In this article, we shall study defects in the crystal structure, sources of defects and their types.

Source of Defects in Crystal Structure:

At absolute zero, crystals tend to have a tendency to have a perfectly ordered arrangement. This arrangement at absolute zero represents the lowest energy state of the crystal. As the temperature increases, there is a change in the orderly arrangement of constituents in the crystal. The defects in crystal structure are basically irregularities in the arrangement of constituent particles. Any deviation from the perfectly ordered arrangement constitutes a defect or imperfection. Such defects which are due to temperature change are referred as thermodynamic defects. The imperfection also may be due to impurities present in solid.

The defects in the crystal due to the irregularities in the arrangement of atoms or ions are called atomic imperfections. They are due to missing or misplaced ions. Such defects are referred as point defects. Point defects are the irregularities or deviations from ideal arrangement around a point or an atom in a crystalline substance

If the deviation from periodicity extends over microscopic regions of crystal then they are called lattice imperfections. If lattice imperfections extend along the line they are called line defects. The line defects are the irregularities or deviations from an ideal arrangement in entire rows of lattice points. They are also called dislocations. If the defects extend along a plane they are called plane defects. These imperfections in crystal lead to modification of some properties of the solid or may give rise to new properties.

Usually a solid consists of an aggregate of a large number of small crystals. These small crystals have defects in them. This happens when the crystallization process occurs at the fast or moderate rate. Single crystals are formed when the process of crystallization occurs at an extremely slow rate. Even these crystals are not free of defects.

Point Defects in Crystal Structure:

Point defects are the irregularities or deviations from ideal arrangement around a point or an atom in a crystalline substance. Point defects can be classified into three types:

Stoichiometric Defects in Crystal Structure:

The compound in which the number of positive and negative ions are exactly in the ratios indicated by their chemical formulae are called stoichiometric compounds. The point defects that do not disturb the stoichiometry of the solid are called stoichiometric defects. They are intrinsic or thermodynamic defects. The electrical conductivity of crystal increases due to this defect. These defects are of two types, vacancy defects and interstitial defects.

Vacancy Defects or Schottky Defects:

When some of the lattice sites are vacant, the crystal is said to have vacancy defect. The defect produced due to vacancies caused by an absence of anions and cations in the crystal lattice of ionic solid is called a Schottky defect. Thus in such defect, one positive ion and one negative ion are missing from their respective positions leaving behind a pair of holes. This defect can also develop when a substance is heated.

Due to this defect, the observed density of crystal is found to be lower than the expected density. This defect is observed in ionic compounds with high coordination number, high radius and with cations and anions have almost equal size. like NaCl, KCl, CsCl, AgBr, KBr, etc.

Vacancy defects increase with the increase in temperature. It can be observed w.r.t. NaCl crystal as given in the table

Interstitial Defects or Frenkel Defects:

When cation and anion from ionic solid leave its regular site and moves to occupy a place between the lattice site (interstitial sites) is called an interstitial defect. This defect associated with the ionic compound is called Frenkel defect or dislocation defect. The ions occupying interstitial sites are called interstitials. The formation of this defect depends on the size of interstitials.

This defect is observed in case of an ionic compound having low coordination number and relatively smaller cations which can fit into interstitial space. This defect is common when the difference in ionic radii of two participating ions is large.

The presence of this defect does not alter the density of the solid. The presence of ions in interstitial sites increases the dielectric constant of the crystal.

Example: In AgCl the defect is observed due to Ag⁺ ions. In ZnS the defect is observed due to Zn⁺⁺ ions. AgBr, AgI

Non-Stoichiometric Defects in Crystal Structure:

There are many compounds in which the ratio of positive and negative ions present in the compound differs from that required by the ideal chemical formula of the compound. Such defects are termed as non-stoichiometric or Berthollide compounds. For example, vanadium oxide is represented as VO_x. The value of x in the crystal of the oxide varies from 0.6 to 1.3. Similarly, iron oxide is represented as Fe_xO. The value of x in the crystal of the oxide varies from 0.93 to 0.95. The balance of positive and negative charge in such compounds is maintained by having extra electrons or extra positive charges. Such defects are called non-stoichiometric defects.

These defects are of two types: (i) metal excess defect and (ii) metal deficiency defect.

Metal Excess Defect:

A metal excess defect due to anionic vacancies:

This defect is likely to be shown by a crystal which shows Schottky defects. A compound may have excess metal ion if the negative ion is absent from its lattice site, leaving a hole which is occupied by an electron to maintain electrical neutrality.

Alkali halides like NaCl and KCl show this type of defect. When crystals of NaCl are heated in an atmosphere of sodium vapour, the sodium atoms are deposited on the surface of the crystal. The Cl^– ions diffuse to the surface of the crystal and combine with Na atoms to give NaCl. This happens with the loss of an electron by sodium atoms to form Na⁺ ions. The released electrons diffuse into the crystal and occupy anionic sites. As a result, the crystal now has an excess of sodium. The anionic sites occupied by unpaired electrons are called F-centres (from the German word Farbenzenter for the colour centre). They impart a yellow colour to the crystals of NaCl. The colour results by excitation of these electrons when they absorb energy from the visible light falling on the crystals. Greater the number of F-centres, greater is the intensity of the colour.

Similarly, the excess of lithium makes LiCl crystals pink and excess of potassium makes KCl crystals violet (or lilac).

Such solids having F- centres are paramagnetic due to the presence of unpaired electrons in the hole.

A Metal excess defect due to the presence of extra cations at interstitial sites:

This defect is due to the presence of extra positive ion at the interstitial site. Electrical neutrality is maintained by the presence of electron at the interstitial site. This defect is likely to be shown by a crystal which shows Frenkel defects.

Zinc oxide is white in colour at room temperature. On heating, it loses oxygen and turns yellow. Now there is an excess of zinc in the crystal and its formula becomes Zn_1+xO. The excess Zn²⁺ ions move to interstitial sites and the electrons to neighbouring interstitial sites. Due to the presence of an electron in interstitial space, the electrical conductivity of ZnO increases on heating.

Metal Deficiency Defect Due to Cationic Vacancies: 

The non-stoichiometric compounds may have metal deficiency due to the absence of the metal ion from its lattice site. The electrical neutrality is maintained by acquiring a higher positive charge by the adjacent ion.

There are many solids which are difficult to prepare in the stoichiometric composition and contain less amount of the metal as compared to the stoichiometric proportion.

This type of defect is mainly shown by transition elements. A typical example of this type is FeO which is mostly found with a composition of Fe_0.95O. It may actually range from Fe_0.93O to Fe_0.96O. In crystals of FeO some Fe²⁺ cations are missing and the loss of positive charge is made up by the presence of the required number of Fe³⁺ ions.

Note:

Both metal excess compounds and metal-deficient compounds act as semiconductors. Due to the presence of electron metal excess compounds work as n-type semiconductors. While metal deficient compounds conduct electricity through positive hole conduction work as p-type semiconductors.

Point Defects due to Presence of Foreign Atoms or Impurity Defects:

This defect occurs when regular cation of a crystal is replaced by some different cation. The different cation may occupy a regular lattice site or interstitial site.

If the impurity cation is substituted in place of regular cation then the defect is known as substitution impurity defect. If the impurity cation occupies interstitial space then the defect is called interstitial impurity defect.

Interstitial impurity in Stainless Steel:

Substitutional impurity in Brass

Substitutional impurity in NaCl

If molten NaCl containing a little amount of SrCl₂ is crystallised, some of the sites of Na+ ions are occupied by Sr²⁺. Each Sr²⁺ replaces two Na+ ions. It occupies the site of one ion and the other site remains vacant. The cationic vacancies thus produced are equal in number to that of Sr²⁺ ions.

Another similar example is the solid solution of CdCl₂ and AgCl.

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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

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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

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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

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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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In this article, we shall study the packing efficiency of different types of unit cells.

In whatever way the constituent particles atoms, molecules or ions are packed, there is always some free space in the form of voids.

The fraction of the total space in the unit cell occupied by the constituent particles is called packing fraction. Mathematically

Packing efficiency is the percentage of total space filled by the constituent particles in the unit cell.

Packing efficiency = Packing Factor x 100

A vacant space not occupied by the constituent particles in the unit cell is called void space.

The fraction of void space = 1 – Packing Fraction
% Void space = 100 – Packing efficiency.

Packing Efficiency of Simple Cubic Crystal Lattice (SCC):

In a simple cubic lattice, the atoms are located only on the corners of the cube. The particles touch each other along the edge as shown. Thus, the edge length (a) or side of the cube and the radius (r) of each particle are related as a = 2r. Thus the radius of an atom is half the side of the simple cubic unit cell.

The volume of the cubic unit cell = a³ = (2r)³ = 8r³

Since a simple cubic unit cell contains only 1 atom.

The packing efficiency of the simple cubic cell is 52.4 %. Thus 47.6 % volume is empty space (void space) i.e. almost half the space is empty. Hence the simple cubic crystalline solid is loosely bonded.

Packing Efficiency of Body Centred Cubic Crystal Lattice (BCC):

In a body-centred cubic lattice, the eight atoms are located on the eight corners of the cube and one at the centre of the cube. Let the edge length or side of the cube ‘a’, and the radius of each particle be r. The particles along the body diagonal touch each other.

∴ AD = r  + 2r + r = 4r   ……. (1)

Thus the radius of an atom is √3/₄ times the side of the body-centred cubic unit cell.

The volume of the cubic unit cell

Since a body-centred cubic unit cell contains 2 atoms

The packing efficiency of the body-centred cubic cell is 68 %. Thus 32 % volume is empty space (void space).

Packing Efficiency of Face Centred Cubic Crystal Lattice (FCC):

In a face-centred cubic lattice, the eight atoms are located on the eight corners of the cube and one at the centre of the cube. Let the edge length or side of the cube ‘a’, and the radius of each particle be r. The particles along face diagonal touch each other.

∴ AB = r + 2r + r = 4r  ……. (1)

Thus the radius of an atom is 1 / _√8   times the side of the face centred cubic unit cell.

The volume of the cubic unit cell

Since a face centred cubic unit cell contains 4 atoms,

The packing efficiency of the face centred cubic cell is 74 %. Thus 26 % volume is empty space (void space).

Summary of the Three Types of Cubic Structures:

Nearest Neighbouring Atoms:

Note:

From the unit cell dimensions, it is possible to calculate the volume of the unit cell. Knowing the density of the metal, we can calculate the mass of the atoms in the unit cell. The determination of the mass of a single atom gives an accurate method of determination of Avogadro constant.

Steps involved in finding the density of a substance:

• Find the volume of the unit cell using formula Volume = a³
• Find the type of cubic cell. Find the number of particles (atoms or molecules) in that type of cubic cell. Let it be denoted by ‘n’
• Find the mass of one particle (atoms or molecules) using formula

Mass of one particle = Molar (Atomic) mass of substance / Avogadro’s number

Mass of one particle = M/N

Where M = Molecular mass of the substance

N = Avogadro’s number = 6.022 x 10^-23 mol^-1.

• Find the mass of each unit cell using formula

Mass of unit cell = Mass of each particle x Number of particles in the unit cell

Mass of unit cell = (M/N) x n = nM/N

• Find the density of the substance using the formula

Steps involved in finding the radius of an atom:

• Find the type of cubic cell. Find the number of particles (atoms or molecules) in that type of cubic cell.  Let it be denoted by ‘n.
• ‘Find molar mass of one particle (atoms or molecules) using formula
• Find the mass of one particle (atoms or molecules) using formula

Mass of one particle = Molar (Atomic) mass of substance / Avogadro’s number

Mass of one particle = M/N

Where M = Molecular mass of the substance

N = Avogadro’s number = 6.022 x 10²³ mol^-1.

• Find the mass of each unit cell using formula

Mass of unit cell = Mass of each particle x  Number of particles in the unit cell

Mass of unit cell = (M/N) x n = nM/N

• Find the length of the side of the unit cell

• Write the relation between a and r for the given type of crystal lattice and calculate r.

Steps involved in finding atomic mass:

• Find the volume of the unit cell using formula Volume = a³
• Find the type of cubic cell. Find the number of particles (atoms or molecules) in that type of cubic cell. Let it be denoted by ‘n’
• Find the value of M/N from the following formula.

Now M/N gives the atomic mass

Science > Chemistry > Solid State > Packing Efficiency of Unit Cell

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In this article, we shall study the structures of Bravais Lattices.

A crystal is a homogenous portion of a solid substance made of a regular pattern of structural units bonded by plane surfaces making a definite angle with each other.

Unit Cell:

A unit cell is the smallest structural repeating unit of crystalline solid (space lattice). When unit cells of the same crystalline substance are repeated in space in all directions, a crystalline solid is formed. The unit cell is represented on paper by drawing lines connecting centres of constituent particles.

The geometrical form consisting only of a regular array of points in space is called lattice or space lattice. It may be defined as an array of points showing how molecules, atoms or ions are arranged in different sites, in a three-dimensional space. Each point at the intersection of lines in the unit cell represents constituent particle viz. molecule or atom or ion. This point of intersection of lines in the unit cell is called lattice point or lattice site.

Note: Each lattice point in the crystal lattice has the same surroundings or environment. Two or more crystalline substance may have the same lattice space. Each and every particle in the array is always represented by a lattice point in the three-dimensional array.

Characteristics of Crystal Lattice:

• The crystal lattice is a regular arrangement of constituent particles of a crystalline solid in three-dimensional space.
• It consists of a large number of unit cells.
• The crystal lattice is defined in terms of properties of the unit cell.
• On break up, it forms numerous unit cells.
• A crystal lattice can be obtained, handled and studied in a laboratory during experiments.
• It is a macroscopic property.

Characteristics of Unit Cell:

• A unit cell is the smallest structural repeating unit of crystalline solid.
• It is the fundamental unit of crystal lattice which possesses all the properties of the crystal.
• Unit cell defines fundamental properties crystal lattice.
• Unit cells are a fundamental unit, hence can not be divided further.
• A unit cell is hypothetical concept Hence it can not be obtained during experiments.
• It is a microscopic property.

Types of Space Lattice:

One Dimensional Lattice:

In this lattice, there exists a set of points arranged at equal distances along.

Two Dimensional Lattice:

In this type, two sides and the angle between them is specified. There are five types of two-dimensional lattice.

Hexagonal lattice is a rhombus with θ = 60°

Bravais Lattices:

Let lengths of three edges of the unit cell be a, b, and c. Let α be the angle between side b and c. Let β be the angle between sides a and c. Let γ be the angle between sides a and b.

French mathematician Bravais said that for different values of a, b, c, and α, β, γ, maximum fourteen (14) structures are possible. These arrangements are called Bravais Lattices.

14 Bravais Lattices:

Sr. No.	Crystal   System	Space Lattice   Type	Diagram	Edge Lengths	Angles	Examples
1	Cubic	Simple   primitive		a = b = c	α = β = γ = 90°	Polonium (Po)
2	Cubic	Body centred		a = b = c	α = β = γ = 90°	Iron (Fe), Rubidium (Rb), Sodium (Na), Titanium (Ti), Tungsten (W), Uranium (U), Zirconium (Zr)
3	Cubic	Face centred		a = b = c	α = β = γ = 90°	Copper (Cu), Aluminium (Al), Nickel (Ni), Gold (Au), Silver (Ag), Platinum (Pt)
4	Tetragonal	Simple   primitive		a = b ≠ c	α = β = γ = 90°	SnO2
5	Tetragonal	Body centred		a = b ≠ c	α = β = γ = 90°	Ti, O2, CaSO4
6	Orthorhombic	Simple   primitive		a ≠ b ≠ c	α = β = γ = 90°	Rhombic sulphur
7	Orthorhombic	Body centred		a ≠ b ≠ c	α = β = γ = 90°	KNO3
8	Orthorhombic	Face centred		a ≠ b ≠ c	α = β = γ = 90°	BaSO4
9	Orthorhombic	End centred		a ≠ b ≠ c	α = β = γ = 90°	MgSO4, 7H2O
10	Monoclinic	simple   primitive		a ≠ b ≠ c	α =γ = 90° and  β ≠ 90°	Monoclinic sulphur
11	Monoclinic	End centred		a ≠ b ≠ c	α =γ = 90° and  β ≠ 90	Na2SO4. 10H2O
12	Triclinic	simple   primitive		a ≠ b ≠ c	α =γ = 90° and  β ≠ 90°	K2Cr2O7, H3BO3
13	Hexagonal	Simple   primitive		a = b ≠ c	α = β = 90°   and γ = 120°	ZnO,   BeO, CoS, SnS
14	Rhombohedral	Simple   primitive		a ≠ b ≠ c	α = β  =  γ ≠ 90°	Calcite, NaNO3, FeCO3

Particle at Corner:

The particle at the corner of the unit cell is shared by 8 unit cells (4 layers below and 4 layers above). Hence each unit cell 1/8 particle.

Particle at Face:

The particle at the face of the unit cell is shared by 2 adjacent unit cells. Hence each unit cell 1/2 particle.

Particle at Edge:

The particle at the edge of the unit cell is shared by 4 unit cells (2 layers below and 2 layers above). Hence each unit cell 1/4 particle.

Coordination Number:

The coordination number of the constituent particle of the crystal lattice is the number of particles surrounding a single particle in a crystal lattice.

More the coordination number more tightly the particles are packed in the crystal lattice. Coordination number is the measure of the hardness of the crystal.

Number of Particles in Unit Cell and coordination Number

Simple Cubic Structure (scc):

From the structure, we can see that there are 8 particles at 8 corners of the unit cell. Each corner particle is shared by 8 neighbouring unit cells. Hence each unit cell contains 1/8 th of the particle at its corner.
Number of corners = 8, Hence number of particles in unit cell  = 1/8 x 8 = 1

Each particle in this structure is directly in contact with four other particles in its plane and one particle each in the layer above and the layer below. Hence the coordination number for simple cubic structure is 4+ 1+ 1 = 6

Body-Centred Cubic Structure (bcc):

From the structure, we can see that there are 8 particles at 8 corners of the unit cell. Each corner particle is shared by 8 neighbouring unit cells. Hence each unit cell contains 1/8 th of the particle at its corner. Number of corners = 8, hence number of particles in unit cell at corners = 1/8 x 8 = 1

At the same time, there is an atom at the centre of the cell, Hence the number of particles in unit cell 1 + 1 = 2

Each particle in this structure is directly in contact with four other particles in the layer above and four particles in the layer below. Hence the coordination number for body centred cubic structure is 4 + 4 = 8

Face Centered Cubic Structure (fcc):

From the structure, we can see that there are 8 particles at 8 corners of the unit cell. Each corner particle is shared by 8 other neighbouring unit cells. Hence each unit cell contains 1/8 th of the particle at its corner. The number of corners = 8. Hence the number of particles in a unit cell at corners = 1/8 x 8 = 1

There are 6 particles at 6 faces of the unit cell. Each face particle is shared by 2 neighbouring unit cells. Hence each unit cell contains 1/2 of the particle at its face.
The number of faces = 6. Hence the number of particles on face = 1/2 x 6 = 3

Hence number of particles in unit cell 1 + 3 = 4

Each particle in this structure is directly in contact with 4 other particles in its layer and with 4 particles in the layer above and 4 particles in the layer below. Hence the coordination number for face centred cubic structure is 4 + 4 + 4 = 12

Notes:

For edge centred particle it is shared by 4 unit cells. Hence each unit cell consists of 1/4 particle The number of atoms per unit cell is in the same ratio as the stoichiometry of the compound. Hence it helps to predict the formula of the compound.

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