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Atomic Structure

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Published in: Chemistry
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Atomic Structure

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  1. S-BLOCK ELEMENTS INTRODUCTION A solid is defined as that form of matter which possesses rigidity and hence possesses a definite shape and a definite volume. Unlike gases and liquids in which the molecules are free to move about and hence constitute fluid state, in a solids the constituent particles are not free to move but oscillate about their fixed positions. CLASSIFICATION OF SOLIDS Crystalline Solids Solids Amorphous Solids (a) Crystalline solids: Crystalline solids have long range order. The long range order means the atoms or ions or molecules are arranged in a regular fashion and this symmetrical arrangement extends throughout the crystal length. (b) Amorphous solids: An amorphous solid differs from a crystalline substance in being without any shape of its own and has a completely random particle arrangements, i.e. no regular arrangement. Example: Glass, A B A Plastic B o o o o Amorphous solid Crystalline solid DISTINCTION BETWEEN CRYSTALLINE AND AMORPHOUS SOLIDS 1. 2. 3. 4. 5. 6. 7. 8. Cr stalline solids The internal arrangement of particles is regular so the ossess definite and re ular eometr The have shar meltin oints There is regularity in the external form when cr stals are formed Crystalline solids give a regular cut when cut with a shar — ed ed knife The have characteristic heat of fusion. Crystalline solids are rigid and their shape is not distorted b mild distortin forces Crystalline solids are regarded as true solids Crystalline solids are anisotropic. This implies that physical properties such as refractive index, conductivity, thermal expansion etc are different in different directions. This is due to orderly arran ement of articles 1. 2. 3. 4. 5. 6. 7. 8. Amor hous solids The internal arrangement of particles is irregular. Thus the do not have an definite eometr . The do not have shar meltin oints There is no regularity in the external form when amor hous solids are formed Amorphous solids give irregular cut. The do not have characteristic heat of fusion. Amorphous solid are not very rigid. These can be distorted b bendin or com ressin forces. Amorphous solids are regarded as super cooled li uids or seudo solids Amorphous solids are isotropic in nature. This implies that various physical properties are same in all the directions. This is because of random arrangement of particles. TYPES OF SOLID Depending upon the nature of the forces holding the structural units — atoms, molecules or ions, solids can mainly be divided into four types. They are: 2. Molecular solid: is a solid in which the constituent particles are atoms or molecules which are held together by weak intermolecular forces i.e. Vander waals forces. Ionic psolid: is a solid that consists of positively and negatively charged ions arranged in a regular manner throughout the crystal. These cation and anion are held together by the strong electrical attraction of opposite chagers.
  2. 3. 4. Covalent solid: is a solid in which the constituent particles are atoms held together by covalent bond to form a large network. Diamond is a three dimensional network solid in which each carbon atom is covalently bonded to four other, to form a crystal which can be considered as an immense molecule. Metallic solid: is a solid which consists of positive cores of atoms immersed in a sea of mobile electrons (metallic bonding). In this type of bonding, the positive charge atomic cores are surrounded by delocalised electrons. UNIT CELL In this topic we would be studying certain properties of a solid which depend only on the constituents of the solid and the pattern of arrangement of these constituents. The smallest amount of the solid whose properties resemble the properties of the entire solid irrespective of the amount taken is called a unit cell. It is the smallest repeating unit of the solid. Any amount of the solid can be constructed by simply putting as many unit cells as required. . Unit cells arranged together In a 3-dimensional space lattice, to specify a unit cell we need the values of three vectors which gives three distances along the three axes and three angles as shown in figure below z c Y x Only four possible ways in which constituent species of a unit cell can be arranged. (a) Primitive or simple (c) Face centred and (b) Body centred (d) End centred So, the total number of possible unit cells are 7 x 4 = 28. Bravais, then went on to predict that out of these 28 possible unit cells, only 14 of them exist in nature. Simple cubic Monoclinic = 900, 900 Tetragonal Triclinic 900 Orthorhombic Rhombohedral 01 = = 900 Hexagonal Primitive = 900 , Y = 1200
  3. CALCULATION OF NUMBER OF PARTICLES PER UNIT CELL 2. 3. 4. The no of atom in a unit cell can be calculated by keeping in view following points An atom at the corners is shared by eight unit cells. Hence the contribution of an atom at the corner to a particular cell = 1/8 An atom at the face is shared by two unit cells. Hence the contribution of an atom at the face to a particular cell = 1/2 An atom at the edge centre is shared by four unit cells in the lattice and hence contributes only 1/4 to a particular unit cell. An atom at the body centre of a unit cell belongs entirely to it, so its contribution = 1 The number of atoms per unit cell is in the same ratio as the stoichiometry of the compound. Hence it helps to predict the formula of the compound Example 1: Solution A metal (atomic mass = 50) has a body centred cubic crystal structure. The density of the metal is 5.96 g/cm3 . Find the volume of unit cell. (No = 6.023 x 1023 atoms mol-I) Mass of the unit cell = mass of one atom x No. of atoms present per unit cell 50 2 x 6.023 x 1 023 Mass Volume of unit cell Density 2.78 x 10 23 cm3 TYPE OF UNIT CELL 1 6.023 x 1 023 5.96 1. 2. A units cell is obtained by joining the lattice points. The choice of lattice points to draw a unit cell is made on the basis of the external geometry of the crystal, and symmetry of the lattice. There are four different types of unit cells. Primitive or Simple cubic (PC/SC) unit cell In a primitive cubic unit cell, all the eight corners of the cube are occupied by atoms and they are not present anywhere else in the cube. If all the eight atoms are of same type, the unit cell is called ideal primitive and if they are not same, it is referred as pseudo primitive or primitive-like. The volume of a cubic unit cell is not completely occupied by atoms. The ratio of volume occupied by the effective atoms to the volume of the unit cell is called packing fraction (PF). The fraction of volume of a unit cell, that is empty is called void fraction (VF). Volume occupied by the effective atoms in a unit cell Packing fraction = Volume of the unit cell (a3) and void fraction = (1 - Packing fraction) 3 1 x —It r 3 (2r)3 0.523 6 Body centered cubic (BCC) unit cell In body centered cubic unit cell, the lattice points are corners and body centre. This implies that the atoms are present at all the corners and at the body centered position and are not present anywhere else in the cube. Since, the body centered atom touches all 8 corner atoms, so the body diagonal (6 a) would be equal to 4r,
  4. where 'a' represents the edge length and 'r' is the radius of atom. The fraction of volume occupied by the atoms is given by 3 3 = 0.6802 8 3. Face centered cubic (fcc) unit cell In FCC unit cell, the lattice sites are corners and face centres. Thus, in face centered cubic unit cell, the atoms are present at the corners and at the face centres and are not present anywhere else in the cube. When all the 14 atoms are of same size, the unit cell is called ideal FCC while if the atoms present at the corners and face centres are different, the unit cell is referred as pseudo FCC or FCC-like. The fraction of volume occupied by the atoms is given by 3 3 = 0.7405 6 (4) Side centered unit cell Atoms are arranged at the centre of only one set of faces in addition to the atoms at the corner of the unit cell. Example 2 : Solution: Suppose you have a face-centred cubic arrangement of X and Y-atoms where X-atoms are at the corners of the unit cell and Y-atom are at the face-centres. (a) (b) (a) What is the formula of the compound? What would be the simplest formula if (i) One of the X-atoms were missing from a corner is each unit cell? (ii) Two X-atoms were missing from the two corners? Number of X-atom at the corners of the unit cell 1 = 8 x — (at the corners) = 1 8 Number of Y-atom at the face-centres of the unit cell 1 6 x — (at the face-centres) = 3 2 Hence, the formula of the compound is XY3 If one-X-atom were missing from a corner in the unit cell, then, total number of X-atom in the unit cell Since, the number of Y atoms is 3, the formula of compound becomes i.e. X7Y24 If 2 X-atoms were missing from two corners, then the number of X-atom 1 1 6 corresponding to a unit cell will be 1 8 Since the number of Y atoms is 3, the formula of compound now becomes i.e. X i.e.
  5. CLOSE PACKING IN CRYSTALLINE SOLIDS In the formation of crystals, the constituent particles (atoms, ions or molecules) get closely packed together. The closely packed arrangement is that in which maximum available space is occupied. This corresponds to a state of maximum density. The closer the packing, the greater is the stability of the packed system. (1) Close packing in two dimensions : The two possible arrangement of close packing in two dimensions. (i) Square close packing : In which the spheres in the adjacent row lie just one over the other and show a horizontal as well as vertical alignment and form square. In this arrangement each sphere is in contact with four spheres. Square packing (ii) Hexagonal close packing : In which the spheres in every second row are seated in the depression between the spheres of first row. The spheres in the third row are vertically aligned with spheres in first row. The similar pattern is noticed throughout the crystal structure. In this arrangement each sphere is in contact with six other spheres. cbse (2) Close packing in three dimensions : In order to develop three dimensional close packing, let us retain the hexagonal close packing in the first layer. For close packing, each spheres in the second layer rests in the hollow at the centre of three touching spheres in the layer as shown in figure. The spheres in the first layer are shown by solid lines while those in second layer are shown by broken lines. It may be noted that only half of the triangular voids in the first layer are occupied by spheres in the second layer. The unoccupied hollows or voids in the first layer are indicated by in figure. Close pacuøtg in ttuee dimensions There are two alternative ways in which species in third layer can be arranged over the second layer, (i) Hexagonal close packing(hcp) : The third layer lies vertically above the first and the spheres in third layer rest in one set of hollows on the top of the second layer. This arrangement is called ABAB . type and 74% of the available space is occupied by spheres. This arrangement is found in Be, Mg, zn, Cd, sc, Y, Ti, zr, Tc, Ru. Hexagonal close packing (hcp) in three dimensions (ii) Cubic close packing (ccp) : The third layer is different from the first and the spheres in the third set of hollows marked 'C' in the first layer. This arrangement is called layer lie on the other ABCABC..... type and in this also 74% of the available space is occupied by spheres. The cubic
  6. close packing has face centred cubic (fcc) unit cell. This arrangement is found in Cu, Ag, Au, Ni, Pt, Pd, co, Rh, ca, Sr. e o dog in thr•• (iii) Body centred cubic (bcc) : This arrangement of spheres (or atoms) is not exactly close packed. This structure can be obtained if spheres in the first layer (A) of close packing are slightly opened up. As a result none of these spheres are in contact with each other. The second layer of spheres (B) can be placed on top of the first layer so that each sphere of the second layer is in contact with four spheres of the layer below it. Successive building of the third will be exactly like the first layer. If this pattern of building layers is repeated infinitely we get an arrangement as shown in figure. This arrangement is found in Li, Na, K, Rb, Ba, Cs, V, Nb, Cr, MO, Fe. öööo pEi.ing in Comparison of hcp, ccp and bcc Property Arrangement of packing Type of packing Available space occupied Coordination number Malleability and ductility TYPES OF VOIDS hcp Close packed AB AB A... 12 Less malleable, hard, brittle ccp Close packed ABC ABC A... 12 Malleable ductile bcc Not close packed AB AB A... 68% 8 and Even in the close packing of spheres, there is left some empty space between the spheres. This empty space in the crystal lattice is called site or void or hole. Voids are of following types . (1) (2) Trigonal void : This site is formed when three spheres lie at the vertices of an equilateral triangle. Size of the trigonal site is given by the following relation, r = 0.155R r = Radius of the spherical trigonal void R = Radius of closely packed spheres Tetrahedral void : A tetrahedral void is formed when an atom fits into the depression formed by three other closest packed atoms (forming an equilateral triangle). Tetrahedral void
  7. (3) (4) In general the number of effective tetrahedral voids in a unit cell is double the number of effective atoms in that unit cell. Let the radius of host atom (forming FCC lattice) be rh while the radius of foreign atom entering a tetrahedral void in an ideal FCC lattice be rf, then the sum of radii of host and foreign atom in terms of edge length (a) is given by 4 OCTAHEDRAL VOID : An octahedral void is formed when three closest packed atoms of one layer (forming an equilateral triangle) is placed over three closest packed atoms of the second layer, their positions being inverted with respect to each other. Each atom touches four other atoms, except the atom diagonally opposite to it. The effective number of octahedral voids in a unit cell is equal to the number of effective atoms in that unit cell. Let the radius of host atom (forming FCC lattice) be rh and the radius of foreign atom occupying octahedral void in an ideal FCC lattice be r; , then the distance between a host atom and a foreign atom is given by where a is the edge length of ideal FCC lattice. Tetrahedral void Number of tetrahedral voids per FCC Octahedral 1 void An octahedral void at the centre of an edge in a FCC unit cell. Cubic void : This type of void is formed between 8 closely packed spheres which occupy all the eight corner of cube. 0.732 The decreasing order of the size of the various voids is, Cubic > Octahedral > Tetrahedral > Trigonal Cubic void Example 3 : Solution: If the close packed cations in an AB type solid have a radius of 75 pm, what would be the maximum and minimum sizes of the anions filling the voids? For closed packed AB type solid L = 0.414 — 0.732 . Minimum value of r- Maximum value of r- 75 pm = 102.5 pm 0.732 0.732 75 pm = 181.2 pm 0.414 0.414 RADIUS RATIO RULES The structure of many ionic solids can be accounted by considering the relative sizes of the cation and anion, and their relative numbers. By simple calculations, we can work out as how many ions of a given size can be in contact with a smaller ion. Thus, we can predict the co-ordination number from the relative size of the ions. RADIUS RATIO Limiting radius ratio, x < 0.155 0.155 < 0.225 0.255 < 0.414 0.414 < 0.732 0.414 < 0.732 0.732 < 0.999 x Co—ordination number 2 3 4 4 6 8 12 Shape Linear Triangular planar Tetrahedral Square planar Octahedral Body centered cubic Ideal FCC Example BeC12 BC13 zns, Beo [PtC14]2 Naa, CaO csa
  8. Example 4 : Solution: Iron crystallises in a body centred cubic structure. Calculate the radius of Fe atom if edge length of unit cell is 286 pm. Edge length, a = 286 pm For BCC, radius of atom, r = —a = — 286 = >< 4 4 = 123.8 pm 4 TYPES OF IONIC STRUCTURES (1) (2) (3) (4) For any ionic solid of the general formula AxBy, in which the molar ratio of cation and anion is x : y, the ratio of co-ordination number of cation to anion would be y : x. Rock Salt Structure : NaCl exhibits this type of structure. In the rock salt structure, Cl- ions exists in FCC pattern and Na+ ions occupy all octahedral voids. There are 4 effective Na+ ions and 4 effective Cl- ions in a unit cell of NaCl. So, the general formula is Na4C14 or NaCl as per the effective ions in a unit cell. The co-ordination number of Na+ ion is 6 and co-ordination number of Cl- ion is also 6. So, the general formula (using co-ordination number of ions) is Na6C16 or NaCl. Halides of all alkali metals except cesium and oxides of all alkaline earth metals except beryllium also represent this structure. AgCl, AgBr & Agl also crystallize in this structure. Other examples include TiO, FeO, NiO etc.. Let the radius of Na+ ion and Cl- ion be represented by rc and ra respectively and 'a' be the edge length of the unit cell, then rc + ra / 2. Rock salt structure Zinc Blende (Sphalerite) Structure : This structure is represented by ZnS. In the zinc blende structure, S2- ions forms FCC lattice and Zn2+ ions occupy alternate tetrahedral voids. The effective Zn2+ and S2 ions in a unit cell are 4 each, so the general formula is Zn4S4 or ZnS. The co-ordination number of both the ions (Zn2+ and S2-) is 4, so the general formula (using co-ordination number of ions) is also Zn4S4 or ZnS. The zinc blende structure is also exhibited by BeO. Let the radius of Zn2+ and S2- ions be rc and ra respectively and edge length of the unit cell be 'a', then 4 Fluorite Structure : This structure is exhibited by CaF2. In the exist as FCC and F- ions occupy all tetrahedral voids. The fluorite structure is also represented by other halides of alkaline earth metals and by ionic compounds of the general formula AB2 like IJ02, Th02, Pb02, HgF2 etc. Let 'a' be the edge length of the unit cell and rc and ra be the radius of Ca2+ and F- ions respectively, then 4 lattice Zinc blende structure structure of CaF2, Ca2+ ions Fluorite structure Antifluorite Structure : This structure is shown by Li20. In the antifluorite structure, 02 ions forms FCC lattice and Li+ ions occupy all tetrahedral voids. The effective number of Li+ and 02- ions in a unit cell are 8 and 4 respectively, so the general formula is Li804 or Li20. Antifluorite structure in just reverse of fluorite structure because the positions of cations and anions are interchanged This type of structure is also exhibited by oxides of other alkali metals like Na20, 1
  9. (5) (6) Cesium Halide Structure This type of structure is exhibited by cesium chloride. Chloride ions exist as primitive cubic and cesium ions ocupy body centered position (cubic void). Overall this structure is referred as BCC-Iike structure. The effective number of Cs+ and Cl- ions are one each, so the general formula is CsCl. The co-ordination number of both the ions (Cs+ and Cl-) is 8, so the general formula using co-ordination number of ions is Cs8C18 or CsCl. This kind of structure is also exhibited by other halides of cesium and by the halides of ammonium like NH4Cl, NH4Br etc. Let the radius of Cs+ and Cl- be rc and ra respectively and the edge length of the unit cell be 'a', then 2 Cesium chloride structure Spinel and Inverse Spinel Structure : Spinel is a mineral (MgA1204). Generally they can be represented as M2+M23+04, where M2+ is present in one-eighth of tetrahedral voids in a FCC lattice of oxide ions and M3+ ions are present in half of the octahedral voids. M2+ is usually Mg, Fe, Co, Ni, Zn and Mn; M3+ is generally Al, Fe, Mn, Cr and Rh. Examples are ZnA1204, Fe304, FeCr204 etc. Many substances of the type M M2 04 also have this structure. In an inverse spinel the ccp is of oxide ions, M2+ is in one-eight of the tetrahedral voids while M3+ would be in one-eight of the tetrahedral voids and one-fourth of the octahedral voids. Example 5 : Prove that void space in fluorite structure per unit volume of unit cell is 0.251. Solution : Cations form cubical closest packing and anions occupying the tetrahedral holes. There are 4 cations and 8 anions per unit cell. Here, face diagonal. where, r + = radius of cation 2NF2r 16 Volume of unit cell = Fraction of volume occupied per unit voume of unit cell 4 3- 3 3 i.e. Packing fraction = 16Uär3+ 2r3 Here in fluorite structure, for tetrahedral hole Radius of anion 0.225 Radius of cation r —{1 + = 0.749 Packing fraction = . Void space = 1 — 0.749 Void fraction = 0.251
  10. IMPERFECTIONS IN A CRYSTAL The discovery of imperfections in an another ideally perfect crystal is one of the most fascinating aspects of solid state science. An ideally perfect crystal is one which has the same unit cell and contains the same lattice points throughout the crystal. The term imperfection or defect is generally used to describe any deviation of the ideally perfect crystal from the periodic arrangement of its constituents. (1) Stoichiometric Defects : Two types of defects may be observed in stoichiometric compounds, called Schottky and Frenkel defects respectively. At absolute zero, crystals tend to have a perfectly ordered arrangement. As the temperature increases the amount of thermal vibration of ions in their lattice sites increases and if the vibration of a particular ion becomes large enough, it may jump out of its lattice site. (i) Schottky Defects: A vacancy at a cation site is mostly accompanied by a vacancy at a nearby anion site. Such paired cation-anion vacancies are referred as Schottky defect. In NaCl the existence of two vacancies, one due to a missing Na+ ion and the other due to a missing Cl- ion in a crystal of NaCl, is shown in figure. The crystal, as a whole remains neutral because the number of missing positive and negative ions is the same. Thus a Schottky defects consists of a pair of holes in the crystal lattice. Na+ missing Cl- missing (ii) Frenkel Defects : When an ion (cation or anion) leaves its lattice point and occupies some interstitial space, the defect is called Frenkel defect. This defect also preserves the electrical neutrality of the cr stal and the densit of the cr stal also remains unaltered as shown in fi ure for the cr stal of A Br. Frenkel Defects in a Crystal Remark : If small size atoms or ions occupy the vacant interstial sites, the defect is called interstitial defect. Thus, Frenkel defect is a hybrid type of defect arising from the combination of Schottky defect and interstitial defect. 2 Non-Stoichiometric Defects : The defects discussed so far do not disturb the stoichiometry of the crystalline substance. However, a large number of non-stoichiometric inorganic solids are known which contain the constituent elements in non-stoichiometric ratio due to defects in their crystal structures.ln non-stoichiometric or Berthollide compounds the ratio of the number of atoms of one kind to the number of atoms of the other kind does not correspond exactly to the ideal whole number ratio implied by the formula. These defects are of two types: (i) metal excess defect and (ii) metal deficiency defect (I) Metal Excess defect: This may occur in two different ways
  11. (i) F-Centres are the sites where anions are missing and instead electrons are present. They are responsible for colour (F = Farbe = Colour). (ii) Interstitial ions and electrons: Metal excess defects also occur when an extra positive ion occupies an interstitial position in the lattice and electrical neutrality is maintained by the inclusion of an interstitial electron. Their composition may be represented by general formula A1+öX. This kind of metal excess defect is much more common than the first and is formed in crystals which would be expected to form Frenkel defects. Examples include ZnO, CdO, Fe203. (Il) Metal Deficiency Defects: In certain cases, one of the positive ions is missing from its lattice site and the extra negative charge is balanced by some nearby metal ion acquiring two charges instead of one. There is evidently, a deficiency of the metal ions although the crystal as a whole is neutral. This type of defect is generally found amongst the compounds of transition metals which can exhibit variable valency. Example 6 : Solution: Sodium chloride is doped with 10-3 mol percent of SrC12. Find the concentration of cation vacancies. The charge on Na+ ion is +1 and that on Sr2+ ion is +2. Substitution of 1 Na+ ion with 1 Sr2+ ion causes 1 Na+ ion to leave the solid. The electrical neutrality in the lattice is thus maintained. Thus, concentration of cation vacancies = concentration of cations of dopped material (i.e. SrC12) Therefore, 1Sr2+ ion causes 1 cationic vacancy. From the question, concentration of Sr2+ ions is 10-3 mol percent. Hence, the concentration of cation vacancies should be equal to 10-3 mol percent. 1 Concentration of cation vacancies = 10 3 6.023 x 1023 X X 100 = 6.023 x 1018

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