Problem 97
Question
The edge length of the unit cell of \(\mathrm{NaCl}\) crystal lattice is \(552 \mathrm{pm}\). If the ionic radius of sodium ion is \(95 \mathrm{pm}\). What is the ionic radius of chloride ion? (a) \(181 \mathrm{pm}\) (b) \(190 \mathrm{pm}\) (c) \(276 \mathrm{pm}\) (d) \(368 \mathrm{pm}\)
Step-by-Step Solution
Verified Answer
The ionic radius of the chloride ion is 181 pm.
1Step 1: Understanding the Problem
We're given the edge length of an NaCl unit cell and the ionic radius of the sodium ion. We need to find the ionic radius of the chloride ion. NaCl has a face-centered cubic (fcc) crystal lattice, where each edge of the cube contains one Na⁺ and one Cl⁻.
2Step 2: Finding the Relationship in FCC Lattice
In an fcc lattice like NaCl, the edge of the cube (aa) is equal to twice the sum of the radii of Na⁺ and Cl⁻: \(a = 2(r_{\mathrm{Na^+}} + r_{\mathrm{Cl^-}})\).
3Step 3: Plugging in the Values
Substitute the given values into the equation: the edge length \(a = 552 \, \mathrm{pm}\) and the radius of Na7179\(r_{\mathrm{Na^+}} = 95 \, \mathrm{pm}\)717. Therefore, the equation becomes: 71\(552 = 2(95 + r_{\mathrm{Cl^-}})\)717.
4Step 4: Solving for the Radius of Chloride Ion
Rearrange the equation: 717\(552 = 190 + 2r_{\mathrm{Cl^-}}\)717. Simplify to find the radius of the chloride ion: 717\(2r_{\mathrm{Cl^-}} = 552 - 190\)717, which gives: 717\(2r_{\mathrm{Cl^-}} = 362\)717. Finally, solving for 717\(r_{\mathrm{Cl^-}}\)717, we have: 717\(r_{\mathrm{Cl^-}} = \frac{362}{2} = 181\, \mathrm{pm}\)717.
Key Concepts
Face-Centered Cubic (FCC) LatticeNaCl Crystal StructureUnit Cell Edge Length
Face-Centered Cubic (FCC) Lattice
The face-centered cubic (fcc) lattice is a common arrangement of atoms within a crystalline structure. This structure is particularly significant in certain material sciences, including metal crystallography and ionic compounds like sodium chloride (NaCl). In an fcc lattice, atoms are arranged in such a way that each cube face has an atom at the center, in addition to the atoms at each corner of the cube.
This arrangement allows for efficient packing and is characterized by:
This arrangement allows for efficient packing and is characterized by:
- Atoms touching each other along the face diagonal, rather than along the cube edge.
- A coordination number of 12, meaning each atom touches 12 other atoms.
- A highly symmetric structure, which optimizes space filling.
NaCl Crystal Structure
Sodium chloride, commonly known as table salt, has a crystal structure that follows the face-centered cubic arrangement. However, its fcc lattice is distinct because it's not just composed of a single type of atom. NaCl consists of two different ions: sodium ions (1+Na^+1+) and chloride ions (1+Cl^-1+).
One unique aspect of the NaCl structure is that:
One unique aspect of the NaCl structure is that:
- Each sodium ion (1+Na^+1+) is surrounded by six chloride ions (1+Cl^-1+) and vice versa, leading to a 6:6 coordination number.
- Sodium and chloride ions alternate along the crystal axis, maintaining charge balance and lattice stability.
- The entire structure can be thought of as face-centered cubic in terms of arrangement of ions, but featuring ionic interactions rather than metallic or covalent bonds.
Unit Cell Edge Length
The edge length of a unit cell is a fundamental measurement in crystallography that describes the size of the smallest repeating unit of a crystal lattice. For NaCl, this is particularly significant because it directly relates to the size and arrangement of both sodium and chloride ions.
Important points about unit cell edge length include:
Important points about unit cell edge length include:
- It defines the distance between two identical points within the lattice, such as two chloride ions or two sodium ions.
- Determines the packing efficiency and density of the crystal.
Other exercises in this chapter
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