A mineral having the formula AB2 crystallises in the cubic closest - p...
A CN=8 Cubic coordination
B CN=4 Tetrahedral coordination
AB2 type crystal structure (CaF2):
Blue – Ca
Yellow- F
Co ordination number of A(For example Ca)= 8 (White one is surrounded by 8 neighbours)
Co ordination number of B(For example F)= 4 (yellow one is surrounded by 4 neighbours)
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A mineral having the formula AB2 crystallises in the cubic closest - p...
Solution:
Given: A mineral having formula AB2 crystallizes in cubic closest - packed lattice with A-atoms occupying the lattice points.
To find: Fraction of tetrahedral sites occupied by 6-atoms
Explanation:
- In cubic closest-packed lattice, there are 4 tetrahedral voids and 8 octahedral voids per unit cell.
- Since the formula of the mineral is AB2, there are 2 B-atoms for every A-atom. Therefore, the number of tetrahedral sites occupied by B-atoms = 2 x number of A-atoms.
- Let the number of A-atoms in the unit cell be x. Then, the number of B-atoms in the unit cell is 2x.
- Since the A-atoms occupy the lattice points, they do not occupy any of the tetrahedral or octahedral voids.
- Let the fraction of tetrahedral sites occupied by B-atoms be y.
- Then, the number of B-atoms occupying tetrahedral sites = y x 4x = 4y x^2
- Since each tetrahedral site is surrounded by 4 atoms, the total number of atoms surrounding the tetrahedral sites = 4 x 4 = 16.
- Therefore, the number of 6-atom molecules surrounding the tetrahedral sites = (16/6) x (4y x^2) = 32/3 y x^2
- Since the total number of atoms in the unit cell = x + 2x = 3x, the fraction of tetrahedral sites occupied by 6-atoms = (32/3 y x^2)/(6x) = 16/9 y
To find the value of y, we need to use the fact that the mineral crystallizes in cubic closest-packed lattice.
- In cubic closest-packed lattice, each sphere is surrounded by 12 nearest neighbors arranged in the form of an octahedron.
- Therefore, the number of octahedral voids = number of spheres.
- Let the radius of the A-atom be rA and the radius of the B-atom be rB.
- Since the A-atoms occupy the lattice points, the distance between two adjacent A-atoms = 2rA.
- The radius of the octahedral void = rO = (sqrt(2)/2) x 2rA = sqrt(2) x rA
- The distance between the centers of two adjacent octahedral voids = 4rO = 4 x sqrt(2) x rA.
- Therefore, the length of the edge of the unit cell = 4 x sqrt(2) x rA.
- The distance between the center of an octahedral void and the center of the closest A-atom = rA + rO = (1 + sqrt(2)) x rA.
- Therefore, the distance between the centers of two adjacent tetrahedral voids = 2 x (1 + sqrt(2)) x rA.
- Let the distance between two adjacent B-atoms occupying the tetrahedral voids be d.
- Then, d^2 = (2 x (1 + sqrt(2)) x rA)^2 + (2rB)^2
- Since the mineral crystallizes in cubic closest-packed lattice, the distance between two adjacent B
A mineral having the formula AB2 crystallises in the cubic closest - p...
(d) AB2 crystallises as cubic close packed (ccp).
Thus, A2+ is surrounded by eight B- and B- is surrounded by four A2+ .
Thus, coordinatio n number of A2+ = B
coordination number of B- = 4
Number of atoms in tetrahedral in small cube of unit cell of atom B = 4
Thus, fraction occupied = 4/4 = 1 (100%)