Courses for Kids
Free study material
Offline Centres
Store Icon

In a unit cell, atoms A, B, C and D are present at corners, face-centre, body-centre, and edge centre respectively. If atoms touching one of the plane passing through two diagonally opposite edges are removed, then formula of the compound is:

Last updated date: 20th Jun 2024
Total views: 403.8k
Views today: 12.03k
403.8k+ views
Hint:We know that unit cell is the smallest repeating unit of the crystal lattice. We can call the building block of a crystal a unit cell. We have to know that a unit cell can be a face-centered cubic cell (FCC), body-centered cubic cell (BCC), and primitive cubic cell.

Complete step by step answer:
We have to know that in a unit cell, an atom's coordination number is the number of atoms it is touching.
The face-centered cubic (fcc) has a coordination number of twelve and the number of atoms per unit cell is four.
The body-centered cubic (bcc) has a coordination number of eight and comprises two atoms per unit cell.
The simple cubic has a coordination number of six and comprises one atom per unit cell.
The edge-centered cubic comprises four atoms per unit cell.
In the given unit cell,
Atom A is present at the corners.
Atom B is present at the face-centres.
Atom C is present at the body-centres.
Atom D is present at the edge-centres.
Initially, the atoms effectively present in the lattice are given as follows,
For an atom A, $8 \times \dfrac{1}{8}$
For an atom B, $6 \times \dfrac{1}{2}$
For an atom C, $1$
For an atom D, $12 \times \dfrac{1}{4}$
After we remove atoms lying on the plane that passes through two diagonally opposite edge,
Four corner atoms,
Two face centered atoms,
One body centered atoms
Two edge centered atoms are removed.
The atom (A) left at the corner would be $8 - 4 = 4$
Therefore, contribution of atoms=$4 \times \dfrac{1}{8} = \dfrac{1}{2}$
The atoms (B) left at the face centre would be $6 - 2 = 4$
Therefore, contribution of atoms=$4 \times \dfrac{1}{2} = 2$
The atoms (C) left at the body center would be $1 - 1 = 0$
The atoms (D) left at the edge center would be $12 - 2 = 10$
Therefore, contribution of atoms=$10 \times \dfrac{1}{4} = 2.5$
Let us now take the ratio of all the four atoms \[A:B:C:D\]
Ratio of \[A:B:C:D\]=\[\dfrac{1}{2}:2:0:2.5\]
Let us multiply through by the number two.
Ratio of \[A:B:C:D\]=\[1:4:0:5\]
On simplifying, we get the formula of the compound as $A{B_4}{C_0}{D_5}$.
Therefore, the formula of the compound is $A{B_4}{D_5}$.
Therefore, Option (D) is correct.

Note: We must remember that the cubic, tetragonal, orthorhombic, hexagonal, monoclinic, triclinic, and rhombohedral are seven simple crystal structures. They vary in their arrangement of crystallographic axes and angles. We can relate hexagonal closed packed systems to face-centered cubic cells. An example of a compound that has face-centered cubic lattice is sodium chloride. Examples of other compounds that have face-centered cubic structures are lithium fluoride, lithium chloride, Sodium fluoride, potassium fluoride, potassium chloride etc.