STRUCTURE OF FULLERENCES :

(1) A fascinating discovery was the synthesis of spherical carbon-cage molecules called fullerences. The discovery of fullerene was awarded the noble prize in chemistry (1996). Fullerenes   were first prepared by evaporation of graphite using laser.

(2) Fullerences are sooty material so formed consists of C₆₀ with small amount of C₇₀ and other fullerences containing an even number of carbon up to 350

(3) Fullerences have a smooth structure and unlike diamond and graphite, dissolved in organic solvent like toluene.

(4) C₆₀ is the most stable fullerene. It has the shape of a football and called buckminsterfullerene

(5) C₆₀ consists of fused five and six membered carbon rings

(6) Six membered rings surrounded by alternatively by hexagons and pentagons of carbon.

(7) Five membered rings are surrounded by five hexagons carbon rings.

(8) There are 12 five –membered rings

(9) There are 20 Six –membered rings

(10) In fullerenes all the carbon sp2 hybridised each carbon formed three sigma bond and the fourth electron delocalized to formed pi bond .

(11) All the carbon atoms are equivalent but all C-C bond are not equivalent.

(12)  In the structure C-C bonds of two different bond length occur at the fusion of two six membered rings the bond length is C-C = 135.5 pm and at the fusion of five and six membered rings C-C bond length is 146.7 pm.

(13) There are both single and double bonds

(14) The smallest fullerenes are C₂₀.

(15) Thermodynamically the most stable allotrope of carbon is considered to be graphite. This is due standard enthalpy of formation of graphite is taken zero .while enthalpy of formation of diamond and fullerenes are 1.90 KJ/Mole and 38.1 KJ/Mole respectively.      

REDOX TITRATIONS:

LAW OF EQUIVALENT WEIGHT:

No of equivalent of A reacted = No of equivalent of B reacted = No of equivalent of C formed = No of equivalent of D formed

RELATION BETWEEN NORMALITY (N) AND MOLARITY (M):

SOME COMMON OXIDISING AGENT:

SOME COMMON REDUCING AGENT:

ILLUSTRATIVE EXAMPLE (1): Calculate the moles of KMnO₄ required to reacting with 180 gm of Oxalic acid (H₂C₂O₄). Also calculate the volume of CO2 at STP produce in the reaction? (K=39, Mn=55, Of=16)

SOLUTION:

ILLUSTRATIVE EXAMPLE (2): Calculate the weight of K₂Cr₂O₇ which reacts with KI to liberate 254 gm of I₂ ?.(Cr=51.9961, I=127)

SOLUTION:

ILLUSTRATIVE EXAMPLE (3): Calculate volume of 0.05 M KMnO₄ required to react with 50 ml of 0.1M H₂S in acidic medium , given H₂S oxidized to SO₂?.

SOLUTION:

ILLUSTRATIVE EXAMPLE (4): Calculate the mass of Fe₃O₄ required to react completely with 25 ml of 0.3 M K₂Cr₂O₇?.

SOLUTION:

ILLUSTRATIVE EXAMPLE (5): Calculate the concentration of H₂O₂ if 20 ml H₂O₂ Solution react with 10 ml of 2M KMnO₄ in acidic medium ?

SOLUTION:

ILLUSTRATIVE EXAMPLE (6): Calculate the moles of KCl which  required to produce 10 ml of Cl₂ when reacted with KClO₃ ?

SOLUTION:

ILLUSTRATIVE EXAMPLE (7): Calculate the moles of KMnO₄ required for Oxidation of 1.25 moles Cu₂S.

SOLUTION:

ILLUSTRATIVE EXAMPLE (8): Calculate the Molarity of H₂O₂ if 11.2 ml H₂O₂ require 30 ml of 0.5 M K₂Cr₂O₇ for its Oxidation . also calculate the volume of strength of H₂O₂.

SOLUTION:

ILLUSTRATIVE EXAMPLE (9): 696 gm of Fe₂O₃ and FeO reacts completely with 158 gm of KMnO₄ in acidic medium . Calculate the composition of mixture.

SOLUTION:

ILLUSTRATIVE EXAMPLE (10): 829 gm of K₂Cr2O₇ and H₂C₂O₄ reacts  completely with 7/3 moles of K₂Cr₂O₇ .
(1) Calculate the moles of each in mixture.
(2) Calculate the moles of NaOH required to react with above mixture.

SOLUTION:

ILLUSTRATIVE EXAMPLE (11): 50ml of KMnO₄ is mixed with excess of KI the I₂ liberated require 30 ml of 0.1M Na₂S₂O₃ solution calculate the Molarity of KMnO₄ solution.

SOLUTION:

ILLUSTRATIVE EXAMPLE (12): 50 Cm³ of 0.04 M K₂Cr₂O₇ in acidic medium oxidized a sample  of H₂S gas to Sulphur . The volume of 0.03 M KMnO₄ required to Oxidize the same amount of H₂S gas to Sulphur, in acidic medium is.

SOLUTION:

ILLUSTRATIVE EXAMPLE (13): What volume of 0.4 M Na₂S₂O₃ would be required to react with the I₂ liberated by adding excess of KI to 50 ml of 0.2 M CuSO₄? (Ans= 25 ml).

SOLUTION: Try yourself .......
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BORON:

Boron is metalloids of 13^th group and it does not occur in Free State. The major ores of boron are a small number of borate (boron oxide) minerals, including

PREPARATION OF BORON:

(1) From Borax: Boron may be obtained by treating borax with hot concentrated HCl, or H₂SO_4,  igniting the boric acid H₃BO₃ to give the oxide B₂O₃ and finally reduced with Na, K, Al, Mg.

Step-(1) conversion of borax into boron oxide:

Step-(2) conversion of Boron oxide into boron:

(2) From Colemanite:

Physical properties:

(1) It is a non-metal. Boron occurs in two different allotropic forms Amorphous and   Crystalline

(2) Amorphous boron has not been obtained in the pure state.

(3) Crystalline boron is a black powder, extremely hard with a metallic appearance but with very low electrical conductivity.

Chemical properties:

(1) Reaction with non oxy acids:

(2) Reaction with Oxy acids:

(3) Reaction with Oxy water:

(4) Reaction with Base:

(5) Reaction with Metals:

(6) Reaction with Non-metals

 (7) Reaction with Ammonia:

Structure of Borazine:

BORIC ACID (H3BO3) and its structural features:

Orthoboric acid, H₃BO_3, commonly known as boric acid, and Metaboric acid, HBO₃ , are two common oxy acids of  boron . Orthoboric acid naturally found in volcanic steam vents called Suffioni. 

STRUCTURE OF BORIC ACID:

(1) In dilute Solution Boric acid exist as monomeric form;

(2) When concentration of the solution of acids  is very high then boric acid exist as polymeric metaboric and many more ions.

(3) In Solid State It exist sp3 hybridised BO3-  2D Sheet .

 GENERAL PROPERTIES:

(1) H₃BO₃ is soluble in water and behaves as weak monobasic acid. It does not donate protons but rather it accepts OH^- .Therefore it acts as a Lewis acid [B(OH)₃] .It is not a proton donor because it accept lone pair or hydroxyl ion from water.

(2) Since B(OH)₃ only partially reacts with water to form H₃O⁺ and [B(OH)₄]^- it behaves as a weak acid. Thus it cannot be titrated satisfactorily with NaOH as a sharp end point is not obtained.  

(3)  B(OH)₃ (Boric acid ) does not  titrated even strong alkali like NaOH but If certain polyhydroxy compounds such as glycerol, mannitol or sugar are added to the titration mixture then B(OH)₃ behaves as a strong monobasic acid and hence can be titrated with NaOH and end point is diluted using phenolphthalein as indicator.

(4) The added compound must be a cis-diol to enhance the acidic proprieties. In this way the cis-diol forms very stable complexes with [B(OH)₄]¯ formed in forward direction above, thus effectively removing it from solution. Hence reaction proceeds in forward direction (Le-Chatelier principle.)

PHYSICAL PROPERTIES:

CHEMICAL PROPERTIES:

USES:

Related Questions:

What is structure of solid Ortho Boric acid ?

Why Boric acid become strong acid in the presence of cis 1,2-diol or 1,3-diol ?

What is the structure of trimetaboric acid and trimetaborate ion?  

What is basicity of "Boric acid" ?

Why Boric acid exist in solid state ?

Reference Books:

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BORAX BEAD TEST:

When Borax is heated on loop of platinum wire then loop of platinum wire swells up due to loss of water of crystal of borax, if it is further heated then a transparent glass of Sodium metaborates (NaBO₂) and boric anhydride (B₂O₃) is obtained which  is called borax Bead. 

When a solution of given salt is heated on bead , then volatile part of salt displaced by B₂O₃  and corresponding metaborates salt is formed which gives Characteristic colour on oxidizing and reducing flame and colour of the bead is noted in hot and cold for each type of flame .

Chemical reactions:

Illustrative example of copper:

In oxidising flame;

In Hot flame: green colour transparent glassy bead appear

In Cold: Blue colour transparent glassy bead appear

In reducing flame;

IMPORTANT NOTE:
In this bead test colour of bead of metaborates of given basic radicals are formed provided  the metal cation must contain at least one unpaired electron , Orthoborates are also formed  but they are not responsible for bead colour.

SUMMARY OF BORAX BEAD TEST:

ILLUSTRATIVE EXAMPLE (1): Which of the following do not respond borax bead test?

     (1) Nickel Salts               (2) Copper Salts             (3) Cobalt Salts      (d) Aluminium Salts

ILLUSTRATIVE EXAMPLE (2): the “Borax Bead “Contains:

       (1)  NaBO₃                    (2) NaBO₂                       (3) B₂O₃                 (d) Na₂ B₄O₇.10H₂O  

ILLUSTRATIVE EXAMPLE (3): This Cation gives a colourless bead in Borax Bead Test:

       (1) Mg²⁺                        (2) Ca²⁺                            (3) Cu⁺¹                  (4) Cu⁺² 

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