Hi,
I was wondering if anyone could provide me the necessary information to solve this problem.
For the following elements:
CdSe
CsI
Li2O
KBr
NaCl
ZnS(I)
I've been asked to find the
-Unit Cell of Larger ion
-Type of Hole where Smaller Ions are found
-# of Cations per unit cell
-# of Anions per unit cell
If anyone could walk me through this one i'd greatly appreciate it.
More information needed to respond. Please repost with more data. Thanks for asking.
Sure! I can help you walk through these calculations.
To solve this problem, we need to understand the concept of unit cells, crystal structures, and the arrangement of ions in them.
First, let's discuss unit cells. A unit cell is the smallest repeating structural unit in a crystal lattice. In general, there are seven types of unit cells, but for this problem, we'll focus on the most common ones: simple cubic (SC), body-centered cubic (BCC), and face-centered cubic (FCC).
Now let's move on to the given elements and determine the type of unit cell for each one:
1. CdSe: The larger ion is Cd, which has a radius of 112 pm. To determine the unit cell, we need to calculate the ionic radii ratio (r+/r-). Since Se has a radius of 184 pm, the ratio is approximately 0.61. Based on this ratio, CdSe adopts a zinc blende structure, which is a FCC unit cell.
2. CsI: The larger ion is Cs, with a radius of 167 pm, and the smaller ion is I, with a radius of 206 pm. The ionic radii ratio is approximately 0.81. Based on this ratio, CsI adopts a caesium chloride (CsCl) structure, which is a BCC unit cell.
3. Li2O: The larger ion is Li, with a radius of 76 pm, and the smaller ion is O, with a radius of 140 pm. The ionic radii ratio is approximately 0.54. Based on this ratio, Li2O adopts an antifluorite structure, which is a FCC unit cell.
4. KBr: The larger ion is K, with a radius of 152 pm, and the smaller ion is Br, with a radius of 196 pm. The ionic radii ratio is approximately 0.78. Based on this ratio, KBr adopts a NaCl structure, which is a FCC unit cell.
5. NaCl: The larger ion is Na, with a radius of 102 pm, and the smaller ion is Cl, with a radius of 181 pm. The ionic radii ratio is approximately 0.56. Based on this ratio, NaCl adopts a NaCl structure, which is a FCC unit cell.
6. ZnS(I): The larger ion is Zn, with a radius of 131 pm, and the smaller ion is S, with a radius of 184 pm. The ionic radii ratio is approximately 0.71. Based on this ratio, ZnS adopts a zinc blende structure, which is a FCC unit cell.
Now that we know the crystal structures, we can determine the type of hole where smaller ions are found:
- In a FCC structure, the smaller ions are found in tetrahedral holes or octahedral holes.
- In a BCC structure, the smaller ions are found in octahedral holes.
- In a zinc blende structure, the smaller ions are found in tetrahedral holes.
The number of cations and anions per unit cell can be determined using the stoichiometry of the compound. For example, in NaCl, the unit cell contains one Na+ ion and one Cl- ion.
I hope this explanation helps you understand how to solve this problem. Let me know if you have any further questions!