Solid States

The stability and existence of solid-state favours lower temperatures as it depends on 2 rival forces.

When the temperature is low, under these circumstances the thermal energy of molecules ( energy which tends to move them faster and makes them apart; sometimes results in a change of state) is sufficiently low and intermolecular forces (forces b/w molecules, ions, atoms that tend to keep them closer) are high enough to make molecules bring in a closer contact that they attach oneself tightly to another and occupy fixed position (in solid-state).

• Isotropicity (equal values of physical properties) or anisotropicity (unequal values of symmetrical properties) is the property that deals with distribution of inherent physical property( eg. - mechanical strength, refractive index) of any substance to the different molecular axes in accordance with the symmetry of the molecules of that substance.

• Crystalline solids are anisotropic in nature i.e. they show different values of physical properties like electrical resistance or refractive index in different directions, unlike the amorphous solids. This is primarily because of their uneven arrangement of particles in different directions (x, y and z axes). It depends upon the symmetry of the molecules which are arranged in a different way in all the planes (3-dimensional x, y and z planes ).

This figure above shows an example of a simple 2-dimensional arrangement of 2 different types of atoms in a crystal. Here the CD direction row there is 2 types of atoms whereas in AB direction row is made of only one type of atoms – this arrangement will result in anisotropicity and different values for physical properties in different rows of directions.

• Quartz glass is an amorphous or non-crystalline solid because it lacks in a long-range order i.e. a regular long pattern of constituent particles in three dimensions which periodically repeats itself throughout the whole crystal and have a definite geometrical shape which is the case for other 3 options (i) Graphite (C), (iii) Chrome alum and (iv) Silicon carbide (SiC).

• Instead of that quartz glass has a short-range order i.e. the repetitive pattern falls through a short distance only and the regular pattern is scattered and they are disarrayed in nature (resemblances that of a liquid) unlike quartz itself which is a crystalline solid.

• Antiferromagnetism is the case where the magnetic moment of the atoms or molecules occur in accordance with spins of their electrons (usually in low temperatures) align themselves with a regular pattern with the neighbouring spins oriented in opposite directions cancelling each other’s magnetic moments which is option (iv) .

• Options (i) and

(ii) are simply the case for ferromagnetism when the substance is placed in a magnetic field the domains (substance particles grouped in different regions) orient themselves in the direction of the magnetic field which results in strong magnetism.

• Option (iii) is clearly the case of ferrimagnetism where the domains (tiny magnets) are aligned in parallel and anti-parallel orientation in unequal numbers which result in very weak magnetism.

Since, quartz glass in an amorphous solid having short range order of constituents. Hence value of refractive index is same in all direction, can be measured and not be equal to zero.

• Amorphous solids usually have definite values of various physical properties like mechanical strength or refractive index, electrical conductivity etc.(isotropic not anisotropic) through all the directions. This is due to lack of long-range repetitive arrangement or order unlike crystalline solids; arrangement of particles is indefinite in different directions (x,y & z planes) which results in an overall equivalent arrangement in all directions, therefore giving birth to isotropicity and the value of physical properties will be same along all directions.

• Other 3 options (i) On heating they may become crystalline at certain temperature , (ii) They may become crystalline on keeping for long time and

(iii) Amorphous solids can be moulded by heating are possible cases for amorphous solids.

• Crystalline solids usually have sharp and well-defined melting points because they have a regular pattern of constituent particles which repeats itself periodically throughout the crystal i.e. long-range order. Therefore they have same distance between them or they have same kind of adjacent neighbours and this regularity in crystal lattices create equal environments for all the particles. Thus the intermolecular forces among them are evenly equal and same amount of thermal energy will be needed to break each interaction b/w particles (atoms, molecules or ions) at the same time.

• Options (i) a regular arrangement of constituent particles observed over a short distance in the crystal lattice and (iii) same arrangement of constituent particles in different directions are entirely untrue for crystalline solids.

• And option (iv) different arrangement of constituent particles in different directions cannot explain the phenomena of sharp melting points of crystalline solids because this is in the side of contrary.

• Iodine molecules are a form of non-polar constituent particles which form crystals and held together by London force (or induced dipole-dipole interaction) which is a weakly dispersion force. It is possible when 2 adjacent atoms of molecules come in a position where a temporary dipole is formed, it is the weakest intermolecular force.

Other 3 options (ii) dipole-dipole interactions (iii) covalent bonds and (iv) coulombic forces are not the case of iodine molecules because they require polarity or ionic entities to occur.

• Diamond is a crystalline solid of non-metal Carbon and it has covalent bonding throughout the crystal. Between two adjacent atoms of the crystal there is a strong and directional interaction which results in a covalent bond which helps the atoms to held strongly at their positions. Therefore resulting in a very high melting point which indicates that diamond is a network solid.

The network structure of diamond

• The other 3 options (i) SO₂ (Solid) (ii) I₂ and (iv) H₂O (Ice) are examples of molecular crystalline solids

• in which SO₂ (Solid) is a type of polar molecular solid (strong dipole-dipole interaction b/w S and O results in stronger polar covalent bonds).

• I₂ can be classified as non-polar molecular solid (formed by weakest dispersion force, London force due to temporary induced dipole-dipole interaction.

• And at last H₂O (Ice) is hydrogen-bonded molecular solid (polar covalent strong interaction b/w H and O reinforces the formation of hydrogen bond.

• I₂ (s) is a non-polar molecular crystalline solid obviously not an electrical conductor i.e. they are non-conductors of electricity because it is formed by weakly dispersion forces; London forces and there is no entity (like ions or ) to conduct electricity.

H₂O (s), which is hydrogen-bonded molecular crystalline solid also do not conduct electricity due to stronger hydrogen-bonding between adjacent water molecules there are no free ions left to conduct electricity.

• On the other hand option (A) Mg (s), which is metal and eventually has electrical conductivity properties because it has free and mobile electron (valence shell) which are cast evenly throughout the crystal.

• And in the case of (B) TiO (s) which behaves like a metal also can conduct electricity because the d-electrons are partially filled and leads to electrical conductivity.

• Although the component ions of the ionic solids are not free for movement in the solid-state as they are held together by strong electrostatic forces ; they become free to move about in the molten state or aqueous state (when they get dissolved in water because) they can overcome coulombic attraction in that state and hence they have a certain amount of electrical conductivity which is high in the molten state.

• Other 3 options (ii) Brittle nature (iii) Very strong forces of interactions and (iv) Anisotropic nature are 3 important characteristics of ionic solids

• Graphite has an exceptional but typical structure which leads to conductance.

• In this structure, the carbon atoms are arranged in multiple layers and each C atom is covalently bonded to another 3 adjacent C atoms residing in the same layer.

• And the 4^th valence electrons which were left out of the C atoms are present among different layers and are therefore free for movement. These free electrons result in graphite in a good conductor.

The other 3 options, (i) lone pair of electrons (iii) cations and (iv) anions are not possible for neutral carbon atoms present in graphite.

• TiO₃ behaves as conductor or insulator mainly because of the energy gap variability between valence band and conduction band. This gap changes in accordance with temperature; as the temperature increases so decreases the gap and it leads to more conductive nature which is opposite in case of lowering of temperatures.

• Options (i) TiO (ii) SiO₂ and (iv) MgO are incorrect .

• TiO which behaves like a metal also can conduct electricity because the d-electrons are partially filled and leads to electrical conductivity.

• (ii) SiO₂ (quartz) is a network -solid which is covalent in nature and always behave as an insulator.

•

• And (iv) MgO is an ionic crystalline solid which means it acts as an insulator in the solid-state and conducts electricity in the molten or aqueous state.

•

• CrO₂ behaves like a metal also can conduct electricity. Because Cr is a transition metal and has d-electrons (electrons present in d-orbitals) which are partially filled and leads to electrical conductivity just like metals having free electrons.

(i) SiO₂ (quartz) is a network solid which is covalent in nature and always behave as an insulator ( no free electrons to conduct electricity).

And (ii) MgO is an ionic crystalline solid which means it acts as an insulator in the solid-state( as there is no free movement for ions) and conducts electricity in the molten or aqueous state(ions overcome electrostatic attractions and are free to move).

(iii) SO₂(s) is a polar molecular crystal and therefore is formed by stronger dipole-dipole interactions which make them non-conductors of electricity, unlike metals.

The lattice site in a pure crystal cannot be occupied by electrons; because in each lattice point of a pure crystal (no impurity) there is only space for either molecule or atom or ions which are essentially joined together in straight lines and leads to a specific geometrical shape for each crystal. And they are also arranged in stoichiometric ratio which is fixed in proportion.

• But electrons can only occupy when there is a vacancy or defect in the corresponding crystal, i.e. crystal is no more in pure form it has impurities.

Hence, options (i) molecule, (ii) ion and (iv) atom are not correct because they can occupy the lattice site of a pure crystal.

• Graphite cannot be classified as an ionic solid because it does not have the characteristic of an ionic solid-

• it does not have ions as constituent particles rather neutral C atoms.

• Even in water, it does not form any sort of ions for conductivity.

• Whereas, graphite belongs to the classes (i) conducting solid (having multiple-layered structures and 3 fold bonding it is left with free electrons for conduction.

• (ii) network solid (carbon atoms are covalently bonded and form network type crystalline structure) and

• (iii) covalent solid ( constituent C atoms are covalently bonded with each other in multiple layers).

• In Frenkel defect which is also known as dislocation defect, the smaller ions (usually the cations) get dislocated from its normal lattice site to an interstitial site, therefore resulting a vacancy defect in the normal site (vacancy for cation) and simultaneously an interstitial defect in the new location of the cation.

• On the other hand (ii) Schottky defect (iii) Vacancy defect (iv) Metal deficiency defect do not necessarily deal with cations.

• In (ii) Schottky defect the number of missing cations and anions are equal in order to maintain electrical neutrality in the crystal and thus decreases density.

•

• (iii)Vacancy defect occurs in case of any kind of vacancy in the crystal.

• And last (iv) Metal deficiency defect is not even a stoichiometric defect like the other options given. It is rather a non- stoichiometric defect that deals with less amount of metal as compared to the ideal stoichiometric proportion (loss of positive charge).

•The Schottky defect is basically a kind of vacancy defect with the tendency to maintain electrical neutrality, therefore same no.s of cations are anions are missing from the crystal lattice.

• Other 3 options are incorrect.

• (i) some of the cations move from their lattice site to interstitial sites – this occurs in the case for Frankel defect only.

• some lattice sites are occupied by electrons- this is the case of metal excess defect.

• (iv) some impurity is present in the lattice – this is observed in crystals with impurity defects.

• P-type semiconductors itself have positive charge carriers (holes) which are relatively free to move from one site to another (although the holes do not move actually , it is the electrons that move to holes and create a new hole in its original position).

• But at last it is neutral because there are fixed acceptors ( atoms ) who accept the electrons and become negative ; therefore a neutrality is worked out between the holes (positively charged) and the accepting atoms (negatively charged).

• Hence, options (i) , (iii) , (iv) are in correct.

• In order to build a n-type semiconductor from silicon, it has to be doped with electron-rich impurities (element having more valence electrons).

• As group 14 element Si has 4 valence electrons, therefore to get an extra electron carrier for the semiconductor it has to be doped with elements of valency 5 i.e. group 15 elements such as P or As.

• Four out of five electrons of group 15 element (P or As) will be forming four covalent bonds with the four silicon atoms which are adjacent.

• The fifth valence electron of the group 15 element now becomes an extra and gets freed and delocalised. These delocalised electrons enhance the conductivity of doped silicon.

Hence, other 3 options (i) 2 (ii) 1(iii) 3 are incorrect, because their doping would not form any n type semiconductor.

• We know that ,if n be the number of atoms in a crystal,

then the number of tetrahedral voids must be 2n.

• And for a face centred unit cell -

(i)there are 8 Corner atoms i.e. (8 X1/8) atoms per unit cell=1atom and

B. face-centred atom number is 6 i.e.( 6 X 1/2) atoms per unit cell= 3 atom (because each atom at the face centre is used by to adjacent unit cells)

Hence no. Of tetrahedral voids for face centred unit cell = 2(4) = 8.

Hence other 3 options (i) , (iii) and (iv) cannot be correct.

• AgBr(s) has CCP(cubic close-packed) type crystals and shows both of the Schottky defect and the Frenkel defect.

• this is mainly due to the intermediate sizes of the radius of Ag⁺ and Br^- ions for which they are able to show defect with cation dislocation in the lattice (Frankel ) and the missing of equal no.s of cations and anions (Schottky) as well.

• AgBr(s) crystals do not show (C) Metal excess defect or (D) Metal deficiency defect hence other options (ii) (C) and (D) (iii) (A) and (C) (iv) (B) and (D) are all incorrect.

• In hcp (hexagonal close packing), the third layer and the first layer resembles each other in arrangement of the spheres and is able to cover all the tetrahedral voids.

• Since the pattern of spheres is repetitive in alternative layers, the stacking for hcp may be described as "A-B-A-B-A-B."

• The atoms in a hexagonal closest packed structure efficiently occupy 74% of space while 26% is empty space which is one of the efficient packings.

• The arrangement of ccp (cubic close packing )also efficiently fills up 74% of space( Similar to hexagonal closest packing).

• In ccp the 2nd layer of the spheres is placed on half of the depressions of the first layer and the third layer is completely different than that of the first two layers and is stacked in the depressions of the second layer, thus it covers all of the octahedral voids.

• The spheres in the third layer are not in line with those in layer 1, and the pattern of spheres does not repeat until a fourth layer is added. The fourth layer is the same as the first layer, so the arrangement of layers is "A-B-C-A-B-C."

• In a body-centred cubic arrangement of lattices, the constituent atoms are located on the 8 corners of the cube with one atom at the centre of the cube.

Let the edge length of the cube be ‘a’ (ED or BC in the fig) and radius of each constituent spherical particle be ‘r’ and the arrangement is such as that atom at the centre is in touch with other two atoms diagonally.

• Hence the body diagonal AF will be equal to = r +2r + r = 4r (1)

In the right angle ∆ EFD,

FD² or b²⁼ FE² + ED² ( by Pythagoras’s theorem ) = a² + a² = 2a²

Or, FD or b = √2a

• In the right-angled triangle, ∆ AFD

AF² or c²= AD² + FD² = a² + b² = a² + 2a² (as b²= 2a² ) =3a²

Or, AF or c =√3a (2)

Hence, from equations (1) and (2) we can get ,

AF = 4r =√3a a

Or, r = √3a/4

And a = 4r/√3

• Then, the volume of 1 cubic unit cell is a³ = (4r/√3) = 64r³/ 3√3

• Since in body centred arrangement , there are 2 atoms per unit cell

Volume occupied by the atoms = 2 X 4/3 π r³ = 8/3 π r³

Since, Packing efficiency = Volume occupied by two spheres in the unit cell X 100 %

Total volume of the unit cell

=8/3 π r³X 100 % = 68.04 % = 68%_appx.

64r³/ 3√3

Therefore , Packing efficiency in a body centred cubic lattice is 68 %

So, empty space or void space will be = (100 -68 )% = 32 %

• Hexagonal close-packed crystal structure forms only when the centres of spheres in the 3^rd layer are vertically above the centres of first layer (A) as the 2^nd layer is placed in only one type of depressions of the 1^st layer( at each junction of spheres ).

• This depression gives rise to voids which by placing new layer of spheres i.e. the 3^rd one (also A) can be covered well.

• This third layer has to resemble the first layer because layer 1 and layer must be in the same alignment and thus the pattern of spheres is repeated in alternate layers. That is why, this pattern is often written as ABAB .......pattern.

Unlike in ccp which is the actual case for (iv).

• The coordination number is 12 – it is true as in hcp each sphere in a layer is surrounded by 6 spheres in the same plane , 3 spheres above it and 3 spheres below it – therefore making a total 12 which is the coordination number for this arrangement.

• (iii) It has 74% packing efficiency – also true as it resembles with ccp in efficiency it can be calculated in accordance with ccp structure ;

If the unit cell edge length= a and face diagonal AC=b

Then in right angled triangle ∆ ABC ,AC² or b² =BC²+AB²= (a²+a²⁾=2a²

Or , b=√2a

if r is the radius of the sphere, then b=r +2r+r=4r=√2a

or, a=4r/√2=2√2r (i.e. r=a/2√2)

it is known that each unit cell in ccp/hcp arrangement has effectively four spheres.

Hence , the total volume occupied by four sphere is = to 4×(4/3)πr³ and

Therefore, volume of the cube = a³ = (2√2r)³.

Therefore, packing efficiency= (volume occupied by 4 spheres in unit cell/ total volume of the unit cell )×100℅

= 4×(4/3)πr³/(2√2r)³ ×100%=74%

• Tetrahedral voids of the second layer are covered by the spheres of the third layer –

this is also a true fact for hcp , because When a sphere of the second layer(B) is placed above the void of the first layer (A) always tetrahedral void is formed this is because a tetrahedron forms when the centres of four spheres are joined and this tetrahedral voids can be totally covered by the placement of a 3^rd layer (A) above them.

•

In NaCl, the Cl^- ions are arranged in ccp arrangement where Cl^- ions are present at the corners as well as the center of the face of the cube. The Na⁺ ions are located such that each Na⁺ ion is surrounded with six chloride ions. Na⁺ ions occupy all the octahedral voids. Each chloride ion is also surrounded by six Na⁺ ions and the stoichiometry is 1:1. Thus the coordination number for both species is the same.

In CaF₂ arrangement, the Ca²⁺ ions are arranged in ccp arrangement where Ca²⁺ ions are placed at the corners as well as the center of the face of the cube. Fluoride ions occupy tetrahedral voids and in way that each ion is surrounded by four calcium ions. The coordination number of F^- is 4. There are two tetrahedral sites available for each calcium ion, and thus the F^- ions occupy all eight tetrahedral voids. However, the stoichiometry is 1:2, and the coordination number of calcium ion is 8, therefore the coordination number for these species is not the same.

In Na₂O arrangement, the O^2- ions are arranged in ccp arrangement where O^2- ions are placed at corners as well as the center of the face of the cube. Na⁺ ions occupy all tetrahedral voids surrounding O^2- ion. Each sodium ion is surrounded by 4 oxide ions and each oxide ion is surrounded by 8 sodium ions. The coordination number of Na⁺ is 4 and O^2- is 8, therefore the coordination number of these species is not the same.

In Zinc blende ZnS arrangement, the S^2- atoms are in ccp arrangement, where each sulphideion is surrounded by four zinc ions. There are eight tetrahedral voids and four sulphide ions occupy half of the tetrahedral holes. Each sulphide ion itself is surrounded by four zinc ions, thus making both of their coordination numbers 4.

Thus the correct answer is (i).

Two dimension close packed structures, where rows of identical spherical molecules are stacked on top of each other, can be done in two ways, square and hexagonal. In square close packed structure, the second row is stacked just above the first row, and the spheres are aligned horizontally and vertically as the way mentioned above. If we observe a sphere in this arrangement, it is surrounded by four spheres which are in direct contact with it. Joining of the centers of the four spheres also forms a square, thus giving the term square close packed structure. Thus by definition of coordination number, a sphere in this structure is 4.

The correct number is (iii).

The process of increasing the conductivity of intrinsic semiconductors by addition of appropriate amounts of suitable impurity is calling doping. Dislocation, Schottky, and Frenkel defects are stoichiometric defects, these are vacancy defects and do not affect the electronic conductivity of the solid. Addition of dopant induces an electronic defect in the parent intrinsic solid which can be positive or negative. The correct answer is (iv).

When Silicon is doped with an electron-rich impurity, for example, an element from Group 15 like Phosphorous and Arsenic, which has 5 valence electrons, some of the electrons will occupy some of the lattice sites in the silicon crystal. Four out of the five will form covalent bonds with the four neighbouring silicon atoms, while the remaining electron becomes delocalized and increases the conductivity of the silicon. The increase in the conductivity is caused due to increase in negative charge and thus the doped silicon is called n-type semiconductor. The correct answer is (ii).

Every element has some magnetic property due to the property of the electrons present. On the basis of their magnetic properties, substances can be classified as paramagnetic, diamagnetic, ferromagnetic, antiferromagnetic and ferrimagnetic.

Paramagnetic substances are weakly attracted by magnetic field and are magnetized in a magnetic field in the same direction. Option (i) is a true statement.

Ferromagnetic substances are those substances which are strongly attracted to a magnetic field and can be magnetized permanently. The domains of the substance when placed in a magnetic field all get oriented in one direction, that is, the direction of the magnetic field and persist even when the field is removed. Thus option (ii) is not true.

Antiferromagnetic substances show the presence of domains like ferromagnetic substances but they are all oppositely oriented and cancel out each other’s magnetic moment. Option (iii) is correct.

Diamagnetic substances are those which are weakly repelled by a magnetic field. This happens because all the electrons in these substances are paired and no electron is unpaired. Pairing out electrons cancels the magnetic moment. Option (iv) is correct.

Ionic solids are a type of crystalline solids, where the constituent particles are ions. When voids are generated during formation of crystal structure, anions, which are the bigger ions (because of the repulsion of the electrons in the shells) form the close packed structure and smaller ions, the cations occupy the voids. Option (i) is correct.

Continuing the previous statement, not all tetrahedral or octahedral voids are occupied, if the smaller ions are small enough, they occupy the tetrahedral voids, otherwise they fit in octahedral voids. Option (ii) is correct.

The fraction of occupied tetrahedral voids to octahedral voids depends on the chemical formula of the substance and it is not necessary that all voids are occupied. Option (iii) is correct.

The fraction of octahedral or tetrahedral voids does not depend on the radii of the ions occupying the voids but the chemical formula of the substance. Option (iv) is incorrect.

A ferromagnetic substance is defined as a substance which is strongly attracted by a magnetic field. In the unmagnetised substance, the domains are randomly oriented but when the substance is placed in a magnetic field, the domains are oriented in the direction of the magnetic field. Option (i) is correct.

Diamagnetic substances are those which are weakly repelled by the magnetic field as their domains get oriented in the opposite direction of the magnetic field. Option (ii) is incorrect.

When domains are oriented randomly, there is no net magnetic behavior shown by the substance. All substances have some amount of magnetic properties associated to it and are affected by the magnetic field Option (iii) and (iv) are incorrect.

Packing efficiency is defined as the total space occupied by the constituent particles in a unit cell of a crystal lattice. Packing efficiency of each unit cell can be calculated.

For fcc,

Let us consider an fcc unit cell cube with edge length a and the face diagonal which connects the centers of the particles at the edges and the face AC = b.

In ΔABC,

AC² = b² = BC² + AB²

= a² + a² = 2a²

b = a

If r is the radius of the sphere, then b = 4r =

Or,

Each unit cell in in ccp has 4 spheres. Volume of each sphere is ³. So, volume of 4 spheres is 4 × ³ and the volume of the cube is a³ or [³.

Packing efficiency is then calculated as

Packing efficiency =

Now we calculate the packing efficiency of bcc.

In body-centered cubic cell, the atoms are placed at the corners of the cube with one atom in the center of the cube. The central atom will be in touch with the other two diagonally arranged atoms.

In ΔEFD, FD² = EF² + ED²

EF = ED = a, which is the side of the cube. FD = b.

b = a

In ΔAFD, let the diagonal AF = c.

c² = a² + b² = a² + 2a² = 3a².

c = a.

The length of the diagonal of the cube is equal to 4 times of the radius of the sphere as the three spheres are along the diagonal.

Therefore, a = 4r

a = and r = a.

This structure contains two spherical atoms and the volume is calculated as 2 × ³

The volume of the cube is a³. Substituting a gives []³.

Calculating packing efficiency,

Packing efficiency = = 68%

Now we calculate the packing efficiency of simple cubic cell.

In this arrangement, the atoms are located only at the corners of the cube and the edges of the spheres touch each other. The side of the cube, a, is equal to twice the radii r of the sphere due to this structure. This can be written as a = 2r.

Volume of the cube is (side)³ i.e. a³ = 8r³

A simple cubic cell contains only one atom, so the volume will be

Packing efficiency is calculated as

Packing efficiency =

= = 52.4%.

From these results, it is concluded that the correct answer is (ii).

Frenkel defect is a stoichiometric point defect which occurs in the lattice arrangement of crystalline solids. Frenkel defect is shown by ionic solids where a smaller ion in the lattice, usually a cation, is dislocated and it occupies another site. The site where the ion leaves creates the vacancy defect and the new site of placement is the interstitial defect. This defect is thus known as dislocation defect. The correct answer is (i).

In arranging the layers of two dimensional close packed structures, the cubic close packed structure is formed when the third layer of spheres are placed above the second layer in a manner that its spheres cover the octahedral voids, where this third layer is not aligned with either the first or the second layer. Only when the fourth layer is placed on top, the fourth layer aligns with the first, forming an …ABCABC… arrangement. If we consider a ccp unit cell, we see that there are atoms present at the corners of the cube and also at the center of the faces of the cube. If we divide the cube equally in eight parts, we get eight cubes with atoms at alternate corners and each small cube has four atoms. If we joined the corners together, a tetrahedron would be formed. So the eight cubes would make eight tetrahedral voids and one unit cell of the ccp structure makes eight tetrahedral voids. The correct answer is (iv).

Referring explanation to Question 33, edge lengths can be calculated in terms of the radius of the spheres.

In case of fcc structure, since the atoms are located at the corners and the center of the face of the cube, the edge length can be calculated from the diagonal formed by joining the radii of the three spheres.

In ΔABC,

AC² = b² = BC² + AB²

= a² + a² = 2a²

b = a

If r is the radius of the sphere, then b = 4r =

Or,

In case of bcc structure, the atoms are placed at the corners of the cube with one atom in the center of the cube. The central atom will be in touch with the other two diagonally arranged atoms.

In ΔEFD, FD² = EF² + ED²

EF = ED = a, which is the side of the cube. FD = b.

b = a

In ΔAFD, let the diagonal AF = c.

c² = a² + b² = a² + 2a² = 3a².

c = a.

The length of the diagonal of the cube is equal to 4 times of the radius of the sphere as the three spheres are along the diagonal.

Therefore, a = 4r

a = .

In case of simple cubic cell, the atoms are located only at the corners of the cubic unit cell with the spheres touching the edges of the adjacent spheres. Thus the edge length is equal to twice the radius of the sphere, 2r.

The correct option is (i).

Solids have a wide range of electrical conductivities, and can be categorized into metals, insulators, and semiconductors. Metals are solids with conductivities between 10⁴ and 10⁷ ohm^-1 m^-1, with metals having conductivity in the order of 10⁷ ohm^-1 m^-1 being good conductors. Insulators are solids with low conductivities and their conductivities range from 10^-20 ohm^-1 m^-1 to 10^-10 ohm^-1 m^-1. Semiconductors are solids with conductivities in the intermediate range, between 10^-6 and 10⁴ ohm^-1 m^-1. The conductivity of metals is many degrees higher than insulators, while semiconductors have higher conductivity than insulators but lower than metals. The correct answer is (i).

All real structures are three dimensional. Hexagonal close packed structure is formed from three dimensional close packing from two dimensional hexagonal close packed layers placed on top of each other. Here, the second layer is aligned on the first layer such that the spheres of the second layer fits into the depressions i.e. the spaces between the spheres of the first layer. Not all the triangular depressions of the first layer is covered by the spheres of the second layer. Option (i) and (ii) are correct statements. If we observe this structure further, we see the formation of tetrahedral and octahedral voids. When a sphere of the second layer is located above the triangular void of the first layer, the shape of the void formed when the centers of these four spheres are joined is a tetrahedron and is known as tetrahedral void. In other places in the structure, the triangular voids of the second layer lie above the triangular voids of the first and the triangular shapes do not overlap, forming a void surrounded by six spheres and is known as octahedral void and joining the centers of these spheres forms an octahedron. Now in the case of hcp structures, the spheres are alternately aligned, as in the spheres of the third layer are aligned with the spheres of the first layer. From these statements, option (iii) and (iv) are incorrect.

Antiferromagnetic substances are solids which show domain structure similar to ferromagnetic substances but these domains are oppositely oriented, canceling out each other’s magnetic moment. They do not get oriented by the magnetic field because they are unaffected by it. Hence (iii) and (iv) are correct.

Vacancy defect is a type of stoichiometric defect, where some of the lattice sites in the solid remain vacant. It results in the decrease in the density of the substance. Option (i) is correct.

Interstitial defect is also a type of stoichiometric defect where constituent particles such as atoms and ions occupy the interstitial spaces. This defect increases the density of the substance. Option (ii) is correct.

Impurity defect is a type of point defect, which is caused by the presence of an impurity in a crystal. For example, ionic impurities in ionic solids which have a different valence than the main constituent, cause formation of vacancies. It affects the density of the substance, thus option (iii) is incorrect.

Frenkel defect is a stoichiometric point defect seen in ionic solids. This defect causes both vacancy and interstitial defects. This occurs when a smaller ion, usually a cation is displaced from its original position to an interstitial site, causing a vacancy defect at the original position and an interstitial defect at the new position. Frenkel defect occurs in ionic solids where there is a huge size difference between cations and anions. It does not change the density of the solid. Hence (iv) is incorrect.

Metals conduct electricity through the movement of electrons in solid as well as molten state. The conductivity is dependent on the number of valence electrons. Electrical conductivity of a substance depends on the ability of the solid to flow electrons from the valence band to a conduction band. In a crystal lattice, the atomic orbitals of the atoms overlap and form molecular orbitals with very close values in energy, and form a band. A partially filled valence or an overlapping valence band are featured in metals. The band gap, even if negligible can be determined in metals using UV-Vis spectrometry. Options (i), (ii) and (iv) are correct.

Semiconductor crystals doped with electron-deficient impurities such as Group 13 elements B, Al or Ga which contain only 3 valence electrons. The fourth missing electron spot is called the electron vacancy or electron hole. An electron from the neighbouring atom can come and fill up the vacancy, leaving a vacancy in its original position. On application of an electric field, the holes will appear to move in the direction opposite to the direction of the electrons. The dislocated electrons move towards the positive plate, the holes move towards the negative plate. These type of semiconductors are known as p-type semiconductors, p being positive. Option (iii) is impossible to occur, hence (i), (ii) and (iv) are correct statements.

Silicon doped with electron-rich impurity is an n-type semiconductor, where n stands for negative because of the excess of delocalized electrons. The increase in conductivity is because of negatively charged electrons. Hence, (i) is incorrect while (ii) and (iii) are correct. Electron vacancies are formed in p-type semiconductors where the vacancies are formed due to addition of electron-deficient impurities. Option (iv) is incorrect.

When excess of KCl ions are heated, the colour of the substance appears violet or lilac. This phenomenon can be explained by metal excess defect due to anionic vacancies. When crystals of KCl are heated in an atmosphere of potassium vapour, the K⁺ atoms are deposited on the surface of the crystal. The Cl^– ions diffuse to the surface of the crystal and combine with K⁺ ions to give KCl. This happens by loss of electron by potassium atoms to form K⁺ ions. The released electrons diffuse inside the crystal and occupy anionic sites. The crystal ends up with an excess of potassium. The anionic sites occupied by the unpaired electrons are called F-centres. They impart lilac or violet colour to the crystals of KCl. The colour results due to excitation of these electrons when they absorb energy from the visible light falling on the crystals. The correct options are (i) and (iv).

NaCl crystals are ccp structures, and a property of ccp structures is that they contain 4 atoms per unit cell

The number of tetrahedral voids is twice the number of atoms per unit cell, and number of octahedral voids is equal to the number of atoms per unit cell.

Hence, NaCl contains 4 × 2 = 8 tetrahedral voids and 4 octahedral voids. The correct answer is (ii) and (iii).

Amorphous solids are solid substances that lack a defined shape, that is, the opposite of crystalline solids. Unlike crystals, the atoms/molecules of amorphous solids are not arranged in a repeating, defined pattern, ie. they have a short range order, where a regular and periodically repeating pattern is observed over short distances only. Examples include glass and wax. Crystalline solids have a sharp melting point but amorphous solids soften over a range of temperatures and can be easily molded and shaped. Amorphous solids have a tendency to flow, though very slowly. Therefore, sometimes these are called pseudo solids or super cooled liquids. Glass windows, for example, become slightly thicker at the bottom than the rest over a long period of time because glass moves downwards slowly.

There are two types of semiconductors, intrinsic and extrinsic semiconductors. Intrinsic, as shown in the question, are perfect crystals of the semiconductors like silicon. These have too low of conductivity to be of practical use. The conductivity is increased by “doping”, adding a suitable impurity in appropriate amount. The doped semiconductors are extrinsic semiconductors. Silicon belongs to group 14 of elements in the periodic table and contains four valence electrons. When it is doped with a group 15 element like P or As, which contains five valence electrons, they occupy some of the lattice sites in silicon the crystal. Four out of five electrons in P or As are used in the formation of four covalent bonds with the four neighbouring silicon atoms. The fifth electron becomes delocalized which leads to increase in conductivity. Silicon doped with electron-rich impurity is called n-type semiconductor because the conductivity increase is due to negatively-charged electrons. Therefore, the semiconductors doped with As and P are n-type semiconductors.

Hence, option (i) and (iii) are n-type semiconductors.

Ferrimagnetic substances are materials that have domains of atoms of opposing magnetic moments which are unequal in number, therefore causing spontaneous magnetization, that is, they show behaviour just like ferromagnets below the Curie Temperature and show no magnetic order above the Curie Temperature, as in, they become paramagnetic. So, option (i) is a correct statement.

Option (ii) is incorrect as ferrimagnetic substances do become paramagnetic on heating.

Antiferromagnetic substances are materials with equal populations of atoms of opposing magnetic moments, and their magnetic moments are cancelled by each other. Option (iii) is incorrect.

Ferromagnetic substances, in unmagnetized state have little to no net magnetic field. They become aligned on application of an external magnetic field, and remain aligned even on the removal of the field, creating a magnetic field of their own extending into space around the material, thus creating a "permanent" magnet. Option (iv) is correct.

Thus, (i) and (iv) are the correct option.

Amorphous solids have irregular arrangement of atoms and molecules, with short range order. They are characterised as isotropic, that is, because of their irregular arrangement, the refractive index is the same in all directions when measured. Amorphous solids also do not have a defined heat of fusion because they do not have a sharp melting point. They also can soften over a long range of temperatures so they are known as supercooled liquids. Quartz glass is a very pure form of amorphous silica, as it does not contain any impurities added to normal glass in order to lower its melting temperature. Being amorphous, it does not have a defined crystalline structure, so option (i) is a feature not shown by quartz glass.

Quartz glass’ refractive index is not zero and same in all directions. Option (ii) is a feature of quartz glass.

Because it is amorphous, it does not have a fixed melting point and no definite heat of fusion. Option (iii) is not a feature of quartz glass.

As amorphous solids are also known as supercooled liquid, Option (iv) is a feature of quartz glass.

Thus, option (i), (iii) are the correct option.

Crystalline solids are classified on the basis of intermolecular forces working in them with four categories – molecular, ionic, metallic and covalent solids. The constituent particles of molecular solids are molecules, further subdivided on the basis of the bond types.

According to the properties of these solids, only (iv) I₂ is a molecular solid and the other options are covalent solids.

Thus, (i), (ii), (iii) are the solutions for this question.

In solids, the constituent particles are closely packed, leaving the minimum vacant space between them. They are called voids and they are shaped particularly in crystalline solids. Close packing in two dimensions can be of two types; square close-packed and hexagonal close-packed. Stacking of two dimensional layers one above the other creates three dimensional structures. When 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, the arrangement generated is simple cubic lattice. A simple cubic has an atom at each corner of the cube. A body-centered cubic (bcc) unit cell has an atom at each of its corners and also one atom at its body center. In face centered cubic (fcc) the particles are arranged at the corners and also at the center of each of the faces. The formation of voids can be explained diagrammatically, where tetrahedral voids are formed when the sphere of the second layer is above void of first layer, a tetrahedral void is formed and in 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 kind of voids surrounded by six spheres are octahedral voids. So according to the structures of the unit cells, option (ii) and (iii) cannot have octahedral voids.

1: Structures of hcp and fcc lattice formed after stacking of two dimensional sphere structures and top, side and bottom views of the same. Image obtained from Wikipedia under Creative Commons license. By en:User:Twisp - Own work, Public Domain.

Thus, hcp (hexagonal closed packing) and fcc (face-centred cubic cell) arrangements show the octahedral voids.

Thus, option (i), (ii) are the correct option.

Frenkel Defect is a stoichiometric defect, which are point defects in crystal structure which do not affect the stoichiometry of the solid. Stoichiometric defects show vacancy as well as intrinsic defects. Frenkel Defect is shown by ionic solids, where the smaller ion (usually cation) is dislocated from its normal site to an interstitial site. It creates a vacancy defect at its original site and an interstitial defect at its new location. Frenkel defect is also called dislocation defect and it does not change the density of the solids.

Thus, the correct option is (i) and (ii).

Interstitial defect occurs when some constituent atoms or molecules occupy an interstitial site in the lattice. This defect increases the density of the substance it occurs in. Vacancy defect occurs when sites in the lattice are vacant, and this reduces the density of the solid. Frenkel Defect is when the smaller ion (usually cation) is dislocated from its normal site to an interstitial site. It creates a vacancy defect at its original site and an interstitial defect at its new location. This does not change the density. Schottky defect is like a vacancy defect in ionic solids and this also decreases the density of the solid.

Thus, vacancy defect and Schottky defect decrease the density of solid. The correct options are (ii) and (iv).

The state of various substances to categorize as solid, liquid or gaseous depends on the nature of the constituent particles and the interactions and bonding between them. Liquids and gases are categorized as fluids because The liquid and gases have a property to flow i.e the molecules can move past and tumble over one another freely. The molecules in these two substances are free to move about unlike solids who are rigidly placed in their position and can only oscillate.

Solids are characterized by short intermolecular distance and strong intermolecular forces. Their constituent particles which can be atoms, molecules or ions have fixed positions unlike liquids or gases and can only oscillate about their mean positions. This happens because of the short inter- and intramolecular bond lengths and strong intermolecular forces. It takes a large amount of energy to break these forces and to make solids compressible. Hence, solids are incompressible.

Crystalline solids or crystals have long range order, that is, they have repeating pattern of identical unit cells over a long distance over the entire crystal. Despite this property, crystals are usually not perfect because of deviations or “defects”. The crystalline solids usually consist of a large number of small crystals, each having a definite shape. So, defects in the entire structure may either happen because constituent particles may not get the time to arrange themselves in the correct order during formation or inclusion of impurities in the crystal lattice. The imperfections caused by impurities in lattice are of two types; point defects which occur due to deviation from arrangement of an atom and line defects which occur due to deviation from arrangement of rows of atoms in structure.

This phenomenon can be explained by metal excess defect due to anionic vacancies. It is a type of non-stoichiometric defect and commonly seen in alkali halides. When table salt crystals are heated, it forms an atmosphere of sodium vapour and the Na⁺ atoms are deposited on the surface of the crystal. The Cl^– ions diffuse to the surface of the crystal, combine with Na⁺ atoms and give NaCl. 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 (German word Farbenzenter for colour centre). They impart yellow colour to the crystals of NaCl because of excitation of these electrons when they absorb energy from the visible light falling on the crystals. Hence, NaCl sometimes appears yellow at high temperatures.

Figure 2: Diagrammatic representation of "F-center.” The blue spheres are Cl^- ions and red spheres are Na⁺ ions.

Iron(II) oxide or ferrous oxide is the inorganic compound with the formula FeO. This compound also refers to a family of related non-stoichiometric compounds, which are typically iron deficient with compositions ranging from Fe_0.84O to Fe_0.95O. This happens because of metal deficiency defect, a non-stoichiometric defect. In crystals of FeO, some Fe²⁺ cations are missing and the loss of positive charge is made up by the presence of required number of Fe³⁺ ions. Three Fe²⁺ ions are replaced by two Fe³⁺ ions to make up for the loss of positive charge. Hence, FeO is not formed in stoichiometric composition.

When zinc oxide crystals are heated or flamed with a blowtorch in the presence of air, it becomes yellow coloured on heating, and on removal of heat, goes back to white colour. This happens due to the phenomenon of metal excess defect due to presence of extra cations at interstitial sites, which is a non-stoichiometric defect. On heating ZnO, oxygen leaves as 2O₂ leaving behind Zn²⁺ and 2 electrons. The formula is given below:

Zn²⁺ and the 2 electrons move to the interstitial sites of the crystal which provides excess electrons in the crystal lattice of ZnO. Also there is excess of zinc in the crystal and its formula becomes Z_n1+xO. The excess Zn²⁺ ions move to interstitial sites and the electrons to neighbouring interstitial sites. When light falls on these crystals, these electrons absorb a part of the light in the visible region and hence impart a yellow colour to the ZnO.

Figure 3: Metal Excess Defect due to excess cations at interstitial sites, shown in ZnO crystal. The blue spheres are O_2, the red spheres are Zn^2+, the green spheres are the excess Zn ions and electrons.

Increasing the temperature of a semiconductor causes the free electron to get more energy and crosses the energy gap to the conduction band from the valence band, as the gap between the valence band and conduction band is small. Band gap energy between the conduction and valence bands is smaller compared to insulators, which allows for electrons to be excited across the band gap, allowing for conductivity. As for comparison with metals, they would decrease in conductivity as temperature increases and a semiconductor would increase in conductivity as temperature increases. Hence, electrical conductivity of semiconductors increase with rise in temperature.

Figure 4: Simple diagram of semiconductor band structure, showing a few bands on either side of the band gap. Image obtained from Wikipedia, used under Creative Commons license: https://creativecommons.org/licenses/by-sa/4.0/deed.en

Germanium, which is a group 14 element, can be doped with a group 13 element like Gallium, which contains only three valence electrons. The missing fourth valence electron is called electron hole or electron vacancy. An electron from a neighbouring atom can come and fill the hole, but in doing so it would leave an electron hole at its original position. In this case, it would appear as if the electron hole has moved in the direction opposite to that of the electron that filled it. Under the influence of electric field, electrons would move towards the positively charged plate through electronic holes, while it appears as if electron holes are positively charged and are moving towards negatively charged plate. These type of semiconductors are called p-type semiconductors and this is the mechanism to increase the conductivity of germanium with doping.

Figure 5: Doping of Germanium crystal with trivalent Gallium impurity, causing formation of electron hole, leading to p-type semiconductor.

In ccp structure, each unit cell contains atoms at all the corners and at the centre of all the faces of the cube. Each atom at a corner is shared between eight adjacent unit cells, four unit cells in the same layer and four unit cells of the upper (or lower) layer. Only 1/8^th of a particle actually belongs to a particular unit cell. The atom at the faces is also shared with the adjacent unit cell, both sharing half of the atom i.e. only 1/2 of each atom belongs in the unit cell.

To calculate number of atoms per unit cell of ccp,

In a FCC

There are 8 corner atoms × 1/8 atom per unit cell = 8 x 1/8 = 1 atom

There 6 face-centered atoms × 1/2 atom per unit cell = 6 x 1/2 = 3 atoms

Total atoms in unit cell = 1 + 3 = 4 atoms.

The number of tetrahedral voids is twice the number of atoms and octahedral voids is same as the number of atoms.

In cubic closed packed (ccp) lattice, number of atoms per unit cell = 4.

Number of tetrahedral voids present = 2 x number of atoms per unit cell = 2 x 4 = 8.

Given that metal atoms (M) occupy 1/3^rd of present tetrahedral voids, M = 1/3 x 8 = 8/3

So the ratio of M and N in a unit cell is M:N = 8/3:4 = 2/3:1

Hence, the formula is M₂N_3.

Continuous slow heating and cooling of amorphous substances over long periods of time causes small portions of the substance to turn crystalline. Crystalline solids have a sharp melting point. On the other hand, amorphous solids soften over a range of temperatures and can be moulded and manipulated. On heating they become crystalline at some temperature. For example, some glass objects from ancient civilisations are found to become milky in appearance because of some crystallization in their structure.

(i)-c , (ii)-a , (iii)-d , (iv)-b

(i) Simple vacancy defect: when some lattice sites are absent from the crystal creating voids, it is called vacancy defect. Since some ions or atoms leaves the site it results in decrease in density.

(ii) Simple Interstitial defect: when some atoms or molecules leaves their sites and move to interstitial sites, it is called Interstitial defect. Since there is no net movement of atoms from the crystal, density remains same.

(iii) Frenkel defect: In this type of defect, the smaller ion (normally cation) moves to interstitial sites leaving vacancy space in its original site. Thus, like simple interstitial defect density remains same. This defect is shown by ionic solids like ZnS, AgCl, AgBr etc.

(iv) Schottky Defect: Unlike interstitial defect, it is vacancy defect in which both cations and anions leaves their crystal lattice. To maintain electrical neutrality both leaves in same number. Like simple vacancy defect, density decreases in this type of defect. Ex. NaCl, KCl, AgBr etc..

(i)-(b), (c) ,(ii)-(c),(d) , (iii)-(c), (e) , (iv)-(a), (d)

(i): for primitive cubic unit cell, a=b=c and it has one atom in each corner. So 1/8^th atoms contributes in each cell. So, from all 8 corners =1.

(ii): for body centered cubic cell, a=b=c and being a body centered cell it has one atom in each corner and 1 atom in the body center. So contribution from all corners is and 1 atoms from body center atom.

(iii): for face centered cubic cell, a=b=c and it has one atom in each corner and 1 atom in each body face which is shared among 2 cells so contribution from one face atom is 1/2. Net contribution from all face atoms is =3. Total atoms in face centred is therefore 3+1= 4 atoms.

(iv): In end centered orthorhombic unit cell a≠b≠c since all sides are not equal.

Total contribution of atoms present at the corners of the cubic cell =

Total contribution of atoms present at end center of the cubic cell = .

(i)-c , (ii)-a , (iii)-b

(i) Impurity Defect: This type of defect arises in crystal lattice due to presence of impurity or foreign particles in lattice. When Sr²⁺ is added to NaCl, Sr²⁺ removes to Na⁺ ions and occupy one site and leaves other vacant. Thus this cause cation vacancy.

(ii) Metal excess defect: When NaCl is heated in atmosphere of sodium vapour, sodium gets deposited on the surface of lattice. The inside Cl^- diffuses to combine with outer sodium ion and thus combines with it with release of an electron by sodium. As a result crystal has now excess of sodium ions and the sites occupied by unpaired electron is called F centers.

(iii) Metal Deficiency defect: It is caused due to cation vacancy created by replacement of some valent ions by its higher charge ions. Example FeO with Fe³⁺. Where Fe²⁺ and Fe³⁺ both are present in lattice.

(i)-d , (ii)-c , (iii)-b , (iv)-a

(i) Mg in solid state shows electronic conductivity due to presence of free electrons, hence it is known as electronic conductors.

(ii)MgCl₂ shows electrolytic conductivity in molten state due to presence of electrolytes (Mg²⁺ and Cl^-), thus shows electrolytic conductivity.

(iii) When silicon is doped with phosphorus which contains 5 electrons. 4 electrons are used in bonds and 1 electron left delocalized and thus shows conductivity in electric field. Since the conductivity is due to electron which is negatively charged it is called N type semiconductor.

(iv) When germanium is doped with boron which contains 3 electrons, it generates one hole due to lack of electron. This hole acts as a positive charge and helps in conductivity when placed in electric field. Since the conductivity is due to positive charge it is called P type semiconductors.

(i)-c , (ii)-a , (iii)-d , (iv)-b

(i) In square close packing each sphere is in contact which nearby 4 spheres. Thus it has a coordination number 4.

(ii) In hexagonal close packing in 2 dimension each sphere is in contact which nearby 6 spheres. Thus it has a coordination number 6. This arrangement creates triangular voids.

(iii) In hexagonal close packing in 3 dimension, there is repetition of pattern of sphere in alternate layers to form ABAB kind of pattern.

(iv) Cubic close packing in 3 dimensions has repetition pattern of spheres in every fourth layer thus forming ABCABC kind of pattern.

In simple cubic unit cell, atoms are present only at the corners and one atom is shared by 8 adjacent cells, so contribution of 1 atom in a cell is 1/8. Since there is 8 corners so total number of atoms present is =1.

So, Assertion and reason both are correct statements and reason is correct explanation for assertion.

Graphite is a good conductor of electricity because three of its four electrons are involved in bonding and 1 electron is free for conduction. But in case of diamond all four electrons are involved in bond formation. So no electron is available for conduction hence diamond is bad conductor of electricity. Graphite has layered structure where layers are held by weak van der wall force, that’s why it has soft in nature. In case of diamond, it has hard & compact 3D structure with strong forces acting gives it hard and brittle structure.

So, Assertion and reason both are correct statements but reason is not correct explanation for assertion.

Octahedral voids are present in the edge centres which is shared among 4 adjacent cells and one octahedral void is present in body centre. So total octahedral voids are.

So, assertion is correct and reason is wrong.

The FCC has maximum packing efficiency of 74% which in case of BCC is 68% and Simple cubic has 52.4%. FCC has coordination number 12. Assertion and reason is correct but reason is not correct explanation of assertion.

The conductivities of semiconductors lies in between the conductors and insulators and of range 10^–6 – 10⁴ ohm^–1m^–1. But intermediate conductivity is due to small energy gap between valance and conduction band which is zero in case of conductors and very large in case of insulators. It has filled valance band.

So assertion is correct and reason is wrong.

In cubic close packing, each cube consists of eight small cubic components as shown in figure. Since atoms are present in corners and face centres, total atoms are:

Total atoms= atoms at face centres and atoms at corners

=8× (1/8) + 6× (1/2)=1+3 = 4

Total number of atoms = 4.

Now octahedral voids are present at Edge centres and body centre. (refer figure)

Total number of octahedral voids =

Hence proved. 4 octahedral voids are present in cubic close packing.

In cubic close packing, each cube consists of eight small cubic components as shown in figure. Since atoms are present in corners and face centres, total atoms are:

Total atoms= atoms at face centres and atoms at corners

==1+3 = 4

Total number of atoms = 4.

Considering single small cubic unit, tetrahedral void in present in body centre. Since here we have cubic structure of 8 small cubic unit, total tetrahedral voids = 8× 1 = 8.

Lets consider single cubic lattice consists of 8 small cubic units, 2 tetrahedral voids are present at each diagonal each equal distance apart from corner. There are 4 diagonal in lattice so total number of tetrahedral voids = 4× 2 =8.

Hence proved. There are 8 tetrahedral voids in CCP.

The conductivity of semiconductors like silicon or germanium can be increased by adding certain amount of impurity or foreign particles in it, this is called doping. The doping can be done either by adding electron rich compound (Example phosphorus) or electron deficit compound (example boron etc). The doped semiconductors are called extrinsic semiconductors and has conductivity greater than pure semiconductor called intrinsic semiconductors. This impurities causes electronic effect in them. When doped with electron rich compound, an unpaired electron becomes delocalized. These delocalized electrons increases the conductivity of doped silicon due to negatively charged electrons when placed in electric field. Silicon doped with electron rich impurity is called n type semiconductor. Similarly when silicon is doped with electron deficit impurity, a delocalized hole is generated with acts as positive charge ion as increases conductivity of silicon when placed in electric field. Silicon doped with electron deficit impurity is called p type semiconductor.

For 1→ O there is 0.93 Fe

Let 100→ O^2- is present in sample. To maintain neutrality Fe²⁺ and Fe³⁺ should be present in such a ratio that it counterbalances the negative charge.

Let number of Fe²⁺ be X

number of Fe³⁺ be 93-X

since total positive charge must be equal to total negative charge as compound is overall neutral,

2X+3(93-X)=2× 100

2X+279-3X=200

X=79

Fe²⁺ =79

Fe³⁺ =93-79=14

Fraction of Fe²⁺ is = 0.849.

This is metal deficiency defect as iron present in sample is in less amount than required for stoichiometric composition.