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In a solid \(\mathrm{AB}\) having \(\mathrm{NaCl}\) structure, A atoms occupy the corners of the cubic unit cell. If all the facecentred atoms along one of the axes are removed, then the resultant stoichiometry of the solid is (a) \(\mathrm{AB}\) (b) \(\mathrm{A}_{2} \mathrm{~B}_{2}\) (c) \(\mathrm{A}_{2} \mathrm{~B}_{3}\) (d) \(\mathrm{A}_{3} \mathrm{~B}_{4}^{2}\)

Short Answer

Expert verified
The stoichiometry of the resultant solid is (c) \( \mathrm{A}_{2} \mathrm{~B}_{3} \).

Step by step solution

01

Understand the NaCl Structure

In a typical NaCl structure, A atoms (e.g., Na) are located at the corners and B atoms (e.g., Cl) are at the face centers. There are 8 corner atoms and 6 face-centered atoms.
02

Calculate Contribution of Corner Atoms

In a cubic cell, each corner atom contributes \( \frac{1}{8} \) to the unit cell. Therefore, the contribution of all corner atoms is \( 8 \times \frac{1}{8} = 1 \). Hence, there is 1 full A atom in the unit cell.
03

Calculate Contribution of Face-centered Atoms

Each face-centered atom contributes \( \frac{1}{2} \) to the unit cell. Typically, all 6 B atoms contribute \( 6 \times \frac{1}{2} = 3 \) full B atoms in the unit cell.
04

Remove Face-centered Atoms Along One Axis

If face-centered atoms along one axis (say the z-axis) are removed, this results in the removal of 2 B atoms, one from each face. So, now only 4 face-centered atoms remain.
05

Calculate New Contribution of Face-centered Atoms

The remaining 4 face-centered B atoms contribute \( 4 \times \frac{1}{2} = 2 \) full B atoms to the unit cell.
06

Determine New Stoichiometry

With 1 full A atom from the corners and 2 full B atoms from the remaining face centers, the stoichiometry becomes \( \text{A}_1 \text{B}_2 \), or simply \( \text{A}_2 \text{B}_3 \) when adjusted for smallest whole numbers.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

NaCl Structure
The NaCl structure is a classic example of ionic crystals, where sodium (Na) and chloride (Cl) ions form a repeating pattern. This pattern mimics the arrangement found in an actual NaCl cube, reflecting its alternating ion placement.
In an NaCl structure, each Na ion is surrounded by six Cl ions, and each Cl ion is similarly surrounded by six Na ions. This arrangement forms a face-centered cubic lattice.
It is important to note that in this structure:
  • "A" atoms (like Na) occupy the corners of the cube.
  • "B" atoms (like Cl) are located at the centers of each face of the cube.
Understanding this pattern is essential when studying ionic solids and their properties.
Unit Cell
To fully grasp the structure of a solid, we focus on its unit cell. A unit cell is the smallest repeating unit in a crystal lattice. It contains the entire example of the lattice's geometrical and symmetrical properties.
In the NaCl structure, the unit cell is cubic, and it embodies the complete symmetry and formula of the entire solid. The unit cell helps in determining the formula unit and stoichiometry.
In the case of NaCl:
  • Each corner of the unit cell contains an "A" atom, contributing \( \frac{1}{8} \) of an atom to the unit cell.
  • Face-centered "B" atoms contribute \( \frac{1}{2} \) to the unit cell.
  • Overall, in the whole unit cell, the combination of these contributions leads to one full "A" and typically three full "B" atoms.
Manipulations in the unit cell, like removing atoms along an axis, affect the overall stoichiometry.
Stoichiometry
Stoichiometry in chemistry refers to the relationship between the quantities of reactants and products in chemical reactions. In solid-state chemistry, especially with the NaCl structure, it deals with the ratio of different atoms in a unit cell.
The traditional stoichiometry for NaCl is a ratio of 1:1, meaning equal numbers of "A" and "B" atoms form the stable crystalline structure. However, altering the unit cell, such as removing specific face-centered "B" atoms, can impact this balance.
For example, if two face-centered atoms (one from each face along an axis) are removed, it disrupts the 3 full "B" atoms typical in a complete NaCl unit cell. Now, only two full "B" atoms remain, while the "A" atom count stays the same.
Ultimately:
  • The resulting stoichiometry turns to \( \mathrm{A}_1 \mathrm{B}_2 \), often written as \( \mathrm{A}_2 \mathrm{B}_3 \) for clarity and balance.
  • This subtle yet significant change demonstrates how stoichiometry reflects variations in unit cell composition and contributes to the physical and chemical properties of the solid.

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Most popular questions from this chapter

Ice crystallizes in hexagonal lattice. At a given temperature, the density of ice is \(0.92 \mathrm{gcm}^{-3}\). The volume of unit cell is \(1.3 \times 10^{-22} \mathrm{~cm}^{3} .\) The number of \(\mathrm{H}_{2} \mathrm{O}\) molecules per unit cell is

Experimentally it was found that a metal oxide has. formula \(\mathrm{M}_{0.98}\) O. Metal \(\mathrm{M}\), is present as \(\mathrm{M}^{2+}\) and \(\mathrm{M}^{3+}\) in its oxide. Fraction of the metal which exists as \(\mathrm{M}^{3+}\) would be (a) \(6.05 \%\) (b) \(5.08 \%\) (c) \(7.01 \%\) (d) \(4.08 \%\)

\(\mathrm{M}_{2} \mathrm{X}\) have a structure closely related to that of fluorite. It is actually reverse of fluorite structure hence called antifluorite structure. In such a structure (a) \(\mathrm{F}^{-}\)ions occupy all the 8 octahedral voids (b) each \(\mathrm{F}^{-}\)is surrounded by \(4 \mathrm{Ca}^{2+}\) in tetrahedral arrangement (c) larger cations occupy the position of \(\mathrm{F}^{-}\)ions and smaller anions that of \(\mathrm{Ca}^{2+}\) ions. (d) smaller cations occupy the position of fluoride ions and larger anions that of \(\mathrm{Ca}^{2+}\) ions

What is the contribution of the atom present at the edge centre to the unit cell? (a) \(\frac{1}{8}\) (b) \(\frac{1}{2}\) (c) \(\frac{1}{4}\) (d) \(\frac{1}{3}\)

Which of the following is the correct relation between enthalpy of vaporization (Joule/mole) and boiling point of a liquid? (a) \(\Delta \mathrm{H}=1.5 \mathrm{P}_{\mathrm{b}}\) (b) \(\Delta \mathrm{H}=77 \mathrm{~T}\) (c) \(\Delta \mathrm{H}=88 \mathrm{~T}_{\mathrm{b}}^{\mathrm{b}}\) (d) \(\Delta \mathrm{H}=0.5 \mathrm{~T}_{\mathrm{b}}\)

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