What is the effect of temperature and pressure on crystal structure of Caesium Chloride?

Effect of temperature on crystal structure:

Increase of temperature decreases the coordination of number, e.g. upon heating to
760 K, the CsCl type crystal structure having coordination 8:8 changed to NaCl type crystal structures having coordination 6:6.

Effect of pressure on crystal structure:

Increase of pressure increases the Co – ordination number during crystallization e.g. by applying pressure, the NaCl type crystal structure having 6:6 coordination number changes to CsCl type crystal having coordination number 8:8

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CAESIUM CHLORIDE (CsCl) STRUCTURE:

SODIUM CHLORIDE (ROCK SALT TYPE) STRUCTURE:

What fraction of edge is not covered by atoms in FCC unit cell ?

CCP or FCC has two lattice point corner as well as face centred:

Suppose 'r' be the radius of sphere and 'a' be the edge length of the cube As there are 4 sphere in FCC unit cell

(1) Relation between radius (r) and side (a)

In FCC, the corner spheres are in touch with the face centred sphere. Therefore, face diagonal AD is equal to four times the radius of sphere AC = 4r

But from the right angled triangle ABC:

This is the relation between radius (r) and edge length (a)

So we can find the edge length in term of radius is

 (a) =2√2 r

Edge uncovered can be calculated as : 

          2√2 r- 2r = 0.83 r

Percentage edge uncovered : (0.83r/a)×100=29.27 %

What is the number of atoms on one unit cell of HCP?

ANALYSIS OF HCP UNIT CELL:

Number of effective atoms in HCP unit cell (Z):

Lattice point:  corner- total 12 carbon contribute 1/6 to the unit cell

Lattice point:   face- total face 2 contribute ½

Lattice point: body centre- total atom 3 (100% contribution)

Find the ratio of Fe+3 and Fe+2 in a non Stoichiometric oxide of iron , formulated as Fe_0.90 S_1.0 .

This is a non Stoichiometric compound formed due to variable valency of transition elements.   In the Compound Fe_0.90 S_1.0  both Fe+3 and F+2 ions present,  if number of  Fe+2 ions are  x then number of Fe+3 ions are (0.90-x).

We can find the value of x by oxidation state method as :

If the total number of atoms per unit cell in an hcp structure and a bcc structure gets halved, then ratio of percentage voids in hcp and bcc structures is

In HCP ; Packing efficiency:

Atoms occupy:74%

Empty space(voids): 28% 

In BCC; Packing efficiency:

Atoms occupy: 68% 

Empty space(voids): 32% 

According to the given condition if the total number of atoms per unit cell in a hcp structure and a bcc structure get halved. Then packing efficiency in HCP and BCC respectively given as:

In HCP ; Packing efficiency:

Atoms occupy: 74/2 = 37%

Empty space(voids): (100 - 37) = 63%

In BCC; Packing efficiency:

Atoms occupy: 68/2 = 34%

Empty space(voids): (100 - 34) = 66%

Then new ratio of their voids = 63%/66% = 21/22