Sodium oxide \(\left(\mathrm{Na}_{2} \mathrm{O}\right)\) adopts a cubic structure with Na atoms represented by green spheres and O atoms by red spheres. $$ \begin{array}{l}{\text { (a) How many atoms of each type are there in the unit cell? }} \\ {\text { (b) Determine the coordination number and describe the }} \\ {\text { shape of the coordination environment for the sodium }} \\\ {\text { ion. }} \\ {\text { (c) The unit cell edge length is } 5.550 \text { A. Determine the den- }} \\ {\text { sity of } \mathrm{Na}_{2} \text { O. }}\end{array} $$

To find the number of atoms of each type, we need to know their positions within the unit cell. In a cubic unit cell, there are eight corners, six faces, and one center. Sodium (Na) atoms are located on corners and faces, while Oxygen (O) atoms are located in the center. Each corner atom is shared by eight cubes, so each contributes 1/8th to the unit cell. There are 8 corner Na atoms, contributing 8 * 1/8 = 1 Na atom. Each face atom is shared by two cubes, so each contributes 1/2 to the unit cell. There are 6 face Na atoms, contributing 6 * 1/2 = 3 Na atoms. The Oxygen atom in the center of the unit cell entirely belongs to the unit cell. So there is 1 O atom. The unit cell contains a total of 1 Na + 3 Na + 1 O = 4 Na atoms and 1 O atom.

The coordination number is the number of nearest neighbors of an atom. In the case of Na2O, each sodium ion is surrounded by Oxygen ions and vice versa. Each face-centered Na atom is surrounded by four O atoms in the same plane and one O atom above and below the plane (six nearest neighbors in total). The coordination number for sodium ions is therefore 6. The shape of the coordination environment is defined by the arrangement of the nearest neighboring atoms. In this case, the six oxygen ions surrounding each sodium ion form an octahedron. Thus, the coordination environment is octahedral.

To determine the density of Na2O, we need to find the mass of the unit cell divided by its volume. The mass of one unit cell can be found using the molecular weight of Na2O: M(Na2O) = 2 * M(Na) + M(O) = 2 * 22.990 g/mol + 16.00 g/mol = 61.980 g/mol. The volume of the unit cell can be calculated using the edge length, which is given as 5.550 A (1 A = 10^{-10} m): Volume = edge_length^3 = (5.550 * 10^{-10} m)^3 = 1.70378 * 10^{-28} m^3. Now we can calculate the density using mass and volume: Density = (Mass of one unit cell) / (Volume of one unit cell). Density = (61.980 g/mol) / [1.70378 * 10^{-28} m^3 * (1 mol / (6.022 * 10^{23} unit cells))]. Density = 2.269 g/cm^3. So, the density of Na2O is approximately 2.269 g/cm^3.