Q1. What is the difference between Schottky and Frankel defect?
Ans1.
| Sr.No. | Schottky defect | Frankel defect |
| 1. | It decreases the density of the crystal. | It does not decrease the density of the crystal. |
| 2. | It occurs in compounds with high Co-ordination number. | It occurs in compounds with low number. |
| 3. | It occurs in compounds in which cations and anions are of similar size. Examples: NaCl, KCl, KBr, CsCl. | It occurs in compounds in which cations and anions differ in their size to a large extent. Examples: ZnS, AgCl, AgBr, Agl. |
Q.-2. A metal crystallises into two cubic phases, fcc and bcc, whose unit
lengths are 3.5 and 3.0Ao respectively.
Calculate the ratio of the densities of fcc and bcc.
Ans-2. d α Z / a3,
For FCC, Z = 4, For BCC, Z= 2, Then
dfcc / dbcc = ( Zfcc / a3 fcc ) x (a3 bcc / Z bcc)
= (4 / 2) x ( 3 / 3.5)3 = 1.26
Q.3 In corundum, oxide ions are arranged in hcp arrangement and the aluminium ions occupy 2/3 of the octahedral voids.
What is the formula of corundum.
Ans-3. Number of oxide Ions = 8 x (1/8 )= 1 per unit cell
Number of Al ions = 2 / 3
formula Al2/ O3 or Al2 O3
OR
Rank of hcp = 6 = No. of octahedral voids.
No. of Al ions = 6 x (2 / 3) = 4
Formula Al4O6 or Al2O3.
Q4. Explain why?
(i) Conductivity of metals decreases with increase in temperature.
(ii) Conductivity of semiconductors increases with increase in temperature.
A-4. (i) => The conductivity of metals is due to the migration of free mobile electrons under the influence of applied potential difference.
=> This migration of electrons is hindered to some extent by the lattice vibrations.
=> At low temperature, lattice vibrations are quite insignificant, and as such metals are excellent electric conductors at low temperature.
=> But with the rise of temperature lattice vibrations increases due to thermal energy and as such migration of electrons is hindered.
=> Therefore, electrical conductivity of metals decreases with rise in temperature.
(ii) => Electrons and holes produced by the ionisation or defects contribute to the
electronic conduction of semiconductors.
=> Unlike metals, the conductivity of semiconductors increases with increase in temperature.
This can be explained as follows.
=> In semiconductors electrons are bound rather tightly to local centers at room temperature.
=> When temperature is raised these electrons are freed and are now able to move through the crystal.
=> The higher the temperature, the greater the number of electrons freed.
=> Due to greater number of free electrons the conductivity increases even though lattice vibrations offer more resistance at the higher temperature.
Q-5. Given that for Fe, a=286 pm ;d=7.86g/cm3. Find the type of the cubic lattice to which the crystal of iron belongs to. Also calculate the radius of Fe atom.
A-5. d =Z X M/a3 N0
Z= (d x a3 x N0 ) / M
= [ ( 7.86 gcm-3) X ( 286 x 10-10 cm )3 x (6.02 x 1023) mol-1] / 55.85 gmol-1
= 2
Since the no. of the atoms in the unit cell is 2 , it is therefore Body - Centered Cubic Structure.
For the body centered cubic structure , the radius of the atom.
Q-6. Chromium metal crystallizes with a body- centered cubic lattice . the length of the unit cell edge is found to be 287 pm . calculate the atomic radius.
What would be the density of chromium in g/cm3?
A-6. For the body centered cubic structure , the radius of the atom .
r = √3 / 4 x a = √3 / 4 x 287 pm = 124.27 pm
THUS, THE ATOMIC RADIUS OF CHROMIUM=124.27 pm
Mass of an atom of chromium = atomic mass of Cr / Avogadro`s no. = 51.99 g / 6.02 x 1023.
No. of atoms in one unit cell = 2
Volume of the unit cell = (287 pm)3 = (287 x 10-10cm)3
Density of Cr = mass of unit cell/ volume of the unit c
= 2 x 51.99/ (287 x 10-10 cm)3 x 6.02 x 1023 = 7.32 g/cm3
Q. 7. In the mineral spinal; having the formula MgAl2O4. The oxide
ions are arranged in CCP, Mg2+ ions occupy the tetrahedral voids.
While Al3+ ions occupy the octahedral voids.
(i) What percentage of tetrahedral voids is occupied by Mg2+ ions ?
(ii) What percentage of octahedral voids is occupied by Al3+ ions ?
Ans-7. According to the formula, MgAl2O4. If there are 4 oxide ions, there will be 1 Mg2+ ions and 2 Al3+. But if the 4 O2– ions are ccp in arrangement,
there will be 4 octahedral and 8 tetrahedral voids.
(i) Percentage of tetrahedral voids occupied by Mg2+ = (1 / 8) × 100
= 12.5%
(ii) Percentage of octahedral voids occupied by Al3+ = (2 / 4) × 100
= 50%.
Q. 8. Analysis shows that nickel oxide has the formula NiO.98O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions ?
Ans-8. NiO.98O1.00 Let Ni2+ be x and Ni3+ be 0.98 – x
Total charge on compd. is equal to zero.
[2 (Ni2+) + 3 (Ni3+) – 2 (O2–)] = 0
[2 x + 3 (0.98 – x) – 2] = 0
x = 0.94
Therefore Ni2+ % = ( 0.94 / 0.98 ) × 100 = 96%
Ni3+ = 4%
Q. 9. How many unit cells are present in a cube shaped ideal crystal of NaCl of mass 1 gm ?
Ans-9. Mass of 1 unit cell = volume × density
= a³ × d
= (a³ x M x Z ) / ( N0 x a³ )
= (58.5 x 4) / ( 6.023 x 1023)
No. of unit cells in 1 gm = 1/M
= (6.023 × 1023 ) / (58.5 × 4)
= 2.57 × 1021
No comments:
Write comments