A certain metal crystallizes in a face-centered cubic arrangement and has a density of a density and atomic radius of 12.41 g/cm3 and 134.5 pm, respectively. Which element below is it?
http://en.wikipedia.org/wiki/Rhodium
To identify the element based on the given information, we need to compare the density and atomic radius with the known values of elements.
First, let's understand the face-centered cubic (FCC) arrangement. In the face-centered cubic lattice, the atoms are arranged in a pattern where there is an atom at each of the corners of the unit cell and one atom in the center of each face. This arrangement results in a coordination number of 12, which means that each atom is in contact with 12 neighboring atoms.
Now, let's find the element by comparing the density and atomic radius:
1. Atomic Radius: The given atomic radius is 134.5 pm. Comparing this value with the periodic table, we find that the element with an atomic radius close to this value is cesium (Cs), which has an atomic radius of 167 pm. However, this does not match exactly, so we'll continue to the next comparison.
2. Density: The given density is 12.41 g/cm3. Comparing this value with the periodic table, we find that the element with a density close to this value is tungsten (W), which has a density of 19.25 g/cm3. However, this does not match exactly either.
Since neither the atomic radius nor the density matches exactly, we can use the discrepancy in the values to identify an element that could fit both criteria.
One way to approach this is to consider atomic radii that are close to 134.5 pm, such as those of the elements above and below cesium (Cs) in the periodic table. The element that satisfies both criteria is Barium (Ba), which has an atomic radius of 135 pm and a density of 3.51 g/cm3.
Therefore, the element with a face-centered cubic arrangement, a density of 12.41 g/cm3, and an atomic radius of 134.5 pm is Barium (Ba).