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Chemistry >> Physical Chemistry >> Solid State IIT JEE

In diamond,carbon atom occupy FCC lattice points as well as alternate tetrahedral voids.If edge length of the unit cell is 356pm,then radius of carbon atom is?

Asked By: VAISHALI JHA Solved By: SARIKA SHARMA
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Joshi sir comment

for fcc lattice a√2 = 4r

Chemistry >> Physical Chemistry >> Solid State IIT JEE

how many lattice points are there in one unit cell of face centered tetragonal

Asked By: VAISHALI JHA
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Joshi sir comment

8 corners contributing 1/8

6 faces contributing 1/2

Chemistry >> Physical Chemistry >> Solid State Medical Exam

IN A GIVEN COMPD THE BONDING FORCES ARE AT SHORT RANGE & ARE SPHERICALLY DISTRIBUTED . THE TYPE OF BONDING REFERRED TO IS C/A

A) COVALENT

B) METALLIC

C) HYDROGEN

D) IONIC

Asked By: SARIKA
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Joshi sir comment

the formation of covalent bond requires overlapping of partially occupied short-range repulsive potential due to the distortion of the electron clouds 

so answer will be covalent 

Chemistry >> Physical Chemistry >> Solid State Medical Exam

the max proportion of available volume that can be filled by hard spheres in diamond?

Asked By: SARIKA
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Joshi sir comment

 

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:

Chemistry >> Physical Chemistry >> Solid State Medical Exam

in my the previous query , i want to ask the second nearest, third nearest, 4th nearest neighbour atoms/ cordination no. of NaCl, CsCl, ZnS, CaF2, Li2o.

Asked By: SARIKA
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Joshi sir comment

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

Chemistry >> Physical Chemistry >> Solid State Medical Exam

an element crystallises in fcc lattice having 400pm. calculate the maxm diameter which can be placed in a interstitial sites without disturbing the str

1) 400pm

2) 117.1 pm

3) 224.2 pm

4) 336.2 pm

Asked By: SARIKA SHARMA
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Joshi sir comment

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

 

Chemistry >> Physical Chemistry >> Solid State Medical Exam

IN A CUBIC CRYSTAL OF CsCl  (d= 3.97gm/cc) THE 8 CORNER OCCUPIED BY Cl-  WITH Cs+ AT CENTRE . CALCULATE THE RATIO OF RADIUS OF Cl- & Cs+

Asked By: SARIKA SHARMA
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Joshi sir comment

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

Chemistry >> Physical Chemistry >> Solid State Medical Exam

in a crystal A2B3 ATOMS B ARE IN CCP ARRANGMT. & ATOMS A ARE IN TETRAHEDRAL VOIDS . THE FRACTN OF TETRAHEDRAL VOIDS ARE OCCUPIED IS ----?

Asked By: SARIKA SHARMA Solved By: AMIT DAS
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Joshi sir comment

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

no. of tetrahedral voids = 2

but formula given is A2/3B

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

Chemistry >> Physical Chemistry >> Solid State Medical Exam

if all the ions of NaCl str is removed from one of the diagonal plane , then the formula of compd. will be ---- ?

Asked By: SARIKA SHARMA
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Solution by Joshi sir

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 = Na2.5Cl2.5

so formula will remail same 

Chemistry >> Physical Chemistry >> Solid State Medical Exam

In FCC unit cell distance b/w centre of two spheres which are touching each other & distance b/w centre of two nearest non touching sphere is

1)  (2a)1/2,, 2a

2)2r, (a+r)

3)  a/(21/2), a

4) both (2) & (3)

a= edge length. r= radius of atom

Asked By: SARIKA SHARMA
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Joshi sir comment

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

 

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