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DIBORANE-PREPARATION:

Boranes are hydride of Boron and diborane is famous borane. It is gas and is highly inflammable in air and poisonous Diborane is used for preparing substances such as high energy fuel and propellants.

Other Methods:

STRUCTURE OF BORAZINE OR BORAZOLE:

(1) Borazine is an inorganic compound with the chemical formula (B3N3H6).and also called borazole. It is a cyclic compound, containing the 3-(BH) units and 3- (NH) units alternate.
(2) Borazine formed by reaction of B2H6 and NH3 in the ratio of 1:2 at room temperature.

(3) Borazine is isoelectronic and isostructural with benzene. For this reason borazine is known as “inorganic benzene”. Like benzene, borazine is a colourless liquid

(4) In Borazine molecule, nitrogen is more electro negative than the boron. Nitrogen acquires partial negative charge and boron acquires partial positive charge and back bonding take place between boron and nitrogen.
(5) As compared with Benzene, Borazol is less stable and more reactive toward Electrophilic Aromatic Substitution reactions due to high polarity of molecule.
(6) Borazine is a highly polar molecule due to high Electronegativity difference between Boron and Nitrogen.
(7) The pi bonds in borazine are highly polarized than pi bonds in benzene. \Thus borazine is more nucleophillic (Negative) hence more reactive than benzene.

STRUCTURE OF DIBORANE :

(1)  B2H6 contains 4-Terminal H are bonded by Sigma bond and  remaining 2-H are bridging hydrogen’s and of these are broken then dimer become monomer.
(2) Boron undergoes sp3 hybridisation 3 of its sp3 hybridised orbitals contain one( e¯) each and fourth sp3 hybrid orbital is vacant.
(3) 3-(Three) of these sp3 hybrid orbitals get overlapped by s orbitals of 3-hydrogen atoms.
(4) One of the sp3 hybrid orbitals which have been overlapped by s orbital of hydrogen gets overlapped by vacant sp3 hybrid orbital. Of 2nd Boron atom. And it’s vice versa.

(5) By this two types of overlapping take place 4 (sp3– s) overlap bonds and 2(sp2 – s – sp3) overlap bonds.
(6) H is held in this bond by forces of attraction from B and This bond is called 3 centred two electron bonds (3C-2e bond) . Also called Banana bonds. Due to repulsion between the two hydrogen nuclei, the delocalised orbitals of bridges are bent away from each other on the middle giving the shape of banana.

(7) The two bridging hydrogens are in a plane and perpendicular to the rest four hydrogen..

ILLUSTRATED EXAMPLE (1): In Diborane
(A) 4 bridged hydrogens and two terminal hydrogen are present
(B) 2 bridged hydrogens and four terminal hydrogen are present
(C) 3 bridged and three terminal hydrogen are present
(D)None of the above
ILLUSTRATED EXAMPLE (2): Which one of the following statements is not true regarding diborane?
(A) It has two bridging hydrogens and four perpendicular to the rest.
(B) When methylated, the product is Me4B2H2.
(C) The bridging hydrogens are in a plane and perpendicular to the rest.
(D ) All the B–H bond distances are equal
ILLUSTRATED EXAMPLE (3): The structure of diborane (B2H6) contains
(A) Four (2C–2e–) bonds and two (2C–3e–) bonds
(B) Two (2C–2e–) bonds and two (3C–2e–) bonds
(C) Four (2C–2e–) bonds and four (3C– 2e–) bonds
(D )None of these
ILLUSTRATED EXAMPLE (4): The molecular shapes  of diborane is shown:
Consider the following statements for diborane:
1. Boron is approximately sp3 hybridised
2. B–H–Bangle is 180°
3. There are two terminal B–H bonds for each boron atom
4. There are only12 bonding electrons available
Of these statements:
(A ) 1, 3 and 4 are correct                  (B) 1, 2 and 3 are correct
(C) 2, 3 and 4 are correct                    (D) 1, 2 and 4 are correct

STRUCTURE OF DIAMOND :

(1) Each carbon is linked to another atom and there is very closed packing in structure of Diamond.
(2) Density and hardness is very much greater for diamond because of closed packing in diamond due to sp3 hybrid and are tetrahedrally arranged around it. And C-C distance is 154pm

(3) Diamond has sharp cutting edges that's why it is employed in cutting of glass.
(4) Diamond crystals are bad conductor of electricity because of absence of mobile electron.
(5) 1 carat of diamond = 200 mg.
(6) Diamond powder if consumed is fatal and causes death in minutes.

STRUCTURE OF GRAPHITE:

(1) In Graphite Carbons are sp2 hybridised out of the four valence electrons, three   involved in (sp2-sigma) covalent bonds form hexagonal layers and fourth unhybridised p– electron of each carbon forms an extended delocalized p-bonding with carbon atoms of adjacent layers
(2) Each carbon is linked with 3 carbons and one carbon will be left and form a two dimensional shed like structure.

(3) Distance between two layers is very large so no regular bond is formed between two layers. The layers are attached with weak vander waal force of attraction.


(4) The carbon have unpaired electron so graphite is a good conductor of current.

(5) The C-C bond length within a layer is 141.5 pm while the inter layer distance is 335.4 pm shorter than that of Diamond (1.54 Å).
(6) Due to wide separation and weak interlayer bonds, graphite is sift , greasy and a lubricant character and low density.
(7) Graphite marks the paper black so it is called black lead or plumbago and so it is used in pencil lead.
(8) Composition of pencil lead is graphite plus clay .the percentage of lead in pencil is zero .
 (9) Graphite has high melting point so it is employed in manufacture of crucible.
(10) Graphite when heated with oxidizing agents like alkaline KMnO4 forms mellatic acid 
                                                 (Benzene hexa carboxylic acid).
(11) Graphite on oxidation with HNO3 gives acid i.e. known as Graphite acid C12H6O12

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 13th 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 H2SO4,  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, H3BO3, 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)4formed 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:





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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