In an ionic compound `A^(+)X^(-)`, the radii of `A^(+) and X^(-)` ions ar 1.0pm and 2.0om, respectively. The volume of the unit cell of the crystal AX will be:

Options

A. `27"pm"^(3)`
B. `64"pm"^(3)`
C. `125"pm"^(3)`
D. `216"pm"^(3)`

Answer

To find the volume of the unit cell of the ionic compound \( A^+X^- \), we will follow these steps:

### Step 1: Identify the radii of the ions
We are given:
- Radius of cation \( A^+ \) = 1 picometer (pm)
- Radius of anion \( X^- \) = 2 picometers (pm)

### Step 2: Calculate the radius ratio
The radius ratio \( \frac{r_{A^+}}{r_{X^-}} \) can be calculated as follows:
\[ \text{Radius ratio} = \frac{r_{A^+}}{r_{X^-}} = \frac{1 \, \text{pm}}{2 \, \text{pm}} = 0.5 \]

### Step 3: Determine the type of lattice
Since the radius ratio \( 0.5 \) lies between \( 0.414 \) and \( 0.732 \), we can conclude that the cations \( A^+ \) will occupy the octahedral voids in the face-centered cubic (FCC) lattice formed by the anions \( X^- \).

### Step 4: Understand the arrangement in the unit cell
In an FCC lattice:
- Anions \( X^- \) are located at the corners and face centers of the cube.
- The edge length \( a \) of the unit cell can be determined by the arrangement of the ions.

### Step 5: Calculate the edge length of the unit cell
The edge length \( a \) can be calculated using the formula:
\[ a = 2r_{X^-} + 2r_{A^+} \]
Substituting the values:
\[ a = 2(2 \, \text{pm}) + 2(1 \, \text{pm}) = 4 \, \text{pm} + 2 \, \text{pm} = 6 \, \text{pm} \]

### Step 6: Calculate the volume of the unit cell
The volume \( V \) of the cubic unit cell is given by:
\[ V = a^3 \]
Substituting the edge length:
\[ V = (6 \, \text{pm})^3 = 216 \, \text{pm}^3 \]

### Final Answer
The volume of the unit cell of the crystal \( AX \) is \( 216 \, \text{pm}^3 \).

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