Chapter 8 Chemical Bonding Concepts I
There are basically 3 types of chemical bonding, ionic, covalent (or molecular) and metallic bonding. In all 3 cases, valence electrons are important in the chemical bonds.
Ionic bonds are due to electrostatic attraction. Ionic bonds are formed between metals and nonmetals or between cations and anions. For binary ionic compounds, the nonmetal is typically more electronegative and “hogs” the bonding electrons. A good indicator of the stability of ionic compounds is their lattice energy of the ionic compound. Lattice energy is the energy required to separate completely 1 mole of a binary ionic compound into its gaseous ions, as governed by this equation: E el
= (ĸQ
1
Q
2
)/d
K = 8.98 x 10
9
JmC
-2
; Madelung => 1.747 binary-divalent; 1.64 binary-univalent
MgCl
2(s)
+ lattice energy → Mg +2
(g)
+ 2Cl -
(g)
Q
1
and Q
2
are charges on the ions; d is the distance between the ions. The charge values dominate in this equation, followed by the distance. For example, MgO would have a larger lattice energy than NaCl ( 3795 kJ/mol vs 788 kJ/mol) because the Q values for
MgO would be of magnitude 4 as opposed to 1 for NaCl. For MgO and CaO, distance between the charges would determine the larger lattice energy because the Q values would be of the same magnitude. For CaO it is 3414kJ/mol. The Mg-O bond distance is shorter than the Ca-O bond distance (for Mg-O, d = 2.03Å, Ca-O, d = 2.47Å).
Experimental values equation = E el
= M {
(ĸQ
1
Q
2
)/d } (1 – 1/n); n is Born exponent
Covalent bonds are formed mainly between nonmetals by sharing the bonding electrons.
Metallic bonds are formed between atoms in a metal, like the atoms in iron metal. The valence electrons move between atoms. Hence, metals typically are good conductors of electricity.
Lewis Dot Symbols - These symbols are used to show how the valence electrons play a role in these types of bonds, especially covalent bonds. In a Lewis dot symbol, dots are used to represent the valence electrons
; the element’s chemical symbol has a minimum of 1 dot on any of 4 sides or a maximum of 2 dots, depending on how many valence electrons the element has.
For Sodium atom: 1 valence electron
 
Na
.
For Chlorine atom: 7 valence electrons
 
Cl

∙∙
For Oxygen atom: 6 valence electrons
  Ö ∙
∙
Lewis Dot Structure for Covalent and Ionic Bonds
In a Lewis dot structure of a compound, the atoms try to obey the octet rule (8 dots surrounding each atom to give the electronic stability of the noble gases) or the duet rule for the hydrogen atom. In the dot structure for ionic compounds, the dots surround the nonmetal atom with square brackets enclosing the dots and the nonmetal atom and a charge sign at the upper right of the right bracket.

.. .. ..
Ex. Na

+ ∙Cl
 
Na + [
 C l

] = Na

Cl:
∙∙ ∙∙ ∙∙
The dash line represents a single( or sigma, σ) bond(2 dots.)
Ex. H

+

H

H

H

H

H
Ex.

Ö : + : Ö :
 
Ö::Ö
  
Ö = Ö

( = is a double bond, consists a σ and pi bond)
Ex.

N

+

N
  
N

N
  
N

N

(

is a triple bond, consists a
σ
and 2 pi bonds)
∙ ∙
A triple bond is shorter but stronger than a double bond, which is in turn shorter but stronger than a single bond(For this course, S , O, N, C form multiple bonds!!).
Drawing Lewis Dot Structures-(AB n
),( AB n
E m
)
1.
Total the number of valence electrons
2.
Connect the surrounding atoms to the central atom with single bonds(2 dots per bond).
3.
Distribute the remaining dots so that the surrounding atoms obey the octet rule.
4.
Place any leftover electron(s) on the central atom.
5.
If there are not enough electrons for the atoms to obey the octet rule, try forming multiple bonds (double or triple bonds).
Draw Lewis dot structures for CN
-
, PCl
3
, CH
2
Cl
2
, BrO
3
-
Ex. CN Carbon has 4 valence electrons(e), Nitrogen has 5 + 1 more to account for the charge on the cyanide ion. Total is 4 + 5 +1 = 10e. Connect C & N by 1 single bond, leaving 8e. Put 4 of the remaining 8 on C & 4 on N. If neither atom obeys the octet rule, try forming multiple bonds. For ions, enclose structure in square brackets and put ion charge outside, upper right of right bracket.
: C: + :N: + 1e

[

C

N

]
-
∙
Chemical bonds that are not between atoms of the same element or between 2 different elements that have the same electronegativity should have a bond polarity because of the difference in electronegativities. For example, NaCl bond has a bond polarity because the Cl atom is more electronegative than Na. The negative end of the polarity should be on Cl and the positive end should be on Na.
+→
Na

+ 
Cl

The polarity is indicated thus: Na

Cl
If the difference in electronegativity between the elements of a bond is less than 0.5, the the bond is nonpolar covalent (NPC); if it is between 0.5 and 2.1, then the bond is polar covalent (PC); if it is greater than 2.1, then the bond is ionic (I).
0.5 < 0.6 - 2.1 ≤ 2.1
NPC < PC ≤ I

If the Lewis dot structure of an ion or compound is symmetric, the sum of the bond polarities should be zero.
Ex. CO
2

Ö = C = Ö

(molecule is nonpolar)
CH
4
(nonpolar)
If the structure is asymmetric, the sum of the bond polarities will impart a net polarity on the structure.
Ex. CO  C  O  (polar)
NH
3
(polar)
Formal Charge (F.C.) - The charge that an atom in a molecule would have if all atoms had the same electronegativity.
F.C. = Group# - 0.5(#bonding e) – (#nonbonding e)
For CO
 C  O 
F.C. for Carbon = 4 – 0.5(6) – 2 = -1
F.C. for Oxygen = 6 – 0.5(6) – 2 = +1 (aberration, oxygen should have the –1 F.C.)
For CO
2

Ö = C = Ö

F.C. for each Oxygen = 6 – 0.5(4) – 4 = 0
F.C. for carbon = 4 – 0.5(8) – 0 = 0
In a molecule or ion, the sum of the formal charges of the atoms is equal to the net charge on the molecule or ion.
Resonance Structures - Ocasionally, a Lewis dot structure can have several representations that differ by the arrangement of the valence electrons.
These 2 structures of ozone are valid:

Ö

Ö = Ö

and

Ö = Ö

..

Both structures are called resonance structures of ozone and they are equivalent and contribute equally to the real structure of ozone. The true structure of ozone is a resonance hybrid of the 2 structures above.

Use of Formal Charges
When resonance structures of a molecule or ion are not equivalent, one can use formal charges to choose the structure that contributes most to the real structure of that molecule or ion. Consider the 3 resonance structures of the cyanate ion (NCO )
∙∙ ∙∙ ∙∙
[ : N

C

O

]

[

N

C

O

]

[

N

C

O

]
-
..
..
..
The 1 st
structure gives O a F.C. of +1, C a F.C. of 0, N a F.C. of –2. The 2 nd
structure gives O F.C. of 0, C (0), N (-1). The 3 rd
structure gives the F.C. of –1 to oxygen, C (0) and
N
(0) which makes more sense because of oxygen’s high electronegativity value, as compared to carbon and nitrogen.
Exceptions to the Octet Rule
Molecules with an odd number of electrons (radicals)
NO, ClO
2
, NO
2
Molecules with Less than an Octet About a Central Atom (Boron, Beryllium)
BF
3
, BeCl
2
; Ofcourse Hydrogen obeys the duet rule.
Molecules with more than an Octet About the Central Atom (The central atom has d orbitals).
PCl
5
, SF
6
Bond Enthalpy
Bond enthalpies (

H associated with the breaking of chemical bonds in gaseous state) can be used to approximate

H rxn
of chemical reactions when

H
 f
values are not available. One must draw Lewis structures to solve this type of problem.
CH
4
+ 2O
2

CO
2
+ 2H
2
O

H rxn
= [4D(C-H) + 2D(O=O)] – [2D(C=O) + 4D(O-H)]
= [4(413) + 2(495)] - [2(799) + 4(463)]
= [1652 + 990] - [1598 + 1852]
= 2642 - 3450
= -808kJ
The true value is –802.34kJ

Chapter 9 CHEMICAL BONDING CONCEPTS II
Molecular Geometry and Bonding Theories
We can use Lewis dot structures and Valence-shell electron-pair repulsion (VSEPR) models to construct 3-dimensional shapes of molecules and ions ( ABnEm ). Electron
Domains(ED) are areas on the central atom of a molecule or ion where electrons are localized. Valence Bond theory provides hybridization of central atom.
An electron domain can be a pair of nonbonding electrons, a single bond, a double bond or a triple bond of the central atom . In VSEPR models, nonbonding domains occupy more space than triple bonds , which in turn occupy more space than double bonds , which in turn occupy more space than single bonds.
The idea is that the electron domains of a molecule try to orient themselves into a molecular geometry(MG) that minimizes electron domain repulsions. There are 5 fundamental electron domains from which one derives molecular geometries.
#ED ED Geometry #BD #NBD M Geometry SHAPE & Angle Hybrid
2 linear 2 0 linear CO
2
sp
● ─●─ ●
180°
3 trigonal planar 3 0 trigonal
:
BF
3
120° sp
2
trig planar 2 1 bent
:
O
3
<120° sp
2
4 tetrahedron 4 0 tetrahedron
> <
CH
4
109.5° sp
3
tetrahedron 3 1 trig. Pyramidal
/|\
NH
3
107.5° sp
3

tetrahedron 2 2 bent
:
H
2
O 104.5° sp
3
#ED ED Geometry #BD #NBD M Geometry SHAPE & Angle Hybrid
5 trig.bipyramidal 5 0 trig. Bipyramidal PCl
5
90°,120°,180° sp 3 d
trig. Bipyramidal 4 1 seesaw(disphenoid)
\/
SF
4
120°,180° sp
3 d
trig. Bipyramidal 3 2 T-shape
┬
BrF
3
<180° ,90° sp 3 d
trig. bipyramidal 2 3 linear
XeF
2
180° sp 3 d
6 octahedral 6 0 Octahedron
*
SF
6
90°, 180° sp
3 d
2
octahedral 5 1 square pyramid
*
IF
5
<90°,180° sp
3 d
2
octahedral 4 2 square planar
:
•
:
XeF
4
90°,180° sp 3 d 2

Hybridization
The central atom is thought to mix its s, p and/or d orbitals in order to form more stable bonds with the surrounding atoms. This is called hybridization. The number of electron domains (E.D.) indicates what type of hybridization. Two electron domains means sp, 3
E.D. means sp
2
, 4 E.D. means sp
3
, 5 E.D means sp
3 d( OR dsp
3
), 6.E.D. sp
3 d
2
( OR d
2 sp
3
)
Single(sigma) Bonds or (

) and pi bonds(

)
Single bonds are formed when 2 s orbitals combine or when 2 p orbitals combine end to end. Pi bonds are formed when p orbitals or d orbital combine side by side.
A double bond has one single(sigma) bond and one pi bond. A triple bond has one single bond and two pi bonds.
Molecular Orbital Theory(MO)
While VSEPR models explain shape of molecules and ions, molecular orbital theory explains spectroscopic aspects of molecules, such as excited states, bond strengths. In
MO theory, atomic orbitals combine to form molecular orbitals. The number of molecular orbitals formed is equal to the number of starting atomic orbitals.
The atomic orbitals combine to form bonding molecular orbitals and an equal number of antibonding molecular orbitals. The bonding orbitals are at lower energies than their respective antibonding orbitals.
Atomic orbitals of similar energy and symmetry are the most likely to combine to form molecular orbitals. Molecular orbitals are designated as either sigma(

) or pi(

). Just like atomic orbitals, one can write electronic configurations for the molecular orbitals of a molecule and calculate bond orders. Bond order indicates that the bond between 2 atoms is a single, double or triple bond . A bond order of 1 means single bond, etc..
Molecular Orbitals of Diatomic Molecules
H
2
, B
2
, N
2
, O
2
Ex. The atomic orbitals of 2 hydrogen atoms combine to form an equal number of molecular orbitals. Two atomic 1s orbitals form two molecular orbitals. One molecular orbital is bonding(

1s
); the other molecular orbital is antibonding(

*
1s
).

M.O. configuration He
2
: (

1s
) 2 (

1s
* ) 2
.
_↓↑

*
1s
1s
 ↓ ↓ 
1s
 
1s
The above diagram is the molecular orbital energy diagram for H
2
molecule. The

1s molecular orbital is called the H ighest O ccupied M olecular O rbital ( HOMO ) for the obvious reason that it is the highest molecular orbital that is occupied by electrons. The

*
1s
molecular orbital is called the L owest U noccupied M olecular O rbital( LUMO ).
Just like atomic orbitals, one can write electronic configurations for the molecular orbitals of a molecule with the lowest energy orbital written first, etc. Hund’s rule is also obeyed for degenerate molecular orbitals (

orbitals).
For H
2
the electronic configuration is:

1s
2
Only the orbitals occupied by electrons are written in ascending energy. The 2 superscript indicates the number of electrons in that orbital.
The BOND ORDER for H
2
is = 0.5(# of bonding electrons - # of antibonding electrons).
For H
2
the bond order is 0.5(2 – 0) = 1. This means that the H atoms of the H
2
molecule are connected by a single bond.
Theoretically, a diatomic molecule or ion with a bond order greater that 0 should exist.
Molecular orbital configuration sequence for H
2
to N
2
.

1s

1s
*

2s

2s
* π
2Px
π
2Py

2Pz
π
2Px
* π
2Py
*

2Pz
*
.
Molecular orbital configuration sequence for O
2
to Ne
2
.

1s

1s
*

2s

2s
*

2Pz
π
2Px
π
2Py
π
2Px
* π
2Py
*

2Pz
*
.

Li
2
? M.O. configuration (

1s
) 2 (

1s
* ) 2 (

2s
) 2 (

2s
* ) 0
___

*
2s
LUMO
2s
 
2s
 
2s
HOMO
↑↓

*
1s
1s
 ↓ ↓ 
1s
 
1s

1s

1s
*

2s

2s
* π
2Px
π
2Py

2Pz
π
2Px
* π
2Py
*

2Pz
*
For H
2
to N
2
.
__ σ*
2Pz
π*
2Px
___ ___π*
2Py
2Px,y,z ≡ ≡2Px,y,z
___
σ
2Pz
π
2Px
___ ___π
2Py
___

*
2s
2s___ ___ 2s

___

2s
___

*
1s
1s___ ___ 1s
___

1s

1s

1s
*

2s

2s
*

2Pz
π
2Px
π
2Py
π
2Px
* π
2Py
*

2Pz
*
For O
2
to Ne
2
.
___ σ*
2Pz
π*
2Px
___ ___π*
2Py
2Px,y,z ≡ ≡2Px,y,z
π
2Px
___ ___π
2Py
___σ
2Pz
___

*
2s
2s _ _ 2s

___

2s
___

*
1s
1s____ ___ 1s
___

1s
--------------------------------------------------------------------------------------------------
IGNORE CHAPTER 5 LECTURE HEREIN
Chapter 5: Ideal Gases and Ideal Gas Laws
The gaseous state is one of the states of matter. Gases have many properties. Gases can occupy any shape , have high entropy (disorder) values and are very compressible .
Mixtures of gases are called solutions because gaseous mixtures are homogenous . The air we breath is a solution of-among other things- Nitrogen , oxygen and carbon dioxide gases.
One important property of gases is pressure. Pressure is the force per unit area (Newton per meter squared ). Pressure measurements can be in units of pascal(pa), torr, mmHg, atm(atmosphere).
1.01325 x 10
5
pa = 1atm = 760 torr = 7

Open- End Manometer Pgas = 760 - 12 = 748mmHg
Gas Laws
Boyle’s Law
- V

1/P (constant n & T)
Charles’s Law
- V

T (constant n & P)
Avogadro’s Law
- V

n (constant P & T)
V

(nT)/P
V = R(nT)/P
PV = nRT Ideal Gas Law
R is gas constant = 0.0821 (L-atm)/mol-K
OR
8.314 J/mol-K
V = Volume is always in liters.
T = Temperature is always in Kelvin
P = Pressure is usually in atmospheres (atm)(1 atm = 760 torr OR mmHg).
Standard Pressure & Temperature (STP):
1 atmosphere for pressure (1 atm)
273° K For temperature
Given 3 variables in the ideal gas equation, the 4 th
value can be calculated.
Ex. If a gas occupies V = 250ml, P = 1.30 atm, T = 31

C, n = ?
n = (PV)/RT n = (1.3atm)(0.25L)
(0.0821 L-atm/mol-K)(304K) n = 0.013 mole.
At standard temperature and pressure (STP, 1 atm, 0

C), 1 mole of every gas occupies a volume of 22.4 liters (molar volume).
The ideal gas equation can be used to derive the gas laws of Boyle and Charles at two different conditions and make calculation.
P
1
V
1
= n
1
RT
1
P
2
V
2
= n
2
RT
2
If n is constant, then P
1
V
1
= T
1
P
2
V
2
= T
2
What if n & T are constant?, (n & P, n & V)?

Using the ideal gas equation, one can relate the gas density(g/L) to its molar mass(M) .
PV = (g/ M) RT g/V = PM/RT = density = d
Ex. What is density of carbon tetrachloride gas at 714 torr and 125

C?
M = 154g/mol
Change 714 torr to atm(714 torr/760 torr = 0.939 atm)
Change 125

C to K(273 + 125 = 398 K) d = (0.939atm x 154g)/(0.0821 x 398 K) = 4.43 g/L
Partial Pressures of Gases in a Mixture
The total pressure of a mixture of gases is equal to the sum of the partials pressures.
P t
= P
1
+ P
2
+ P
3
….
P t
= ( n
1
+ n
2
+ n
3
….)RT/V
Use mole fractions ( X
1
= n
1
/ n t
) to calculate partial pressures(P
1
).
P
1
= X
1
P t
Ex. A mixture of gases has 2 moles O
2
, 7 moles N
2
, 0.4 mole H
2
in 4 liter container.
What is the total pressure of the mixture and what is the partial pressure of H
2
in mmHg?
K = 273°
Amount of n = 2+7+0.4 = 9.4moles
Pt = (9.4moles)(0.0821)(273K)/4liters = 52.7atm = 52.7 x 760mmHg = 4 x 10
4 mmHg
P(H
2
) = (0.4/9.4)(4 x 10 4 mmHg) = 1702 mmHg
Sometimes in an experiment at a certain temperature, a gas can be collected over a body of water. The total pressure over the water is due to the gas and the water vapor from the body of water. The partial pressure from the water vapor at that temperature can be looked up in a chemistry textbook.
Root Mean Square Speed of a gas (u)- This is close in value to the average speed(
OR rate) of the gaseous molecules.
_______ u =

(3RT/ M ); u is in m/s; M is molecular mass in kilograms ; In this case R =
8.314J/mol-K
Graham’s Law of Effusion r
1
/ r
2
=

( M
2
/ M
1
)

Ex. What is molecular mass of a homonuclear diatomic gas that effuses at 0.355 times the rate of O
2
? r x
/ r
2
= 0.355 =

( M
2
/ M x
) =

(32g/ M x
)
0.126 = 32g/ M x
M x
= 32g/0.126 = 254g/mol

I
2
Chapter 10 Intermolecular Forces, Liquids & Solids
Intermolecular Forces- Special forces of attraction that exist between molecules or between molecules and ions. These forces are generally weaker than ionic or covalent bonds and of electrostatic nature. There are four basic intermolecular forces.
Ion-Dipole Force - This force is between an ion and a polar molecule. Examples of such attractions are found in solutions of ions of alkali metal-halide salts and water(i.e. water and KBr or NaCl..)
Dipole-Dipole Forces - These are electrostatic attractions between polar molecules. The positive end of molecule X is attracted to the negative ends of nearby molecules. The negative end of X is attracted to the positive ends of nearby molecules (H
2
S, CH
3
Cl).
London Dispersion Forces - This attraction force comes about because of the induced distortion of the electron cloud of an atom or molecule by the electron cloud of a nearby atom or molecule. This distortion causes the atom or molecule to have (for an instant) a positive end and a negative end. In that instant, the positive and negative ends of an atom/molecule are attracted to the respective opposite ends of nearby atoms/molecules.
An example is attractive forces between atoms of liquid Neon.
Hydrogen Bonding - Attraction between hydrogen atom of one molecule and an electronegative atom on another nearby molecule (Usually, Oxygen in water, Nitrogen in ammonia or Fluorine in HF). Hydrogen bonding accounts for the high boiling point of water as well as the stability of the double helix of our DNA. Intermolecular forces that bind molecules together are called cohesive forces ; those that bind a molecule to a surface are called adhesive forces .
Some Properties of liquids
Viscosity - Resistance to flow. Cooking oil is more viscous than water.

Surface Tension - Liquids like water and Mercury can form spherical droplets. The spheres arise because of an inward attraction on the surface molecules by the inner molecules. The energy required to increase the surface area of the surface of a liquid at a given temperature is surface tension.
Phase Changes
Energy is associated with a phase change ( freezing/melting or boiling/condensing or subliming/depositing ).
Gas
Solid Liquid
Heating Curve - Indicates the amount of energy in several stages needed to heat a substance from solid state to gaseous state and vice versa. For a substance, one uses its heat capacity in the solid state, heat capacity in the liquid state, heat capacity in the gaseous state and its heat of fusion and its heat of vaporization and sum up the heats.
Solid__________________│___ liquid
_____________│___ gas __________
.q = mCp∆T H fus q = mCp∆T H vap q =mCp∆T
For water: Heat capacity for ice is 2.09 J/g-K
Heat capacity for liquid water is 4.184 J/g-K
Heat capacity for water vapor is 2.09 J/g-K
Heat of fusion is 6.01 kJ/mole at 0

C
Heat of vaporization is 40.67 kJ/mol at 100

C
Calculate heat required to change 1 mole of ice at -25

C to 125

C
AB:

H = (1mol)(18 g/mol)(2.09 J/g-K)(25 K) = 940J = 0.940kJ
BC:

H = (1mol)(6.01k J/mol) = 6.01 kJ
CD:

H = (1mol)(18g/mol)(4.184 J/g-K)(100 K) = 7531J = 7.531 kJ
DE:

H = (1mol)(40.67 kJ/mol) = 40.67 kJ
EF:

H = (1mol)(18 g/mol)(2.09 J/g-K)(25 K) = 940.5 J = 0.941 kJ

H total
= 0.94 kJ + 6.01 kJ + 7.531 kJ + 40.67 kJ + 0.941 kJ = 56.1 kJ
Critical Temperature & Pressure-
A gas at a given temperature liquefies (changes to liquid form) when enough pressure is applied. As the temperature increases, the pressure must increase also. But there comes a

high enough temperature at which beyond that temperature, no amount of pressure can change the gas to liquid. That temperature is called the critical temperature and the minimum liquefaction pressure at that temperature is called the critical pressure (for water they are 374

C & 217.7atm).
Vapor Pressure- The vapor pressure of a liquid at a given temperature is the pressure the vapor of the liquid exerts when the liquid phase and the gaseous phase of the liquid are in equilibrium(or when condensation and vaporization are in dynamic equilibrium). At 1 atmosphere , the boiling point of the liquid is the temperature at which the gaseous phase of a liquid is in equilibrium with atmospheric pressure( normal boiling point ).
For a given set of liquids at a given temperature, the liquid with the highest vapor pressure is said to be the most volatile (Vapor pressure vs. temperature graph).
A B C D
1000-
800-
p 600-
(mmHg) 400-
200-
l l l l l l
40 80 120 160 200 240 T( o C)
Phase Diagram- The phase diagram of a substance shows the equilibrium relationships of the 3 states of that substance in a pressure vs. temperature graph. The triple point of this graph is the pressure-temperature point at which all 3 states of the substance are in equilibrium. Most phase diagrams have a Y shape and are divided into 3 parts(solid, liquid and gas).
What phase exists at the point labeled b ?

Solids (unit cells) Solids can be crystalline (well defined regular arrangement of the ions, atoms or molecule) or can be amorphous (ion, atoms or molecules have an irregular arrangement). Crystalline solids are made of repeating units called unit cells .
There are main 3 types of unit cubic cells, Simple(Primitive) cubic cell( Scc ), Face centered cubic( Fcc ), and Body-centered cubic cell ( bcc ). Iron is made up of repeating
Bcc unit cells in which an atom is located at the center,1/8 of an atom is wedged at each of the 8 corners of the unit cube.
So, (1/8 x 8) + ( 1 ) = 2 atoms in Bcc unit cell
An atom that has an Scc unit cell has 1/8 of an atom wedged at the 8 corners only.
So, (1/8 x 8) = 1 atom in a Scc unit cell .
An atom that has an Fcc unit has 1/8 of an atom wedged at the 8 corners and half of an atom resting at the inside of the 6 faces of the unit cell.
So, (1/8 x 8) + (6 x ½) = 4 atoms in a Fcc unit cell .
For the three basic unit cells, a is edge length of the unit cell.
│← a →│
Scc => a = 2r Bcc => a = (4/√3)r Fcc => a = (√8) r
The r is the radius of the atom(s) that occupy the unit cell.
Ex.
Crystalline iron has a body centered cell unit with the sides measuring 2.8664 angstroms.
Calculate the density of iron metal.
Volume = a
3
= (2.8664 x 10
-8 cm)
3
= 2.36 x 10
-23 cm
3
Bcc => 2 Fe atoms in unit cell
Mass = 2 x 55.8 amu x 1.66 x 10
-24 g/ amu = 1.85 x 10
-22
g
Density = 7.84 g/ cm
3
Ex. Gold(Au, 197g/mol) has an Fcc unit cell lattice and has a density of 19.3 g/ cm
3
.
Calculate the radius a gold atom.
Fcc => a = (√8) r

Remember, a is edge length of cubic cell.
Given density, we need mass => volume => a => r
4 x 197 amu x 1.66 x 10
-24
g/amu = 1.31 x 10
-21
g
V = M / D = 1.31 x 10
-21
g / 19.3 g/cm
3
= 6.78 x 10
-23
cm
3
.a = (6.78 x 10
-23
cm
3
)
1/3
= 4.08 x 10
-8
cm
.r = a/√8 = 1.44 x 10
-8
cm