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Showing posts with label THE SOLID STATE. Show all posts
Showing posts with label THE SOLID STATE. Show all posts

Thursday, May 6, 2021

How to calculate packing efficiency of cubic unit cell of diamond ?

Diamond structure is ZnS type structure  in which carbon atoms forms  a face centred cubic (FCC/CCP) lattice as well as  four out of eight (50%) or alternate tetrahedral voids are occupied by carbon atoms. Every atom in this structure is surrounded tetrahedrally by four other. No discrete molecule can be discerned (identified) in diamond .the entire crystal is giant molecule a unit cell of which is shown as below.

Note : Only those atoms which form four covalent bond produce a repeated 3D structure using only covalent bonds.

Lattice of Diamond is ZnS type structure.

(1) C- form FCC/CCP (4-atoms)

(2) C- atoms present at the (50%) alternative tetrahedral voids (4-atoms)

(3) Total Number of one lattice unit is eight (8) hence molecular formula of diamond is (C8) (i.e. Z= 8)

(4) Number of C-C bond in lattice cell is = 4×4= 16

(5) Number of C-C bond per carbon atom is 16/8=2

(6) The distance between two Corbin atom is dC-C = a√3/4 and the radius of carbon atom = dc-c/2 = rc = a√3/2x4

(8) Packing efficiency (PE = π√3/10= 0.34 or 34%):

(9) Voids = 66 %

The edge length of unit cubic cell of diamond is 356.7 pm and this cubic unit cell contains 8 carbon atoms calculate (A) Distance between two carbon atoms, assuming them to spheres in contact (B) Radius of carbon atoms (C) Fraction of the total volume that is occupied by the carbon atoms.

(A) Distance between two carbon atoms:

(B) Radius of carbon atoms:

(C) Fraction (Packing efficiency) of the total volume that is occupied by the carbon atoms:

Related Questions:

In diamond, carbon atoms occupy FCC/CCP lattice point as well as alternate tetrahedral voids. If edge length of unit cell is 3.56pm, the radius of carbon atom is ?

We know that in diamond structure the distance between two carbon atoms is

Related Questions:

Saturday, April 17, 2021

Crystal lattice Structure of Oxides:

Related Questions:

Corundum (Al2O3) type Structure:

This type structure is exhibited by oxides of trivalent metals for examples Fe2O3, Al2O3, Cr2O3, Mn2O3 etc.

General formula is M2O3:

In Corundum  Structure:

O-2 = ions forms HCP unit cell:

M+3 = Occupied  2/3 (66.66%) of octahedral voids (OV): and 33.33% OV are empty and TVs are 100% empty.

In Hexagonal close packing (HCP): Formula determination:

O-2 = HCP = O6

M+3=2/3 x 6 = M4

=M4O6  or  M2O3

Related Questions:

Rutile (TiO2) type Structure:

This type of structure is represented by TiO2 and other example are includes MnO2 , SnO2 , MgF2, NiFetc

In Rutile structure:

O-2  = ions forms hexagonal cubic packing (HCP).

M+4 =Tetravalent cations are occupied 50 % (1/2) of octahedral voids (OV) while all the tetrahedral voids (TO) are empty.

HCP= Lattice points (LP) = O-2= 6

M+4 ions = OV (50%) =6 x 1/2 = 3

M3O6 = MO2

It means each HCP unit cell certain three (M+4 ) and  six (O-2 )ions

Related Questions:

Pervoskite (ABO3) type structure:

ABQ3 is a proto type solid where A is a divalent cation  and B is a tetravalent cation.

A+2 = Divalent cations

B+4 =Tetravalent cations

In Pervoskite structure:

A+2 = placed at corner of cubic unit cell

O-2 = placed at all the face centre

That means A+2 and O-2 ions combinelly form CCP/FCC like structure  and

B+4 = Tetravalent cations occupy the central octahedral voids (25% of total OV) i.e centre of the centre of the unit cell so general formula is ABO3

Other examples are BaTiO3, SrTiO3

Related Questions: