217 - Chemistry Questions Answers

In Diamond, there is 8 lattice points per unit cell.  The maximum radius that a hard sphere can have is , r and a are in their usual meaning, Hence the packing fraction is:

second, third and fourth coordination number means to count the no. of atoms of either Cl or Na, having second, third and fourth closest distance

for NaCl         first coordination number =    6   (6 faces)            distance (a/2)

                 second coordination number =  12   (12 edges)        distance (a/√2)

                       third coordination number =    8   (8 corners)        distance (a√3/2)

                    fourth coordination number =    6    ( 6 centre)          distance (a)

similarly try for others

according to the given conditions

along the face of a side 4x = 400√2             or  x = 100√2                 here x is radius of the atom forming lattice

and along an edge 2x+2y = 400                      here y is interstitial radius

so y = 200-x = 200-100√2  = 100 (2-1.414) = 100(0.586) = 58.6 pm

so diameter = 117.2

use the formula d = Mn/N₀a³

calculate a?                here n = 1

now 2r ^- + 2r ⁺ = a √3

and 2r ^- = a

divide these equations for getting ratio

B atom forms ccp, let No. of atoms of B = 1

no. of tetrahedral voids = 2

but formula given is A_2/3B

so % of voids occupied = (2/3)*100/2 = 33.34 %  

Normal formula of compound is Na₄Cl₄

now on removing all the atoms from 1 body diagonal, 

lost atoms = 4 Cl at corner, 2 Na from edge centre, 2 Cl from face centre, 1 Na from body centre

so loss in Cl = 4*1/8 + 2*1/2 = 1.5

and loss iin Na = 2*1/4 + 1*1 = 1.5

so new formula = Na_2.5Cl_2.5

so formula will remail same 

along diagonal of a face 4r = a√2   or  r = a√2/4 so distance b/w centre of two touching spheres is 2r = a√2/2

similarly distance b/w two nearst non touching sphere = a (along edge)

so 3) will be the option

the question is incomplete, after ONE EIGHTH, i guess, it will be one eighth of the tetrahedral void are occupied by B

so let no. of O = 1

then no. of octahedral voids = 1 and half of this = 1/2

similarly tetrahedral voids = 2 so 1/8 of it = 2*1/8 = 1/4

so formula = A_.5B_.25O₁  or A₂B₁O₄

if my assumption for question is correct then answer would be 3)     here B is divalent positive ion.

NaCl    fcc structure   6 for Na and 6 for Cl

ZnS      In ZnS , S forms the face centered cube and Zn occupies half of the tetrahedral holes..The unit cell can be devided into eight equal small cubes,each with 1 Zn and 4 S ions so the coordination no will be 4 for Zn and 4 for S

CsCl  bcc structue  8 for Cs and 8 for Cl

CaF₂   fcc structure of Ca with 100% filled tetrahedral voids by F, so coordination no. of Ca is 8 and that of F is 4 

Li₂O  anti fluorite structure, fcc structure of O with 100% filled tetrahedral voids by Li so coordination of O is 8 and that of Li is 4 

its simple because question is based on volume and gas covers the complete volume of the vessel in which it is placed.