THE SOLID STATE ( MARKS-4)
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1 THE SOLID STATE ( MARKS-) PREPARED BY : APSINGH BHADOURIA PGT (CHEM), KV KHATKHATI Crystalline solid [] In a crystalline solid the particles (atoms, molecules or ions) are arranged in a regular and repetitive three dimensional arrangements [2] These solids have sharp melting point [] These solids are anisotropic, ie their physical properties such as electrical conductivity, refractive index, thermal expansion etc have different values in different directions Examples: All the metallic elements like iron, copper and silver; Non metallic elements like sulphur, phosphorus and iodine and Compounds like sodium chloride, zinc sulphide and naphthalene Amorphous solid [] In amorphous solid the particles (atoms, molecules or ions) are arranged in an irregular and non repetitive three dimensional arrangements [2] Rapidly solidified liquids are amorphous substances, eg Glass, rubber etc [] These solids are generally isotropic, ie physical properties are same in all directions [] These solids do not have sharp melting point and boiling point ie they melt gradually over a temperature range What makes a glass different from a solid such as Quartz? Under what conditions quartz could be converted into glass? In glass, amorphous silica ( SiO 2 ) is present SiO tetrahedral have an irregular arrangement In quartz, crystalline silica ( SiO 2 ) is present SiO tetrahedral have a regular arrangement When quartz (SiO 2 ) is melted and the melt is cooled very rapidly, quartz converted into glass Why glass pans fixed to windows or doors of old building are invariably found to be slightly thicker at the bottom than at the top? Due to fluidity property, the glasses flows down very slowly and make the bottom portion slightly thicker Why solids are rigid in nature? The constituent particles in solid have fixed position and can only oscillate about their mean position, for which solids are rigid Name the factors which determine the stability of a substance [] Intermolecular forces tend to keep the molecules or constituent particles closer [2] Thermal energy tends to keep them apart by making them move faster Characteristic properties of solid [] They have definite mass, volume and shape [2] The inter molecular distances are short [] The inter molecular forces are strong [] Their constituent particles have fixed positions and can only oscillate about their mean positions [5] They are incompressible and rigid Why glass is considered as a super cooled liquid? Amorphous solids have a tendency to flow Since glass is an amorphous solid, so it is called super cooled liquid or pseudo solid Why some glass objects from ancient civilizations are found to become milky in appearance? Glass becomes crystalline at some temperature For which glass objects from ancient civilizations become milky in appearance because of some crystallization
2 Types of solid Types of solid [] Molecular solids (a) Non-polar (b) Polar (c)hydrogen bonded Constituent particles Attractive forces Examples Physical nature Molecules Dispersion or London forces Ar, CCl, H 2, I 2, CO 2 Soft Dipole-dipole interactions Hydrogen bonding HCl, SO 2 H 2 O ( ice) [2] Ionic solids Ions coulombic or electrostatic NaCl, MgO,ZnS, CaF2 Hard but brittle Soft Hard Electrical conductivity Insulator Insulator Insulator Insulator in solid state but conductors in molten state and in aqueous solutions Melting point Very low Low Low High [] Metallic solids Positive ions in a sea of delocalised electrons Metallic bonding Fe, Cu, Ag, Mg Hard but malleable and ductile Conductors in solid state as well as in molten state Fairly high [] Covalent or network solid Atoms Covalent bonding SiO 2 ( quartz), SiC, Diamond, AlN Hard Insulator Very high Graphite Soft Conductor Space lattice or crystal lattice A regular three dimensional arrangement of points in space is called space lattice or crystal lattice The points represent the constituent particles of the crystal Unit cell An unit cell is the smallest portion of the crystal lattice When it is moved repeatedly a distance equal to its own dimension along each direction, a three dimensional crystal lattice is generated Types of unit cell Unit cells are of two types: [] Primitive unit cell It is also called simple unit cell, which has particles as points only at its corners - (Eight lattice points) a a a [2] Non-Primitive unit cell / centered unit cell- In this type of unit cells, particles as points are present not only at the corners but also at some other positions [Particles as points are located at the corners and also at the centre of the cube] Nine lattice points Body - centered [ Particles as points are located at the corners and also in the centre of each face ] Fourteen lattice points Face-centred End-centred- Particles as points are located at the corners and laso at the centres of any two opposite faces - Ten lattice points Parameters of a unit cell Depending upon the symmetry of the axial distance(a,b,c)and also the axial anglebetween the edge ( ) the various crystals can be divided into seven systems Further the Z seven crystal systems on the basis of unit cells present, classified into fourteen different types of lattices, called Bravis lattices Y [] Cubic [-Primitive, Face-centred,Body-centred] --NaCl, Zinc blende and Cu [2] Tetragonal - [ Primitive,Body-centred]- SnO 2, TiO 2, CaSO, White tin X [] Orthorhombic-[Primitive,FC,End-centred & BC]-KNO, BaSO,Rhombic sulphur [] Hexagonal - [Primitive]- Graphite, ZnO, CdS [5] Rhombohedral or Trigonal - [ Primitive ] - Calcite ( CaCO ), HgS ( cinnabar ) [6] Monoclinic- [Primitive,End-centred]-Na 2 SO 0H 2 O, Monoclinic sulphur [7] Triclinic -[ Primitive ]- K 2 Cr 2 O 7, CuSO 5H 2 O, H BO
3 /8 CONTRIBUTION PER UNIT CELL FULL CONTRIBUTION /2 CONTRIBUTION
4 Rank of unit cell ( Z ) The number of particles as points in a unit cell is known as rank To calculate rank, these points should be noted:- - A point at each corner of unit cell is counted as - The points on an edge are counted as 8 - The points at each face are counted as - The points within the unit cell is conted as 2 The rank of unit cell of :- [] Primitive / simple cubic = 8 X 8 = [2] Body-centred cubic = X 8 + = 2 8 [] Face-centred cubic = 8 X X = [] In hexagonal system,each corner atom shared by 6 hexagonal unit cell,each face shared by 2 unit cell and inside thebody there are three atoms Z= ( X 2 corners) + ( X 2 faces ) + ( inside the body ) = Packing in metallic crystal:- The identical solid spheres can be packed in a number of ways [] In the first layer the spheres are arranged in a hexagonal manner in which each sphere is in contact with six other spheres [2] In the second layer the spheres will fit into the depression of the first layer [] For the third layer, there are two possibilities:- (a) The spheres can be placed in the depression of the second layer ie the third layer is directly above the first and the fourth layer is directly above the second This leads to the arrangement ABABAB This type of arrangement is known as hexagonal closed packed (HCP) structure eg Zinc, magnesium crystallizes in this type of structure [b] Alternatively, the sphere can be placed in the depressions, of the second layer; do not lie directly above the atoms of the first layer ie the spheres in fourth layer lie exactly above the first, fifth above the second, sixth above the third and so on This leads to the arrangement ABCABCABC This type of arrangement is cubic closed packed or face centered cubic arrangement eg Cu,AgAu crystallizes in this type of structure Co-ordination number It is the number of atoms or spheres that surrounds the single sphere / atom in a crystal CNof tetragonal arrangement = CNof tetrahedral arrangement = CNof octahedral arrangement = 6 CNof body centered cubic arrangement = 8 Any close ( tight ) packing having CN = 2 ie hcp and ccp ie fcc having CN = 2 Void / Hole/ Interstices The space which is left in between the closest pack arrangement is called void In close packing two types of voids are created [] Tetrahedral void: It refers to one empty space surrounded by four spheres which lie at the vertices of a regular tetrahedron There are eight tetrahedral voids around each sphere Cordination number = Radius ratio = 0225 The number of tetrahedral voids belonging to a sphere in close packing arrangement Number of tetrahedral voids around a sphere = Number of sphere around a void = 8 = 2 [2] Octahedral void : It refers to one empty space surrounded by six spheres which lie at the vertices of a regular octahedronthere are six octahedral voids around each sphere Cordination number = 6 Radius ratio = 0 The number of octahedral voids belonging to a sphere in close packing arrangement Number of octahedral voids around a sphere = = 6 Number of sphere around a void 6 = Octahedral voids are larger than the tetrahedral voids In FCC there are four octahedral voids and eight tetrahedral voids A cubic solid is made up of two elements P & Q Atoms of Q are at the corners of cube and P at the body centre What is the formula of the compound? What are the cordination number of P & Q There are eight corners which are occupied by Q No of Q per unit cell = 8 X 8 = P is situated at the body centre, So no of P per unit cell = So formula of the compound = PQ CN of P = 8 & CN of Q = 8
5 Relationship between edge length (a) and radius of the sphere ( r ) in unit cell For FCC, a = 2 2 r For BCC, a = r For simple cubic, a = 2 r Packing efficiency Volume occupied by all the spheres in a unit cell Packing efficiency = x 00 Volume of unit cell Packing efficiency, For simple cubic= 526 %, For BCC = 68 %, For FCC = 7 % Calculate the efficiency of packing in case of a metal crystal for Simple cubic, body centred cubic and face centred cubic [] For simple cubic a = 2 r Rank ( Z ) = So volume occupied by one sphere in the unit cell = r Volume of the unit cell = ( 2 r ) = 8r Hence, packing efficiency = r x 00 = 526 % 8r [2] For body centred cubic a = r Rank (z) =2 So, volume occupied by two spheres in the unit cell = 2 x r Volume of unit cell = 2 x r r So packing efficiency = x 00 = 68 % r [] For face centred cubic a = 2 2 r Rank (Z) = So, volume occupied by four spheres in the unit cell = x r Volume of unit cell = So packing efficiency = x r 2 2 r x 00 = 7 % 2 2 r Gold(Atradius= 0nm)crystallizes in afcc unit cell What is the length of the side oftheunitcell? Atomic radius= 0 nm So, length of side a = 2 2 r = 2 2 x 0 = 006 nm Aluminum crystallizes in a cubic closed packed structure Its metallic radius is 25 pm [] What is the length of the side of the unit cell? [2] How many unit cells are there in one cc of aluminium? [] For FCC, a = 2 2 r = 2 2 x 25 = 5 pm, So the edge length of the unit cell = 5 x 0-0 cm [2] Volume of unit cell = a = ( 5 x 0-0 ) cm Therefore, number of unit cells in cc of aluminum = / ( 5 x 0-0 ) cm = 225 x 0 22 Unit cells Radius ratio in ionic crystal The ratio of the radius of the smaller sphere with that of the larger is called the radius ratio Radius ratio in ionic crystal = Radius of cation ( r + ) Radius of anion ( r - ) SNo Types of packing Co-ordination number Radius ratio Example Tetrahedral Void ZnS 2 Octahedral Void NaCl Body centred cubic CsCl High pressure increases the cordination number and high temperature decreases the cordination number Thus high pressure converts NaCl structure into CsCl structure and high temperature converts CsCl into NaCl
6 If the radius of the octahedral void is r and radius of the atoms in close packing is R, derive relation between r and R The figure shows the cross section of an octahedral void In Triangle ABC, BC 2 = AB 2 + AC 2 => (2R) 2 = (r + R) 2 + (r+r) 2 => R 2 = 2 (r + R) 2 B A C => 2 R 2 = (r + R) 2 => r + R = 2 R => r R + = 2 => r R = => 2 - = - r =0 R Ferric oxide crystallizes in a hexagonal closed pack array of oxide ions with two out of every three octahedral holes occupied by ferric ions Deduce the formula of the ferric oxide No of atoms in one unit cell of hcp structure = 6 No of oxide ions per unit cell = 6 No of octahedral voids = 6 Since ferric ions occupy only two out of every three octahedral voids, there fore, no of octahedral voids occupied by ferric ions = ( 2/ ) x 6 = Stoichiometric ratio of Fe + and O 2- is : 6 = 2 : Hence the formula of ferric oxide is Fe 2O Analysis shows that nickel oxide has the formula Ni 098 O 00 What fraction of nickel exist as Ni 2+ and Ni + ions? Let amount of Ni + be x mol Then amount of Ni 2+ is (098 x) Total oxidation number of Ni in the compound is x + 2 (098 x) Oxidation number of oxygen is -2 Since the sum of the oxidation number of all the constituents in a compound is zero => x + 2 (098 x) 2 = 0 => x x 2 = 0 => x = 00 Hence % of Ni + = ( 00 / 098 ) x 00 = 08 % % of Ni 2+ = = 9592 % Relationship between density (d) and the dimension of unit cells Let the edge length of unit cell be a There fore volume of unit cell = a Let no of atoms in unit cell = Z Gm Atomic mass = M There fore mass of one atom = M / N A Where N A = Avogadro s number ie 602 x 0 2 Mass of Z atoms = ZM N A Density of unit cell ( d ) = Mass of unit cell Volume of unit cell = ZM N A a d = ZM a N A Silver crystallizes in FCC lattice If edge length of the cell is 05 x 0 8 cm and density is 05 gm/cm Calculate atomic mass of silver Given data- Edge length ( a ) = 07 X 0 8 cm Density ( d ) = 05 gm/ cm Since silver crystallizes in fcc lattice, so rank, ie the no of silver atoms per unit cell ( z ) = N A = 602 X 0 2 ZM So d = a M = N A d a N A z M = 05 X ( 07 X 0-8 ) X 602 X 0 2 = 07 gm / mol So atomic mass of silver = 07 amu Niobium crystallizes in body-centered cubic structure If density is 855 g / cm, calculate atomic radius of niobium using its atomic mass 9 u Density ( d ) = 855 gm/cc Atomic mass ( M ) = 9 gm/ mol Rank ( Z ) = 2 so d = Z M a a N = Z M a 2 X 9 = a = 609 X 0-2 CC A d N A 855 X 602 X 0 2 In bcc, a = r a = ( 609 X 0-2 ) / = 05 X 0-8 cm r = a X 05 X 0-8 r = = X 0-8 cm = nm
7 Copper crystallises into a fcc lattice with edge length density is in agreement with its measured value of 892 g cm - cm Show that the calculated For fcc, rank ( Z ) = ; Edge length,( a ) = 6 X 0-8 cm ; Molmass ( M )= 65 gm/mol ZM d = X 65 a N d = A ( 6 X 0-8 ) X 602 X 0 2 = 896 gm / cm So, the measured value is nearly equal to the calculated value Crystal Internal irregularities of crystals is known as crystal Line Stoichiometric Vacancy Interstitial Schottky Frenkel Crystal Point Non-Stoichiometric Metal excess Meta-deficiency Colour centre or F-centre or Anion vacancy Impurity Cation vacancy # Single crystals are formed when the process of crystallization occurs at extremely slow rate # Point s are the irregularities or deviations from ideal arrangement around a point or an atom in a crystalline substance # Line s are the irregularities or deviations from ideal arrangement in entire rows of lattice points # Point s do not disturb the stoichiometric of the solid is known as stoichiometric or intrinsic or thermodynamic # Vacancy When some of the lattice sides are vacant, the crystal is said to have vacancy This results the decrease in density of the substance This develops when a substance is heated # Interstitial - When some constituent particles occupy an interstitial site, the crystal is said to have interstitial This increases the density of the substance Schottky [] It occurs when a pair of ions of opposite charge are missing from the ideal lattice [2] The presence of a large number of Schottky in a crystal lowers its density [] This occurs if cation and anion having similar size with high coordination number [] In sodium chloride, there is one schottky for 0 6 ions One cc of sodium chloride contains 0 22 ions Therefore, one cc of sodium chloride possesses 0 6 schottky pair of ions Frenkel [] When an ion leaves its position in the lattice and occupy interstitial site leaving a gap in the crystal ie it creates a vacancy and interstitial [2] This will occur if size of cation is smaller than anion, with low coordination number [] Frenkel s are not found in pure alkali halide Due to larger size of cations, ions can not accommodate in interstitial site [] Frenkel are found silver halide, because silver ions are smaller in size and can get into the interstitial site [5] The Frenkel does not change the density of the solid [6] In silver bromide, both Schottky and Frenkel s are found Metal excess [a] F-Centers ( Farbe: means colour ) [] When there is an excess of metal ions in non-stoichiometric compounds, the crystal lattice has vacant anion site The anion sites occupied by electrons are called F-centre [2] The F-centers are associated with the colour of the compounds Excess of K in KCl makes the crystal violet Excess of Li in LiCl makes the crystal pink [] Solid containing F-centre are paramagnetic, because the electrons occupying the F-centers are unpaired [] When the crystal having F-centers are exposed to light, they become photoconductor
8 5 Kj/mol -6 Kj/mol [b] Metal excess due to presence of extra cation at the interstitial site Zinc oxide is white in colour at room temperature on heating, it looses oxygen and turns yellow Now there is excess of zinc in the crystal and its formula becomes Zn + x O The excess zinc ions move to interstitial site and the electrons to neighboring interstitial site Metal deficiency FeO, mostly found with a composition of Fe 095 O ie range from Fe 09O Fe 096O In crystals of FeO, some Fe 2+ ions are missing and the loss of positive charge is made up by the presence of required number of Fe + ions Electron sea model of metallic bonding [] A metal consists of a lattice of positive ions ( Kernel ) immersed in a sea of valence electrons ( mobile electrons ) [2] The force of attraction between the mobile electrons and the the kernels is known as metallic bond [] The electrical and thermal conductivity of metals can be explained by the presence of mobile electrons in metal The Band Model Of Metallic Bonding The band model of metal is based on molecular orbital theory When a large no of orbital overlap in metal, it results a continuous energy level produced by a large number of molecular orbital is called energy band Coduction band 2p 0 2s 2 2p band ( conduction band ) 2s- band ( valence band ) Valence band [ Insulator ] [ Semi conductor ] [ conductor ] s 2 Conductivity of Insulator s- band ohm - m - Semiconductor ohm - m - Conductor ohm - m - The lowest unoccupied energy band is known as conduction band The highest occupied energy band is known as valence band The energy difference between the top of valence band and the bottom of the conduction band is known as energy gap Impurity ( Doping ) The introduction of s in a particular crystalline solid by the addition of other elements is known as doping Doping increases the conductivity of crystal For example, if we mix strontium chloride ( SrCl 2 ) with sodium chloride, some strontium ( Sr 2+ ) ions occupy the lattice sites of sodium ions ( Na + ) and equal number of sodium ( Na + ) sites remain vacant Such vacancies in the crystal increase the electrical conductivity because certain ions from the neighboring sites can move into these vacant holes In this the number of positive ions are less as compared to negative ions Crystals with such s also act as semiconductor Since the conductivity is due to holes, these are known as P-type semiconductors If NaCl is doped with 0 - mol % of SrCl2 What is the concentration of cation vacancies? The addition of SrCl 2 to NaCl produces cation vacancies equal in number to that of Sr 2+ ions No of moles of SrCl 2 added to one mol of NaCl = 0 - / 00 = 0-5 mol No of holes created in one mole of NaCl = 0-5 X 602 X 0 2 = 602 x 0 8 Semiconductor These are solids whose conductivity lies in between those of conductors and insulators The conductivity of semiconductors increases with increase of temperature Intrinsic semiconductor An insulator capable of conducting electric current at higher temperature or when irradiated with electromagnetic radiations, are known as intrinsic semiconductor This happens because certain covalent bonds are broken and the released electrons are in a position to conduct electric current eg Silicon, Germanium
9 Extrinsic semiconductor These are formed when impurities of certain elements are added (doped) to insulator N-type semiconductors It is obtained by doping group elements with group 5 elements Suppose Si is doped with P with 5 valence electrons, out of 5 valence electrons, only valence electrons are involved in bond formation The fifth electron is not bound any where and can be easily promoted to the conduction band The conduction is thus mainly caused by the movement of electrons P-type semiconductors It is obtained by doping group elements with group elements Suppose Si is doped with Ga which has valence electrons, valence electrons are involved in bond formation with neighboring Si atom A vacancy is left which can be filled by the transfer of a valence electron from a neighboring Si atom The movement of electron into the vacancy leaves behind a hole which carries positive charge Another electron from a neighboring Si atom can move into the hole leaving behind another hole It appears as if the hole has moved through the lattice The movement of positively charged hole is responsible for the conduction of charge Non-stoichiometric cuprous oxide, Cu 2O can be prepared in laboratory In this oxide, copper to oxygen ratio is slightly less than 2: Can you account for the fact that this substance is a p-type semiconductor? [] Since Cu 2O is non-stoichiometric oxide, it contains Cu in two oxidation states, + and +2 [2] Cu 2+ provides an excess of positive charge As a result an electron from a neighboring Cu + is transferred to Cu 2+ [] The transfer of electron leaves behind a hole, which carries an extra positive charge and a negative hole is created [] It appears that the positive hole moves through the lattice, hence it appears as P-type semiconductor 2 6 and 5 compounds Combination of elements of Gr and Gr 5 or Gr 2 and Gr 6 produce compounds which stimulate average valence of four as in Ge or Si 2 6 compounds > ZnS, CdS, CdSe, HgTe 5 compounds > InSn, AlP, GaAs Magnetic properties [] Diamagnetic Diamagnetic substances are the substances which are weakly repelled by a magnetic field The electrons in diamagnetic substances are all paired They do not contain unpaired electrons eg TiO 2, NaCl, C 6H 6,N 2, Zn [2] Paramagnetic Paramagnetic substances are those which are attracted by a magnetic field but they lose their magnetism in the absence of magnetic field These substances have permanent magnetic dipole, due to presence of atoms, molecules or ions containing unpaired electrons eg Cu 2+, Fe +, O 2, NO, CuO, etc Substances containing unpaired electrons are further classified as: (a) Ferromagnetic substances > Ferromagnetic substances are those substances which are strongly attracted by a magnetic field and can be made into permanent magnets These substances show magnetism even in the absence of a magnetic field The large magnetism in these substances is due to the spontaneous alignment of magnetic moment,ie Unpaired electron in the same direction Example: Iron, cobalt, nickel, gadolinium and CrO 2 CrO 2 is used to make magnetic tapes for audio recording (b) Anti-ferromagnetic substances > Anti-ferromagnetic substances are those substances in which equal number of magnetic moments are aligned in opposite directions so as to give zero net moment Example: MnO, MnO 2 and Mn 2O (c) Ferrimagnetic substances Ferrimagnetism is observed when the magnetic moments of the domains in the substance are aligned in parallel and anti-parallel directions in unequal numbers They are weakly attracted by magnetic field as compared to ferromagnetic substances Example: Fe O (magnetite) and ferrites like MgFe 2O and ZnFe 2O These substances also lose ferrimagnetism on heating and become paramagnetic
10 UNIT I - SOLID STATE - HOME ASSIGNMENT By AP SINGH, PGT- CHEMISTRY, KV KHATKHATI How may the conductivity of an intrinsic semiconductor be increased? (CBSE 202) 2 Which stoichiometric increases the density of a solid? (CBSE 202,,09) What are n-type semiconductors? (CBSE 202) What type of semiconductor is obtained when Si is doped with As? (CBSE 200) 5 Which point decreases the density of a solid? (CBSE 200, 09,2008) 6 What is the No of atoms in a unit cell of fcc crystal? (CBSE 2009) 7 Define forbidden zone of an insulator? (CBSE 2008) 8 What is the No of atoms in a unit cell of simple cubic crystal? (CBSE 200) 9 State a feature to distinguish a metallic solid from an ionic solid (CBSE 200) 0 Crystalline solids are anisotropic in nature What does this statement mean? (CBSE 20) Which point in crystal does not change the density of a solid? (CBSE 200,09),2 What type of interaction hold the molecule together in a polar molecule? (CBSE 200) Wrire a distinguishing feature of metallic solid (CBSE 200) What is meant by intrinsic semiconductor? (CBSE 20) 5 What is the No of atoms in a unit cell of bcc crystal? (CBSE 20) 6 Why LiCl acquire pink colour when heated in Li vapour? (CBSE sample 20) 2 Silver crystallises in fcc structure Each side of the unit cell has a length of 09 pm What is the radius of Ag atom? (CBSE 20,0,09) 22 Calculate the packing efficiency of a metal crystal for simple cubic lattice (CBSE 20) 2 Explain how you can determine the atomic mass of an unknown solid if u know the mass density and the dimensions of unit of its crystal (CBSE 20) 2 define the following terms wrt crystalline solid (a) Unit cell (b) Coordination No Give example in each case 25 KF has ccp structure Calculate the radius of unit cell if the side of the cube is 00 pm How many F ions & Octrahedral voids are there in this unit cell (CBSE sample 20) 26 Give reason (CBSE sample 20) (a) Why is Frankel found in AgCl? (b) What is difference between phosphorus doped & gallium doped semiconductors? 27 Gold (Atomic mass = 97 u, Atomic radius = 0 nm) crystallises in a fcc unit cell Determine the density of Gold (CBSE sample 20) 28 Classify each of the following as being either a p- type or a n-type semiconductor Give reason (a) Si doped with In (b) Si doped with P (CBSE sample 20) Copper crystallises with fcc unit cell if the radius of Copper atom is 278 pm, calculate thedensity of copper metal (Atomic Mass of Cu = 655 & Avagadro No = 602x 0 2 mol - ) (CBSE 202, 0) 2 Iron has a bcc unit cell with the cell dimension of pm Density of iron is 787 g/cc use this information to calculate & Avagadro s No (At Mass of Fe = 56) (CBSE 202) The density of copper metal is 895 g/cc, if the radius of Cu atom is 278 pm, is the Cu unit cell simple cubic, bcc or fcc? (CBSE 202, 0) Silver crystallises in fcc structure if the edge length of the unit cell is 07 x 0-8 cm & density of the crystal is 05 g/cc, calculate the atomic mass of Silver (CBSE 200, 09) 5 Niobium (Nb) crystallises in bcc structure If its density is 855 g/cc, calculate the atomic radius of Nb (atomic mass of Nb = 9 u) (CBSE 2008) 6 The well known mineral fluorite is chemically calcium fluoride It is known that in one unit cell of this mineral there are Ca 2+ ions and 8F ions and that Ca 2+ ions are arranged in fcc lattice The F ions fill all the tetrahedral holes in the fcc lattice of Ca 2+ ions The edge of unit cell is 56 x 0-8 cm in length the density of solid is 8 g/cc Use this information to calculate the Avagadro s number (Molar mass of CaF 2 = 7808 g/mol) (CBSE 200) 7 Aluminium crystallises in a cubic close-packed structure Radius of the atom in the metal is 25 pm (i) What is the length of the side of the unit cell? (ii) How many unit cells are present in cm of Aluminium? (CBSE 20) 7 Sodium crystallises in a bcc unit cell calculate the approximate No of unit cells in 92 g of Sodium ( CBSE Sample,)
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