Solid States Notes PDF

Solid States General Properties y Solid is that state of a matter in which Concept Ladder constituents are firmly bound because of strong forces. Matter can exist in one of y They have definite mass, shape and volume. three main sta...

Solid States General Properties y Solid is that state of a matter in which Concept Ladder constituents are firmly bound because of strong forces. Matter can exist in one of y They have definite mass, shape and volume. three main states : solid, y They are incompressible, rigid and have liquid, or gas. strength. Three phases of water y They have close packed arrangement of Ice (solid), Water (liquid) particles. and Steam (gas) y They have high density but very slow diffusion rate. y They can only have vibrational motion as constituents have fixed positions. Types of Solids Solids are classified into 2 types, crystalline and amorphous. Rack your Brain Crystalline Solids Why density of ice is lower than y In crystalline solids, constituents are arranged water? in a definite or proper order which repeats itself over long distances. y They generally have definite geometry having flat faces and sharp edges. y They have sharp melting points. y They are considered true solids. y They give clean cleavage with fixed cleavage planes. Concept Ladder y They show anisotropic behaviour which means different physical properties in different Isotropic : Properties of a material are identical in all directions because of orderly arrangement of directions. constituents. Anisotropic : Properties y They are incompressible. For example, CaF2, of a material depend on ZnS, diamond, quartz, NaX. the direction; for example, y All elements and compounds are of this type. wood. In a piece of wood, you can see lines going in Amorphous Solids one direction; this reaction y In these type of solids, constituents are not is referred to as “with the arranged in a regular or orderly manner for Solid States grain”. long range. 1.  y They do not have sharp melting points. y They are pseudo solids. Concept Ladder y They show isotropic behaviour which is same as physical properties in all directions because There exists strong of irregular arrangement. electrostatic force in ionic y They do not give clean cleavage. They have crystalline solids like NaCl irregular cut. because of attraction E.g., glass, plastic, and rubber between anions and cations. Terms related to crystalline solids Crystal: This is a homogeneous part of solid substance which is made by regular pattern of structural units which are bonded by plane surface making definite angles with each other. Crystal lattice or space lattice: It is regular arrangement of constituent particles (atom, ions etc.) of a crystal in 3-D space. Face: It is plane surface of crystal. Rack your Brain Edge: It is formed because of intersection of two adjacent faces. What is interfacial angle between two adjacent faces of a cube? Interfacial angle: It is angle between perpendiculars of two intersecting faces. Q.1 Why is glass of window panes of very old buildings found to be thicker at the bottom than at the top and why is it milky? A.1 Glass is amorphous solid which is a supercooled liquid of high viscosity and hence possesses fluidity. Due to presence of this property it is thicker at the bottom than at the top. Milkiness of glass is due to fact that it undergoes heating during the day and cooling at night, i.e., annealing over several years. as a result, it acquires some crystalline character.. Types of crystalline solids Nature of constituent particles and binding forces of crystalline solids are generally classified as shown in table given below. Solid States 2. Types of Crystalline Solids Solid angle When three or more than three edges are intersect, a solid angle is formed. Bragg’s Equation Max von Laue identified possibility of diffraction of X-rays by crystals as order of wavelength of X-rays is compatible to inter- atomic distances present in a crystal. y Bragg’s equation gives a simple relationship between wavelength of X-rays and distance Concept Ladder between planes in crystal and angle of reflection. The equation can be given as: Due to their bonding, nλ = 2d sin θ metallic solids have Here, delocalised electrons. λ = Wavelength of X-rays These free electrons can n = Order of reflection it is taken generally move around, therefore as 1. they can conduct θ = Angle of incident light electricity. But in ionic d = Distance between 2 layers of crystals solids electrons are not free to move so they can’t Solid States conduct electricity. 3.  y For a given number of lattice planes, value of ‘d’ is fixed so possibility of getting maximum reflection depends only on q. If we increase q gradually several positions will be observed at which there will be maximum reflection. Applications y Bragg’s observation has proved to be highly useful in determination of structures and dimensions of ionic crystalline solids. y It also explains many properties of X-rays. Rack your Brain y Equation helped in construction of an X-ray spectrometer to describe crystalline structure What is the shape of unit cell iof of crystals as in case of face-centered cubic NaCl crystalline structure? structure of NaCl. Unit Cell y Unit cell is the smallest unit or three- dimensional portion of the space lattice which when repeated and again in different directions gives rise to the complete space lattice. y It has characteristics properties of a, b, c Previous Year’s Questions (edge distance) and a, b, g, (angles). y It is smallest geometrical figure which has The number of carbon atoms per properties of a crystal in a lattice. unit cell of diamond is y A crystal can have infinite number of unit [NEET-2013] cells. (1) 6 (2) 1 Solid States (3) 4 (4) 8 4. Types of Unit Cell They are of four types, namely, simple, face- centred, end-centred and body-centred. Simple or primitive or basic unit cell: In this, lattice points (particles) are present only at Concept Ladder corners. Face-centred unit cell: In this, lattice points or In cube of any crystal particles are present not only at corners but also A-atom placed at every at centre of each face. corners and B-atom placed Body-centred unit cell: In this, particles are at every centre of face. The present at all the corners and body centre of unit formula of this compound cell. is AB3. End-centred unit cell: In this, particles are present at all the corners as well as at centre of 2 opposite faces. Types of Symmetry in Crystals A crystal can have centre of symmetry, axis of symmetry and plane of symmetry. Rack your Brain y Plane of symmetry (POS): It is an imaginary plane which passes from centre of crystal What is the symmetry of PCl5 and divides it into two equal part. (mirror molecule? images of each other). y Centre of symmetry: It is an imaginary point which separates surface of crystal which is at equal distances in both the directions by Solid States drawing any line through it. 5.  y A crystal generally has one centre of symmetry. y Axis of symmetry is imaginary straight line Concept Ladder which on rotation of crystal gives same appearance more than a single time. y It is 2-fold, 3- fold, 4-fold and 6-fold type A sphere has all kinds of respectively. symmetry of point, axis and plane. Crystal Systems There are 7 types of crystal systems and fourteen bravais lattices as given in table below Crystal Systems and Bravais Lattices Solid States 6. Symmetry in Crystals (a) Rectangular Plane of Symmetry (b) Diagonal Plane of Symmetry (c) Axis of Four Fold Symmetry (d) Centre of Symmetry Mathematical Analysis of Cubic System Atomic Radius (r) Previous Year’s Questions y Atomic radius: It is half the distance between two nearest neighbour atoms in crystal. For orthorhombic system axial y It’s expressed in terms of length of edge ratios are a ≠ b ≠ c and the axial angles are (a) Of unit cell of a crystal. a [AIPMT] y In simple cubic r = (1)       90 2 a (2)       90 y In face-centred cubic (FCC) r = 2 (3)     90,   90 3a (4)       90 y In body-centred cubic (BCC) r = 4 Solid States 7.  No. of Atoms Per Unit Cell/ Unit Cell Content (Z) y It is total number of atoms contained in a Concept Ladder unit cell. nc ne nf ni A point lying at the corner Z= + + + of a unit cell is shared 8 4 2 1 equally by eight unit cells Where; nc = no. of atoms at corner position and therefore, only one- ne = no. of atoms at edge center eighth portion of each of nf = no. of atoms at face center position such a point belongs to the ni = no. of atoms present at the center given unit cell. 1 y In SCC, Z = 8 × =1 8 Every corner atom is shared by surrounding 1 Rack your Brain unit cells so it accounts for of an atom. 8 y In a FCC structure, Z = 3 + 1 = 4 Find out the two nearest atoms 1 in FCC and BCC crystalline Eight(8) corner atoms contribute of an 8 structures and also find their atom. radii? which means only of one atom per unit cell. Each of 6 face-centred atoms are shared by two adjacent unit cells such that only half of one face- centred atom is contributed as its share that is, Previous Year’s Questions 1 6× 3 (atoms per unit cell) = The number of atoms in 100 g of a 2 fcc crystal with density d = 10 g/ total of 4 atoms per unit cell. cm3 and cell edge equal to 100 pm, y In a BCC structure, Z = 1 + 1 = 2. Each of eight is equal to corner atoms will be contributing only 1 atom [AIPMT] per unit cell. (1) 2 × 10 25 (2) 1 × 1025 Central atom has contribution of only one (3) 4 × 10 25 (4) 3 × 1025 atom per unit cell. Solid States 8. Coordination Number (C. No.) y It is equal to the number of nearest neighbours atom, that is, touching particles present around a species in a crystal. Its larger value Concept Ladder shows closer packing. y The values depend upon structure of crystal. In ionic crystal, the number S.C : Coordination number is 6 of oppositely charged FCC : Coordination number is 12 ions surrounding each BCC : Coordination number is 8 ion is known to be its coordination. NaCl ; CsCl ; ZnS ; CaF4 ; Na2O 6:6 8:8 4:4 8:4 4:8 Density of Lattice Matter (d) y Density of lattice matter is ratio of mass per unit cell to total volume of unit cell and it is given as: Z × Atomic weight d= Previous Year’s Questions N0 × Volume of unit cell (a3 ) Here, A metal crystallises with a face N0 = Avogadro number centred cubic lattice. The edge a3 = Volume of the unit cell is 408 pm. The Z = Number of atoms in unit cell diameter of the metal atom is a = Edge length in cm [AIPMT-2012] d = Density in g/ml or g/cm3 (1) 288 pm (2) 408 pm (3) 144 pm (4) 204 pm Solid States 9.  Packing Fraction y ⚫ It is defined as the ratio of volume Concept Ladder 4  occupied by spheres  r 3  in a unit cell to 3  A crystal may have a the total volume (a3) of that unit cell. Fraction nubmer of planes or of volume that is empty is called void fraction. axis of symmetry but it 4 3 possesses only one centre Z× πr of symmetry. P.F. = 3 a3 y In a SCC, 4 3 πr P.F. = 3= 0.52 = 52% (2r)3 Rack your Brain Why the closest packed structures have high value of coordination? Here, a = 2r; % void = 48% y In a body-centred cubic structure, 4 3 2r P.F.  3  0.68  68% 3  4r    Previous Year’s Questions  2 The intermetallic compound LiAg crystallizes in cubic lattice in which both lithium and silver have coordination number of eight. The crystal class is [AIPMT] (1) face-centred cube (2) simple cube 4r Here, a = ; % void = 32% (3) body-centred cube Solid States 3 (4) none of these 10. y In a face-centred cubic structure, Concept Ladder y In fcc, the nearest neighbour of a corner atom is the face centred atom and thus, atom on each corner has 4 4 3 nearest neighbours. 4× πr y It also possesses four P.F. = 3= 0.74 = 74%  4r  3 face centred atoms   in planes below and  2 3 above it.  4r  Here, a =    2 % void = 26% Solid States 11.  Interstitial Voids Interstitial voids are spaces left after hexagonal Concept Ladder close packing (hcp) and cubic close packing (ccp). The spaces or voids are of the following types: Voids are terms as Trigonal void: It is a vacant space touching three the gaps between the spheres that is, two-dimensional void formed constituent particles in a when three spheres are in same plane having closed packed structure. corners are at corners of triangle. Close packing in solids can be generally done in three Tetrahedral voids: Vacant space created when ways : 1D close packing, 2D each sphere of second layer rests on vacant close pakcing and 3D close space created by three spheres (of first layer) packing. touching each other is called a tetrahedral void. Octahedral voids: They are formed by combination of two triangular voids of first and second layers. They are so called because they are enclosed between six spheres, centres of which occupy corners of a regular octahedron. Location and number of voids Rack your Brain (1) Tetrahedral void: They are located at body diagonals, two in each body diagonal at one What is the coordination number fourth of the distance from each end. for tetrahedral void? Number of Tetrahedral voids = 8 (2) Octahedral voids: These are located at middle of cell edges and at centre of cubic unit cell. 1 Total number of Octahedral voids = × 12 + 1 = 4 4 Size of voids: Voct = 0.414 × r Previous Year’s Question Vtetra = 0.214 × r The number of octahedral void(s) Vtri = 0.115 × r per atom present in a cubic close- packed structure is where r is radius of the biggest sphere. [AIPMT-2012] (1) 1 (2) 3 Solid States (3) 2 (4) 4 12. Radius Ratio Radius ratio is radius of octahedral void Concept Ladder to radius of sphere forming close packed arrangement. The packing density data (1) For stability of ionic compounds, each cation reveals that close packing must be surrounded by maximum no, of anions of atoms in cubic structure and vice-versa. follow the order, FCC > BCC > SCC, i.e., more closely (2) The maximum no. of opposite charged ions packed atoms are in FCC surrounding another ion is known as coordination structure. number. Since, ionic bonds are non- directional arrangement of ions in crystal is determined by sizes. Ratio of radius of cation to anion is known as radius ratio, i.e., Radius of the cation(r + ) Radius Ratio = Rack your Brain Radius of the anion(r − ) If the radius of the bromide ion (4) If radius ratio is greater, then size of cation is 0.182 nm, how large a cation is larger and therefore greater is its coordination can fit in each of the tetrahedral number. hole? (5) Relationships between radius ratio and the coordination number and structural arrangement are called radius ratio rules. Q.2 The two ions A+ and B- have radii 88 and 200 pm respectively. In the close packed crystal of compound AB, predict the coordination nubmer of A+. A.2 r r(A  )   88 pm  0.44 r r(B ) 200 pm  It lies in the rage 0.414 to 0.732 Hence, the coordination number of A+ = 6 Solid States 13.  Packing of Constituents in Crystals Crystal constituents can be packed in two dimensions and 3-Dimensions. Close Packing in Two Dimensions In this, close packing arrangement in two dimensions are as follows: Previous Year’s Questions Square close packing y In this, each sphere is in contact with four In crystals of which one of following other spheres. ionic compounds would you expect y Voids form a square in this type of packing. maximum distance between y About 52.4% space is occupied by the spheres. centres of cations and anions? [AIPMT] (1) CsI (2) CsF (3) LiF (4) LiI Q.3 The edge length of a unit cell of a metal having molecular mass 75 g/mol is 5Å which crystallizes in a cubic lattice. If the density is 2 g/cc, then find the radius of the metal atom. A.3  Z M a 3  N0   a 3  N0 2  (5  108 )3  6.02  1023 or Z   2 M 75 This shows that the metal has body-centred cubic lattice. For BCC lattice, 3 1.732 r a  5Å  2.165Å 4 4 Solid States 14. Hexagonal close packing y It is more densely packed than square closed Concept Ladder packing. y In this, voids are triangular. y In this, 60.4% space is occupied by spheres. In 2D hexagonal close pakcing, their is less free space between sphere than 2D square close packing. Thus hexagonal close pakcing is denser than square close packing. Rack your Brain At what angle the HCP structure can be rotated so that it appears same as original one? Packing in Three Dimensions This packing is of following three types: Previous Year’s Questions Hexagonal close packing y Atoms are present at center and at corners of A compound is formed by cation two hexagons placed parallel to each other, C and anion A. The anoins form 3 more atoms are placed in a parallel plane hexagnoal close packed (hcp) midway between the two planes. lattice and the cations occupy 75% y It generally has a six-fold axis of symmetry. of octahedral voids. The formula of y In this, packing gives an arrangement of the compound is layers as AB AB, that is, odd number layers [NEET-2019] are similar and so are even number layers. (1) C4A3 (2) C2A3 Solid States (3) C3A2 (4) C3A4 15.  Concept Ladder In both hexagonal close packing (hcp) and cubic close packing (ccp) the coordination number of Cubic close packing spheres remian twelve. y Sphere in fourth layer will correspond to that in first layer and give rise to ABC, ABC, type of packing. y It has 3-fold axis of symmetry which pass through the diagonal of the cube. y Here coordination number is 12. Examples: Cu, Ag, Au, Ni, Pt Previous Year’s Questions Structure of a mixed oxide is cubic close packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. The formula of the oxide is [AIPMT-2012] (1) ABO2 (2) A2BO2 (3) A2B3O4 (4) AB2O2 Solid States 16. 17. Solid States  Body-centred cubic packing y In this, each sphere is in contact with 8 Concept Ladder spheres: 4 in lower layer and four in upper layer. y Ionic solids have y It is possible when spheres in first layer are consituents particles slightly opened that is none of spheres is as ions. These are touching the other. Examples: Li, Na, K, Rb, formed by arrangement Cs, Ba of cations and anions by strong Coulombic Structure of Some Ionic Solids forces and also are hard Rock salt (NaCl) type and brittle in nature. y Cl– has a close cubic packing (ccp) type y Ionic solids acts as an structure that is face centred cubic type (fcc). insulator in its solid y In this, Na+ occupies octahedral spaces. states where as they y Na+ and Cl– have coordination number 6. acts as conductors in y No. of formula units per unit cell are 4. r + its molten and aqueous y Theoretically, Na should be 0.414 but it is states. rCl − 0.525. Examples: LiX, NaX, KX, AgCl, AgBr, NH4Cl. CsCl type y In this, Cl– ions are at corners of cube and Cs+ ions are in cubic void. y Coordination number of both Cs+ and Cl– is 8. y In this, number of formula units per unit cell is 1. r + Rack your Brain y Theoretically, Cs should be 0.732 but it is 0.93. rCl − In NaCl type structure, the larger Examples: CsX, TiCl, TiBr, NH4Cl, NH4Br atoms form _______ arrangement and smaller atoms fill all ______ Zinc Blend (ZnS type) voids. y In this, S2– occupies CCP while Zn2+ ions occupy alternate tetrahedral voids. y Only half of total voids are occupied. y In this, coordination no.of both Zn2+ and S2– is 4. y No. of formula units in a unit cell is 4. Examples: ZnS, CuCl, CuBr, CuI, AgI, BeO Solid States 18. Fluorite structure (CaF2 type) y In this, Ca2+ occupies CCP and F– occupies all Concept Ladder tetrahedral voids. y Also, coordination no. of Ca2+ is 8 while for F– it is 4. The crystal in which all the y In this, no. of formula units per unit cell is lattice points are occupied Examples: CaF2, BaCl2, BaF2, SrF2. by the component particles or groups of particles is Antifluorite structure (Na2O type) knonw as ideal crystal. y Here negative ions (O2–) occupy ccp while With change in cations (Na+) occupy all the tetrahedral voids. temperature, molecular y Co-ordination no. of Na+ is 4 while for O2–. motion increases, which It is 8. causes deviation from y In this, no. of formula units per unit cell is 4. ordered arrangement and Examples: Na2O, Li2O, K2O. gives rise to a defect or imperfection in the crystal. Normal spinel structure (AB2O4) They have a general formula AB2O4. A Bivalent Cation (Mg+2) B Trivalent Cation (Al3+) y MgAl2O4 is spinel crystal. y In this, Mg2+ occupy tetrahedral voids while Rack your Brain oxide ions occupy CCP. Aluminium occupies octahedral voids. Coordination number of fluorine y Ferrites [ZnFe2O4] can also have structure. in calcium chloride is? y They are used in telephones, memory loops of computers as magnetic material. Structure of Fe3O4 (Magnetite) y In Fe3O4, Fe3+ and Fe2+, they are present in 2:1 ratio. y In this, oxide ions are in CCP. Fe2+ occupies Previous Year’s Questions octahedral voids where as Fe3+ occupies octahedral and tetrahedral voids. Formula of nickel oxide with metal y MgFe2O4 also has this kind of structure deficiency defect in its crystal is Ni0.98O. The crystal contains Ni2+ Imperfections In Solids and Ni3+ ions. The fraction of nickel y Deviation from perfectly ordered arrangement existing as Ni2+ ions in the crystal is constitutes a defect or imperfection. [NEET-2019] y The defects are also called thermodynamic Solid States (1) 0.96 (2) 0.04 defects as a number of the defects depend (3) 0.50 (4) 0.30 on temperature. 19.  y Crystals may possess additional defects because of presence of impurities. Imperfection not only modifies properties of solids but also gives rise to new properties. Electronic Imperfection Electrons are present in fully occupied lowest energy states but at very high temperatures Previous Year’s Questions few electrons may occupy higher energy states depending upon the temperature. Which one of following elements y For example, in crystals of pure silicon or silicon should be doped so as to germanium some electrons are released give p-type of semiconductor? thermally from covalent bonds at temperature above 0 K. These electrons are free to move [AIPMT] in crystal and are responsible for electrical (1) Selenium (2) Boron conductivity. This type of conduction is called (3) Germanium (4) Arsenic as intrinsic conduction. y Electron deficient bond formed by release of an electron is known as hole. In presence of an electric field the positive holes move in a direction opposite to that of the electrons and conduct electrically. Q.4 Why stoichiometric defects are also called intrinsic defects? A.4 Stoichiometric defects are called because they do not change the stoichiome- try of the crystal (Schottky defect and Frenkel defect). They are called intrinsic defects because it is due to the deviation from regular arrangement of atoms or ions within the crystal and no external substance is added.. Q.5 CaCl2 will introduce Schottky defect if added to AgCl crystal. Explain. A.5 Two silver ions will be replaced by one calcium ions to maintain electrical neu- trality. So, a hole is created at the lattice site for every Ca2+ ion introduced.. Solid States 20. Atomic Imperfection Compounds in which number of irregularities Concept Ladder present in arrangement of atoms or ions are called atomic imperfections. It is of two types: In Schottky defect the Stoichiometric defects: Compounds in which density of crystal decreases. number of positive and negative ions are The crystal begins to exactly in ratio indicated by the chemical formula conduct electricity to small are known as stoichiometric compounds for extent by ionic mechanism. example, NaCl. These solids show following types of defects: (a) Schottky defect: Schottky defect is created when same number of negative and positive ions are missing from respective positions Rack your Brain leaving behind a pair of holes. Explain why ZnO becomes yellow on heating. Previous Year’s Questions The correct statement regarding y Schottky defect more common in ionic defects in crystalline solids is compounds with high coordination number [NEET-2015] and where size of negative and positive (1) Frenkel defects decrease the ions is almost equal. density of crystalline solids y This defect decreases density of crystals but (2) Frenkel defect is a dislocation maintain neutrality e.g., NaCl, CsCl, KCl, KBr. defect In case of NaCl, there is nearly 106 Schottky (3) Frenkel defect is found in pairs per cm3 at room temperature. halides of alkaline metals (4) Schottky defects have no effect Solid States on the density of crystalline solids. 21.  (b) Interstitial defect: Interstitial defect is caused due to the presence of ions in the normally Concept Ladder vacant interstitial sites in the crystal. Generally, cations are (c) Frenkel defect: Frenkel defect is created smaller than anions, hence when an ion leaves its correct lattice site and it is more common to find occupies an interstitial site. This defect is the cations occupying the interstitial sites. common in ionic compounds which have low coordination number and in which there is large difference in size between negative and positive ions. Due to this defect, neutrality density remains same but dielectric constant of the medium increases. For example, ZnS, AgCl, AgBr, AgI etc. Rack your Brain State whether the density of solid remains same or not for Frenkel defect. Previous Year’s Questions Non-stoichiometric defect: In Non-stoichiometric defect many compounds in which ratio of The appearance of colour in solid positive and negative ions differs from what is alkali metal halides is generaly due required by the ideal formula of the compound. to [AIPMT] Such compounds are called non-stoichiometric (1) interstitial positions compounds, for example, VOx. In this type of (2) F-centres compounds, a balance of positive and negative (3) Schottky defect charges is always maintained by having extra (4) Frenkel defect Solid States electrons or extra positive charge. 22. These defects are explained below: (a) Metal excess defects due to anion vacancies Concept Ladder y Compound may have excess metal ions if a negative (–ve) ion is absent from its lattice Defects due to interstitial site, leaving a hole which is engaged by an cations is shown by crystals electron to maintain electrical neutrality. which are likely to exhibit Frenkel defect. An excess y Holes occupied by electrons are called positive ion is located in F-centres and are responsible for colour of the interstitial sites. compounds, examples are, 1. Excess of Na in NaCl makes crystal appear yellow. 2. Excess of Li in LiCl makes it pink. 3. Excess of K in KCl makes it violet. y Greater the number of F-centres, greater will Rack your Brain be intensity of colour. This kind of defect is found in crystal which is likely to possess Write some difference between Schottky defect. stiochiometric and non- stiochiometric defects. Previous Year’s Questions Ionic solids, with Schottky defects, contain in their structure [AIPMT] (1) cation vacancies only (2) cation vacancies and interstitial (b) Metal excess defects cations y This defect occurs if an extra positive ion is (3) equal number of cation and present in an interstitial site. anion vacancies y Electrical neutrality is sustained by presence (4) anion vacancies and interstitial of an extra electron in interstitial site. anions. Solid States 23.  y The type of defect is shown by crystals which are likely to show Frenkel defects, for example, Concept Ladder yellow colour of Zn. When extra positive ions occupy interstitial site to maintain electrical neutrality, some extra electron occupy some other interstitial sites. (c) Metal deficiency due to cation vacancies Rack your Brain y Non-stoichiometric compounds can have metal deficiency due to absence of a metal What will be the colour of KCl ion from their lattice site. crystal mixed in the atmosphere y The charge is balanced by an adjacent ion K? having high positive charge. y These type of defect is generally shown by compounds of transition metals, for example, FeO, FeS and NiO. Previous Year’s Questions If NaCl is doped with 10-4 mole % of SrCl2. Calculate the concentration of cation vacancies? (NA = 6.02 × 1023 mol-1) [AIPMT] (1) 6.02 × 10 mol 16 -1 (2) 6.02 × 1017 mol-1 (3) 6.02 × 1014 mol-1 Solid States (4) 6.02 × 1015 mol-1 24. Magnetic Properties of Solids Diamagnetic substances: Concept Ladder They are weakly repelled by magnetic field and do not have any unpaired electron, i.e., all The temperature at which paired electrons. a feromagnetic substance loses its ferromagnetism and attains paramagnetism only is known as Curie temperature. For Iron the For example, NaCl, Zn, Cd, Cu+ , TiO2. Curie temperature is 1033K y They act as electrical insulators. and Ni it is 629K. Paramagnetic substances: They are attracted by magnetic field and have unpaired electrons. They lose magnetism in absence of magnetic field. Rack your Brain For example, Mention some application based 1. Metal oxides like CuO, VO2 etc. on ferromagnetic and anti- 2. Transition metals for example, Cr, Mn, Ni, Co, ferromagnetic substances. Fe etc. Ferromagnetic substances: They are attracted by magnetic field and show permanent magnetism even in absence of magnetic field. Some examples are, Fe, Co, Ni, CrO2 (used in the audio and video tapes) Fe3O4 etc. y Ferromagnetism arises because of Concept Ladder spontaneous alignment of magnetic moments in same direction. A material is diamagnetic if it tends to move out of a magnetic field and paramagnetic if it tends to Solid States y When we go above curie point/curie move into a magnetic field. temperature ferromagnetism doesn’t exist. 25.  Anti-ferromagnetic substances: They are expected to possess paramagnetism Concept Ladder or ferromagnetism on basis of unpaired electrons but in real have zero net magnetic moment, some Solids are classified examples are, MnO, MnO2, Mn2O3, FeO, Fe2O3. into three groups Anti-ferromagnetism occurs when number of viz – conductors, parallel magnetic moments is equal to number semiconductors and insulators. Conductivity of of anti-parallel magnetic moments. This results metal ranges from 10-7 – 104 in a net zero magnetic moment. ohm m, for semi conductor from 10-6 – 104 ohm m and for insulators from 10-10 – 10-20 ohm m. Ferrimagnetic substances: In case of ferrimagnetic substances, there are unequal number of parallel and anti-parallel magnetic moments which lead to net magnetic Rack your Brain moment, for example, Fe3O4, ferrites. Name the process of introducing an impurity into semi-conductors to enhance their conductivity. Ferrimagnetic, anti-ferromagnetic and ferromagnetic solids change into paramagnetic substance at a particular temperature. For example, ferrimagnetic Fe3O4 on heating to 850 K becomes paramagnetic. This is because of Previous Year’s Questions alignment of spins in one direction on heating. If we mix a pentavalent impurity Curie Temperature in a crystal lattice of germanium, The ferromagnetic substance has a what type of semiconductor formation will occur? characteristic temperature above which no [AIPMT] ferromagnetism is seen. It is known as curie (1) n-type semiconductor temperature. (2) p-type conductor (3) Both (1) and (2) (4) None of these Solid States 26. Electrical Properties of Solids Piezoelectricity: It is electricity produced when Concept Ladder mechanical force is applied on polar crystals due to displacement of ions. A piezoelectric crystal acts like a mechanical electrical transducer. They y Diode is used as a are used in record players. rectifier which is formed by combinaiton Pyroelectricity: It is the electricity produced of n-type and p-type when some polar crystals are heated. e.g., LiNbO3. semiconductors. Ferroelectricity: In few piezoelectric crystals, y Transistors can be ‘npn’ dipoles are permanently polarized even in or ‘pnp’ type which absence of electric field. However, on applying are formed by making electric field, direction of polarization changes. sandwitch of a layer of The phenomenon is known as ferroelectricity p-type semiconductor due to analogy with ferromagnetism. between two n-type semiconductors (i.e., npn). Barium titanate, sodium potassium tartrate (Rochelle salt) and potassium dihydrogen phosphate are some of polar crystals which exhibit ferroelectricity. Anti-ferroelectricity: In few crystals, dipoles align in such a way that they alternately point up and down in a manner that crystal does not possess any net dipole moment, for example, lead zirconate. Previous Year’s Questions On doping Ge metal with a little Superconductivity: Superconductivity defined of In or Ga, one gets when it offers no resistance to flow of electricity. [AIPMT] (1) p-type semiconductor There are no substances which are having super (2) n-type semiconductor conductance at room temperature. (3) insulator (4) rectifier Solid States 27.  y Superconductors: They are widely used in building super magnets, electronic power transmission, etc. Examples: YBa2Cu3O7, Nb3 Ge alloy, La1.25 Ba0.15 CuO4 , (TMTSF)2 PF6 (TMTSF stands for Definition tetra methyl tetra Selena fulvalene). The type of semiconductor y Kammerlingh Onnes observed this formed by doping of impurities phenomenon at 4K in mercury. to any substance is known as Semiconductors: They are electronic conductors extensive semiconductors. which have electrical conductivity in range of 104– 107 Ω–1 cms, for example, Sn, Ge, Si (grey only), SiC, Cu2O. y Pure substances which are semiconductors which are known as intrinsic semiconductors, for example, Si, Ge. y In case of semiconductors, if conductivity is due to impurities, they are known as extrinsic Rack your Brain semiconductors. y ⚫ Addition of impurities in to a semiconductor What are the group numbers is known as doping. Example includes, when of elements which form phosphorous and arsenic (5th group element) semiconducters substances? are doped in silica (4th group element) n-type of conductance is seen. When a group 3rd element (for example, Ga) is Doped, p-type of conductance is observed.. Q.6 Despite long-range order in the arrangement of particles why are the crystals usually not perfect? A.6 Crystallization process will be faster so that particle may not get enough time to arrange in perfect order. That is why crystals have the long- range arrange- ment of particles but not perfect.. Q.7 Why is FeO (s) not formed in stoichiometric composition? A.7 The composition of Fe2+ and O2- ions is not 1:1 it is 0.95:1. This is obtained if Solid States and only if a small number of Fe2+ ions are replaced by two-third of Fe3+ in OH sites.. 28. Q.8 Gold crystallizes in an FCC unit cell. What is the edge length of unit cell (r = 0.144 mm)? A.8 r = 0.144 nm a  2 2r  2  1.414  0.144 nm  0.407 nm Q.9 Analysis shows that a metal oxide’s empirical formula is M0.98O1.00. Calcu late the percentage of M2+ and M3+ ions in the crystal. A.9 Let the M2+ ion in crystal be x M3+ = 0.98 – x Since, total charge on compound must be zero, So, 2x + 3(0.98 – x) – z = 0 Or x  0.94 0.94 % of M2   100  97.9% 0.98 % of M3  100  97.9  2.1% Q.10 Explain how electrical neutrality is maintained in compounds showing Frenkel and Schottky defect. A.10 The compound showing Frenkel defect, ions just get displaced within lattice, while in compounds showing Schottky defect,have equal number of cations and anions are removed from lattice. So it’s electrical neutrality is maintained in both cases.. Q.11 The electrical conductivity of a metal decreases with rise in temperature while that of a semi-conductor increases. Explain. A.11 In metals with increase of temperature, kernels electrons start vibrating at faster rate and thus offer resistance to own of electrons. Then conductivity of metals decreases. In case of semi-conductors (14th group elements), with increase of temper- Solid States ature, more electrons can shift from valence band to conduction band. Then conductivity of semiconductors increases. 29.  Q.12 What type of substances would make better permanent magnets – ferromagnetic or ferromagnetic ? Why ? A.12 Ferromagnetic substances make better permanent magnets. This is because metal ions of a ferromagnetic substance are grouped into small regions known as domains. Each domain acts as small magnet and get oriented in direction of magnetic field in which it is placed. This persists even in absence of mag- netic field. Q.13 In a crystalline solid, the atoms A and B are arranged as follows : (a) Atoms A are arranged in ccp array. (b) Atoms B occupy all the octahedral voids and half of the tetrahedral voids. What is the formula of the compound ? A.13 No. of A (ccp) = 4 No. of B  octahedral voids  Tetrahedral 2 8 4 2 8 A:B so, formula of the compound is AB2 4:8 1:2 AB2 Q.14 A unit cell consists of a cube in which X atoms are at the corners and Y atoms are at the face centres. If two atoms are missing from two corners of the unit cell, what is the formula of the compound? A.14 Total contribution of ‘X’ atoms from 6 corners  1 6  3 8 4 Number of atoms of Y from face centres = 3 3 \x:y= : 3 = 3 : 12 or 1 : 4 4 Solid States Hence, formula is XY4. 30. Q.15 Calculate the number (n) of atoms contained within (a) simple cubic cell (b) a body centred cubic cell (c) a face centred cubic cell. A.15 (a) The simple cubic jnit cell has 8 atoms at eight corners. Each aotm is shared by 8 unit cells. 1  n  8 1 8 (b) The body centred cubic (BCC) cell consists of 8 atoms at the corners and one atom at centre.  1  n  8    1  2  8 (c) The face centred cubic (FCC) unit cell consists of 8 atoms at the eight corners and one atom at all faces. This atom at the face is shared by two unit cells. 1  1  n  8  6    4 8  2 Q.16 Calculate number of atoms in a cubic unit cell having one atom on each corner and tow atoms on each body diagonal. A.16 There are total 4 body diagonals and there are 2 atoms at each body diagonal. Hence nubmer of atoms from 4 diagonals = 8 Number of atoms from 8 corners = 1 \ Total number of atoms in this unit cell = 8 + 1 = 9 Q.17 In compound atoms of element Y forms ccp lattice and those of element X occupy 2/3rd of tetrahedral voids. What is the formula of the compound ? Solid States 31.  A.17 No. of Y atoms per unit cell in ccp lattice = 4 No. of tetrahedral voids = 2 × 4 = 8 No. of tetrahedral voids occupied by X = 2/3 × 8 = 16/3 Therefore, Formula of the compound = X16/3Y4 = X16Y12 = X4Y3 Q.18 Potassium crystallizes in a body centred cubic lattice. How many unti cells are present in 2g potassium? (At. mass of K = 39) A.18 A bcc unit cell has 2 atoms per unit cell 2 \ Mole of potassium = 39 2  6.022  1023 Number of atoms of K  39 Number of unit cells  2  6.022  10  1.54  1022 23 39  2 Q.19 A solid A+B- has NaCl type close packed structure. If the anion has a radius of 241.5 pm, what should be the ideal radius of the cation? Can a cation C+ having radius of 50 pm be fitted into the tetrahedral hole of the crystal A+B-? A.19 As A+B- has NaCl structure, A+ ions will be present in octahedral voids. Ideal radius of cation will be equal to radius of the octahedral void becuase in that case, it will touch the anions and arrangement will be close packed. Hence, Radius of octahedral void = rA+ = 0.414 × rB– = 0.141 × 241.5 pm = 100.0 pm Radius of tetrahedral void = 0.225 × rB– = 0.225 × 241.5 pm = 54.3 pm As radius of cation C+(50 pm) is smaller than size of tetrahedral void and can be placed into tetrahedral voids (but not exactly fitted into it). Solid States 32. Summary y In NaCl, there are nearly 106 Schottky pairs per cm3 at room temperature (As in 1 cm3 there are nearly 1022 ions so there is one schottky defect per 1016 ions.) y Combination of ‘p’ and ‘n’ type semiconductors are used to make electronic components e.g., Diode. y On increasing temperature of CsCl structure the coordination number changes from 8–8 to 6–6. While on increasing pressure in NaCl structure Co No increases from 6–6 to 8–8. y The production of frenkel or schottky defects is an endothermic process. y He has HCP structure while rest of inert gases have C.C.P structure. y The most unsymmetrical crystal system is Tri clinic as a ≠ b ≠ c and α ≠ β ≠ γ ≠ 90°. y At the highest temperature at which super conductivity was called as 23 K in case of alloys of Niobium. y d–spacing: It is a distance between two parallel planes in a cubic lattice. a d= h + k2 + l 2 2 Here, a = edge length h, k, l = miller indices of plane. y Atomic Radius → and Edge Length (a) a In a simple cubic unit cell r = 2 In face-centred cubic cell (FCC) r = a 2 3a In body-centred cubic cell (BCC) r = 4 y ⚫ No. of Atoms in a Unit Cell/ Unit Cell Content (Z) nc nf ni Z   3 2 1 Here nc = 3, nf = 6 , ni = 1 Solid States 33.  y Density of Lattice Matter (d) Z × Atomic weight d= N0 × Volume of unit cell (a3 ) Here d = Density Z = Number of atoms N0 = Avogadro number a3 = Volume a = Edge length y Packing Fraction 4 3 Z× πr P.F = 3 a3 y Bragg’s Equation nλ = 2d sinθ Solid States 34.