Final answer:
The ionic radius of H⁻ in NaH is calculated using the given cell edge length of 4.880 Å. The density of NaH can be calculated using the molar mass and volume of the unit cell.
Step-by-step explanation:
To calculate the ionic radius of H⁻, we can use the fact that NaH has the same crystal structure as NaCl. In a NaCl-like structure, the cations touch each other along the cell edges, and the anions are located in the octahedral holes.
The radius of the chloride ion, Cl¯, is given as 181 pm. Since NaCl has a 1:1 stoichiometry, the chloride ion's radius is equal to half of the cell edge length.
Therefore, the cell edge length of NaH is twice the ionic radius of H⁻. Using the given edge length of 4.880 Å, we can calculate the ionic radius of H⁻ as half of 4.880, which is 2.440 Å.
To calculate the density of NaH, we need to know the molar mass and volume of the unit cell. The molar mass of NaH can be calculated as the sum of the atomic masses of Na and H.
The volume of the unit cell can be calculated as the cube of the cell edge length. Once we have the molar mass and volume, we can use the formula density = mass/volume to calculate the density of NaH.