Due to Frenkel defect, the density of the ionic solids
does not change
decreases
increases
may increase or decrease
The dimensions of a unit cell of a crystal are a = 0.397, b = 0.387, c = 0.504 and α = β = 90°, γ = 120° the crystal system is
cubic
hexagonal
orthorhombic
rhombohedral
Sodium atom crystallizes in bcc lattice with cell edge a = 4.29 Å, the radius of sodium atom is
18.6 Å
1.86 Å
0.186 Å
37.2 Å
Which is mismatched for NaCl crystal ?
Coordination number = 6.6
Edge of unit cell =(r+ + r-)
Crystal structure = fcc
The structure of ionic compound A+B identical to NaCl. If the edge lenght is 400pm and cation radius is 75pm, the radius of anion will be
100 pm
125 pm
250 pm
325 pm
The length of unit cell edge of a body-centred cubic metal crystal is 352 pm. The radius of metal atom is
162.4 pm
152.4 pm
142.4 pm
156.pm
If NaCl is doped with 10-3 mol% of SrCl2, the number of cation vacancies will be
6.023 × 1018
1 × 10-3
6 × 1012
6.023 × 1023
A.
6.023 × 1018
Given, concentration of SrCl2 = 10-3 mol % concentration is in percentage so that take total 100 mL of solution
Number of moles of NaCl = 100-3 moles of SrCl2
Moles of SrCl2 is very negligible as compare to total moles so percentage always taken as 100 so that,
1 mole of NaCl is dipped with = mole of SrCl2
= 10-5 mole of SrCl2
So, cation vacancies per mole of NaCl = 10-5 mol
Since, 1 mole = 6.023 × 1023 particles
So, cation vacancies per mole of NaCl
= 10-5 × 6.022 × 1023
= 6.022 × 1018
Match List I with List II and select the correct answer using the given codes.
List I (Crystal system) | List II (Example) |
A.Cubic | 1. TiO2 |
B. Tetragonal | 2.Graphite |
C. Hexagonal | 3. K2Cr2O7 |
D. Triclinic | 4.ZnS |
A B C D
2 3 4 1
A B C D
1 4 3 2
A B C D
3 2 1 4
A B C D
4 1 2 3