Ammonium chloride, crystallizes in a body centered cubic lattice with edge length of unit cell equal to 387pm. If the size of `Cl^(-)` ion is 181pm, the size of `NH_(4)^(+)` ion would be:

Options

A. (a) 116pm
B. (b) 154pm
C. (c) 174pm
D. (d) 206pm

Answer

To find the size of the \( NH_4^+ \) ion in ammonium chloride (\( NH_4Cl \)), which crystallizes in a body-centered cubic (BCC) lattice, we can follow these steps:

### Step 1: Understand the BCC Structure
In a body-centered cubic lattice, there are atoms located at the eight corners of the cube and one atom at the center of the cube. The body diagonal of the cube connects two opposite corners and passes through the center atom.

### Step 2: Identify the Given Data
- Edge length of the unit cell (\( a \)) = 387 pm
- Radius of \( Cl^- \) ion = 181 pm

### Step 3: Calculate the Body Diagonal
The body diagonal (\( d \)) of a cube can be calculated using the formula:
\[ d = \sqrt{3} \times a \]
Substituting the value of \( a \):
\[ d = \sqrt{3} \times 387 \text{ pm} \]

### Step 4: Set Up the Equation for the Body Diagonal
In the BCC structure, the body diagonal can also be expressed in terms of the radii of the ions. The body diagonal consists of:
- Radius of \( NH_4^+ \) ion (let's denote it as \( r_{NH4} \))
- Diameter of \( Cl^- \) ion (which is \( 2 \times r_{Cl} \))

Thus, the equation becomes:
\[ d = r_{NH4} + 2 \times r_{Cl} + r_{NH4} = 2 \times r_{NH4} + 2 \times r_{Cl} \]

### Step 5: Substitute Known Values
Now substituting the known values into the equation:
\[ \sqrt{3} \times 387 = 2 \times r_{NH4} + 2 \times 181 \]

### Step 6: Solve for \( r_{NH4} \)
First, calculate \( \sqrt{3} \times 387 \):
\[ \sqrt{3} \approx 1.732 \implies 1.732 \times 387 \approx 670.82 \text{ pm} \]
Now substituting this back into the equation:
\[ 670.82 = 2 \times r_{NH4} + 362 \]
Subtracting 362 from both sides:
\[ 670.82 - 362 = 2 \times r_{NH4} \]
\[ 308.82 = 2 \times r_{NH4} \]
Now divide by 2:
\[ r_{NH4} = \frac{308.82}{2} \approx 154.41 \text{ pm} \]

### Step 7: Final Answer
Thus, the size of the \( NH_4^+ \) ion is approximately \( 154 \text{ pm} \).

### Summary
The size of \( NH_4^+ \) ion is approximately \( 154 \text{ pm} \).

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