Carbohydrate is stored in the body as glycogen and in plant as starch.
(iii) Carbohydrates such as cellulose which is the primary components of plant cell wall, is used to make paper, furniture (wood) and cloths (cotton)
(iv) Simple sugar glucose serves as an instant source of energy.
(v) Ribose sugars are one of the components of nucleic acids.
(vi) Modified carbohydrates such as hyaluronate (glycosaminoglycans) act as shock absorber and lubricant.

(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) The organic compound (A) which is sodium salt of phenolic acid is sodium salicylate.
(ii) (A) on heating with soda lime gives Compound (B) phenol. Phenol gives violet colouration with neutral perchloride.

(iii) Phenol on treatment with C₆H₅COCl in the presence of NaOH gives on ester 'C':
C₆H₅OH + C₆H₅COCl \(\overset { NaOH }{ \longrightarrow } \underset { (C) }{ { C }_{ 6 }{ H }_{ 5 }OCO{ C }_{ 6 }{ H }_{ 5 }+HCl } \)

Compound	Compound Name	Formula
A	Sodium benzoate'
B	Phenol	C6H5OH
C	Phenyl benzoate	C6H5OCO C6H5

Ethers are a class of organic compound in which an oxygen atom is connected to two alkyl/aryl groups Ethers can be considered as the derivatives of hydrocarbon in which one hydrogen atom is replaced by an alkoxy (-OR) or an aryloxy (-OAr ) group. The general formula of aliphatic ether is C_nH_2n+2O.
Classification:

(i) CH₃CN (Methyl cyanide) on reduction gives CH₃CH₂NH₂ (ethylamine) because addition of hydrogen takes place at ≡ CN
\({ CH }_{ 2 }-C\equiv N\xrightarrow [ 4H ]{ { LiAH }_{ 4 } } { \underset { Ethylamine\\ (Primary\quad amine) }{ { CH }_{ 3 }-{ CH }_{ 2 }{ NH }_{ 2 } } }\)
Whereas CH₃NC (Methyl isocyanide) on reduction with LiAlH₄ gives secondary amine
\({ CH }_{ 3 }-\underset { \overset { | }{ H } }{ N } -{ CH }_{ 3 }\)

In methyl cyanide, -CN group is attached to alkyl group and by reduction it gives a primary amine wherease in methyl isocyanide -NC group is attached to alkyl group and by reduction it gives a secondary amine.
(ii) (CH₃)₂ NH (Secondary amine) requires 2 moles of CH₃I to give a quaternary salt (crystalline product) whereas (CH₃)₂N (tertiary amine) require only one mole of CH₃I to give the same quaternary salt. It is due to the number of alkyl groups present in amines.
\(\underset { Secondary \ anmine }{ { ( }{ CH }_{ 3 })_{ 2 }-NH+{ 2CH }_{ 3 }I } \rightarrow { { [(CH }_{ 3 })_{ 4 }N] }^{ + }\underset { Tetramethy\\ ammonium \ iodide }{ { I }^{ - } } \)
(iii) Nitration of aniline with cone. HNO₃ results in the formation of m-nitro aniline because nitric acid is a strong acid. It protonates aniline forming anilinium ion C₆H₅NH3+ because of positive charge on nitrogen, it is meta directive and -NO₂ group is substituted at meta position.

(iv) p-Toluidine contains a methyl group which has +I effect (electron withdrawing group and due to this, p-toluidine is a stronger base than p-nitro aniline in which the nitro group is less reactive and it deactivate the benzene rin make it a less basic

An organic compound A (C₂H₆O₂) liberates hydrogen with metallic sodium.

(ii) Compound A when heated with anhydrous zinc chloride ultimately gives B (C₂H₄O)

(iii) Compound A when heated with conc. phosphoric acid gives C (C₄H₁₀O₃).

(iv) Compound A on oxidation with acidified K₂Cr₂O₇ gives compound D (CH₂O₂).

Compound	Compound Name	Formula
A	Ethylene glycol	\(\overset { { CH }_{ 2 }OH }{ \underset { { CH }_{ 2 }OH }{ | } } \)
B	Acetaldehyde	CH3-CHO
C	Diethylene glycol	\(\overset { HCOOH }{ \underset { HCOOH }{ + } } \)
D	Formic acid

(i) Compound A is phenol. (C₆H₅OH)
Phenols on reaction with metallic Na compound (B) ie., Sodium Phenoxide
\(\underset { (A) }{ { 3C }_{ 6 }{ H }_{ 5 }OH+2Na } \longrightarrow \underset { (B) }{ { 2C }_{ 6 }{ H }_{ 5 }ONa+{ H }_{ 2 } } \) 
(ii) (B) is heated with CO₂ at 400K under pressure, to give (C). This is decomposed by dil.HCl, and (D) is formed

Compound	Compound Name	Formula
A	Phenol	C6H5OH
B	Sodium Phenoxide	C6H5ONa
C	Sodium Salicylate
D	Salicylicacid

Electrophoresis:
(i) When electric potential is applied across two platinum electrodes dipped in a hydrophilic sol, the dispersed particles move toward one or other electrode.
(ii) This migration of sol particles under the influence of electric field is called electrophoresis or cataphoresis.
(iii) If the sol particles migrate to the cathode, then they posses positive (+) charges, and if the sol particles migrate to the anode then they have negative charges(-).
(iv) Thus from the direction of migration of sol particles we can determine the charge of the sol particles.
(v) Hence electrophoresis is used for detection of presence of charges on the sol particles

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

(i) The formation of rust requires both oxygen and water. Since it is an electrochemical redox process, it requires an anode and cathode in different places on the iron.
(ii) The iron surface and a droplet of water on the surface form a tiny galvanic cell.
(iii) The region enclosed by water is exposed to low amount of oxygen and it acts as the anode.
(iv) The remaining area has high amount of oxygen and it act a cathode. So an electro chemical cell is formed. Corrosion occurs at the anode i.e., in the region enclosed by the water.

Anode (oxidation)Iron dissolve in the andoe region:
\({ 2Fe }_{ (s) }\rightarrow { 2Fe }_{ (aq) }^{ 2+ }+{ 4e }^{ - }\) E^o = 0.44V
The electrons move through the iron metal from the anode to the cathode area where the oxygen dissolved in water, is reduced to water.
Cathode (reduction):
The reaction of atmospheric carbon dioxide with water gives carbonic acid which furnishes the H⁺ ions for reduction.
\({ O }_{ 2(g) }+{ 4H }_{ (aq) }^{ + }+{ 4e }^{ - }\rightarrow { 2H }_{ 2 }O(l)\)  E^o = 1.23 V
The electrical circuit is completed by the migration of ions through water droplet.
The overall redox reactions is,
\({ 2Fe }_{ (s) }+{ O }_{ 2(g) }+{ 4H }_{ (aq) }^{ + }\rightarrow { 2Fe }_{ (aq) }^{ 2+ }+{ 2H }_{ 2 }O(l)\)  E^o = 0.444 + 1.23 = 1.67V
The positive emf value indicates that the reaction is spontaneous
Fe²⁺ ions are further oxidised to Fe³⁺ which on further reaction with oxygen to form rust.
\({ 4Fe }_{ (Aq) }^{ 2+ }+{ O }_{ 2(g) }+{ 4H }_{ (aq) }^{ + }\rightarrow { 4Fe }_{ (aq) }^{ 2+ }+{ 2H }_{ 2 }O(l)\)
\({ 3Fe }_{ (Aq) }^{ 3+ }+{ 4{ H }_{ 2 }O }(l)\rightarrow { Fe }_{ 2 }{ O }_{ 3 }.{ H }_{ 2 }{ O }_{ (s) }+{ 6H }_{ (aq) }^{ + }\)

Given:
E₁ = 0.5 V
Cr³⁺+ 3e^- ➝ Cr_(s).......(1)
E₂ = 0.41 V
Cr³⁺+ e^- ➝ Cr²⁺  .....(2)
The required reaction is,
Cr²⁺+ 2 e^- ➝ Cr_(s)
Then,
Formula:
\({ E }_{ 3 }=\frac { 3{ E }_{ 1 }+{ E }_{ 2 } }{ 2 } \)
Solution:
= \(\frac { 3(0.5)+(0.41) }{ 2 } =\frac { 1.5+0.41 }{ 2 } \)
= 0.955 V
E₃ = 0.955 V

Initial concentration of
AgNO₃ = 1M= IN
Quantity of current 0.75 F
Formula:
1Faraday = 1equivalent mass
Solution:
For IF current In AgNO₃ will be liberated.
For 0.75 F current 0.75 N AgNO₃ will be liberated
The concentration of AgNO₃ remaining
= 1.0 - 0.75 = 0.25 N
ஃ The concentration of AgNO₃ remaining

Degree if dissociation \(\alpha =\sqrt { \frac { K_{ a } }{ C } } ,C\alpha =\sqrt { { K }_{ a }.C } \)
∴ \(pOH=\log\frac { 1 }{ \sqrt { { K }_{ b }.C } } =\log\frac { { 10 }^{ 3 } }{ \sqrt { 1.75\times { 10 }^{ -5 }\times .1 } } \)
\(=\log\frac { 1 }{ \sqrt { 1.75\times { 10 }^{ -6 } } } =\log\frac { { 10 }^{ 3 } }{ \sqrt { 1.75 } } \)
\(=3-\frac { 1 }{ 2 } \log1.75\)
\(=3-\frac { 1 }{ 2 } \times 0.2430=2.8785\)
pH = 14 - pOH
∴ pH = 14 - 2.8785 = 11.1215
Alternating,
\(\left[ { H }^{ - } \right] =\sqrt { { K }_{ b }\times C } \)
\(=\sqrt { 1.75\times { 10 }^{ -5 }\times 0.1 } =\sqrt { 1.75\times { { 10 }^{ -6 } } } \)
= 1.322 x 10^-3
∴ pOH = - log [H⁺] = - log (1.323 x 10^-3)
3 - 0.1216 = 2.8784
∴ pH = 14 - pOH = 14 - 2.8784 = 11.1216

HCI ⟶ H⁺ + Cl^-. HCI is a strong acid.
[H⁺] from HCI is very much greater than [H⁺] from water which is 1 x 10^-7 M.
∴ [H⁺] = [HCI] = 0.001 M
∴ pH = -log (0.001) = 3.0
∴ That is acidic solution.

Degree if dissociation \(\alpha =\sqrt { \frac { { K }_{ B } }{ C } , } C\alpha =\sqrt { { K }_{ a }.C } \)
∴ \(pH=\log\frac { 1 }{ \sqrt { { K }_{ a }.C } } =\log\frac { 1 }{ \sqrt { 1.8\times { 10 }^{ -5 }\times .1 } } \)
= \(\log\frac { 1 }{ \sqrt { 1.8\times { 10 }^{ -3 } } } =\log\frac { { 10 }^{ 3 } }{ \sqrt { 1.8 } } \)
\(=3-\frac { 1 }{ 2 } \log1.8\)
\(=3-\frac { 1 }{ 2 } \times 0.2553=3-0.1276=2.8724\)
Alternating,
\(\left[ { H }^{ + } \right] =\sqrt { { K }_{ a }\times C } \)
\(=\sqrt { 1.8\times { 10 }^{ -5 }\times 0.1 } =\sqrt { 1.8\times { 10 }^{ -6 } } \)
= 1.341 x 10^-3
∴ -log [H⁺] = -log (1.341 x 10^-3)
= 3 - 0.1277 = 2.8723.

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

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) As per the position in the periodic table, the element A is phosphorus as below:
(ii) Phosphorus reacts with dry chlorine to form phosphorus trichloride as below
\(\underset { (A) }{ { P }_{ 4 } } +6{ Cl }_{ 2 }\longrightarrow \underset { (B) }{ { 4PCl }_{ 3 } } \)
Hence, compound B is phosphorus trichloride.
(iii) PCl₃ on hydrolysis gives phosphorus acid H₃PO₃, which is dibasic acid,
\(\underset { (B) }{ { PCl }_{ 3 } } +3{ H }_{ 2 }O\longrightarrow { 3HCl }+\underset { (C) }{ { H }_{ 3 }{ PO }_{ 3 } } \)  
Hence, compound C is phosphorus acid.
(iv) H₃PO₃ on heating undergo auto oxidation and reduction producing phosphoric acid md phosphine as below:
\(\underset { (c) }{ { 4H }_{ 3 }{ PO }_{ 3 } } \overset { \Delta }{ \longrightarrow } { PH }_{ 3 }+\underset { (b) }{ { 3H }_{ 3 }{ PO }_{ 4 } } \)
Phosphoric acid is a tribasic acid. Hence, compound D is phosphoric acid.
Thus A = Phosphorus
B = Phosphorus trichloride
C = Phosphorus acid
D = Phosphoric acid

(i) In this process a mixture of air and hydrochloric acid is passed up a chamber containing a number of shelves, pumice stones soaked in cuprous chloride are placed.
(ii) Hot gases at about 723 K are passed through a jacket that surrounds the chamber.
\(4 \mathrm{HCl}+\mathrm{O}_{2} \frac{400^{\circ} \mathrm{C}}{\mathrm{Cu}_{2} \mathrm{Cl}_{2}} 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{Cl}_{2} \uparrow\)
The chlorine obtained by this method is dilute and is employed for the manufacture of bleaching powder. The catalysed reaction is,
\(2 \mathrm{Cu}_{2} \mathrm{Cl}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{Cu}_{2} \mathrm{OCl}_{2}\\ \quad \quad \quad \quad \quad \quad \quad \text { cuprous oxychloride } \)
\(\mathrm{Cu}_{2} \mathrm{OCl}_{2}+2 \mathrm{HCl} \rightarrow 2 \mathrm{CuCl}_{2}+\mathrm{H}_{2} \mathrm{O}\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad\text { Cupric chloride } \)
\(2 \mathrm{CuCl}_{2} \rightarrow \mathrm{Cu}_{2} \mathrm{Cl}_{2}+\mathrm{Cl}_{2}\\ \quad \quad \quad \quad \text { cuprous chloride }\)

(a) NaCl + MnO₂ + 4H₂SO₄ \(\longrightarrow \) MnCl₂+ 4NaHSO₄ + 2H₂O + Cl₂
(b) 6XeF₄ + 12H₂O \(\longrightarrow \) 4Xe + 2XeO₃ + 24HF + 3O₂

(a) Rhombic Sulphur (α - Sulphur):
(a) It is yellow in colour.
(b) Its melting point is 385.8K and specific gravity is 2.06
(c) It is stable form of sulphur at room temperature.
(d) It is formed on evaporating the solution of sulphur in CS₂.
(e) It in insoluble in water, readily soluble in CS₂ and dissolves to some extent in benzene, alcohol and ether.
(b) Monoclinic sulphur \(\left( \beta -sulphur \right) \):
(a) Its melting point is 393K and specific gravity is 1.98
(b) It is prepared by melting rhombic sulphur in a dish and cooling, till crust is formed. Two holes are made in crust and remaining liquid is powered out. On removing crust, colourless needle - shaped crystals of β - sulphur is formed.
(c) Monoclinic sulphur is stable above 369K and below 369K α - sulphur is stable.
(d) At 369K both forms are stable and this temperature is called transition temperature.
(e) Both rhombic and monoclinic sulphur have S₈ molecules, these are packed to give different crystal structure S₈ form is puckered and has crown shape.
Several other modifications containing 6-20 sulphur atoms per ring are synthesised 

(f) In Cyclo-S₆ the ng adopts chair form.

(g) At elevated temperatures (～1000K), S₂ is dominant species and is, paramagnetic like O₂

Crystal defects are classified as follows
(i) Point defects
(ii) Line defects
(iii) Interstitial defects
(iv) Volume defects
Point defects are further classified as follows

(i) The colour develops because of the presence of electrons in the 8 anionic sites.
(ii) These electron absorb energy from the visible region of radiation and get excited. 
(iii) For example when crystals of NaCl are heated in an atmosphere of sodium vapours, the sodium atoms get deposited on the surface of the crystal and the deposited Na atoms.
(iv) During this process, the Na atoms on the surface lose electrons to form Na⁺ ions 
(v) These electrons get excited by absorbing energy from the visible light and impart yellow colour to the crystals.

DIAMOND	GRAPHITE
C is sp3 hybridised.	C is sp2 hybridised.
Three dimensional, tetrahedral structure.	Two dimensional, sheet like structure.
Crystalline, transparent with extra brilliance.	Crystalline, opaque and shiny substance.
It is hard with high density and high melting point.	It is soft with low density and high melting point.
Bad conductor of and electricity.	Good conductor of heat and electricity.

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.

(i) The impure nickel is heated in a stream of carbon monoxide at around 350 K.
(ii) The nickel reacts with the CO to form a highly volatile nickel tetracarbonyl.
(iii) The solid impurities are left behind 
\({ Ni }_{ (s) }+4{ CO }_{ (g) }\longrightarrow { Ni(CO) }_{ 4(g) }\)
(iv) On heating the nickel tetracarbonyl around 460 K, the complex decomposes to give pure metal.
\({ Ni(CO) }_{ 4(g) }\longrightarrow { Ni }_{ (s) }+{ 4CO }_{ (g) }\)

Valance bond theory helps us to visualize the bonding in complexes. However, it has limitations as mentioned above. Hence Crystal Field Theory to explain some of the properties, like colour, magnetic behavior, etc., This theory I was originally used to explain the nature of bonding in ionic crystals. Later on, it is used to explain the properties of transition metals and their complexes. The salient features of this theory are as follows.
(i) Crystal Field Theory (CFT) assumes that the bond between the ligand and the central metal atom is purely ionic. i.e. the bond is formed due to the electrostatic attraction between the electron rich ligand and the electron deficient metal.
(ii) In the coordination compounds, the central metal atom/ion and the ligands are considered as point charges (in case of I charged metal ions or ligands) or electric dipoles (in case of neutral metal atoms or ligands).
(iii) According to crystal field theory, the complex formation is considered as the following series of hypothetical steps.
Step 1: In an isolated gaseous state, all the five d orbitals of the central metal ion are degenerate. Initially, the ligands form a spherical field of negative charge around the metal. In this filed, the energies of all the five d orbitals will increase due to the repulsion between the electrons of the metal and the ligand.
Step 2: The ligands are approaching the metal atom in actual bond directions. To illustrate this let us consider an octahedral field, in which the I central metal ion is located at the origin and the six ligands are coming from the +x, -x, +y, -y, +z  and -z directions as shown below.

As shown in the figure, the orbitals lying along the axes dx²-y² and dz² orbitals will experience strong repulsion and raise in energy to a greater extent than the orbitals with lobes directed between the axes (d_xy, d_yz, and d_zx). Thus the degenerate d orbitals now split into two sets and the process is called crystal field splitting.
Step 3: Up to this point the complex formation would not be favored. However, when the ligands approach further, there will be an attraction between the negatively charged electron and the positively charged metal ion, that results in a net decrease in energy. This decrease in energy is the driving force for the complex formation.
Crystal field splitting in octahedral complexes: During crystal field splitting in octahedral field, in order to maintain the average energy of the orbitals (barycentre) constant, the energy of the orbitals d_x2-_y2 and d _z2 (represented as e_g orbitals) will increase by 3/5 \({ \triangle }_{ o }\) while that of the other three orbitals d_{xy '} d_yz and d_zx (represented as t_2g orbitals) decrease by 2/5  \({ \triangle }_{ o }\) , Here, \({ \triangle }_{ o }\)   represents the crystal field splitting energy in the octahedral field.

(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.