A certain form of lead has a cubic closest packed structure with an edge length of 492 pm. Calculate the value of the atomic radius and the density of lead.

Interpretation:

        The atomic radius and density of lead cubic close packing structure has to be determined.

Concept introduction:

        In packing of atoms in a crystal structure, the atoms are imagined as spheres and closely packed in a regular pattern. The two major types of close packing of the spheres in the crystal are – hexagonal close packing and cubic close packing. Cubic close packing structure has face-centered cubic (FCC) unit cell.

       In face-centered cubic unit cell, each of the six corners is occupied by every single atom. Each face of the cube is occupied by one atom.

       Each atom in the corner is shared by eight unit cells and each atom in the face is shared by two unit cells. Thus the number of atoms per unit cell in FCC unit cell is,

                     8   ×  18  atoms in corners  +  6  ×  12 atoms in faces  =   1 +  3  =  4 atoms                         The edge length of one unit cell is given by                                                              a  =   2R 2                                               where  a  =  edge length of unit cell                                                              R =  radius of atom

Explanation

Calculate the atomic radius of lead.

given data:  edge length,    a   =  492 pm  =  492  ×  10-12 m                                                    a   =   2R 2

                                        R  =   a22     =   492  ×  10-12 m2  ×  1.414  =    174  × 10-12 m      =    174 pm                                                         

              The edge length value for the FCC unit cell is given. The edge length value is related to the radius of the atom by the equation, a   =   2R 2 . The given value for ‘a’ is plugged and the equation is rearranged in order to get atomic radius ‘R’ of lead.

Calculate the volume of the unit cell.

  given data :  edge length ,  a    =   492 pm  =  492  ×  10-12 m            volume of unit cell  a3   =   (492  ×  10-12 m)3                                                             =    1.19  × 10-22cm3                                                           =    11.9 × 10-23 cm3

             The edge length value for the FCC unit cell is given. Cubic value of the edge length of the unit cell gives volume of the unit cell.

Calculate the mass of the unit cell.

                         Average mass of one Pb atom   =    atomic mass of PbAvogadro number                                                                 =    207