Select right expression for determinig packing fraction (P.F.) of NaCl unit cell (assume ideal ) , if ions along an edge diagonal are absent :

Options

A. `P.F.=((4)/(3)pi(r_(+)^(3)+r_(-)^(3)))/(16sqrt2 r_(-)^(3))`
B. `P.F.=((5)/(2)pi(r_(+)^(3)+4r_(-)^(3)))/(16sqrt2 r_(-)^(3))`
C. `P.F.=((4)/(3)pi((5)/(2)r_(+)^(3)+4r_(-)^(3)))/(16sqrt2 r_(-)^(3))`
D. `P.F.=((4)/(3)pi((7)/(2)r_(+)^(3)+r_(-)^(3)))/(16sqrt2 r_(-)^(3))`

Answer

To determine the packing fraction (P.F.) of the NaCl unit cell, assuming ideal conditions and that ions along the edge diagonal are absent, we can follow these steps:

### Step 1: Understand the Structure of NaCl
NaCl crystallizes in a face-centered cubic (FCC) structure. In this structure:
- Na⁺ ions occupy octahedral voids.
- Cl⁻ ions are arranged in an FCC lattice.

### Step 2: Identify the Number of Ions in the Unit Cell
- **Na⁺ Ions**:
- There is 1 Na⁺ ion at the body center.
- There are 12 Na⁺ ions at the edge centers, each shared by 4 unit cells, so the contribution from the edge centers is \( \frac{12}{4} = 3 \).
- Since ions along the edge diagonal are absent, we have 2 edge centers missing, reducing the effective number of Na⁺ ions to \( 3 - 2 = 1 \).
- Therefore, the total effective number of Na⁺ ions is \( 1 + 1 = 2 \).

- **Cl⁻ Ions**:
- There are 8 Cl⁻ ions at the corners, each shared by 8 unit cells, contributing \( \frac{8}{8} = 1 \).
- There are 6 Cl⁻ ions at the face centers, each shared by 2 unit cells, contributing \( \frac{6}{2} = 3 \).
- Therefore, the total effective number of Cl⁻ ions is \( 1 + 3 = 4 \).

### Step 3: Calculate the Volume of Ions
Let \( R^+ \) be the radius of Na⁺ and \( R^- \) be the radius of Cl⁻. The volume of each ion can be calculated using the formula for the volume of a sphere:
- Volume of Na⁺ in the unit cell:
\[ V_{Na^+} = \frac{5}{2} \times \frac{4}{3} \pi R^+^3 \]
- Volume of Cl⁻ in the unit cell:
\[ V_{Cl^-} = 4 \times \frac{4}{3} \pi R^-^3 \]

### Step 4: Calculate the Total Volume of the Unit Cell
The edge length \( a \) of the NaCl unit cell can be derived from the geometry of the FCC structure. The relationship between the edge length and the ionic radii is given by:
\[ a = 2\sqrt{2}R^- \]
Thus, the volume of the unit cell is:
\[ V_{cell} = a^3 = (2\sqrt{2}R^-)^3 = 16\sqrt{2}R^-^3 \]

### Step 5: Calculate the Packing Fraction
The packing fraction (P.F.) is defined as the ratio of the total volume of the ions to the total volume of the unit cell:
\[ P.F. = \frac{V_{Na^+} + V_{Cl^-}}{V_{cell}} \]
Substituting the volumes calculated:
\[ P.F. = \frac{\left(\frac{5}{2} \times \frac{4}{3} \pi R^+^3\right) + \left(4 \times \frac{4}{3} \pi R^-^3\right)}{16\sqrt{2}R^-^3} \]

### Step 6: Simplify the Expression
Taking \( \frac{4}{3} \pi \) common and simplifying:
\[ P.F. = \frac{\frac{4}{3} \pi \left(\frac{5}{2} R^+^3 + 16 R^-^3\right)}{16\sqrt{2}R^-^3} \]

### Final Expression for Packing Fraction
The final expression for the packing fraction of NaCl, considering the absence of ions along the edge diagonal, is:
\[ P.F. = \frac{5R^+^3 + 32R^-^3}{96\sqrt{2}R^-^3} \]