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 2^nd 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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