Test: Solid State - 1 - Chemistry MCQ

Diamond and Graphite, Two allotropes of carbon are covalent network solids which differ in the bonding geometry of the carbon atoms.

In diamond, the bonding occurs in the tetrahedral geometry, while in graphite the carbons bond with each other in the trigonal planar arrangement. This difference accounts for the drastically different appearance and properties of these two forms of carbon.

In a face-centered cubic system, the interionic distance is given by r_c + r_a = a/2

r_c = Radius of the cation
r_a = Radius of the Anion.

Dry ice, or solid carbon dioxide, is a perfect example of a molecular solid. The van der Waals forces holding the CO₂ molecules together are weak enough that dry ice sublimes, i.e. it passes directly from the solid to the gas phase, at -78^oC.

Every crystal system does not have all the four types of unit cells (simple, face centered, end centered, and body centered). But orthorhombic crystal class has all four types of unit cells- primitive, face centered, end centered and body centered.

Molecular crystals are substances that have relatively weak intermolecular binding, such as Dry Ice (solidified carbon dioxide), solid forms of the rare gases (e.g., argon, krypton, and xenon), and crystals of numerous organic compounds.

The density of the element will be same in both the cases as the two structures will have the same coordination number and hence the same packing fraction.

Molar volume =N4/3πr³ where N is Avogadro number ,

Now Molar volume of KCl is 37.46cm³ So N4/3πR³ = 37.46 cm³

On solving this gives R=2.458 x 10^-8 cm

And for NaCl by using the same formula it gives R=2.22 x 10^-8 cm

Now the ratio of edge length must be equal to ratio of radii as both belong to same crystal system so KCl/NaCl = 2.458/2.22 = 1.108

Given, cell is fcc, So Z =  4
Edge length, a = 404 pm = 4.04 x 10^-8 cm
Density of metal, d = 2.72 g cm^-3
N_A = 6.02 x 10²³ mol^-1
Molar mass ofg the metal, M =?
We know that

No of atoms in a CCP unit cell = similar to FCC = 4

Therefore no of B atoms present = 4

No of A atom = tetrahedral void and octohedral void(4+2) = 6

The formula of the compound  = A₃​B_2​

In ZnS (zinc blends), Zn²⁺ ions are present in alternate tetrahedral voids. 

Given:
- Na crystallizes in fcc lattice
- Edge length of the lattice = 500 pm
- Mass of the Na particle = 4.151 × 10⁻²³ g
To Find:
The density of the lattice
Formula:
Density (ρ) = Mass (m) / Volume (V)
Explanation:
1. The fcc (face-centered cubic) lattice consists of a unit cell with atoms at each of the eight corners and one atom in the center of each face.
2. The edge length of the fcc unit cell can be calculated using the formula: a = 4r / √2, where "a" is the edge length and "r" is the atomic radius.
3. In the case of Na, the atomic radius is not given directly, but we can calculate it using the edge length.
4. The atomic radius (r) can be calculated as: r = a √2 / 4.
5. Substituting the given edge length (500 pm) into the formula, we can calculate the atomic radius of Na.
6. Once we have the atomic radius, we can calculate the volume of the unit cell using the formula: V = a³.
7. The mass of the Na particle is given as 4.151 × 10⁻²³ g.
8. Now, we can calculate the density of the lattice using the formula: ρ = m / V.
Calculation:
1. Calculate the atomic radius of Na:
r = (500 pm) √2 / 4 = 353.55 pm = 3.5355 Å
2. Calculate the volume of the unit cell:
V = (500 pm)³ = 125 × 10⁶ pm³ = 125 × 10⁻²⁴ cm³
3. Calculate the density of the lattice:
ρ = (4.151 × 10⁻²³ g) / (125 × 10⁻²⁴ cm³) = 33.208 g/cm³ ≈ 33.2 g/cm³
Conclusion:
The density of the Na lattice is approximately 33.2 g/cm³.

The centre atom will be considered as whole.

The face centre atom will be considered as half..

The corner atoms will be considered 1/8.

A Schottky defect is a type of point defect in a crystal lattice named after Walter H. Schottky. In non-ionic crystals it means a lattice valency defect. In ionic crystals, this point defect forms when oppositely charged ions leave their lattice sites, creating vacancies.

 Na₂SO₄ and Na₂SeO₄ are different substances possessing similar formula, they are called isomorphous systems.

In ionic compounds as internuclear distance incrreases, melting point decreases so as we move down the group, melting point decreases. So, as r_c + r_a  is greater for RbBr than for NaF so NaF have higher melting point.