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Page 1 Numerical unit cell edge length of . The radius of sodium atom is __________ (Nearest integer) JEE Main 2023 (Online) 13th April Evening Shift Q.1. Sodium metal crystallizes in a body centred cubic lattice with Q.2. An atomic substance A of molar mass 12 g mol -1 has a cubic crystal structure with edge length of 300 pm. The no. of atoms present in one unit cell of A is ____________. (Nearest integer) JEE Main 2023 (Online) 6th April Evening Shift Cubic, tetragonal, orthorhombic, hexagonal, rhombohedral, monoclinic, triclinic Q.4. A metal M crystallizes into two lattices :- face centred cubic (fcc) and body centred cubic (bcc) with unit cell edge length of 2.0 and 2.5A o respectively. The ratio of densities of lattices fcc to bcc for the metal M is ____________. (Nearest integer) JEE Main 2023 (Online) 11th April Morning Shift Given the density of A is Q.3. Number of crystal systems from the following where body centred unit cell can be found, is ____________. Page 2 Numerical unit cell edge length of . The radius of sodium atom is __________ (Nearest integer) JEE Main 2023 (Online) 13th April Evening Shift Q.1. Sodium metal crystallizes in a body centred cubic lattice with Q.2. An atomic substance A of molar mass 12 g mol -1 has a cubic crystal structure with edge length of 300 pm. The no. of atoms present in one unit cell of A is ____________. (Nearest integer) JEE Main 2023 (Online) 6th April Evening Shift Cubic, tetragonal, orthorhombic, hexagonal, rhombohedral, monoclinic, triclinic Q.4. A metal M crystallizes into two lattices :- face centred cubic (fcc) and body centred cubic (bcc) with unit cell edge length of 2.0 and 2.5A o respectively. The ratio of densities of lattices fcc to bcc for the metal M is ____________. (Nearest integer) JEE Main 2023 (Online) 11th April Morning Shift Given the density of A is Q.3. Number of crystal systems from the following where body centred unit cell can be found, is ____________. cell is __________. (Nearest integer) Given: Molar mass of Fe and O is 56 and 16 g mol -1 respectively. JEE Main 2023 (Online) 30th January Evening Shift Q.6. A metal M forms hexagonal close-packed structure. The total number of voids in 0.02 mol of it is _________ × 1021 (Nearest integer). JEE Main 2023 (Online) 29th January Evening Shift Answer key & Explanation 1. Ans. Correct answer is 17 Explanation In a body-centered cubic (BCC) lattice, the relationship between the edge length (a) and the atomic radius (r) is given by: Given the unit cell edge length (a) of sodium metal as 4 Å: JE .5. Iron oxide FeO, crystallises in a cubic lattice with a unit cell edge length of 5.0 Å. If density of the FeO in the crystal is 4.0 g cm-3, then the number of FeO units present per unit E Main 2023 (Online) 1st February Evening Shift Q Page 3 Numerical unit cell edge length of . The radius of sodium atom is __________ (Nearest integer) JEE Main 2023 (Online) 13th April Evening Shift Q.1. Sodium metal crystallizes in a body centred cubic lattice with Q.2. An atomic substance A of molar mass 12 g mol -1 has a cubic crystal structure with edge length of 300 pm. The no. of atoms present in one unit cell of A is ____________. (Nearest integer) JEE Main 2023 (Online) 6th April Evening Shift Cubic, tetragonal, orthorhombic, hexagonal, rhombohedral, monoclinic, triclinic Q.4. A metal M crystallizes into two lattices :- face centred cubic (fcc) and body centred cubic (bcc) with unit cell edge length of 2.0 and 2.5A o respectively. The ratio of densities of lattices fcc to bcc for the metal M is ____________. (Nearest integer) JEE Main 2023 (Online) 11th April Morning Shift Given the density of A is Q.3. Number of crystal systems from the following where body centred unit cell can be found, is ____________. cell is __________. (Nearest integer) Given: Molar mass of Fe and O is 56 and 16 g mol -1 respectively. JEE Main 2023 (Online) 30th January Evening Shift Q.6. A metal M forms hexagonal close-packed structure. The total number of voids in 0.02 mol of it is _________ × 1021 (Nearest integer). JEE Main 2023 (Online) 29th January Evening Shift Answer key & Explanation 1. Ans. Correct answer is 17 Explanation In a body-centered cubic (BCC) lattice, the relationship between the edge length (a) and the atomic radius (r) is given by: Given the unit cell edge length (a) of sodium metal as 4 Å: JE .5. Iron oxide FeO, crystallises in a cubic lattice with a unit cell edge length of 5.0 Å. If density of the FeO in the crystal is 4.0 g cm-3, then the number of FeO units present per unit E Main 2023 (Online) 1st February Evening Shift Q We can now solve for the radius (r) of the sodium atom: Now, we can approximate the numerical value: 2. Ans. Correct answer is 4 Explanation Given: Page 4 Numerical unit cell edge length of . The radius of sodium atom is __________ (Nearest integer) JEE Main 2023 (Online) 13th April Evening Shift Q.1. Sodium metal crystallizes in a body centred cubic lattice with Q.2. An atomic substance A of molar mass 12 g mol -1 has a cubic crystal structure with edge length of 300 pm. The no. of atoms present in one unit cell of A is ____________. (Nearest integer) JEE Main 2023 (Online) 6th April Evening Shift Cubic, tetragonal, orthorhombic, hexagonal, rhombohedral, monoclinic, triclinic Q.4. A metal M crystallizes into two lattices :- face centred cubic (fcc) and body centred cubic (bcc) with unit cell edge length of 2.0 and 2.5A o respectively. The ratio of densities of lattices fcc to bcc for the metal M is ____________. (Nearest integer) JEE Main 2023 (Online) 11th April Morning Shift Given the density of A is Q.3. Number of crystal systems from the following where body centred unit cell can be found, is ____________. cell is __________. (Nearest integer) Given: Molar mass of Fe and O is 56 and 16 g mol -1 respectively. JEE Main 2023 (Online) 30th January Evening Shift Q.6. A metal M forms hexagonal close-packed structure. The total number of voids in 0.02 mol of it is _________ × 1021 (Nearest integer). JEE Main 2023 (Online) 29th January Evening Shift Answer key & Explanation 1. Ans. Correct answer is 17 Explanation In a body-centered cubic (BCC) lattice, the relationship between the edge length (a) and the atomic radius (r) is given by: Given the unit cell edge length (a) of sodium metal as 4 Å: JE .5. Iron oxide FeO, crystallises in a cubic lattice with a unit cell edge length of 5.0 Å. If density of the FeO in the crystal is 4.0 g cm-3, then the number of FeO units present per unit E Main 2023 (Online) 1st February Evening Shift Q We can now solve for the radius (r) of the sodium atom: Now, we can approximate the numerical value: 2. Ans. Correct answer is 4 Explanation Given: Avogadro's number First, calculate the volume of the unit cell Next, calculate the mass of the unit cell using the given density. Density is mass over volume, so mass Then, calculate the number of moles in one unit cell. The molar mass is mass over number of moles, so Finally, calculate the number of atoms in one unit cell. The Avogadro constant is the number of atoms per mole, Rounding to the nearest integer, we get 4 atoms. So, there are 4 atoms present in one unit cell of substance A 3. Ans. Correct answer is 3 4. Ans. Correct answer is 4 Explanation Page 5 Numerical unit cell edge length of . The radius of sodium atom is __________ (Nearest integer) JEE Main 2023 (Online) 13th April Evening Shift Q.1. Sodium metal crystallizes in a body centred cubic lattice with Q.2. An atomic substance A of molar mass 12 g mol -1 has a cubic crystal structure with edge length of 300 pm. The no. of atoms present in one unit cell of A is ____________. (Nearest integer) JEE Main 2023 (Online) 6th April Evening Shift Cubic, tetragonal, orthorhombic, hexagonal, rhombohedral, monoclinic, triclinic Q.4. A metal M crystallizes into two lattices :- face centred cubic (fcc) and body centred cubic (bcc) with unit cell edge length of 2.0 and 2.5A o respectively. The ratio of densities of lattices fcc to bcc for the metal M is ____________. (Nearest integer) JEE Main 2023 (Online) 11th April Morning Shift Given the density of A is Q.3. Number of crystal systems from the following where body centred unit cell can be found, is ____________. cell is __________. (Nearest integer) Given: Molar mass of Fe and O is 56 and 16 g mol -1 respectively. JEE Main 2023 (Online) 30th January Evening Shift Q.6. A metal M forms hexagonal close-packed structure. The total number of voids in 0.02 mol of it is _________ × 1021 (Nearest integer). JEE Main 2023 (Online) 29th January Evening Shift Answer key & Explanation 1. Ans. Correct answer is 17 Explanation In a body-centered cubic (BCC) lattice, the relationship between the edge length (a) and the atomic radius (r) is given by: Given the unit cell edge length (a) of sodium metal as 4 Å: JE .5. Iron oxide FeO, crystallises in a cubic lattice with a unit cell edge length of 5.0 Å. If density of the FeO in the crystal is 4.0 g cm-3, then the number of FeO units present per unit E Main 2023 (Online) 1st February Evening Shift Q We can now solve for the radius (r) of the sodium atom: Now, we can approximate the numerical value: 2. Ans. Correct answer is 4 Explanation Given: Avogadro's number First, calculate the volume of the unit cell Next, calculate the mass of the unit cell using the given density. Density is mass over volume, so mass Then, calculate the number of moles in one unit cell. The molar mass is mass over number of moles, so Finally, calculate the number of atoms in one unit cell. The Avogadro constant is the number of atoms per mole, Rounding to the nearest integer, we get 4 atoms. So, there are 4 atoms present in one unit cell of substance A 3. Ans. Correct answer is 3 4. Ans. Correct answer is 4 Explanation 5. Ans. Correct answer is 4 Explanation 6. Ans. Correct answer is 36 ExplanationRead More
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FAQs on JEE Mains Previous Year Questions (2021-2024): Solid State - Chemistry for JEE Main & Advanced
| 1. What is the basic concept of solid state in JEE? |  |
Ans. In JEE, the solid state refers to the study of crystalline solids, their structures, properties, and behavior. It involves understanding the arrangement of atoms, ions, or molecules in a three-dimensional lattice structure.
| 2. How are unit cells important in the study of solid state for JEE? | |
Ans. Unit cells are the building blocks of crystalline solids and play a crucial role in determining the overall properties of the solid. They help in understanding the symmetry, density, and other characteristics of the solid state.
| 3. What are the different types of defects in solid state as per the JEE syllabus? | |
Ans. In the context of JEE, the different types of defects in the solid state include point defects (vacancies, interstitials), line defects (dislocations), and surface defects (grain boundaries). These defects can significantly impact the properties of solids.
| 4. How does band theory of solids explain the electrical conductivity of materials in JEE? | |
Ans. The band theory of solids in JEE explains the electrical conductivity of materials by describing the energy bands formed by the overlapping atomic orbitals. Conductors have partially filled bands, while insulators have fully occupied bands with a large energy gap.
| 5. Can you explain the concept of superconductivity in the context of solid state for JEE? | |
Ans. Superconductivity is a phenomenon where certain materials exhibit zero electrical resistance when cooled below a critical temperature. In the solid state, this behavior is attributed to the formation of Cooper pairs of electrons that move without resistance through the lattice structure.
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