The Solid State

(a) The unit cell of CsCl has bcc arrangement of ions in which each ion has eight oppositely charged ions around it in the nearest neighbours as
shown below :
(b) In bcc, coordination number of atom is 8
(c) In an unit cell, a comer is shared in eight unit cells and a face centre is shared between two adjacent unit cells.
(d) In NaCI unit cell; Hence, a, c, d are correct.

Atom A occupies all corner and face center of unit cell and atom B in all octahedral void. As all atoms from face center from one axis is removed, hence number of atom A in unit cell remains 3 and B as 4.


after removing atoms



Formula of compound is

In ZnS, (sulphide ions) are present at fee positions giving four sulphide ions per unit cell. To comply with 1 : 1 stoichiometry, four ion must be present in four alternate tetrahedral voids out of eight tetrahedral voids present. In NaCI, ions are present in octahedral voids
while in ions are present in all its tetrahedral voids giving the desired 2:1 stoichiometry. In ions occupies fee positions and all the tetrahedral voids are occupied by fluoride ion

Contribution of circle from comer of square
Effective number of circle per square

Area occupied by circle
Also, diagonal of square where L = side of square.
Packing fraction

(A) The empirical formula of the compound is
(C) The ratio of bond length to the cubic unit cell edge length is

Step 1: Understanding the Structure of Meso-Butane-2,3-Diol
Meso-butane-2,3-diol is a compound with two hydroxyl groups (-OH) located on the second and third carbon atoms of a butane chain. It also contains two methyl groups (-Me) attached to the first and fourth carbon atoms.
Step 2: Analyzing the Conformation
The molecule can adopt different conformations based on the relative positioning of the substituents (-OH and -Me). The goal is to find the conformation with the lowest steric hindrance and torsional strain.
Step 3: Staggered Conformation
In a staggered conformation, the atoms or groups attached to the carbon atoms are positioned as far apart from each other as possible, minimizing repulsion. This is generally the most stable conformation for alkanes and related compounds.
Step 4: Anti Conformation of -OH and -Me Groups
In the most stable conformation, the bulky -OH and -Me groups should be positioned in an **anti** arrangement (opposite to each other), so they do not experience direct steric clashes. The -OH groups, as well as the -Me groups, are positioned as far apart as possible.
Step 5: Conclusion
The conformation that corresponds to the most stable conformation of meso-butane-2,3-diol is the one where both the -OH groups are anti to each other, and the -Me groups are also anti to each other, minimizing steric hindrance and maximizing stability.

Step 1: Understanding the Close-Packed Structure
The structure is a close-packed arrangement, which suggests we are dealing with a face-centered cubic (FCC) structure or cubic close packing (CCP) structure.
In FCC, ions are packed in such a way that the packing efficiency is maximum.
Step 2: Packing Efficiency in FCC Structure
In a face-centered cubic structure, the packing efficiency is approximately 63%, meaning 63% of the total volume is occupied by the ions.
Step 3: Calculating the Packing Fraction
The packing fraction is defined as the ratio of the packing efficiency to 100:

 Packing fraction = Packing efficiency / 100 

Substituting the packing efficiency:

 Packing fraction = 63 / 100 = 0.63 

However, in this case, the packing fraction is 0.63, which suggests that the cation and anion sizes differ, and this lower packing fraction reflects this discrepancy.
Step 4: Considering the Ionic Size Ratio
When the cation is significantly smaller than the anion, the packing efficiency can decrease from the typical FCC value of 74%. In this case, the packing fraction is given as 0.63, which is consistent with this situation.
Final Result
The packing fraction for the given close-packed structure is approximately 0.63.

The correct option is (B): 35

The correct options are (A),(B) and (D)
Given, and




Packing efficiency FCC>BCC>SC
Packing efficiency unit cell x>y>z
In the FCC unit cell,




and


density of x =
density of y =
density of x > density of y

Step 1: Understand the lattice type and parameters.
A cubic close packed (ccp) lattice is the same as a face centered cubic (fcc) lattice. In an fcc lattice, atoms are located at each corner and the centers of all the faces of the cube. The relationship between the edge length and the atomic radius is: Step 2: Calculate atomic radius . Given , Step 3: Number of atoms per unit cell in fcc (ccp) lattice is 4. Step 4: Calculate the volume of the unit cell. Step 5: Calculate mass of atoms in one unit cell. Atomic mass Mass of one atom in grams: Mass of 4 atoms in unit cell: Step 6: Calculate density . % Final answer Density

Step 1: Understand the lattice type and parameters.
A cubic close packed (ccp) lattice is the same as a face centered cubic (fcc) lattice. In an fcc lattice, atoms are located at each corner and the centers of all the faces of the cube. The relationship between the edge length and the atomic radius is: Step 2: Calculate atomic radius . Given , Step 3: Number of atoms per unit cell in fcc (ccp) lattice is 4. Step 4: Calculate the volume of the unit cell. Step 5: Calculate mass of atoms in one unit cell. Atomic mass Mass of one atom in grams: Mass of 4 atoms in unit cell: Step 6: Calculate density . % Final answer Density

We use the formula for density of a unit cell: Where: - = number of atoms per unit cell in CCP = 4
- = molar mass = 105.6 g/mol
- = Avogadro's number =
- = edge length = 400 pm = Now substitute: Wait! That’s too high. Let's check unit conversion: Correcting volume: Therefore, density =