In a AB unit cell (Rock salt type) assuming `A^(+)` forming fcc :

Answer

To solve the problem regarding the AB unit cell (rock salt type) with \( A^+ \) forming a face-centered cubic (FCC) structure, we will follow these steps:

### Step 1: Understand the Structure of Rock Salt
The rock salt structure consists of two types of ions: cations \( A^+ \) and anions \( B^- \). In this case, \( A^+ \) ions form an FCC lattice.

### Step 2: Determine the Number of Ions in the Unit Cell
In an FCC unit cell:
- There are 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom to the unit cell. Therefore, the contribution from the corners is:
\[ 8 \times \frac{1}{8} = 1 \text{ atom} \]
- There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) of an atom to the unit cell. Therefore, the contribution from the faces is:
\[ 6 \times \frac{1}{2} = 3 \text{ atoms} \]
- Total contribution of \( A^+ \) ions in the unit cell:
\[ 1 + 3 = 4 \text{ atoms} \]

### Step 3: Identify the Position of \( B^- \) Ions
The \( B^- \) ions occupy the octahedral voids in the FCC lattice. In a rock salt structure, there are 4 octahedral voids per unit cell, which means there are 4 \( B^- \) ions as well.

### Step 4: Coordination Numbers
- The coordination number of \( A^+ \) (cations) is 6, as each \( A^+ \) ion is surrounded by 6 \( B^- \) ions.
- The coordination number of \( B^- \) (anions) is also 6, as each \( B^- \) ion is surrounded by 6 \( A^+ \) ions.

### Step 5: Nearest Neighbors
- The nearest neighbors of \( A^+ \) are 6 \( B^- \) ions.
- The nearest neighbors of \( B^- \) are 6 \( A^+ \) ions.

### Step 6: Second Nearest Neighbors
- The second nearest neighbors of \( A^+ \) ions include additional \( A^+ \) ions located at the edges of the cube. There are 12 such \( A^+ \) ions, which gives a coordination number of 12 for the second nearest neighbors.

### Step 7: Packing Fraction
The packing fraction for an FCC structure is given by:
\[ \text{Packing Fraction} = \frac{\text{Volume of atoms in the unit cell}}{\text{Volume of the unit cell}} \]
For FCC, the packing fraction is approximately 74%, which does not match the given option of \( \frac{\sqrt{3}}{8\pi} \) (approximately 66.18%).

### Summary of Findings
1. Nearest neighbors of \( A^+ \) is 6 \( B^- \) (True).
2. Nearest neighbors of \( B^- \) is 6 \( A^+ \) (True).
3. Second neighbors of \( A^+ \) is 12 \( A^+ \) (True).
4. Packing fraction of the AB crystal is not \( \frac{\sqrt{3}}{8\pi} \) (False).

### Conclusion
The correct statements are:
- Nearest neighbor of \( A^+ \) is 6 \( B^- \).
- Nearest neighbor of \( B^- \) is 6 \( A^+ \).
- Second neighbor of \( A^+ \) is 12 \( A^+ \).
- The packing fraction statement is incorrect.