Which of the following expression is correct for packing fraction of NaCl if the ions along with face are diagonally removed?

Options

A. `(13/3 pi r_(-)^3 + 16/3pi r_(+)^3)/(8(r_(+) + r_(-))^3)`
B. `(13/3 pi r_(-)^3 + 4/3 pi r_(+)^3)/(8(r_(+) + r_(-))^3)`
C. `(16/3 pi r_(-)^3 + 13/3 pi r_(+)^3)/(8(r_(+) + r_(-) )^3)`
D. `(4/3 pi r_(-)^3 + 13/3 pi r_(+)^3)/(8(r_(+) + r_(-) )^3)`

Explanation

To find the correct expression for the packing fraction of NaCl when ions along the face diagonal are removed, we can follow these steps:

### Step 1: Understand the Crystal Structure of NaCl
NaCl crystallizes in a face-centered cubic (FCC) structure. In this structure, each unit cell contains:
- Corner atoms: 8 corners × (1/8 contribution per corner) = 1 Na atom
- Face-centered atoms: 6 faces × (1/2 contribution per face) = 3 Cl atoms

Thus, the total number of atoms in the NaCl unit cell before any removal is:
- Total Na atoms = 1
- Total Cl atoms = 3
- Total atoms = 1 Na + 3 Cl = 4 atoms

### Step 2: Remove Atoms from the Face Diagonal
When we remove atoms along the face diagonal, we need to account for the contribution of the removed atoms:
- Each face diagonal has 2 Cl atoms (each contributing 1/8 to the unit cell).
- The contribution of the removed Cl atoms is: 2 × (1/8) = 1/4 Cl atoms.

### Step 3: Calculate the Net Number of Atoms
After removing the Cl atoms:
- Net Cl atoms = 4 (initial) - 1/4 = 3.75 Cl atoms
- The number of Na atoms remains unchanged at 4.

### Step 4: Calculate the Volume of the Unit Cell
The edge length (a) of the unit cell can be expressed in terms of the radii of the ions:
- \( a = 2(r^+ + r^-) \)
Where \( r^+ \) is the radius of the Na+ ion and \( r^- \) is the radius of the Cl- ion.

The volume of the unit cell (V_cell) is:
\[ V_{cell} = a^3 = [2(r^+ + r^-)]^3 = 8(r^+ + r^-)^3 \]

### Step 5: Calculate the Volume of Atoms in the Unit Cell
The volume occupied by the ions in the unit cell is:
- Volume of Na atoms = Number of Na atoms × Volume of one Na atom
- Volume of Cl atoms = Number of Cl atoms × Volume of one Cl atom

The volume of one ion is given by:
\[ V_{ion} = \frac{4}{3} \pi r^3 \]

Thus, the total volume occupied by the ions is:
\[ V_{occupied} = (3.75 \times \frac{4}{3} \pi r^-^3) + (4 \times \frac{4}{3} \pi r^+^3) \]

### Step 6: Calculate the Packing Fraction
The packing fraction (PF) is defined as the ratio of the volume occupied by the atoms to the total volume of the unit cell:
\[ PF = \frac{V_{occupied}}{V_{cell}} \]

Substituting the volumes we calculated:
\[ PF = \frac{(3.75 \times \frac{4}{3} \pi r^-^3) + (4 \times \frac{4}{3} \pi r^+^3)}{8(r^+ + r^-)^3} \]

### Step 7: Simplify the Expression
This simplifies to:
\[ PF = \frac{13}{3} \pi r^-^3 + \frac{16}{3} \pi r^+^3 \div 8(r^+ + r^-)^3 \]

### Final Answer
The correct expression for the packing fraction of NaCl when ions along the face diagonal are removed is:
\[ PF = \frac{13}{3} \pi r^-^3 + \frac{16}{3} \pi r^+^3 \div 8(r^+ + r^-)^3 \]