Problem 24 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

CsBr has bcc structure with edge length \(4.3\). The shortest inter ionic distance between \(\mathrm{Cs}^{+}\)and \(\mathrm{Br}^{-}\)is (a) \(3.72\) (b) \(1.86\) (c) \(7.44\) (d) \(4.3\)

Step-by-Step Solution

Answer

(a) 3.72

Step 1: Determine Initial Information

CsBr has a body-centered cubic (bcc) structure with an edge length of 4.3 units. The bcc structure means there is one atom at each corner of the cube and one atom in the center.

Step 2: Identify Atoms in the Structure

In a bcc lattice, the central atom (Br-) is at the center of the cube, and the Cesium ion (Cs+) is positioned at one of the corners. Therefore, the shortest distance between \({\text{Cs}}^+\) and \({\text{Br}}^-\) ions is half of the body diagonal of the cube.

Step 3: Calculate the Body Diagonal

The formula for the body diagonal in a cube with edge length \( a \) is \( \sqrt{3}a \). Substituting the given edge length of 4.3, we have: \( \text{Body diagonal} = \sqrt{3} \times 4.3 \approx 7.44 \).

Step 4: Find Half of the Body Diagonal

Since the shortest interionic distance is half of the body diagonal, calculate \( \frac{7.44}{2} = 3.72 \).

Step 5: Determine the Correct Option

The calculated shortest interionic distance of 3.72 units matches option (a). Therefore, the correct option is (a).

Key Concepts

Body-Centered Cubic Structure

In a body-centered cubic (BCC) structure, the arrangement of atoms is quite efficient and is one of the simplest forms of crystal structures in materials. In this structure, each cube contains one atom at each of the eight corners plus one atom in the very center of the cube. This central atom is 'surrounded' by the corner atoms, which gives the structure its name "body-centered."
The BCC is an important structure type because it is common in several metals and alloys. This arrangement allows for high packing efficiency, though not as high as the face-centered cubic (FCC) structure. For those learning about crystal lattices, understanding BCC is essential because it sets the groundwork for grasping more complex structures.
When visualizing BCC, think of the middle atom as being surrounded by the corner atoms, all aligned along the cube's diagonals. Every diagonal in the cube will pass through two atoms, one at the start and one at the center, emphasizing the three-dimensional nature of the structure. This plays an important role in determining characteristics like the shortest interionic distances, as seen in the interionic distance calculation.

Interionic Distance Calculation

Interionic distance calculation in crystals like the body-centered cubic structure is crucial to understanding the arrangement and space between ions within the lattice. In the BCC structure specifically, the shortest distance between ions is not directly along the edges of the unit cell but along the body diagonal.

To calculate the interionic distance in a BCC structure like CsBr (Cesium Bromide), one must first determine the body diagonal of the cube. The formula for the body diagonal of a cube is \( \sqrt{3}a \), where \( a \) is the edge length. For CsBr with edge length 4.3 units, the body diagonal is calculated as \( \sqrt{3} \times 4.3 \approx 7.44 \) units.
Since the shortest distance is half of this body diagonal (from the corner Cs ion to the center Br ion), the calculation becomes \( \frac{7.44}{2} = 3.72 \) units. Understanding this calculation helps us determine key physical properties of ionic compounds, such as ionic sizes and potential for ion movement within the lattice.

Edge Length in Crystal Lattices

The edge length in crystal lattices plays a significant role in determining various properties of the material, including density, stability, and packing efficiency. The edge length is essentially the side of the cube in cubic lattice structures such as the body-centered cubic (BCC) structure.
For instance, in a BCC structure, the atoms at the corners and the center contribute differently to each unit cell's effective volume and mass. The edge length can be used not only to calculate the body diagonal, as previously demonstrated, but also to find other important distances and sizes within the lattice.
Knowing the edge length allows us to calculate the volume of the unit cell using the formula \( a^3 \) (since it is a cube). In addition, it provides a foundational value needed to compute atomic packing factors, which measure how tightly atoms are packed within the lattice. With an edge length of 4.3 units in CsBr, these calculations become directly linked to understanding the material's structure and the spacing of its ions.