The number of `Cl^(-)` ions present around each `Na^(+)` ion in NaCl crystal lattice is

Options

A. 3
B. 4
C. 8
D. 6

Answer

To determine the number of `Cl^(-)` ions present around each `Na^(+)` ion in the NaCl crystal lattice, we can follow these steps:

### Step 1: Understand the Structure of NaCl
NaCl crystallizes in a face-centered cubic (FCC) structure. In this structure, the chloride ions (`Cl^(-)`) form a close-packed arrangement.

**Hint:** Recall that in a face-centered cubic (FCC) lattice, the ions are arranged in a specific geometric pattern.

### Step 2: Identify the Position of Sodium Ions
In the NaCl crystal lattice, sodium ions (`Na^(+)`) occupy the octahedral voids of the chloride ion lattice. Each octahedral void is surrounded by six chloride ions.

**Hint:** Remember that octahedral voids are formed by the arrangement of ions in a cubic lattice.

### Step 3: Count the Number of `Cl^(-)` Ions Around Each `Na^(+)` Ion
Since each `Na^(+)` ion is located in an octahedral void, it is surrounded by six `Cl^(-)` ions. This means that for every sodium ion, there are six chloride ions surrounding it.

**Hint:** Visualize the octahedral arrangement to see how the six `Cl^(-)` ions are positioned around the `Na^(+)` ion.

### Step 4: Conclusion
Thus, the number of `Cl^(-)` ions present around each `Na^(+)` ion in the NaCl crystal lattice is **6**.

**Final Answer:** 6 `Cl^(-)` ions are present around each `Na^(+)` ion in the NaCl crystal lattice.

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