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1. Compute the minimum cation-to-anion radius ratio for a coordination number of 6. Solution This problem asks us to show that the minimum cation-to-anion radius ratio for a coordination number of 6 is 0.414 (using the rock salt crystal structure). Below is shown one of the faces of the rock salt crystal structure in which anions and cations just touch along the edges, and also the face diagonals. From triangle FGH, GF = 2rA and  Since FGH is a right triangle FH = GH = rA + rC  (GH)2 +
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  1.   Compute the minimum cation-to-anion radius ratio for a coordination number of 6. Solution This problem asks us to show that the minimum cation-to-anion radius ratio for a coordination number of 6 is 0.414 (using the rock salt crystal structure). Below is shown one of the faces of the rock salt crystal structure in which anions and cations just touch along the edges, and also the face diagonals. From triangle  FGH  , GF   = 2 r  A  and  FH   = GH   = r  A  + r  C  Since  FGH   is a right triangle ( GH  ) 2 + (  FH  ) 2 = (  FG ) 2  or ( r  A + r  C ) 2 + ( r  A + r  C ) 2 = ( 2 r  A ) 2  which leads to r  A  + r  C  = 2 r  A 2  Or, solving for r  C / r  A      r  C r  A      22    1            0 . 414   2.   Calculate the atomic packing factor for the NaCl crystal structure in which cation-anion radius ratio is 0.414. Solution This problem asks that we compute the atomic packing factor for the rock salt crystal structure when r  C / r  A  = 0.414. From Equation 3.2 APF = V  S  V  C   With regard to the sphere volume, V  S  , there are four cation and four anion spheres per unit cell. Thus, V  S   = (4)43  r  A3      + (4)43  r  C3      But, since r  C / r  A  = 0.414 V  S   = 163  r  A3 1     ( 0 . 414 ) 3   = (17.94) r  A3   Now, for r  C / r  A  = 0.414 the corner anions in Table 12.2 just touch one another along the cubic unit cell edges such that V  C   = a 3  = 2 ( r  A   r  C )   3    2 ( r  A   0 . 414 r  A )   3  ( 22 . 62 ) r  A3  Thus APF = V  S  V  C  = ( 17 . 94 ) r  A3 ( 22 . 62 ) r  A3 = 0.79    3.   Magnesium oxide has the rock salt crystal structure. Calculate the theoretical density. Solution We are asked to calculate the theoretical density of MgO. This density may be computed using Equation (12.1) as AOMg  +   N V  A An C       Since the crystal structure is rock salt, n'   = 4 formula units per unit cell. Using the ionic radii for Mg 2+  and O 2-  from Table 12.3, the unit cell volume is computed as follows:   33-2O+2Mg3 )140.0(2)072.0(22+2 nmnmr r    aV  C    cellunit cm107.62=cellunit nm 0.0762= 3 23-3   Thus,    units/molformula106.022cell/unit cm107.62 g/mol)16.00+g/mol1cell)(24.3units/unitformula(4   23323-      = 3.51 g/cm 3   4.   A circular specimen of MgO is loaded using a three-point bending test. If the specimen radius is 3.5 mm, the support point separation distance is 45 mm, and flexural strength of 105 MPa,  predict whether or not you would expect the specimen to fracture when a load of 580 N is applied. Justify your prediction. Solution (a) This portion of the problem asks that we determine whether or not a cylindrical specimen of MgO having a flexural strength of 105 MPa and a radius of 3.5 mm will fracture when subjected to a load of 580 N in a three-point bending test; the support point separation is given as 45 mm. Using Equation 12.7b we will calculate the value of  ; if this value is greater than   fs  (105 MPa), then fracture is expected to occur. Employment of Equation 12.7b yields      =  FL   R 3   MPa194= N/m10194= )105.3()( 10)580( = 26333 )45(   mm N   Since this value is greater than the given value of   fs  (105 MPa), then fracture is predicted.

unit 11 (1).pdf

Jul 23, 2017
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