216 - Chemistry Questions Answers

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.

v = {3RT/M}^1/2

so according to the given condition 

{3RT/32}^1/2 = {3R300/16}^1/2

or T/32 = 300/16

or T = 600

so t = 600-273 = 327⁰C