What is the coordination number of `Cs^(+)` in `CsCl`?

Answer

To determine the coordination number of \( Cs^+ \) in \( CsCl \), we can follow these steps:

### Step 1: Understand the Structure of \( CsCl \)
- \( CsCl \) crystallizes in a body-centered cubic (BCC) structure. In this structure, the \( Cs^+ \) ions occupy the body center of the cube, while the \( Cl^- \) ions are located at the corners of the cube.

### Step 2: Identify the Positions of Ions
- In a BCC lattice, there is 1 ion at the center and 8 ions at the corners. Each corner ion is shared among 8 adjacent cubes, contributing \( \frac{1}{8} \) of an ion to each cube.

### Step 3: Calculate the Total Number of \( Cl^- \) Ions Surrounding \( Cs^+ \)
- Since there are 8 corner positions in the cubic structure, the total contribution of \( Cl^- \) ions in contact with \( Cs^+ \) is:
\[ 8 \times \frac{1}{8} = 1 \text{ (from corners)} \]
- However, in the case of \( CsCl \), the coordination number refers to how many \( Cl^- \) ions surround a single \( Cs^+ \) ion.

### Step 4: Determine the Coordination Number
- In the BCC structure, each \( Cs^+ \) ion is surrounded by 8 \( Cl^- \) ions located at the corners of the cube. Therefore, the coordination number of \( Cs^+ \) in \( CsCl \) is:
\[ \text{Coordination number of } Cs^+ = 8 \]

### Final Answer
The coordination number of \( Cs^+ \) in \( CsCl \) is **8**.

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