Close Packed Structures Video Lecture | Additional Study Material for Class 12

hexagonal close packing HCP let us place
that layer over the second layer here
are two types of arrangements are
possible one placing the third layer of
spheres on the tetrahedral voids of the
second layer in this type of arrangement
the third layer spheres are exactly as
same as the first layer spheres so the
type of arrangement is a B a B type and
the structure is called ex agonal closed
pattern HCP in this type each sphere is
in contact with 12 other spheres six
from the same layer three from the
bottom layer and three from the top
layer so the coordination number is 12
the total volume occupied by the sphere
is 74% example be daryl IAM mg magnesium
CD cadmium
CEO cobalt Zedan zinc TI titanium TL
tellurium etc
cubic close packing CCP of face centered
cubic FCC structure to placing the third
layer of spheres on the octahedral voids
of the second layer in this type of
arrangement the third layer is not the
repetition of the first layer so this
type of arrangement is called ABC ABC
type of arrangement in this type each
sphere is surrounded by trowels fears
six from the same layer three from the
top layer and three from the bottom
layer so the coordination number is 12
the total volume occupied by the spheres
is 74% example al aluminium CU kapa au
gold
P be lead PD palladium ni nickel and CA
calcium body-centered cubic lattice
this arrangement is obtained when the
spheres in the first layer or slightly
opened up that is none of the sphere
touches other in the same layer now
place second layer of spheres on the
holes of the first layer third layer is
repetition of the first layer so this
type of arrangement is a B a B type in
this type each sphere is in contact with
eight other spheres fall from the top
layer and fall from the bottom layer so
the coordination number is eight the
total volume occupied by the spheres is
68% example na sodium K potassium RB
rubidium CR chromium mo molybdenum W
tungsten etc calculating number of
tetrahedral voids in C CB or FCC the
unit cell is divided into eight small
cubes each cube has four atoms which are
arranged in a tetrahedron so one
tetrahedral void in each cube and eight
in total but we know that there are four
atoms per unit cell therefore the number
of tetrahedral voids is twice the number
of atom in CCP or FCC structure radius
ratio of tetrahedral wide are wide by
our sphere is equal to zero point two
two five
calculating number of octahedral voids
in CCP or FCC the body centered is
surrounded by six atoms which are
present on face centers which generate
an octahedral void along with one at the
body center there are 12 octahedral
voids at the
twelve edge centers but only one-fourth
of it belongs to one unit cell therefore
the number of octahedral holes is full
one body center and three edge centers
12 into 1 by 4 is equal to 3 so the
number of octahedral voids is equal to
number of atoms in FCC or CCP structure
radius of octahedral void are void by a
sphere is equal to zero point four one
four