Solid `A^(+)B^(-)` has a bcc structure .If the distance of closest apporach between two atoms is 173 pm ,the edge length of the cell is:

Options

A. 2 Å
B. 6Å
C. 5 Å
D. 3 Å

Answer

To find the edge length of a body-centered cubic (BCC) solid \( A^{+}B^{-} \) given the distance of closest approach between two atoms (173 pm), we can follow these steps:

### Step 1: Understand the BCC Structure
In a BCC structure, atoms are located at the corners of the cube and one atom is located at the center of the cube. This means that the body diagonal of the cube contains two atomic radii plus the radius of the atom at the center.

### Step 2: Identify the Relationship Between Edge Length and Atomic Radius
The body diagonal \( d \) of a cube can be expressed in terms of the edge length \( a \):
\[ d = \sqrt{3}a \]
In a BCC structure, the body diagonal is equal to four times the atomic radius \( R \):
\[ d = 4R \]
Thus, we can equate the two expressions:
\[ \sqrt{3}a = 4R \]

### Step 3: Relate the Given Distance to Atomic Radius
The problem states that the distance of closest approach between two atoms is 173 pm, which is equal to \( 2R \):
\[ 2R = 173 \text{ pm} \]
From this, we can find \( R \):
\[ R = \frac{173}{2} = 86.5 \text{ pm} \]

### Step 4: Substitute \( R \) into the Equation
Now, substitute \( R \) back into the equation relating \( a \) and \( R \):
\[ \sqrt{3}a = 4 \times 86.5 \text{ pm} \]
Calculating the right side:
\[ \sqrt{3}a = 346 \text{ pm} \]

### Step 5: Solve for Edge Length \( a \)
Now, solve for \( a \):
\[ a = \frac{346 \text{ pm}}{\sqrt{3}} \approx \frac{346}{1.732} \approx 200.5 \text{ pm} \]

### Step 6: Convert to Angstroms
To convert picometers to angstroms (1 Å = 100 pm):
\[ a \approx \frac{200.5 \text{ pm}}{100} = 2.005 \text{ Å} \approx 2 \text{ Å} \]

### Final Answer
Thus, the edge length of the cell is approximately:
\[ \boxed{2 \text{ Å}} \]