For cubic structures of simple atoms (such as metals),
    there are 3 different possibilities :

    a)  simple cube:  cell edge a =2r, because the atoms
        touch each other along the edge.  Therefore, there
        are 6 nearest neighbors.          r = a/2

    b)  fcc:  face diagonal = 2^1/2a = 4r, because
        3 atoms touch each other along the FACE diagonal.
                  h = (a² + a²)^1/2 = (2a²)^1/2 = 2^1/2a = 4r
        There are 12 nearest neighbors.      r = 2^1/2a/4 

    c)  bcc:  body diagonal = 3^1/2a = 4r, because
       3 atoms touch each other along the BODY diagonal.
        Therefore, there are 6 nearest neighbors.
                               r =  3^1/2a/4

 
 

USEFUL EQUATION:     rho = Z*M / V*N,
        where rho = the density of the material in g/cm³
                 Z  = number of atoms (particles) in the cell
                 M = molecular (atomic) weight in g/mol or amu
                 V = volume of the unit cell = a³, in cm³
                 N = Avogadro's number =
                                6.022 x 10²³  particles/mol

  CONVERSIONS:  1 pm = 1 x 10^-12 m = 1 x 10^-10  cm
                                        = 1 x 10^-2 Angstrom
              1 Angstrom =  1 x 10^-8 cm = 1 x 10^-10 m = 100 pm
 

Types of problems:
    1)  Calculate Unit-Cell Dimension from Unit-Cell Type and
          Density:

      Pt crystallizes in a face-centered cubic (fcc) lattice with
      all atoms at the lattice points.  It has a density of
      21.45 g/cm³ and an atomic weight of 195.08 amu.
      Calculate the length of a unit-cell edge.  Compare this
      with the value of 392.4 pm obtained from x-ray
      diffraction.

STRATEGY:  We can calculate the mass of the unit cell from
            the atomic weight.  Knowing the density and the
            mass of the unit cell, we can calculate the volume of
            a unit cell and then the edge length of a unit cell.

    195.08 g Pt  x         1 mol Pt              =3.239 x 10^-22g Pt
      1 mol Pt        6.022  x 10²³Pt atoms         1 Pt atom

   Since there are 4 atoms of Pt per fcc unit cell, the mass
        per unit cell =  4 x 3.239 x 10^-22  = 1.296 x 10^-21 g Pt
                                                                        1 unit cell

       The Volume of the unit cell is:   V = m/d
       V = (1.296 x 10^-21 g Pt/cell) /(21.45 g/cm³)
       V = 6.042 x 10^-23 cm³ /cell

 Since the Volume of the cell = a³,  then a = V^1/3
           a = ( 6.042 x 10^-23 cm³)^1/3 = 3.924 x 10^-8 cm
           a = 3.924 x 10^-10 m = 3.924 Angstrom = 392.4 pm

  This is exactly the number obtained by the x-ray
        diffraction experiment.
 

    2)  Calculate Mass of One Atom from Unit-Cell Dimension
          and Density:
          The unit cell length for Ag was determined to be
          408.6 pm.  Ag crystallizes in a fcc lattice with all
          atoms at the lattice points.  Ag has a density of 10.50
          g/cm³.  Calculate the mass of a Ag atom, and, using
          the known atomic weight (107.87 amu), calculate
          Avogadro's #.

STRATEGY:  We can calculate the unit cell volume.   Then,
        from the density, we can find the mass of the unit cell
        and then the mass of a single Ag atom.   From this, we
        can find Avogadro's number.

 The volume of the cell = a³:    V =  (408.6 x 10^-12 m)³
                                                   V =  6.822 x 10^-29 m³

   The density of silver (in g/m³) is:
        d =10.50 g/cm³ x (1 cm/10^-2m)³ = 1.050 x 10⁷ g/m³

   The mass of a unit cell is:  m = d*V
        m = 1.050 x 10⁷ g/m³  x  6.822 x 10^-29 m³=
        m = 7.163 x 10^-22 g Ag/cell

 Because there are 4 atoms in a fcc cell, the mass of a
    single Ag atom in the cell =
                    7.163 x 10^-22 g Ag/cell /  (4 atoms/cell)
                   = 1.791 x 10^-22 g Ag/atom

   The molar mass (Avogadro's number) = N =

             107.87 g Ag /mol Ag     =  6.023 x 10²³ atoms/mol
           1.791 x 10^-22 g Ag/atom
 

              OR, use the formula:     rho = Z*M / V*N
                                                   N = Z*M/rho*V =

      (4 atoms)(107.87 g/mol)                                             =
  10.50 g/cm³(4.086 Angstroms)³(1x10^-8cm/Angstrom)³

                                          6.023 x 10²³/mol
 

Additional Exercises:
   1)  A heavy metal crystallizes in a simple cubic unit cell
        with an edge length of 3.36 Angstroms.  The density
        of the metal is 13.45 g/cm³
        a)  Calculate the radius of an atom.
        b)  Calculate the volume of 1 atom.
        c)  Calculate the volume of the unit cell.
        d)  Calculate the "empty" volume in the cell.
        e)  Calculate the mass of the unit cell.

   2)  Chromium forms cubic crystals whose unit cell has an
         edge length of 288.5 pm. The density of the metal is
         7.20 g/cm³.  The atomic weight of Cr is 51.996 amu.
         a)  Calculate the number of atoms in a unit cell,
               assuming all atoms are at lattice points.
         b)  What type of cubic lattice does chromium have?
         c)  Calculate the radius of the Cr atom.
         d)  Calculate the empty volume of the cell.
 
 

BAND THEORY OF METALS:
  Very simply, if you have metals, i.e., Na, the atomic
    orbitals overlap to form very closely spaced "bonding"
    and  "antibonding" molecular orbitals.  These constitute
    a nearly continuous band of orbitals that belong to the
    crystal as a whole.

  Because the atoms are very close to one another in the
    solid, electrons can "jump" from their ground-state
    positions to vacant higher energy levels when an
    electric field is applied.

  Also, in some metals, the s and p orbitals really do overlap
    in energies, and therefore, there are even more empty
    orbitals to which the electrons can jump.  According to
    the band theory, the highest energy electrons in these
    metals occupy either a partially-filled band or a filled
    band that overlaps a higher-energy empty band.  The
    band where the electrons can jump to is called the
    conduction band = conductors.

  Crystalline nonmetals, such as diamond, are insulators
    (they do not conduct electricity), because their highest
    energy electrons occupy filled bands of molecular orbitals
    that are separated from the conduction band by a band
    gap that has an energy difference that is too large for
    the electrons to jump to get to the conduction band.

  Semiconductors are materials that have filled bands that
    are only slightly below, but do not overlap with, the
    empty bands.  They do not conduct electricity at low
    temperatures, but a small increase in T is sufficient to
    excite some of the highest energy electrons into the
    empty conduction band.
 

Phase changes:  1)  Liquid-Vapor Equilibrium

    Vapor pressure:  At a given T, a certain number of liquid
molecules possess enough KE to escape from the surface.
      The molecules on the surface do not "feel" the
same attraction as those molecules inside the liquid
(surface ones have no molecules on top of them).

    The gas molecules exert a vapor pressure.  Molecules
with HIGH vapor pressure usually evaporate faster.
Ex.:  diethyl ether and acetone evaporate at a much
        higher rate than does water.

Consider a closed system: place liquid in and stopper.
At first, only molecules with high enough KE escape from
the surface, and traffic is all one way.  But, when
dynamic equilibrium is established. then for every
molecule leaving the surface, a different molecule will go
back to the liquid (called condensation = molecule
becomes trapped by intermolecular forces in the liquid).
The pressure of the vapor is called the "equilibrium vapor
pressure."

KE distribution as a function of T:  Only the molecules with
high KE can escape from the liquid into the vapor sate.

HEAT OF VAPORIZATION AND BOILING POINT:

    Molar heat of vaporization DH_vap = energy required to
    vaporize 1 mole of a liquid.  If the intermolecular attraction
    is strong, it will take more energy to free the molecules from
    the surface of the liquid phase.

The higher DH_vap, the higher the boiling pt. of the liquid:

                          DH_vap(kJ/mol)       boiling pt. (^oC)
    CH₄                     9.2                        -164
     diethyl ether    34.6                          26.0
    benzene            31.0                         80.1
     H₂O                 40.79                         100
 

CLAUSIUS-CLAPEYRON EQUATION:

    ln P = -DH_vap/RT  + C    where ln is natural logarithm,
        R = 8.314 J/K-mol   and C = constant.

This has the form of a straight line:  Y = mX + b
    where Y = ln P ,   X = 1/T and m = -DH_vap/R
      and b is the intercept.

SHOW PLOT!!!!!!   (show negative slope!!)

Another form of the Clausius-Clapeyron:

  ln P₁ = -DH_vap/RT₁  + C

  ln P₂ = -DH_vap/RT₂  + C

Subtracting the 2nd from the 1st, we get:

  ln P₁ - ln P₂ = -DH_vap/RT₁ -[-DH_vap/RT₂]

                        =  (DH_vap/R) [1/T₂ - 1/T₁]

Another form:  ln(P1/P2) = (DH_vap/R)[1/T₂ - 1/T₁]

    ln(P₁/P₂) = (DH_vap/R) [T₁-T₂/T₁T₂]

Ex.:  Calculate the vapor pressure of water at 52^oC if the
vapor pressure at 25^oC is 24.6 torr.  DH_vap = 40.79 kJ/mol

  ln(P₁/P₂) = (DH_vap/R) [T₁-T₂/T₁T₂]
      P₁ = 24.6 torr                 P₂ = ?
      T₁ = 25^oC = 298K          T₂ = 52^oC = 325 K

ln(24.6/P₂)=[40790J/mol/(8.314J/mol-K)][298-325/(298x325)]

_{BOILING POINT:  TEMP. AT WHICH VAPOR PRESSURE
    OF LIQUID IS EQUAL TO EXTERNAL PRESSURE.}

_{CRITICAL TEMPERATURE AND PRESSURE:  EVERY
SUBSTANCE HAS A CRITICAL T ABOVE WHICH A GAS CANNOT
BE MADE TO LIQUEFY NO MATTER HOW MUCH PRESSURE IS
APPLIED  (IT IS ALSO THE HIGHEST TEMP THAT A LIQUID
CAN EXIST).  ALSO, THE CRITICAL PRESSURE IS THE MINIMUM
PRESSURE THAT HAS TO BE APPLIED TO START THE
LIQUEFACTION AT THE CRITICAL TEMP.}
 

2)  Liquid-Solid Equilibrium:
      Melting point of solid = freezing point of liquid = T at
      which solid and liquid phases co-exist in equilibrium
 
      Usually think in terms of 1 atm pressure:  
                    ice ---->water
                         <----

MOLAR HEAT OF FUSION:  DH_fus = ENERGY REQUIRED TO
        MELT 1 MOL OF SOLID

NOTE:  THESE VALUES ARE SMALLER THAN DH_vap BECAUSE
    HERE WE ARE JUST SEPARATING THE SOLID INTO THE
    LIQUID PHASE.  BUT IN ORDER TO FORM VAPOR, WE HAVE
    TO GIVE MOLECULES LOTS OF ENERGY TO GET THE VAPOR
    AWAY FROM THE ATTRACTIVE FORCES OF THE LIQUID.
 

SOLID-VAPOR EQUILIBRIUM:   SUBLIMATION (DEPOSITION) 
               SOLID ---->  GAS
                          <----

DH_sub = energy required to sublime 1 mole of solid.
 
DH_sub = DH_fus  + DH_vap    This holds when all the phase
    changes are at the same temperature.
 

PHASE DIAGRAMS:   1)  GENERAL
                                   2)  WATER
                                   3)  CARBON DIOXIDE 

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