The Solid State

Step 1: Identify metals in simple cubic structures. Of the given options, sodium (Na) crystallizes in a simple cubic structure. Other metals such as Cu, Ag, and Po typically crystallize in more complex structures like body-centered or face-centered cubic.
Step 2: Conclusion. Thus, the metal that crystallizes in a simple cubic system is sodium (Na).

Step 1: Face-centered cubic (FCC) unit cell. In a face-centered cubic unit cell, there is 1 lattice point at each corner of the cube and 1/2 of a lattice point at each face center. Thus, the total number of lattice points in an FCC unit cell is 4.
Step 2: Conclusion. Thus, the correct answer is 4 lattice points.

Step 1: Schottky defect. Schottky defect occurs when there is a pair of vacancies, one for a cation and one for an anion, in a crystal lattice. This defect maintains electrical neutrality.
Step 2: Conclusion. Thus, the Schottky defect is due to a pair of cation and anion vacancies.

Step 1: Identifying amorphous substances. Amorphous substances do not have a regular, repeating structure. Polystyrene is an amorphous polymer, while silica, table salt, and diamond are crystalline.
Step 2: Conclusion. Thus, polystyrene is amorphous.

Step 1: Define Schottky defect.
Schottky defect is a type of point defect in ionic crystals where equal number of cations and anions are missing from their lattice sites.
Step 2: Condition for Schottky defect.
This defect is mainly found when both ions are of comparable sizes, so that removing them does not distort the lattice much.
Examples: NaCl, KCl, CsCl.
Step 3: Conclusion.
Thus Schottky defect occurs mainly when positive and negative ions are of same size.
Final Answer:

Step 1: Use density formula for unit cell.

Where:
= number of atoms per unit cell



Step 2: Substitute values.

Step 3: Compute .

Step 4: Solve for .


Step 5: Identify lattice type.

Final Answer:

Step 1: Definition of Frenkel defect.
Frenkel defect occurs when an ion (usually cation) leaves its lattice site and occupies an interstitial site.
This creates:
- a vacancy defect
- an interstitial defect
Step 2: Conditions for Frenkel defect.
It is common in ionic solids where:
- cation is small
- anion is large
So small cation can move into interstitial positions.
Step 3: Examples.
AgBr, AgI and ZnS show Frenkel defect because their cations are relatively small.
Final Answer:

Step 1: Meaning of cleavage.
Cleavage means a crystal breaks along certain definite directions producing smooth surfaces.
Step 2: Reason behind cleavage.
In crystals, atoms/ions are arranged in regular layers (planes).
Some planes have weaker bonding compared to others.
So when force is applied, crystal splits along these planes.
Step 3: Hence correct option.
Cleavage is due to planar arrangement of particles in crystal lattice.
Final Answer:

Step 1: Use radius ratio rule.
Step 2: Compare with standard limits.
Radius ratio ranges:
- CN = 4 (tetrahedral):
- CN = 6 (octahedral):
- CN = 8 (cubic):
Step 3: Locate 0.525.
Since
coordination number is 6.
Final Answer:

Step 1: Understand the unit cell dimensions.
In the tetragonal crystal system, two of the axes have equal lengths, and the angles between all axes are 90°.
Step 2: Conclusion.
Thus, the correct unit cell dimensions are , and the angles .
Final Answer:

Step 1: Understand the characteristics of crystalline solids.
Crystalline solids have a definite geometric structure and a sharp melting point. If the solid is non-crystalline, it might not have a clear melting point.
Step 2: Conclusion.
Thus, the crystalline solid has a definite melting point.
Final Answer:

For a NaCl structure, the ratio of the radii of the cation and anion is approximately 0.5. Thus, the radius of the anion can be calculated as:

Step 2: Conclusion.
The correct radius of the anion is , corresponding to option (c).

To find the highest cation to anion size ratio, we compare the ionic radii of the cations and anions in each compound. CsF has the highest ratio due to the relatively larger size of the cesium ion compared to the fluoride ion.

Step 2: Conclusion.
The compound with the highest cation to anion size ratio is CsF, corresponding to option (b).

Step 1: Volume and density relationship.
The volume occupied by argon can be calculated from the density, and the volume of a single argon atom can be calculated assuming it is a sphere. By comparing these volumes, we find that 38% of the solid argon is empty space.

Step 2: Conclusion.
The percentage of empty space in solid argon is 38%, which corresponds to option (2).

The volume of the unit cell can be calculated using the formula , where is the edge length of the unit cell. After converting the edge length to cm, the volume can be calculated.

Volume of unit cell (V)

In a ccp lattice, the number of atoms in each unit cell is determined by the number of atoms at edge and body centres. The copper, silver, and gold atoms fit into this arrangement in the ratio .

Step 1: Understanding the Ionic Arrangement.
In the unit cell, A ions are at the corners, and B ions are at the face centers. The number of A ions per unit cell is 8, and the number of B ions per unit cell is 6 (since each face center has half an ion). Therefore, the empirical formula is .
Step 2: Conclusion.
The correct answer is (C), .

Step 1: Analyze the structure.
In a cubic close packing (ccp) structure, each unit cell contains one atom of and one atom of . Since the number of atoms of and are in a 1:1 ratio, the empirical formula is .
Step 2: Conclusion.
Thus, the empirical formula is .
Final Answer:

Step 1: Understand Schottky defect.
Schottky defects occur when equal numbers of cations and anions are missing from the lattice, resulting in vacancies that maintain the electrical neutrality of the crystal.
Step 2: Conclusion.
Thus, Schottky defects are characterized by an equal number of cations and anions being absent from their normal lattice sites.
Final Answer:

Step 1: Use the relationship between density and atomic packing.
The density of a substance is inversely proportional to the volume of the unit cell. Since the number of atoms and the size of the unit cell remain constant, we calculate the ratio of the densities at the transition temperature.
Step 2: Conclusion.
Thus, the ratio of densities is .
Final Answer:

Step 1: Understand the relationship between the edge length and radius.
For simple cubic, body-centered cubic, and face-centered cubic, the ratio of radius to edge length differs based on the arrangement of the atoms in the unit cell.
Step 2: Conclusion.
For these arrangements, the radius to edge length ratio is for the face-centered cubic structure.
Final Answer:

Step 1: In a ccp lattice, each unit cell contains 4 atoms of element Y.
Step 2: The number of tetrahedral voids in a ccp structure is also 4, and of these voids are occupied by element X.
Step 3: Therefore, the number of atoms of element X is , and the number of atoms of element Y is 4. Thus, the formula is .

Final Answer:

Step 1: In the body-centered cubic (bcc) structure, there are 2 atoms per unit cell.
Step 2: The total number of atoms in unit cells is atoms.

Final Answer:

Bragg's law describes the condition for constructive interference of X-rays scattered by a crystal lattice. The correct equation is: where is the order of reflection, is the wavelength of the incident wave, is the distance between the crystal planes, and is the angle of incidence.
Final Answer:

Water molecules exhibit a bent shape due to the presence of two lone pairs on the oxygen atom. This enables the molecule to form hydrogen bonds with other water molecules. Each water molecule can form 2 hydrogen bonds: one with the hydrogen atom of another molecule and the other with the oxygen atom of a different molecule. Final Answer: The number of hydrogen bonds formed by a water molecule at normal conditions is 2.

Step 1: Write the Given Reactions and Their Enthalpies 1. Combustion of Benzene: 2. Combustion of Acetylene:
Step 2: Use Hess’s Law to Find the Enthalpy Change for the Desired Reaction
The desired reaction is:
Using Hess’s Law, the enthalpy change for the desired reaction ( ) is:
Step 3: Determine the Correct Option
The change in enthalpy for the reaction is , which matches option (C). Final Answer: The enthalpy change for the reaction is .