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Copper crystallizes in a face-centered cubic lattice with a unit cell length of $361 pm$ . What is the radius of copper atom in $pm$ ?

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Important Questions on Solid State

EASY

Chemistry>Physical Chemistry>Solid State>Calculations for Standard Cubical Unit Lattice

A compound of formula

A_{2} B_{3}

has the HCP lattice. Which atom forms the HCP lattice and what fraction of the tetrahedral voids are occupied by the other atoms?

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Chemistry>Physical Chemistry>Solid State>Calculations for Standard Cubical Unit Lattice

If 'a' stands for the edge length of the cubic system: simple cubic, body-centred cubic and face-centred cubic, the ratio of radii of the spheres in these systems will be respectively

HARD

Chemistry>Physical Chemistry>Solid State>Calculations for Standard Cubical Unit Lattice

Consider the bcc unit cells of the solids $1$ and $2$ with the position of atoms as shown below. The radius of atom $B$ is twice that of atom $A .$ The unit cell edge length is $50 %$ more in solid $2$ than in $1$ . What is the approximate packing efficiency in solid $2$ ?

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Chemistry>Physical Chemistry>Solid State>Calculations for Standard Cubical Unit Lattice

a

the length of the side of a cube, the distance between the body centered atom one corner atom in the cube will be

MEDIUM

Chemistry>Physical Chemistry>Solid State>Calculations for Standard Cubical Unit Lattice

A given metal crystallises out with a cubic structure having edge length of

361 pm

. If there are four metal atoms in one unit cell, what is the radius of one atom?

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Chemistry>Physical Chemistry>Solid State>Calculations for Standard Cubical Unit Lattice

An element has a face-centered cubic

(fcc)

structure with a cell edge of

a

. The distance between the centres of two nearest tetrahedral voids in the lattice is

MEDIUM

Chemistry>Physical Chemistry>Solid State>Calculations for Standard Cubical Unit Lattice

The total number of octahedral void (s) per atom present in a cubic close packed structure is :

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Chemistry>Physical Chemistry>Solid State>Calculations for Standard Cubical Unit Lattice

A crystal is made up of metal ions

' M_{1}'

and

' M_{2}'

and oxide ions. Oxide ions form a ccp lattice structure. The cation

' M_{1}'

occupies

50 %

of octahedral voids and the cation

' M_{2}'

occupies

12.5 %

of tetrahedral voids of oxide lattice. The oxidation numbers of

' M_{1}'

and

' M_{2}'

are respectively :

EASY

Chemistry>Physical Chemistry>Solid State>Calculations for Standard Cubical Unit Lattice

The packing efficiency of the face centered cubic (FCC), body centered cubic (BCC) and simple/primitive cubic (PC) lattices follows the order

EASY

Chemistry>Physical Chemistry>Solid State>Calculations for Standard Cubical Unit Lattice

A metal crystallizes in a face centred cubic structure. If the edge length of its unit cell is

' a'

, the closest approach between two atoms in the metallic crystal will be

MEDIUM

Chemistry>Physical Chemistry>Solid State>Calculations for Standard Cubical Unit Lattice

The percentage of unoccupied volume in simple cubic cell is

EASY

Chemistry>Physical Chemistry>Solid State>Calculations for Standard Cubical Unit Lattice

The vacant space in bcc lattice unit cell is:

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Chemistry>Physical Chemistry>Solid State>Calculations for Standard Cubical Unit Lattice

The correct order of the packing efficiency in different types of unit cells is

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Chemistry>Physical Chemistry>Solid State>Calculations for Standard Cubical Unit Lattice

A compound is formed by cation

C

and anion

A

. The anions form hexagonal close packed (hcp) lattice and the cations occupy

75 %

of octahedral voids. The formula of the compound is:

HARD

Chemistry>Physical Chemistry>Solid State>Calculations for Standard Cubical Unit Lattice

An element crystallises in a face–centred cubic (fcc) unit cell with cell edge a. The distance between the centres of two nearest octahedral voids in the crystal lattice is:

EASY

Chemistry>Physical Chemistry>Solid State>Calculations for Standard Cubical Unit Lattice

In a compound, atoms of elements

Y

form ccp lattice and those of element

X

occupy

2 / 3 rd

of tetrahedral voids. The formula of the compound can be

MEDIUM

Chemistry>Physical Chemistry>Solid State>Calculations for Standard Cubical Unit Lattice

An element has a body centered cubic (bcc) structure with a cell edge of

288 pm

. The atomic radius is

EASY

Chemistry>Physical Chemistry>Solid State>Calculations for Standard Cubical Unit Lattice

An ionic compound is formed between a metal M and non-metal

Y.

M

occupies half the octahedral voids in the cubic close - packed arrangement formed by

Y,

the chemical formula of the ionic compound is

EASY

Chemistry>Physical Chemistry>Solid State>Calculations for Standard Cubical Unit Lattice

A two-dimensional solid pattern formed by two different atoms

X

and

Y

is shown below. The black and white squares represent atoms

X

and

Y

, respectively. The simplest formula for the compound based on the unit cell from the pattern is

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MEDIUM

Chemistry>Physical Chemistry>Solid State>Calculations for Standard Cubical Unit Lattice

Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of

4.29 Å

. The radius of sodium atom is approximately:

Copper crystallizes in a face-centered cubic lattice with a unit cell length of 361 pm. What is the radius of copper atom in pm?

Important Questions on Solid State

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Copper crystallizes in a face-centered cubic lattice with a unit cell length of $361 pm$ . What is the radius of copper atom in $pm$ ?