X7

Question 1

no. tetrahedral voids in FCC unit cells

Answer 1

8
all belong to unit cell
Z = 8

Question 2

no. octahedral voids in FCC unit cells

Answer 2

13
1 central void belonging to cell
12 shared between unit cells (1/4)
Z = 4

Question 3

no. tetrahedral voids in hexagonal unit cells

Answer 3

10
2 belong to unit cell + 8 shared between unit cells (1/4)
Z = 4

Question 4

no. octahedral voids in hexagonal unit cells

Answer 4

2
all belong to unit cell
Z = 2

Question 5

NaCl rock salt structure
- structure
- Z
- coordination polyhedra

Answer 5

Cl- forms FCC
Z(Cl) = 8(1/8) + 6(1/2) = 4
Na+ fills all octahedral voids
Z(Na) = 12(1/4) + 1 = 4
Z(unit cell) = 4

coordination polyhedra = NaCl6
CN(Na) = 6

Question 6

ZnS zinc blende structure
- structure
- Z
- coordination polyhedra

Answer 6

S2- forms FCC structure
Z(S) = 4
Zn2+ fills 1/2 tetrahedral voids
Z(Zn) = 4
Z(unit cell) = 4

coordination polyhedra = ZnS4
CN(Zn) = 4

Question 7

CaF2 fluorite structure
- structure
- Z
- coordination polyhedra

Answer 7

Ca2+ forms FCC
Z = 4
F- fills all tetrahedral voids
Z = 8
Z(unit cell) = 4 (CaF2)
- there are 4 units of 1 Ca and 2 F

coordination polyhedra = FCa4
CN(F) = 4

Question 8

Na2S anti-fluorite structure
- structure
- Z
- coordination polyhedra

Answer 8

S2- forms FCC
Z = 4
Na+ fills all tetrahedral voids
Z = 8
Z(unit cell) = 4

coordination polyhedra = NaS4
CN(Na) = 4

Question 9

CsCl caesium chloride structure
- structure
- Z
- coordination polyhedra

Answer 9

Cl- forms PC
Z = 8(1/8) = 1
Cl fills central void
Z = 1
Z(unit cell) = 1

coordination polyhedra = ClCs8

Question 10

relationship between unit cell parameter a and atomic radius (R): PC

Answer 10

a = 2R

Question 11

relationship between unit cell parameter a and atomic radius (R): BCC

Answer 11

a = 4R/sqrt(3)

Question 12

relationship between unit cell parameter a and atomic radius (R): FCC

Answer 12

a = 4R/sqrt(2)

Question 13

relationship between unit cell parameter a and atomic radius (R): hexagonal

Answer 13

a = 2R

Question 14

Unit cell volume: PC

Answer 14

a^3 = 8R^3

Question 15

Unit cell volume: BCC

Answer 15

a^3 = (16/sqrt3)(4/3)(R^3)

Question 16

Unit cell volume: FCC

Answer 16

a^3 = 16(sqrt2)R^3

Question 17

Unit cell volume: hexagonal

Answer 17

a^2csin(60)

Question 18

NiAs nickeline structure
- structure
- Z
- coordination polyhedra

Answer 18

As forms hexagonal cell
Z = 8(1/8) + 1 = 2
Ni fills all octahedral voids
Z = 2
Z(unit cell) = 2

coordination polyhedra = NiAs6
CN(Ni) = 6

Question 19

ZnS wurtzite structure
- structure
- Z
- coordination polyhedra

Answer 19

S2- forms hexagonal cell
Zn = 2
Zn2+ fills 1/2 tetrahedral voids
Zn = 4 (1/4) + 1 = 2
Zn(unit cell) = 2

coordination polyhedra = ZnS4
CN(Zn) = 4

Question 20

formula for unit cell mass
- how used to find density

Answer 20

M(unit cell) = (Mcompound x Z)/(Na)

density = mass/volume

Question 21

Diamond structure

Answer 21

analogous to zinc blende (ZnS):
Carbon FCC
Carbon fills 1/2 tetrahedral voids
Z (unit cell) = 8(1/8) + 6(1/2) + 4 = 8

Question 22

C60 fullerene structure

Answer 22

Carbon FCC
- has all octahedral (13) and tetrahedral (8) voids available to be filled

Question 23

K3C60 structure

Answer 23

arises from filling of tetrahedral and octahedral voids in fullerene (C60) FCC structure
Z(C60) = 4
Z(K) = 8 + 12(1/4) + 1 = 12

Question 24

SiO2 beta-cristobalite structure

Answer 24

analogous to Zinc Blende (ZnS) with oxygens bridging the tetrahedral voids:

Si2+ forms FCC and fills 1/2 tetrahedral voids
Z = 8
O2- bridges FCC and tetrahedral Si2+ atoms
Z = 16

Question 25

formula for coverage

Answer 25

no. adsorption sites occupied/ no. adsorption sites available

Question 26

adsorption isotherm

Answer 26

describes change in adsorption (coverage) with pressure x axis = pressure y axis = coverage

Question 27

langmuir isotherm

Answer 27

describes the equilibrium of a monolayer - fractional coverage described by the volume of gas adsorbed (linear increase) - coverage = vol. adsorbed/ vol. saturated x axis = p/v y axis = pressure

Question 28

methods for increasing surface area of a catalyst

Answer 28

- formation of spongy metals (via hydrogenation) - formation of nano particles (e.g. Pt nano-particles formed by H2PtCl6 reduction) - use of non catalytic pores as substrates (prevents clumping and maximises active sites)

Question 29

catalytic oxidation of carbon monoxide on platinum

Answer 29

CO + 1/2O2 = CO2 - to form carbon dioxide (less harmful) faster on 010 over 111 (O prefers 4f over 3f sites)

Question 30

ammonia synthesis

Answer 30

use of Fe BCC catalyst faster on 111 Fe surface facilitated mass production of fertilisers

Question 31

catalytic hydrogenation of alkenes

Answer 31

use of heterogeneous palladium catalyst - addition fo 2 H atoms to form alkanes

Question 32

Brunauer-Emmet-Teller (BET) equation

Answer 32

defines when a monolayer is formed in a physisorption isotherm (linear increase)

Question 33

formula for calculating surface area

Answer 33

surface area = (Vm x Na x cross-section area of adsorbate)/ (mass (sample) x molecular volume)

Question 34

Hydrogen chemisorption

Answer 34

only undergoes molecular chemisorption to 1f undergoes dissociative chemisorption to 1f, 2f and 3f (prefers 111)

Question 35

alkyne chemisorption

Answer 35

only undergoes molecular chemisorption binds 2 2f sites H cannot bind 4f sites (prefers 111)

Question 36

Carbon monoxide chemisorption

Answer 36

undergoes molecular chemisorption with many FCC metals undergoes dissociative chemisorption on noble metals cannot bind 4f sites (prefers 111)

Question 37

oxygen chemisorption

Answer 37

undergoes dissociative chemisorption prefers 4f (prefers 010) 4f > 3f > 2f > 1f