Class-XII Chemistry CHAPTER-I TYPES OF CRYSTALLINE SOLIDS

Class-XII

Chemistry

CHAPTER-I

SOLID STATES

TYPES OF CRYSTALLINE SOLIDS

On the basis of nature of force among constituent particles crystalline solids are divided into four parts.

I]Molecular Solids: i) Non-Polar

ii)Polar

iii)Hydrogen bonded

II]Ionic Solids

III] Metallic Solids

IV] Covalent solids

PROPERTIES OF CRYSTALLINE AND AMORPHOUS SOLIDS

PROPERTY MOLECULAR SOLIDS

IONIC SOLIDS

METALLIC SOLIDS

COVALENT SOLIDS

Constituents Molecules or atoms Positive Ions in a sea of delocalized electrons.

metals Non-metals

Subtypes 1.Non polar 2.Polar 3.Hydrogen bonded

---- -----

Bonding Dispersion or London forces (for Non polar) Dipole-Dipole interactions (for Polar) And Hydrogen bonding for hydrogen bonded

Electrostatic force of attraction

Metallic bonding

Covalent bonding

Melting point Low High Fairly high Very high Conductance Insulators Insulators in

solid state but conductors in

Conductors in solid state as well as in

Insulators (exception: Graphite is

molten state and in aqueous solutions

molten states good conductor of electricity)

Physical Nature and Examples

Polar and non polar are soft but hydrogen bonded are hard in nature H2O, CO2,I2,

Hard but brittle NaCl ,MgO, ZnS,CaF2.

Hard but malleable and ductile Fe ,Cu, Mg, Ag

Hard C(diamond),Quartz , SiC , AlN Soft C(graphite)

INTERSTITIAL VOIDS

In hexagonal close packing (hcp) as well as cubic close packing (ccp) of a crystalline solid

only 74% of the available space is occupied by constituent particles. The remaining 26%

space is vacant and constitutes interstitial voids or spaces.

Types of voids

There are two types of interstitial voids in three dimensional close packing.

i) Tetrahedral voids

ii) Octahedral voids

i) Tetrahedral voids:

In close packing arrangement each sphere in the second layer rests on the hollow

(triangular voids) in the three touching spheres in the first layer. The centres of these four

spheres are at the corners of a regular tetrahedron. The vacant space between these four

touching spheres is called tetrahedral voids.

Radius ratio of Tetrahedral void:

Ratio of the radius of the tetrahedral void to the radius of the sphere is called radius ratio of

Tetrahedral void.

�(��)

�(��)= 0.225

Proving:Let the face length of the cube = a

R and r are the radii of the void and the sphere respectively then considering the right

angled triangle ABC,

the cube = a



Location of Tetrahedral void:

The unit cell of ccp lattice is divided into eight small cubes. The eight tetrahedral voids are

located on the body centres of the small eight cubes in a unit cell of ccp lattice.

of Tetrahedral void:

lattice is divided into eight small cubes. The eight tetrahedral voids are


lattice is divided into eight small cubes. The eight tetrahedral voids are


Number of Tetrahedral voids:

There are eight tetrahedral voids in a unit cell of ccp or fc

If, the number of atoms in a unit cell of ccp or fcc lattice =N

Then, the number of Tetrahedral voids in a unit cell of ccp or fcc lattice =2N

In a close packing, the number of tetrahedral

voids is double the number of sphere

The interstitial void formed by combination of two triangular voids of the first and second

layer is called octahedral void because this is enclosed between six spheres, centres of which

occupy corners of a regular octahedron.

Radius ratio of Octahedral void

Ratio of radius of the octahedral void to the radius of the sphere is called radius ratio of

octahedral void.

�(��)

�(��) =0.414

Proving: Let r and R are the radii of the void and the sphere respectively then considering the

right-angled triangle ABC

There are eight tetrahedral voids in a unit cell of ccp or fcc lattice.



In a close packing, the number of tetrahedral

voids is double the number of spheres.

Octahedral void

interstitial void formed by combination of two triangular voids of the first and second


occupy corners of a regular octahedron.

Radius ratio of Octahedral void


Let r and R are the radii of the void and the sphere respectively then considering the


interstitial void formed by combination of two triangular voids of the first and second



Let r and R are the radii of the void and the sphere respectively then considering the

So, the radius ratio for octahedral void is 0.414.

Location of octahedral voids

In ccp Octahedral voids are located at two positions

i) At body centre: All face centred spheres touch this void.

The void at the centre will be completely inside the unit cell.

ii) At edge centre: at edge centre the void is surrounded by 2 corner spheres and 4

face centred spheres.

OR �

� = 0.414


Location of octahedral voids

Octahedral voids are located at two positions:

: All face centred spheres touch this void.


: at edge centre the void is surrounded by 2 corner spheres and 4



: at edge centre the void is surrounded by 2 corner spheres and 4

The void at the edge centre will not be completely inside 1unit cell. It is shared

among 4unit cells.

Number of Octahedral voids:

There are four octahedral voids in a unit cell of ccp or fcc lattice.


Then, the number of octahedral voids in a unit cell of ccp or fcc lattice =N

In a close packing, the number of octahedral voids is equal to the number of spheres.

[ In ionic solids, the bigger ions (usually anions) form the close packed structure (ccp or fcc )

and the smaller ions (usually cations ) occupy the voids. If the cation is small enough then

tetrahedral voids are occupied then,

��

�_ = 0.225

If the cation is bigger, then octahedral voids are occupied. Then,

��

�_ = 0.414 ]

NCERT QUESTIONS AND ANSWERS [CONTINUE….]

Q.24 Differentiate between: (a) Cubic close-packing and hexagonal close-packing. (b) Unit cell and crystal lattice. (c) Octahedral void and Tetrahedral void.

Ans:

(a) Cubic close packing: When a third layer is placed over the second layer in a manner that the octahedral voids are covered by the spheres, a layer different from the first (A) and second (B) is obtained. If we continue packing in this manner we get the cubic close packing.[ABCABC-Type packing]

Coordination number =12

Hexagonal close packing: When the third layer is placed over the second layer in a way that the tetrahedral voids are covered by the spheres, a 3D close packing is produced where spheres in each third or alternate layers are vertically aligned. If we continue packing in this order we get hexagonal close packing [ABAB -Type packing]]

Coordination number =12

b) Unit cell

It is the smallest portion of a crystal lattice which, when repeated in different directions,

generates the entire lattice, e.g. primitive unit cell.

Crystal lattice

This is the three dimensional arrangement of constituent particles ( each particle is depicted

as a point) in the space which represents how the constituent particles (atoms, ions or

molecules) are arranged in a crystal.

c)Octahedral void – The interstitial void formed by combination of two triangular voids of the first and second layer is called octahedral void because this is enclosed between six spheres, centres of which occupy corners of a regular octahedron. It is a void surrounded by 6 spheres.the radius ratio for octahedral void is 0.414.

Tetrahedral void – In close packing arrangement each sphere in the second layer rests on

the hollow (triangular voids) in the three touching spheres in the first layer. The centres of

these four spheres are at the corners of a regular tetrahedron. The vacant space between

these four touching spheres is called

Tetrahedral voids. it is a void surrounded by 4 spheres.

Radius ratio of Tetrahedral void is 0.225.

Q.25A cubic solid is constituted by two elements X and Y. Atoms of Y occupy the corners of the cube and X is at the centre. Give the compound’s formula. Also, provide the coordination numbers of X and Y.

Ans:

Given:

Atoms of Y occupy the corners of the cube. => Number of Y atoms in a unit cell = 8 x 1/8 = 1 The atom of X occupies the body centre. => Number of Y atoms in a unit cell = 1 Therefore, the ratio of the number of X atoms to the number of Y atoms; X : Y = 1 : 1 Thus the formula of the compound is XY And the coordination number of both the elements is 8.

Q.26Studies demonstrate that the formula of nickel oxide is Ni0.98O1.00. Find the fractions of nickel that exist as Ni2+ and Ni3+ ions.

Ans:

The given formula of nickel oxide is Ni0.98 O1.00. Thus, the ratio of the number of Ni atoms to the number of O atoms is, Ni : O = 0.98 : 1.00 = 98 : 100 Now, Total charge on 100 O2− ions = 100 × (−2) = −200 Let the number of Ni2+ ions = x. So, the number of Ni3+ ions is 98 − x. Now, Total charge on Ni2+ ions = x(+2) = +2x Similarly, total charge on Ni3+ ions = (98 − x)(+3) = 294 − 3x As, the compound is neutral, we can write: 2x + (294 − 3x) + (−200) = 0 ⇒ −x + 94 = 0

⇒ x = 94 Therefore, number of Ni2+ ions = 94And, number of Ni3+ ions = 98 − 94 = 4Thus, the fraction of nickel that exists as NiAnd, the fraction of nickel that exists as Ni

Q.27Find the number of voids in 0.2 moles of a compound forming hexagonal closestructure. What number of these are tetrahedral voids?

Ans:

Given: No. of close-packed particles = 0.2 × 6.022 × 10

Thus, no. of octahedral voids = 1.2044 × 10

And, no. of tetrahedral voids = 2 × 1.2044 × 10

Thus, total number of voids = 1.2044 × 10

Q.28 A compound is made up of two elements A and B. The atoms of element B form ccp and element A’s atoms take up 1/3rd of the tetrahedral voids. Find the formula of this compound.

Ans:

Given: Atoms of element B form ccp, thus no. of atoms = nNo. of oct voids = n No. of td voids = 2n = 2 x n(1/3) = 2n/3Therefore : The formula of the compound is A : B2n/3 : n 2:3 = A2B3

In close packed structures of crystalline solids, only a part of the total available space is

occupied by constituent particle

The percentage of total space filled by the particles is called

Different types of packing arrangements have different packing efficiency.

Packing efficiency in Simple Cubic Lattice (SCC)

A unit cell of simple cubic lattice contains one atom.

ions = 94 − 94 = 4

Thus, the fraction of nickel that exists as Ni2+ = 94/98 = 0.0959 And, the fraction of nickel that exists as Ni3+ = 4/98 = 0.041

Find the number of voids in 0.2 moles of a compound forming hexagonal closestructure. What number of these are tetrahedral voids?

particles = 0.2 × 6.022 × 1023= 1.2044 × 1023

= 1.2044 × 1023

And, no. of tetrahedral voids = 2 × 1.2044 × 1023 = 2.4088 × 1023

Thus, total number of voids = 1.2044 × 1023 + 2.4088 × 1023 = 3.6132 × 10

e up of two elements A and B. The atoms of element B form ccp and element A’s atoms take up 1/3rd of the tetrahedral voids. Find the formula of this compound.

Atoms of element B form ccp, thus no. of atoms = n

No. of td voids = 2n = 2 x n(1/3) = 2n/3

The formula of the compound is A : B

PACKING EFFICIENCY

n close packed structures of crystalline solids, only a part of the total available space is

occupied by constituent particles.

The percentage of total space filled by the particles is called packing efficiency.


Packing efficiency in Simple Cubic Lattice (SCC)

contains one atom.

Find the number of voids in 0.2 moles of a compound forming hexagonal close-packed

= 3.6132 × 1023

e up of two elements A and B. The atoms of element B form ccp and element A’s atoms take up 1/3rd of the tetrahedral voids. Find the formula of this compound.

n close packed structures of crystalline solids, only a part of the total available space is

packing efficiency.


Let the side of a simple cubic lattice is ‘a’ and radius of atom present in it is ‘r’.

Since, edges of atoms touch each other, therefore, a = 2r

Volume of cube =Side3 =a3=(2r)3

Volume of one atom =4/3πr3

Number of atoms in simple cubic unit

Thus, packing efficiency (in %) = ��


=4/3πr�

�� X100

=��

�� X100

=�

� X100

=0.524 X100

=52.4%

Thus, packing efficiency of bcc structure=52.4%

Packing efficiency of body cantered cubic (bcc) structure:

In body cantered cubic unit cell, one atom is present in body centre apart from 4 atoms at its

corners. Therefore, total number of atoms

Let a unit cell of bcc structure with side a.

Let FD (face diagonal) = b and body diagonal AF = c


Since, edges of atoms touch each other, therefore, a = 2r

=8r3

unit cell =1

��

�� ×100

��

�� ×100

Thus, packing efficiency of bcc structure=52.4%



corners. Therefore, total number of atoms present in bcc unit cell is equal to 2.

Let a unit cell of bcc structure with side a.

Let FD (face diagonal) = b and body diagonal AF = c




present in bcc unit cell is equal to 2.

Let the radius of atom present in unit cell = r

Now, in right angled ∆EFD

FD2=ED2+EF2

Or, b2=a2+a2=2a2………………..(i)

Or, b=a√ 2 ……………(ii)

Now, in ΔAFD,

AF2=FD2+AD2

Or, c2=b2+a2

Or, c2=2a2+a2 (from equation (i))

Or, c2=3a2

Or, c=a√ 3

Since c is equal to 4r

Therefore, c = 4r=a√ 3

Or, a=4r/√ 3 …………………..(iii)

Volume of cube =Side3 = a3`

After substituting the value of a from equation (iii) we get

Volume of cube =(4r/√ 3 )3=64r3/3√3

Volume of a sphere =4/3 πr3

Volume of 2 spheres present in bcc structure =2×4/3πr3

Now, packing efficiency (in %) =Volume of 2 spheres in unit cell

Total volume of unit cell X 100

=�×�/��3

��3/3√3X100

=��√�

�� X100

=�√�

�X 100

=�.�� .��

� X100

=0.679 X100 =67.9% = 68%

Thus, packing efficiency of bcc structure=68%

Calculation of pacing efficiency in hcp and ccp structure:

The packing efficiency can be calculated by the percent of space

unit cell.


Since there are 4 atoms in the unit cell of hcp or ccp structure

Therefore, packing efficiency of hcp or ccp structure=

Let the edge length of an unit cell = a

And diagonal AC = b

Now, in ∆ ADC,

AD is perpendicular, DC is base and AC is diagonal

AC2=AD2+DC2

Or, b2=a2+a2

Or, b2=2a2

Or, b=a√ 2

Let r is the radius of sphere, so b=4

Thus, b=4r=a√ 2

Or, a=4r/√ 2 ………(1)

Thus, packing efficiency of bcc structure=68%

alculation of pacing efficiency in hcp and ccp structure:

The packing efficiency can be calculated by the percent of space occupied by spheres present in a

��

�� ×100

Since there are 4 atoms in the unit cell of hcp or ccp structure

Therefore, packing efficiency of hcp or ccp structure=Volume of 4 spheres in unit cell

Total volume of unit cell

Let the edge length of an unit cell = a

AD is perpendicular, DC is base and AC is diagonal

=4r

alculation of pacing efficiency in hcp and ccp structure:

occupied by spheres present in a

cell X 100

Now, volume of cube =Side3=a3

Substituting the value of a from equation (i) we get

Volume of cube =(4r/√ 2 )3=64r3

Now, volume of sphere =4/3πr3

Since oneunit cell of ccp or hcp contains 4 atoms, i.e. 4 spheres

Therefore, volume of 4 atoms, i.e. 4 spheres

Now, packing efficiency (in %) =Volume

=�×�/�� …

��/�√� …X100=

=��√�

��X100=

=�

�√� X100 =

=�.��

� � �.�� 100 =

=0.74 X 100= 74%

Thus, packing efficiency of hcp or ccp structure=74%

Density of Unit CellA unit cell is a three-dimensional structure occupying one, two or more atoms. With the help of dimensions of unit

cells, we can evaluate the density of the unit cell. To do so let’s consider a unit cell of edge length ‘a’, therefore the

volume of the cell will be ‘a3’. Also, density is defined as the ratio of the mass of unit cell and volume of the unit

cell.

So we can write

Mass of unit cell varies with number of atoms “n” and mass of a single atom “m”. Mathematically mass of unit cell

is the product of number of atoms “n” and mass of one atom “m” i.e.

Mass of Unit Cell = number of atoms

Mass of Unit Cell = n × m

Also from quantitative aspect of atoms, mass of one atom can be written in terms of Avogadro Number (N

molar mass of atom ( M), that is,

Mass of one atom (m) = ��

��

Substituting the value of a from equation (i) we get

3/2√2 …………….(ii)

or hcp contains 4 atoms, i.e. 4 spheres

Therefore, volume of 4 atoms, i.e. 4 spheres =4×4/3πr3=4×4/3πr3 ……………….(iii)

Volume of 4 spheres in unit cell

Total volume of unit cell X 100

Thus, packing efficiency of hcp or ccp structure=74%

Cell dimensional structure occupying one, two or more atoms. With the help of dimensions of unit

we can evaluate the density of the unit cell. To do so let’s consider a unit cell of edge length ‘a’, therefore the

’. Also, density is defined as the ratio of the mass of unit cell and volume of the unit


is the product of number of atoms “n” and mass of one atom “m” i.e.

atoms (n) X mass of one atom (m)

Also from quantitative aspect of atoms, mass of one atom can be written in terms of Avogadro Number (N

��

��

……………….(iii)

dimensional structure occupying one, two or more atoms. With the help of dimensions of unit

we can evaluate the density of the unit cell. To do so let’s consider a unit cell of edge length ‘a’, therefore the

’. Also, density is defined as the ratio of the mass of unit cell and volume of the unit


Also from quantitative aspect of atoms, mass of one atom can be written in terms of Avogadro Number (NA) and

https://files.askiitians.com/cdn1/images/2017814-19156147-2038-formula-of-density-of-unit-cell.png

m = ��

��

Volume of Unit Cell = a3

Placing the required values in equation 1 we get

Therefore if we know molar mass of atom “M”, number of atoms “n”, the edge length of unit cell “a” we can

evaluate the density of unit cell. The density of the unit cell is the same as the density of the substance.

At absolute zero, crystals tend to have a perfectly ordered arrangement. This arrangement

corresponds to state of lowest energy. As the temperature increases, the crystals start

deviating from the perfectly ordered arrangement.

Any deviation from the perfectly ordered arrangement c

imperfection.

The defects which arise due to the irregularity in the arrangement of atoms or ions are

called atomic imperfections. The atomic defects caused by missing or misplaced ions are

called point defects.

There are three kinds of point defects.

i)Stoichiometric point defect

ii) Non- Stoichiometric point defect

iii)Impurity defect

i. Stoichiometric point defect:

The compounds in which the numbers of positive and negative ions are exactly in the ratios

indicated by their chemical fo

thatdo not disturb the stoichiometry (the ratio of numbers of positive and negative ions) are

called stoichiometric defects. These are of following types:

a) Vacancy Defects;

When in a crystal some

defect. It is introduced when crystal is heated. Due to this defect the density of the

substance decreases.

Placing the required values in equation 1 we get

if we know molar mass of atom “M”, number of atoms “n”, the edge length of unit cell “a” we can


Imperfections in solids

tend to have a perfectly ordered arrangement. This arrangement


deviating from the perfectly ordered arrangement.

Any deviation from the perfectly ordered arrangement constitutes a defect or



of point defects.

Stoichiometric point defect

Stoichiometric point defect:


indicated by their chemical formulae are called stoichiometric compounds. Thedefects


called stoichiometric defects. These are of following types:

When in a crystal some lattice sites are vacant, the crystal is said to have vacancy


if we know molar mass of atom “M”, number of atoms “n”, the edge length of unit cell “a” we can


tend to have a perfectly ordered arrangement. This arrangement


onstitutes a defect or




rmulae are called stoichiometric compounds. Thedefects


lattice sites are vacant, the crystal is said to have vacancy


b) Interstitial defects;

This type of defect is caused due to the presence of constituent particles in the

normally vacant interstitial sites in the crystal. The interstitial defects result in increase

in density of the substance.

Vacancy and interstitial defects can be shown by non-ionic solids.

Ionic solids show vacancy and interstitial defects as Schottky and Frenkel defects.

c) Schottky defect: This type of defect is created when one positive ion and one negative

ion are missing from their respective positions leaving behind a pair of holes. Thus,

Schottky defect is simply a vacancy defect in ionic solids,

Schottky defects are more common in ionic compounds with high coordination

number, and where the sizes of positive and negative ions are almost equal. For

example, NaCl, KCl, CsCl, AgBr and KBr.

The presence of large number of Schottky defects in crystal results in significant

decrease in its density.

d) Frenkel Defect: This type of defect is created when an ion leaves its correct lattice site

and occupies an interstitial site. Frenkel defect is a combination of vacancy defect and

interstitial defect. It creates a vacancy defect at its original site and interstitial defect at

its new location. It is also called dislocation defect.

It does not have any effect on the density of the solid. Frenkel defects are common in

ionic compounds which have low coordination number, and in which there is large

difference in size between positive and negative ions. For example, ZnS, AgCl, AgBr and

AgI.

ii. Non- Stoichiometric point defect

There are many compounds in which the ratio of positive and negative ions present in the

compound defers from that required by ideal chemical formula of the compound. Such

compounds are called Non-Stoichiometric or Berthollide compounds.

In these compound compositions may vary over a wide range. For example, FeO exists as

non - stoichiometric compound having composition Fe0.93O to Fe0.95O. In these

compounds balance of positive and negative charges is maintained by having either extra

electrons or extra positive charges.

The defects which disturb the stoichiometry of the compound are called non –

stoichiometric defects.

a) Metal excess defects due to anion vacancies

This type of defects are found in crystals which are likely to possess short key defects.

Anion vacancies in alkali halides are produced by heating the alkali halide crystals in an

atmosphere of alkali metal vapours. Under these conditions alkali metals atoms deposit

on the surface of the alkali halide crystal, halide ions move to the surface and combine

with metal ions. The electrons released during conversion of metal atoms into ions diffuse

into the crystal and occupy the sites vacated by anions. The holes occupied by electrons

are called F- centres ( colour centres) and are responsible for the colour of the compound

and many other interesting properties.

b) Metal excess defects due to interstitial cations

Another way in which metal excess defects may occur is, if an extra positiv

in an interstitial site. Electrical neutrality is maintained by the presence of an electron in

the interstitial site. This type of defects are exhibited by the crystals which are likely to

exhibit Frenkel defects.

c) Metal deficiency due to cation vacancies

The non– stoichiometric compounds may have metal deficiency due to the absence of a

metal ion from its lattice site. The charge is balanced by an adjacent ion having higher

positive charge. This type of defects are

metals. For example, non– stoichiometric FeO which is mostly found with a composition

Fe0.95O. In the crystals of FeO some Fe

balanced by the presence of requ

iii. Impurity defect

Metal excess defects due to interstitial cations

Another way in which metal excess defects may occur is, if an extra positiv



e to cation vacancies

stoichiometric compounds may have metal deficiency due to the absence of a


positive charge. This type of defects are generally shown by compounds of transition

stoichiometric FeO which is mostly found with a composition

O. In the crystals of FeO some Fe2+ ions are missing and the loss of positive charge is

balanced by the presence of required number of Fe3+ ions.

Another way in which metal excess defects may occur is, if an extra positive ion is present



stoichiometric compounds may have metal deficiency due to the absence of a


generally shown by compounds of transition

stoichiometric FeO which is mostly found with a composition

ions are missing and the loss of positive charge is

Defects in the ionic solids may be introduced by adding impurity ions. If the impurity ions

have different valency state than that of the host ions, vacancies are created. For

example, addition of SrCl2

occupies Na+ sites and produces cation vacancies equal to the number of the divalent ions

occupying substitutional sites.

NCERT EXERCISES [CONTINUE….]

Q.29 Find the packing efficiency of

(a) simple cubic (b) body-centred cubic (c) face-centred cubic (with the assumptions that atoms are touching each other).

Ans:Refer notes

Q.30 Silver is crystallized to a face4.07 × 10−8 cm and density is 10.5 g cm

Ans:

Given: Edge length, a = 4.077 × 10−8 cmDensity, d = 10.5 g cm−3 The given lattice is of fcc type, Thus the number of atoms per unit cell, z = 4We also know that NA = 6.022 × 10let M be the atomic mass of silver.We know, d = z x M/a3NA => M = d x a3 x Na / z

= 10.5 x 4.077 × 10−8 x 6.022 × 10

Q.31 An element with a density of 2.7 × 10makes a cubic unit cell with an edge length of 405 pm. What kind of cubic unit cell does it have?

Ans:

Given, Density ,d = 2.7 × 103 kg m-3 Molar mass, M = 2.7 × 10-2 kg /molEdge length, a = 405 pm = 405 × 10We know,



to NaCl yields solids solutions where the divalent cation

sites and produces cation vacancies equal to the number of the divalent ions

occupying substitutional sites.

NCERT EXERCISES [CONTINUE….]

Find the packing efficiency of a metal crystal for:

centred cubic (with the assumptions that atoms are touching each other).

Silver is crystallized to a face-centred cubic lattice. Given that the cell has an edge length of cm and density is 10.5 g cm−3. Find its atomic mass.

cm

Thus the number of atoms per unit cell, z = 4 = 6.022 × 1023 / mol

let M be the atomic mass of silver.

x 6.022 × 1023 ) / 4 = 107.13 g /mol

An element with a density of 2.7 × 103 kg m-3 has a molar mass of 2.7 × 10makes a cubic unit cell with an edge length of 405 pm. What kind of cubic unit cell does it have?

kg /mol Edge length, a = 405 pm = 405 × 10-12 m = 4.05 × 10-10 m



to NaCl yields solids solutions where the divalent cation

sites and produces cation vacancies equal to the number of the divalent ions

centred cubic (with the assumptions that atoms are touching each other).

centred cubic lattice. Given that the cell has an edge length of

has a molar mass of 2.7 × 10-2 kg / mol and it makes a cubic unit cell with an edge length of 405 pm. What kind of cubic unit cell does it have?

Avogadro’s number, NA = 6.022 × 1023 mol-1

We also know that d = ( Z x M)/(a3 x NA) => Z = (d x NA x a3)/M = 3.99 ∼4 This means that the cell is a face centred cubic.

Q.32.When solids are heated what kind of defects can arise in it? Find physical property affected by it and the way it is affected.

Ans:

Heating solids produce vacancy defects in the crystal. This means that when the heat is applied to the solid, some ions or atoms leave their lattice site totally, making those sites empty. Thereby, decreasing the solid’s density.

Q33Identify the type of stoichiometric defect present in :

(i) AgBr

(ii) ZnS

Ans:

(i) Frenkel and Schottky defects are present in AgBr. (ii) ZnS contains Frenkel defect.

Q34When a cation of higher valency is put in an ionic solid as an impurity, vacancies are created in the ionic solid. Explain how this happens.

Ans:

When a cation of higher valence is put in an ionic solid, it starts replacing cations of lower valency such that the crystal remains electrically neutral. Thereby, creating some vacant sites. For instance, if Sr2+ is put in NaCl, each Sr2+ ion replaces two Na+ ions. Thereby creating one vacant for every Sr+ ion introduced.

Q35 Explain using an appropriate example of how ionic solids, with anionic vacancies caused due to the metal excess defect, start developing a colour.

Ans:

Let us take an example of NaCl to explain this when NaCl crystals are heated in a sodium vapor atmosphere, sodium atoms are deposited on the crystal’s surface. This causes the Cl– ions to leave their lattice sites to form NaCl with the deposited sodium atoms. In this process, the sodium atoms on the surface lose their electrons to form Na+ ions and the released electrons move into the crystal to fill in the vacant anionic sites. These electrons absorb energy from the incoming visible light and get excited to a higher energy level. Thereby, imparting a yellow colour to the crystals.

Electrical properties of solid

On the basis of electrical conductivity, the solids can be broadly classified into three types:

a. Metal (conductors)

b. Insulators

c. Semiconductors

a). Conduction in metals

A metal is characterized by a band structure in which the highest occupied band (the valency

band), is only partially filled. If the highest energy band (conduction band) i.e. occupied is not

fully filled, electrons may be excited from lower energy level to higher energy level by

supplying a very small amount of energy because vacant orbitals lie just above the occupied

orbitals of highest energy. When the voltage is applied to a metal crystal, electrons are

excited to the unoccupied orbitals in the same band. Thus, a material with a partially filled

energy band is a conductor.

In a metallic crystal the electrons are free to move throughout the crystal. However, the

metal ions are fixed in position and can only vibrate. When a metal is heated the metal ions

start vibrating more vigorously and as a result motion of the valency electrons through the

crystal is hindered. Thus, electricalconductivity of metals decreases with increases in

temperatures.

b). Conduction in insulators

In case of insulators, the highest occupied band is completely filled. The forbidden band just

above the highest filled band is wide. Therefore, the energy gap between the fully filled

valency band and the vacant conduction band is very large and it is not possible to excite the

electrons to the conduction band. Most of the nonmetals are insulators.

c) Conduction in semiconductors

In case of semiconductors, the forbidden band is narrow. The thermal energy available at

room temperature is enough to excite some electrons from the highest occupied band

(valency band) to the next permitted energy band (conduction band).

The electrical conductivity of semiconductors increases with increase in temperature

because with rise in temperature more number of electrons can jump from valency band to

the conduction band. The conduction by pure substances such as silicon and germanium is

called is called intrinsic conduction.

n and p type semiconductors

The conductivity of pure Si and Ge is very low at room temperature and that can be

increased by adding appropriate amount of suitable impurity. The process is called doping.

Doping can be done with an impurity which is electron rich or electron deficit as compared to

the intrinsic semiconductor.

Doping with electron rich impurities

In pure Si each Si atom uses its four valency electrons for the formation of four covalent

bonds with the neighboring Si atoms. When Si is doped with some group-15 element, then

some of the positions in the lattice are substituted by atoms of group-15 element. The group-

15 elements have five valency electrons. After forming the four covalent bonds with Si, one

excess electron is left on them. Since this electron is not involved in bonding it becomes

delocalized and contributes to electrical conduction. Si doped with group-15 element behave

as a n-type semiconductor.

Doping with electron deficit impurities

Electrical conductivity of Si or Ge can also be increased by doping with some group-13

element such as B, Al or Ga. Group-13 elements have only three valency electrons. They

combine with group-14 elements to form an electron deficient bond or electron vacancy or a

hole. These holes can move through the crystal like a positive charge giving rise to electrical

conductivity. Group-14 elements doped with group-13 elements behave as p-type

semiconductors.

12-16 and 13-15 compounds

A large variety of compounds have been prepared by combination of group-13 and group-15

or group-12 and group-16 elements. These compounds possess an average valency of 4 and

behave as semiconductors just like Si and Ge. Some typical compounds of group-13-15 are

AlP, GaAs and InSb. ZnS, CdS, CdSe and HgTe are examples of group-12-16 compounds. The

bonds in these compounds have partially ionic character. The extent of ionic character

depends on the electronegativities of the two elements.

Applications of semiconductors

Semiconductors are widely used in electronic industries

i) A diode is a combination of p- type and n-type semiconductors and is used as a rectifier.

ii) Transistors are pnp or npn ‘Sandwhich’ semiconductors which are used to detect and

amplify radio signals.

iii) The solar cells are also made up of semiconductors which are efficient photo diode.

Magnetic properties of solid

The magnetic properties of substance are due to the magnetic moments associated with

electrons. Each electron in atom possesses magnetic moment due to two reasons.

1. Orbital motion of the electron is along the axis of rotation.

2. Its spin which is directed along the spin axis.

i) Para magnetism. The substance which are weakly attracted by magnetic field are called

paramagnetic substances. These substances have permanent magnetic dipoles due to the

presence of some species (atoms, ions or molecules) with unpaired electrons. The

paramagnetic substances lose their magnetism in the absence of magnetic field. For

example, O2, TiO, VO2 and CuO.

ii) Diamagnetism. Solids can be classified into different types depending upon their behavior

towards magnetic fields. The substance which are weakly repelled by magnetic field are

called diamagnetic substances. For example, H2O, C6H6, TiO2 and NaCl. Diamagnetic

substances have all their electrons paired.

iii) Ferromagnetism. The substances which are strongly attracted by magnetic field re called

ferromagnetic substances. These substances show permanent magnetism even in the

absence of magnetic field. Some examples of ferromagnetic solids are: iron, cobalt, nickel,

gadolinium and CrO2.

In solid state the metal ions of ferromagnetic substances are grouped together into small

regions called domains. Each domain acts as a tiny magnet. In an unmagnetized sample of

ferromagnetic substance the domains are randomly oriented and their magnetic moments

get cancelled. When such a substance is placed in magnetic field all the domains get oriented

in one direction and a strong magnetic effect is produced.

iv) Anti-ferromagnetism. Alignment of domains in opposite direction in a compensatory

manner and resulting in a zero magnetic moment (due to equal number of parallel and

antiparallel magnetic dipoles) gives rise to anti ferromagnetism. For example, MnO, Mn2O3,

and MnO2 are antiferromagnetic.

v) Ferrimagnetism.Alignment of domains in opposite directions resulting in a net magnetic

moment (due to unequal number of parallel and anti-parallel magnetic dipoles) gives rise to

ferrimagnetism. For example, Fe3O4 (where M is some divalent metal such as Mg, Zn, Cu etc.)

are ferrimagnetic in nature.

Ferromagnetic and ferrimagnetic substances change into paramagnetic substances at higher

temperatures due to randomization of spins. Fe3O4 which is ferrimagnetic at room

temperature, becomes paramagnetic at 850 K.

https://youtu.be/nXbxMts3ADk [conductors and insulators]

https://youtu.be/huW5QVdFUVA [semiconductors]

https://youtu.be/5A3vh84eq04 [magnetic properties of solids]

https://youtu.be/huW5QVdFUVA

https://youtu.be/5A3vh84eq04

NCERT EXERCISES;

Q .36. Write notes on the following, providing suitable examples: (a) Ferromagnetism (b) Para magnetism (c) Ferrimagnetism (d) Anti ferromagnetism (e) 12-16 and 13-15 group compounds.

Ans; Refer Notes

Q.37To convert a group 14 element into an n-type semiconductor a suitable impurity is doped into it. To which group does this impurity belong?

Ans:

This impurity should belong to group 15.

Q.38 Which substance would be a better choice to make a permanent magnet, ferromagnetic or ferrimagnetic. Explain your answer.

Ans: Substances like iron, cobalt, nickel, gadolinium and CrO2, are known as the ferromagnetic substances and make better permanent magnets as they are attracted very strongly by the magnetic field. These substances can also be permanently magnetized. The metal ions of ferromagnetic substances, in solid-state are grouped into small regions, which are known as domains. Each domain act as a tiny magnet. Meanwhile, in an magnetized piece of a ferromagnetic substance, the domains are randomly oriented and so the magnetic moments of the domains get cancelled. Nevertheless, when the substance is placed in a magnetic field, all the domains are oriented in the direction of the magnetic field and produces a strong magnetic effect. The ordering of the domains persists even after the removal of the magnetic field. Thus, the ferromagnetic substance becomes a permanent magnet.

Q.39 If the radius of the octahedral void is r and radius of the atoms in close packing is R, derive relation between r and R. Answer

Q.40 Classify each of the following as being either a p-type or an n-type semiconductor: (i) Ge doped with In (ii) B doped with Si. Answer (i) Ge (a group 14 element) is doped with In (a group 13 element). Therefore, a hole will be created and the semiconductor generated will be a p-type semiconductor. (ii) B (a group 13 element) is doped with Si (a group 14 element). So, there will be an extra electron and the semiconductor generated will be an n-type semiconductor.

Q.41 Explain the following terms with suitable examples: (i) Schottky defect (ii) Frenkel defect (iii) Interstitials and (iv) F-centres Answer (i) Schottky defect: Schottky defect is basically a vacancy defect shown by ionic solids. In this defect, an equal number of cations and anions are missing to maintain electrical neutrality. It decreases the density of a substance. Significant number of Schottky defects is present in ionic solids. For example, in NaCl, there are approximately 106 Schottky pairs per cm3 at room temperature. Ionic substances containing similarsized cations and anions show this type of defect. For example: NaCl, KCl, CsCl, AgBr, etc.

(ii) Frenkel defect: Ionic solids containing large differences in the sizes of ions show this type of

defect. When the smaller ion (usually cation) is dislocated from its normal site to an interstitial site, Frenkel defect is created. It creates a vacancy defect as well as an interstitial defect. Frenkel defect is also known as dislocation defect. Ionic solids such as AgCl, AgBr, AgI, and ZnS show this type of defect.

(iii) Interstitials: Interstitial defect is shown by non-ionic solids. This type of defect is created when some constituent particles (atoms or molecules) occupy an interstitial site of the crystal. The density of a substance increases because of this defect.

(iv) F-centres: When the anionic sites of a crystal are occupied by unpaired electrons, the ionic sites are called F-centres. These unpaired electrons impart colour to the crystals. For example, 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 from the crystal to its surface and combine with Na atoms, forming NaCl. During this process, the Na atoms on the surface of the crystal lose electrons. These released electrons diffuse into the crystal and occupy the vacant anionic sites, creating F-centres.

Q.42 In terms of band theory, what is the difference (i) Between a conductor and an insulator (ii) Between a conductor and a semiconductor Answer (i) The valence band of a conductor is partially-filled or it overlaps with a higher energy, unoccupied conduction band. On the other hand, in the case of an insulator, the valence band is fully- filled and there is a large gap between the valence band and the conduction band.

(ii) In the case of a conductor, the valence band is partially-filled or it overlaps with a higher energy, unoccupied conduction band. So, the electrons can flow easily under an applied electric field. On the other hand, the valence band of a semiconductor is filled and there is a small gap between the valence band and the next higher conduction band. Therefore, some electrons can jump from the valence band to the conduction band and conduct electricity.

Question 43 and 44 Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm. (i) What is the length of the side of the unit cell? (ii) How many unit cells are there in 1.00 cm3 of aluminium?

Answer

Q.45

Q.46 and 47

Class-XII Chemistry CHAPTER-I TYPES OF CRYSTALLINE SOLIDS

Documents

Transcript of Class-XII Chemistry CHAPTER-I TYPES OF CRYSTALLINE SOLIDS