The ratio of the radius of cations (r⁺) to the radius of the anion (r^–) is known as the radius ratio of the ionic solid.

The significance of radius ratio:

• It is useful in predicting the structure of ionic solids.
• The structure of an ionic compound depends upon stoichiometry and the size of ions.
• In crystals, cations tend to get surrounded by the largest possible number of anions around it.
• Greater the radius ratio, greater is the coordination number of cations and anions.
• If cations are extremely small and anions are extremely large, then the radius ratio is very small. In such case packing of anions is very close to each other and due to repulsion between anions, the system becomes unstable. Hence the structure changes to some suitable stable arrangement.
• The radius ratio at which anions just touch each other, as well as central cation, is called the critical radius ratio.

Effect of Radius Ratio on Coordination Number:

A cation would fit exactly into the octahedral void and would have a coordination number of six if the radius ratio were exactly 0.414. Similarly, a cation would fit exactly into the tetrahedral void and would have a coordination number of four, if the radius ratio were exactly 0.225.

Let us consider a case in which a cation is fitting exactly into the octahedral void of close pack anions and have the coordination number of six, in this case, the radius ratio is exactly 0.414.

When the radius ratio is greater than this, then the anions move apart to accommodate larger cation. This situation is relatively unstable. If the radius ratio is further increased the anions will move farther and farther apart till to reach a stage at which more anions can be accommodated. Now, the bigger cation moves to bigger void i.e. octahedral void whose coordination number is 8. This happens when the radius ratio exceeds 0.732.

In case of radius ratio becomes less than 0.414, the six anions will not be able to touch the smaller cation. To touch the cation, the anions starts overlapping with each other, which is an unstable situation. Hence smaller cation moves to smaller void i.e. tetrahedral void and coordination number decreases from 6 to 4.

Note: Although a large number of ionic substances obey this rule, there are many exceptions to it.

Radius Ratio of Interstitial Voids:

The vacant space left in the closest pack arrangement of constituent particles is called an interstitial void or interstitial site.

Tetrahedral Voids:

Locating Tetrahedral Voids:

Let us consider a unit cell of ccp or fcc lattice [Fig. (a)]. The unit cell is divided into eight small cubes. Each small cube has atoms at alternate corners as shown. Each small cube has 4 atoms. When joined to each other, they make a regular tetrahedron as shown in the figure. Thus, there is one tetrahedral void in each small cube and eight tetrahedral voids in total in the unit cell. Each of the eight small cubes has one void in one unit cell of ccp structure. The ccp structure has 4 atoms per unit cell. Thus, the number of tetrahedral voids is twice the number of atoms.

Radius Ratio of Tetrahedral Void:

A tetrahedral site in a cube having a tetrahedral void of radius ‘r’ is as shown at the centre of the cube. Let ‘R’ be the radius of the constituent particle of the unit cell. Let ‘a’ be each side of the cube.

Consider face diagonal AB (Right-angled triangle ABC). We have,

AB² = a² + a²

∴  AB² = 2 a²

∴  AB² = √2 a

Now the two-sphere touch each other along face diagonal AB

∴  2R = √2 a

∴  R = √2 a/2

∴  R =  a/√2  ……. (1)

Consider body diagonal AD (Right-angled triangle ABD). We have,

AD² = AB² + BD²

∴  AD² = (√2 a)² + a²

∴  AD² = 2 a² + a² = 3 a²

∴  AD = √3 a

Now the two spheres at the diagonals touch the tetrahedral void sphere along body diagonal AD

∴  2 R + 2r  = √3 a

∴  R + r  = (√3/2)a  ……..  (2)

Dividing equation (2) by (1)

Thus the radius ratio for the tetrahedral void is 0.225

Octahedral Voids:

Locating Octahedral Voids:

Let us consider a unit cell of ccp or fcc lattice [Fig. (a)]. The body centre of the cube, C is not occupied but it is surrounded by six atoms on face centres. If these face centres are joined, an octahedron is generated. Thus, this unit cell has one octahedral void at the body centre of the cube.

Besides the body centre, there is one octahedral void at the centre of each of the 12 edges. [Fig. (b)]. It is surrounded by six atoms, four belonging to the same unit cell (2 on the corners and 2 on face centre) and two belonging to two adjacent unit cells.

Since each edge of the cube is shared between four adjacent unit cells, so is the octahedral void located on it. Only 1/4 th of each void belongs to a particular unit cell.

For cubic close-packed structure:
Number of octahedral void at the body-centre of the cube = 1
12 octahedral voids located at each edge and shared between four unit cells

Thus number of octahedral valve at 12 edges =12 x 1/4 = 3
Thus, the total number of octahedral voids in unit cell = 1 + 3 = 4

In the ccp structure, each unit cell has 4 atoms. Thus, the number of octahedral voids is equal to this number of atoms.

Radius Ratio of Octahedral Void:

A octahedral site in a cube having octahedral void of radius r is as shown at the centre of the cube. Let R be the radius of the constituent particle of the unit cell. Let a be each side of the cube. In this case we can see that a = 2R

Consider Right angled triangle ABC. We have,

AB² = AC² + CB²

∴  (2R + 2r)² = a² + a²

∴  (2R + 2r)² = 2 a²

∴  2R + 2r = √2 a

But from figure a = 2R

∴  2R + 2r = √2 x 2R

∴   R +  r = √2  R

∴   r = √2  R – R

∴   r = (√2  – 1)R

∴   r = (1.414  – 1)R

∴   r = 0.414R

Thus radius ratio for the octahedral void is 0.414

Science > Chemistry > Solid State > Radius Ratio and its Significance

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In this article, we shall study two types of voids formed during hexagonal closed packing namely, a) Tetrahedral voids and b) octahedral voids.

Let us consider a closed pack hexagonal packing in the first layer as shown. There are empty spaces between the particles (sphere) are called voids.

All the voids are equivalent in the first layer they have marked alternately as ‘a’ and ‘b’. The spheres of the second layer can be either placed on voids which are marked ‘a’ or ‘b’ but it is impossible to place spheres on both types of voids simultaneously. When spheres of the new layer are placed on voids marked ‘a’ then voids marked as ‘b; remain unoccupied.

Thus there is no void above ‘a’ in the second layer but there is void above ‘b’ even in the second layer. The void between the first layer and second layer at ‘a’ is tetrahedral void, while the void between the first layer and the second layer at ‘b’ is octahedral void.

T = Tetrahedral void and O = Octahedral void

Tetrahedral Voids:

Wherever a sphere of the second layer is above the void of the first layer (or vice versa) a tetrahedral void is formed. These voids are called tetrahedral voids because a tetrahedron is formed when the centres of these four spheres are joined. These voids have been marked as ‘T’ in the figure.

Octahedral Voids:

At other places, the triangular voids in the second layer are above the triangular voids in the first layer, and the triangular shapes of these do not overlap then the octahedral void is formed. One of them has the apex of the triangle pointing upwards and the other downwards. These voids have been marked as ‘O’ in the figure.

Number of Voids:

The number of these two types of voids depends upon the number of close-packed spheres.

Let the number of close-packed spheres be N, then, the number of octahedral voids generated = N and the number of tetrahedral voids generated = 2N

Characteristics of Tetrahedral Voids:

• The vacant space or void is surrounded by four atomic spheres. Hence co-ordination number of the tetrahedral void is 4.
• The atom in the tetrahedral void is in contact with four atoms placed at four corners of a tetrahedron.
• This void is formed when a triangular void made coplanar atoms (first layer) is in contact with the fourth atom above or below it (second layer).
• The volume of the void is much smaller than that of the spherical particle.
• If R is the radius of the constituent spherical particle, then the radius of the tetrahedral void is 0.225 R.
• If the number of close-packed spheres is N, then the number of tetrahedral voids is 2N.

Characteristics of Octahedral Voids:

• The vacant space or void is surrounded by six atomic spheres. Hence, the coordination number of the tetrahedral void is 6.
• The atom in the octahedral void is in contact with six atoms placed at six corners of an octahedron.
• This void is formed when two sets of equilateral triangles pointing in the opposite direction with six spheres.
• The volume of the void is small.
• If R is the radius of the constituent spherical particle, then the radius of the octahedral void is 0.414 R.
• If the number of close-packed spheres is N, then the number of octahedral voids is N.Characteristics of Voids

Packing Voids in Ionic Solids

We have already seen that when particles are close-packed resulting in either cubic closed packing (ccp) or hexagonal close packing (hcp) structure, two types of voids are generated. The number of octahedral voids present in a lattice is equal to the number of close-packed particles, the number of tetrahedral voids generated is twice this number.

Ionic solids are formed from cations and anions. The charge is balanced by an appropriate number of both types of ions so that net charge on solid is zero (neutral).

In ionic solids, the bigger ions (usually, anions) form the close-packed structure and the smaller ions (usually, cations) occupy the voids. If the smaller ion is small enough then tetrahedral voids are occupied, if bigger, then octahedral voids are occupied.

Not all octahedral or tetrahedral voids are occupied. In a given compound, the fraction of octahedral or tetrahedral voids that are occupied, depends upon the chemical formula of the compound, as can be seen from the following examples.

Example 1:

A compound is formed by two elements X and Y. Atoms of the element Y (as anions) make cubic closed packing (ccp) and those of the element X (as cations) occupy all the octahedral voids. Obtain the formula of the compound.

Solution:

Given that the cubic closed packing (ccp) lattice is formed by the element Y.

The number of octahedral voids generated would be equal to the number of atoms of Y present in it.

Since all the octahedral voids are occupied by the atoms of X, their number would also be equal to that of the element Y. Thus, the atoms of elements X and Y are present in equal numbers or 1:1 ratio.

Therefore, the formula of the compound is XY.

Example 2:

Atoms of element B form hexagonal close packing (hcp) lattice and those of element A occupy 2/3rd of tetrahedral voids. Obtain the formula of the compound formed by the elements A and B.

Solution:

The number of tetrahedral voids formed is equal to twice the number of atoms of element B

Only 2/3rd of these are occupied by the atoms of element A. Hence the ratio of the number of atoms of A and B is 2 × (2/3):1 or 4:3

Hence the formula of the compound is A₄B₃.

Science > Chemistry > Solid State > Tetrahedral Voids and Octahedral Voids

The post Tetrahedral Voids and Octahedral Voids appeared first on The Fact Factor.

In this article, we shall study the different types of close packing of the constituent particles in solids.

The Principle of Packing in Solids:

In a crystal particle (atoms, molecules or ions) occupy the lattice points in the crystal lattice. These particles may be of various shapes and hence mode of packing of these particles will change according to their shape.

Constituent particles are considered rigid incompressible spheres of equal size. The constituent particles try to get closely packed together so that the maximum closely packed structure is attained. The maximum closely packed structure is such that the minimum empty space (void) is left. The closer the constituent particles lie, the greater is the stability of the packed system. By this arrangement, a maximum possible density of the crystal and stability of the crystal is achieved.

Close Packing in One Dimension:

There is only one way of arranging spheres in a one-dimensional close-packed structure, that is to arrange them in a row and touching each other.

In this arrangement, each sphere is in contact with two of its neighbours. The number of nearest neighbours of a particle is called its coordination number. Thus, in a one-dimensional close-packed arrangement, the coordination number is 2.

Close Packing in Two Dimensions:

A two-dimensional close-packed structure can be generated by placing the rows of close-packed spheres side by side horizontally. This can be done in two different ways.

a) Square Close Packing:

The second row may be placed in contact with the first one such that the spheres of the second row are exactly above those of the first row. The spheres of the two rows are aligned horizontally as well as vertically. If we call the first row as ‘A’ type row, the second row being exactly the same as the first one, is also of ‘A’ type. Similarly, we may place more rows to obtain AAA type of arrangement.

In this arrangement, each sphere is in contact with four of its neighbours. Thus, the two-dimensional coordination number is 4. Also, if the centres of these 4 immediate neighbouring spheres are joined, a square is formed. Hence this packing is called square close packing in two dimensions.

b) Hexagonal Close Packing:

In this arrangement, the second row may be placed above the first one in a staggered manner such that its spheres fit in the depressions of the first row.

If the arrangement of spheres in the first row is called ‘A’ type, the one in the second row is different and may be called ‘B’ type. When the third row is placed adjacent to the second in a staggered manner, its spheres are aligned with those of the first row. Hence this row is also of ‘A’ type. The spheres of the similarly placed fourth row will be aligned with those of the second row (‘B’ type). Hence this arrangement is of ABAB type.

In this arrangement, there is less free space and this packing is more efficient than the square close packing. Each sphere is in contact with six of its neighbours and the two-dimensional coordination number is 6. The centres of these six spheres are at the corners of a regular hexagon hence this packing is called two-dimensional hexagonal close packing.

It can be seen in the figure that in these rows there are some voids (empty spaces). These are triangular in shape. The triangular voids are of two different types. In one row, the apexes of the triangles are pointing upwards and in the next layer downwards.

Close Packing in Three Dimensions:

A three-dimensional close-packed structure can be generated by stacking (placing) layers of close-packed spheres in two dimensions one over the other. This can be done in the following ways.

Three-dimensional close packing from a two-dimensional square layer:

In such an arrangement, the second layer is placed over the first layer such that the spheres of the upper layer are exactly above those of the first layer. In this arrangement spheres of both the layers are perfectly aligned horizontally as well as vertically as shown. Similarly, we may place more layers one above the other.

If the arrangement of spheres in the first layer is called ‘A’ type, all the layers have the same arrangement. Thus this lattice has AAA…. type pattern. The lattice thus generated is the simple cubic lattice, and its unit cell is the primitive cubic unit cell.

The coordination number for such arrangement is 6. The volume occupied by particles is 52% i.e. there is 48% void space. Of all metals in the periodic table, only polonium crystalizes in this crystal form.

Three-dimensional close packing from two-dimensional hexagonal close-packed layers:

This close-packed structure can be generated by placing layers one over the other.

All the voids are equivalent in the first layer they have marked alternately as ‘a’ and ‘b’. The spheres of the second layer can be either placed on voids which are marked ‘a’ or ‘b’ but it is impossible to place spheres on both types of voids simultaneously. When spheres of a new layer are placed on voids marked ‘a’ then voids marked as ‘b; remain unoccupied.

Let us take a two-dimensional hexagonal close-packed layer ‘A’ and place a similar layer above it such that the spheres of the second layer are placed in the depressions of the first layer. Since the spheres of the two layers are aligned differently, let us call the second layer as B.

T = Tetrahedral void and O = Octahedral void

Placing the third layer over the second layer can be done in two ways.

It can be observed from the figure that not all the triangular voids of the first layer are covered by the spheres of the second layer.

This gives rise to two different types of holes or voids. Wherever a sphere of the second layer is above the void of the first layer (or vice versa) a tetrahedral void is formed. These voids are called tetrahedral voids because a tetrahedron is formed when the centres of these four spheres are joined. At other places, the triangular voids in the second layer are above the triangular voids in the first layer, and the triangular shapes of these do not overlap. One of them has the apex of the triangle pointing upwards and the other downwards. These voids have been marked as ‘O’ in the figure above.

Covering Tetrahedral Voids (Hexagonal Closed Packing):

Tetrahedral voids of the second layer may be covered by the spheres of the third layer. In this case, the spheres of the third layer are exactly aligned with those of the first layer. Thus, the pattern of spheres is repeated in alternate layers. This pattern is often written as ABAB ……. pattern. This structure is called a hexagonal close-packed (hcp) structure.

This sort of arrangement of atoms is found in many metals like magnesium and zinc.

B) Covering Octahedral Voids (Cubic Closed Packing):

The third layer may be placed above the second layer in a manner such that its spheres cover the octahedral voids. When placed in this manner, the spheres of the third layer are not aligned with those of either the first or the second layer. This arrangement is called ‘C’ type.

Only when the fourth layer is placed, its spheres are aligned with those of the first layer as shown. This pattern of layers is often written as ABCABC ……….. This structure is called cubic close-packed (ccp) or face-centred cubic (fcc) structure. Metals such as copper and silver crystallise in this structure.

Note:

Both these types of close packing are highly efficient and 74% space in the crystal is filled. In either of them, each sphere is in contact with twelve spheres. Thus, the coordination number is 12 in either of these two structures.

Characteristics of Hexagonal Close Packing:

• In this three dimensional arrangement of the unit cell, spheres of the third layer are placed on triangular-shaped tetrahedral voids of the second layer.
• The spheres of the third layer lie exactly above the spheres of the first layer.
• The arrangements of the first layer and third layer are identical.
• The arrangement of hexagonal close packing is represented as ABAB type.
• Packing efficiency is 74%

Characteristics of Cubic Close Packing:

• In this three dimensional arrangement of the unit cell, spheres of the third layer are placed in the positions of tetrahedral voids having apices upward.
• In this arrangement, the spheres of the third layer do not lie exactly above the spheres of the first layer. Actually, the spheres of the fourth layer lie exactly above the spheres of the first layer.
• The arrangements of the first layer and third layer are different. The arrangements of the first layer and fourth layer are identical.
• The arrangement of cubic close packing is represented as ABCABC type.
• Packing efficiency is 74%

Science > Chemistry > Solid State > Packing in solids

The post Packing in solids appeared first on The Fact Factor.

In this article, we shall study to solve problems to calculate the atomic radius, the distance between atoms in the unit cell and to decide the type of crystal structure.

Example – 01:

A naturally occurring gold crystallizes in face centred cubic structure and has a density of 19.3 g cm^-3. Calculate the atomic radius of gold. (Au = 197 g mol^-1)

Given: Density of gold  = 19,3 g cm^-3, Avogadro’s number N = 6.022 x 10²³ mol^-1.Atomic mass of gold = M = 197 g mol^-1.Type of crystal structure = fcc

To Find: Atomic radius of gold =?

Solution:

The number of atoms in the unit cell of a face centred cubic structure is n = 4

Ans: The atomic radius of gold is 144 pm

Example – 02:

Sodium metal crystallizes in the bcc structure with an edge length of unit cell 4.29 x 10^-8 cm. Calculate the atomic radius of sodium metal.

Given: Edge length = a = 4.29 x 10^-8 cm, Type of crystal structure = bcc

To Find: Atomic radius of sodium =?

Solution:

Ans: The atomic radius of sodium is 186 pm

Example – 03:

Niobium crystallizes in bcc structure and has a density of 8.55 g cm^-3. Calculate its atomic radius, if its atomic mass is 93.

Given: Density of niobium = 8.55 g cm^-3, Avogadro’s number N = 6.022 x 10²³ mol^-1.Atomic mass of niobium = M = 93 g mol^-1.Type of crystal structure = bcc

To Find: Atomic radius of niobium =?

Solution:

The number of atoms in the unit cell of body centred cubic structure is n = 2

Ans: Atomic radius of niobium is 143.1 pm

Example – 04:

Sodium metal crystallizes in the body-centred cubic unit cell. If the distance between the nearest Na atom is 368 pm, calculate the edge length of the unit cell.

Given: the distance between nearest Na atom is 368 pm,

To Find: Edge length of unit cell =?

Solution:

The nearest atoms in bcc crystal structure are along the body diagonal of the cube and they are in contact with each other.

2r = 368 pm

∴ r = 184 pm

Ans: The edge length of the cell is 424.9 pm

Example – 05:

Copper crystallizes in fcc structure. If two neighbouring Cu atoms are at a distance of 234 pm, find a) edge length, b) volume of a unit cell. Also, find the distance between the next neighbouring atoms.

Given: the distance between nearest Cu atom is 234 pm,

Solution:

The nearest atoms in fcc crystal structure are along the face diagonal of the cube and they are in contact with each other.

2r = 234 pm

∴ r = 117 pm

a = 331 x 10^-10 m = 3.31 x 10^-8 m

Volume of unit cell = a³ = (3.31 x 10^-8 m)³

Volume of unit cell = (3.31) ³ x 10^-24 m³ = 36.24 x 10^-24 m³

The next neighbouring atom is along the edge of the cell. Thus the distance between next neighbouring atom is ‘a’ i.e. 331 pm

Ans: The edge length = 331 pm, the volume of unit cell = 36.24 x 10^-24 m³, the distance between next neighbouring atom is 331 pm

Example – 06:

A metal occurs as body centred cube and has a density of 7.856 g cm^-3. Calculate the atomic radius of metal. Atomic mass of metal = 58 g/mol

Given: Density of metal = 7.856 g cm^-3, Avogadro’s number N = 6.022 x 10²³ mol^-1.Atomic mass of metal = M = 58 g mol^-1.Type of crystal structure = bcc

To Find: Atomic radius of metal =?

Solution:

Ans: The atomic radius is 125.7 pm

Example – 07:

An element germanium crystallizes in the bcc type crystal structure with an edge of unit cell 288 pm and the density of the element is 7.2 g cm^-3. Calculate the number atoms present in 52 g of the crystalline element. Also, calculate the atomic mass of the element.

Given: Density of germanium = 7.2 g cm^-3, edge length = 288 pm = 288 x 10^-12 m = 288 x 10^-10 cm = 2.88 x 10^-8 cm, Avogadro’s number N = 6.022 x 10²³ mol^-1. Given mass of germanium = 52 g, Type of crystal structure = bcc

To Find: the number atoms present in 52 g of the crystal =? atomic mass of the element = M = ?

Solution:

The number of atoms in the unit cell of bcc structure is n = 2

For bcc there are 2 atoms in a unit cell

Number of atoms in given mass = 2 x 3.023 x 10²⁹ = 6.046 x 10²⁹

Example – 08:

An atom crystallises in fcc crystal lattice and has a density of 10 g cm^-3 with unit cell edge length of 100 pm. Calculate the number atoms present in 1 g of the crystal.

Given: Density = 10 g cm^-3, edge length = 100 pm = 100 x 10^-12 m = 100 x 10^-10 cm = 10^-8 cm, Avogadro’s number N = 6.022 x 10²³ mol^-1. Given mass of crystal = 1 g, Type of crystal structure = fcc

To Find: the number atoms present in 1 g of the crystal

Solution:

The number of atoms in the unit cell of fcc structure is n = 4

For fcc there are 4 atoms in a unit cell

Number of atoms in given mass = Number of atoms in unit cell x Number of unit cells

Number of atoms in given mass = 4 x 10²³

Ans: Number of atoms in given mass = 4 x 10²³

Example – 09:

Aluminium having atomic mass 27 g mol^-1, crystallizes in face centred cubic structure. Find the number of aluminium atom in 10 g of it. Also, find the number of unit cells in the given quantity.

Given: Atomic mass = 27 g mol^-1, Avogadro’s number N = 6.022 x 10²³ mol^-1. Given mass = 10 g, Type of crystal structure = fcc

To Find: the number of aluminium atom in 10 g =? Number of unit cells =?

Solution:

Mass of each atom = Atomic mass / Avogadro’s number

Mass of each atom = 27 g mol^-1/ 6.022 x 10²³ mol^-1 = 4.483 x 10^-23 g

The number of atoms in unit cell of fcc structure is n = 4

Hence mass of unit cell = 4 x 4.483 x 10^-23 g = 1.793 x 10^-22 g

Number of unit cells = Given mass/ mass of unit cell = 10 g/ 1.793 x 10^-22 g = 5.58 x 10²²

Number of aluminium atom = 4 x number of unit cell

Number of aluminium atom = 4 x 5.58 x 10²² = 2.23 x 10²³

Ans: Number of aluminium atoms = 2.23 x 10²³ and number of unit cells = 5.58 x 10²²

Example – 10:

A unit cell of iron crystal has edge length 288 pm and density 7.86 g cm^-3. Find the number of atoms per unit cell and type of crystal lattice. Given the molar mass of iron 56 g/mol.

Given: Density = 7.86 g cm^-3, edge length = 288 pm = 288 x 10^-12 m = 288 x 10^-10 cm = 2.88 x 10^-8 cm, Avogadro’s number N = 6.022 x 10²³ mol^-1. Molecular mass of iron = M = 56 g mol^-1,

To Find: the number of atoms per unit cell and type of crystal lattice

Solution:

Number of atoms per unit cell = 2,

Now body centred cubic structure has 2 atoms in the unit cell. Hence the crystal structure must be bcc

Ans: The crystal structure is bcc

Example – 11:

The edge length of a unit cell of a cubic crystal is 4.3 A°, having atomic mass 89 g/mol. If the density of crystal is 9.02 g cm^-3, find the number of atoms in a unit cell.

Given: Density = 9.02 g cm^-3, edge length = 4.3 A0 = 4.3 x 10^-10 m = 4.3 x 10^-8 cm, Avogadro’s number N = 6.022 x 10²³ mol^-1. Molecular mass = M = 89 g mol^-1,

To Find: the number of atoms in a unit cell.

Solution:

Ans: Number of atoms per unit cell = 5

Example – 12:

The density of silver having atomic mass 107.8 g /mol is 10.7 g cm^-3. If the edge length of a cubic unit cell is 405 pm, find the number of silver atoms in a unit cell and predict its type.

Given: Density = 10.7 g cm^-3, edge length = 405 pm = 405 x 10^-12 m = 405 x 10^-10 cm = 4.05 x 10^-8 cm, Avogadro’s number N = 6.022 x 10²³ mol^-1. Molecular mass = M = 107.8 g mol^-1,

To Find: the number of silver atoms in a unit cell and predict its type.

Solution:

The number of atoms per unit cell = 4,

Now face centred cubic structure has 4 atoms in the unit cell. Hence the crystal structure must be fcc

Ans: The crystal structure is fcc

Example – 13:

A crystalline solid is made up of two elements ‘A’ and ‘B’. Atoms of A are present at the corners and atoms of B are present at face centres. One atom of A is missing from the corner. Find the simplest formula of the solid.

Solution:

Atoms of A are present at 7 corners

Hence the number of atoms of A in the lattice = 7 x 1/8 = 7/8

Atoms of B are present at 6 face centres

Hence the number of atoms of B in the lattice = 6 x 1/2 = 3

Thus the ratio of atoms of A to atoms of B in crystal = 7/8 : 3 = 7 : 24

Hence the simplest formula of the solid is A₇B₂₄

Example – 14:

A solid has a structure in which ‘W’ atoms are located at the corners of the cube, ‘O’ atoms are at the centre of the edges and Na atoms at the centre of the cube. Find the formula of the compound.

Solution:

Atoms of W are present at 8 corners

Hence the number of atoms of W in the lattice = 8 x 1/8 = 1

Atoms of O are present at 12 edge centres

Hence the number of atoms of O in the lattice = 12 x 1/4 = 3

Atoms of Na are present at the centre

Hence the number of atoms of Na in the lattice = 1

Thus the ratio of atoms of W to atoms of O to atoms of Na in crystal = 1:3:1

Hence the simplest formula of the solid is WO₃Na i.e. NaWO₃

Science > Chemistry > Solid State > Numerical Problems on Type of Crystal Structure

The post Numerical Problems on Type of Crystal Structure appeared first on The Fact Factor.

In this article, we shall study to solve numerical problems to find the density of solid, edge length and volume of the unit cell.

Example – 01:

Silver crystallizes in face centred cubic structure. The edge length of a unit cell is found to be 408.7 pm. Calculate the density of silver. (Ag = 108 g mol^-1)

Given: The edge length of the unit cell = a = 408.7 pm = 408.7 x 10^-10cm, Atomic mass of silver = M = 108 g mol^-1, Avogadro’s number N = 6.022 x 10²³ mol^-1. Type of crystal structure = face centred cubic

To Find: Density of silver =?

Solution:

The number of atoms in the unit cell of a face centred cubic structure is n = 4

Ans: Density of silver is 10.51 g/cm³

Example – 02:

Copper crystallizes in face centred cubic structure. The edge length of a unit cell is found to be 3.61 x 10^-8 cm. Calculate the density of copper if the molar mass of copper is 63.5 g mol^-1

Given: The edge length of the unit cell = a = 3.61 x 10^-8 cm, molar mass of copper = M = 63.5 g mol^-1, Avogadro’s number N = 6.022 x 10²³ mol^-1. Type of crystal structure = face centred cubic

To Find: Density of copper =?

Solution:

The number of atoms in the unit cell of a face centred cubic structure is n = 4

Ans: Density of copper is 8.96 g/cm³

Example – 03:

Determine the density of caesium chloride which crystallizes in bcc type structure with an edge length of 412.1 pm. The atomic mass of Cs and Cl are 133 and 35.5 respectively.

Given: The edge length of the unit cell = a = 412.1 pm = 412.1 x 10^-10 cm, atomic mass of Cs = 133, atomic mass of Cl = 35.5, Avogadro’s number N = 6.022 x 10²³ mol^-1. Type of crystal structure = bcc

To Find: Density of caesium chloride =?

Solution:

Volume of unit cell = a³ = (4.121 x 10^-8 cm) ³ = (4.121) ³ x 10^-24 cm³.

Mass of one atom of Cs = Atomic mass of Cs / Avogadro’s number

Mass of one atom of Cs = 133 g mol^-1 / 6.022 x 10²³ mol^-1

Mass of one atom of Cs = 22.09 x 10^-23 g

Mass of one atom of Cl = Atomic mass of Cl / Avogadro’s number

Mass of one atom of Cl = 35.5 g mol^-1 / 6.022 x 10²³   mol^-1

Mass of one atom of Cl = 5.894 x 10^-23 g

Caesium chloride has a body-centred cubic structure such that caesium atom is at centre

and eight chlorine atom at the eight corners of the cube.

Thus there is 1/8 x 8 = 1atom of Cl and 1 atom of Cs.

Thus mass of unit cell of caesium chloride = 22.09 x 10^-23 g + 5.894 x 10^-23 g

Thus mass of unit cell of caesium chloride = 27.98 x 10^-23 g

Density of caesium chloride = mass of unit cell/volume of unit cell

Density of caesium chloride = 27.98 x 10^-23 g /(4.121) ³ x 10^-24 cm³.

Density of caesium chloride = 3.997 g/cm³ = 4 g/cm³

Ans: Density of caesium chloride = 4 g/cm³

Example – 04:

A metal crystallizes as fcc structure and the unit cell has a length of edge 3.72 x 10^-8 cm. Calculate the density of the metal. Given the atomic mass of metal as 68.5 g mol^-1.

Given: The edge length of the unit cell = a = 3.72 x 10^-8 cm, atomic mass of copper = M = 68.5 g/mol, Avogadro’s number N = 6.022 x 10²³ mol^-1. Type of crystal structure = face centred cubic

To Find: Density of metal =?

Solution:

The number of atoms in the unit cell of a face centred cubic structure is n = 4

Ans: Density of metal = 8.83 g/cm³

Example – 05:

Sodium crystallizes in bcc structure. If the atomic radius of sodium is 186 pm. Find a) edge length b) volume of the unit cell and c) density of sodium crystal. The atomic mass of sodium is 23 g mol^-1.

Given: Atomic radius of sodium = r = 186 pm = 186 x 10^-10 cm = 1.86 x 10^-8 cm, atomic mass of copper = M = 23 g/mol, Avogadro’s number N = 6.022 x 10²³ mol^-1. Type of crystal structure = bcc

To Find: Edge length = a =? Volume of unit cell =? Density of sodium =?

Solution:

Volume of unit cell = a³ = (4.30 x 10^-8 cm) ³ = (4.3) ³ x 10^-24 cm³ = 79.4 x 10^-24 cm³

The density of sodium is 0.962 g cm^-3.

Ans: Density of sodium = 0.962 g/cm³

Example – 06:

copper crystallizes in fcc type unit cell. The edge length of a unit cell is 360.8 pm. The density of metallic copper is 8.92 g cm-3. Determine the atomic mass of copper.

Given: The edge length of the unit cell = a = 360.8 pm = 360.8 x 10^-10 cm  = 3.608 x 10^-8 cm, Density of copper  = 8.92 g cm^-3, Avogadro’s number N = 6.022 x 10²³ mol^-1. Type of crystal structure = fcc

To Find: the atomic mass of copper =?

Solution:

The number of atoms in the unit cell of a face centred cubic structure is n = 4

Ans: The atomic mass of copper is 63.07 g mol^-1.

Example – 07:

Silver crystallises in fcc type unit cell. The edge length of a unit cell is 4.07 10^-8 cm. The density of metallic silver is 10.5 g cm^-3. Determine atomic mass of silver.

Given: The edge length of the unit cell = a = 4.07 x 10^-8 cm, Density of silverr = 10.5 g cm^-3, Avogadro’s number N = 6.022 x 10²³ mol^-1. Type of crystal structure = fcc

To Find: Atomic mass of silver =?

Solution:

The number of atoms in the unit cell of a face centred cubic structure is n = 4

Ans: The atomic mass of silver is 106.6 g mol^-1

Example – 08:

Face centred cubic crystal lattice of copper has a density of 8.966 g cm^-3. Calculate the volume of the unit cell. Given the molar mass of copper 63.5 g/mol.

Given: Molar mass of copper = 63.5 g mol^-1, Density of copper = 8.966 g cm^-3, Avogadro’s number N = 6.022 x 10²³ mol^-1. Type of crystal structure = fcc

To Find: Volume of unit cell =?

Solution:

The number of atoms in the unit cell of a face centred cubic structure is n = 4

Ans: The volume of the unit cell is 4.704 x 10^-23 cm³

Example – 09:

Silver crystallizes as fcc structure. If the density of silver is 10.51 g cm^-3. Calculate the volume of a unit cell. Given:  molar mass of silver 108 g/mol.

Given: Molar mass of silver = 108 g mol^-1, Density of silver  = 10.51  g cm^-3, Avogadro’s number N = 6.022 x 10²³ mol^-1. Type of crystal structure = fcc

To Find: Volume of unit cell =?

Solution:

The number of atoms in the unit cell of a face centred cubic structure is n = 4

Ans: The volume of the unit cell is 6,825 x 10^-23 cm³

Science > Chemistry > Solid State > Numerical Problems on Density of Solid

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