Top 1000+ Solid State Short Type Questions And Answers
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Question 91.
An element with molar mass 27 g mol-1 forms a cubic unit cell with edge length 4.05 × 10-8 cm. If its density is 2.7 g cm-3, what is the nature of the cubic unit cell?
Answer:
= 6.022 × 10-2 × 66.43 = 4.0004 = 4
It has Face centred cubic cell/fee.
Question 92.
Examine the given defective crystal :
Answer the following questions :
(i) Is the above defect stoichiometric or non- stoichiometric?
(ii) Write the term used for this type of defect. Give an example of the compound which shows this type of defect.
(iii) How does this defect affect the density of the crystal?
Answer:
(i) It is stoichiometric defect.
(ii) Schottky defect, e.g. NaCl.
(iii) Density of crystal decreases.
Question 93.
Define the following :
(i) Schottky defect
(ii) Frenkel defect
(iii) F-centre
Answer:
(i) Schottky defect: If in an ionic crystal of type A B , equal number of cations and anions are missing from their lattice sites so that the electrical neutrality is maintained, it is called Schottky defect.
(ii) Frenkel defect : If an ion leaves its site from its lattice site and occupies the interstitial site and maintains electrical neutrality, then it is called Frenkel defect.
(iii) F-centre : The centres which are created by trapping of electrons in anionic vacancies
and which are responsible for imparting colour to the crystals are called F-centres. (F = Fabre)
Question 94.
Silver crystallises in fee lattice. If edge length of the unit cell is 4.077 × 10-8 cm, then calculate the radius of silver atom.
Answer:
Given : a = 4.077 × 10-8 cm r = ? for fee lattice
Using formula,
Radius (r) = a22√
r = 4.077×10−82×1.414cm=4.077×10−82.828
∴ r = 1.441 × 10-8 cm
Question 95.
An element crystallizes in a f.c.c. lattice with cell edge of 250 pm. Calculate the density if 300 g of this element contains 2 × 1024 atoms.
Answer:
Given: a = 250 pm = 250 × 10-10 cm
z = 4 (for fee)
M = ? d = ?
Using formula : d = z×Ma3NA
∵ 2 × 1024 atoms of an element have mass = 300g
∴ 6.022 × 1023 atoms of an element have mass
= 300×6.022×10232×1024 = 90.33 g
Now M = 90.33 g
Question 96.
An element crystallizes in a b.c.c. lattice with cell edge of 500pm. The density of the element is 7.5g cm-3. How many atoms are present in 300 g of the element?
Answer:
Given: For b.c.c. structure, z = 2
Edge of the unit cell, a = 500 pm = 500 × 1010 cm
Density d = 7.5 g cm-3
Using the formula,
∵ 282.28 g of the element contains = 6.022 × 1023 atoms
∴ 300 g of the element contains
= 6.022×1023282.28 × 300
= 1806.6282.28×1023 × 1023 = 6.40 × 1023 atoms
Question 97.
If NaCl is doped with 10-3 mol % of SrCl2, what is the concentration of cation vacancies?
Answer:
Concentration of SrCl2 = 10-3 mol% = 10-3/100 mol = 10-5 mol
1 mol of NaCl on doping procuces = 6.022 × 1023 cation vacancies
Therefore, 10-5 mol of NaCl on doping produces = 6.022 × 1023 × 10-5 = 6.022 × 1018 cation vacancies
Question 98.
Silver crystallises in f.c.c. lattice. If edge length of the cell is 4.07 × 10-8 cm and density is 10.5 g cm-3, calculate the atomic mass of silver.
Answer:
Question 99.
(a) Based on the nature of intermolecular forces, classify the following solids: Silicon carbide, Argon
(b) ZnO turns yellow on heating. Why?
(c) What is meant by groups 12-16 compounds? Give an example.
Answer:
(a) Silicon carbide is a covalent or network solid while Argon is a non-polar molecular solid.
(b) ZnO shows metal excess defect due to presence of extra cations, i.e., Zn2+ ions in interstitial sites which on heating changes into yellow due to loss of oxygen.
ZnO⟶ΔZn2++12O2+2e−
(c) Group 12-16 compounds are imperfect covalent compounds in which the ionic character depends on the electronegativities of the two elements, e.g., ZnS, CdS, etc.
Question 100.
(a) Based on the nature of intermolecular forces, classify the following solids: Benzene, Silver
(b) AgCl shows Frenkel defect while NaCl does not. Give reason.
(c) What type of semiconductor is formed when Ge is doped with Al?
Answer:
(a) Benzene — Molecular solid (non-polar) Silver — Metallic solid
(b) Due to intermediate radius of AgCl, the size of Ag+ is smaller than larger Na+ ion of NaCl so it can easily occupy interstitial spaces and shows Frenkel defect.
(c) p-type semiconductor is formed when Ge is doped with Al.
Question 101.
(a) Based on the nature of intermolecular forces, classify the following solids: Sodium sulphate, Hydrogen
(b) What happens when CdCl2 is doped with AgCl?
(c) Why do ferrimagnetic substances show better magnetism than antiferromagnetic substances?
Answer:
(a) Sodium sulphate — Ionic solid
Hydrogen — Molecular solid (non-polar)
(ib) Cd2+ ion is dipositive and therefore addition of one Cd2+ ion results in the loss of two Ag+ ions from the lattice. But out of 2 holes obtained, one is occupied by Cd2+ ion and one left empty. Hence, addition of CdCl2 results in an impurity defect with cation vacancy.
(c) In ferrimagnetism, domains /magnetic moments are aligned in opposite direction in unequal numbers while in antiferromagnetic substances, the domains align in opposite direction in equal numbers so they cancel magnetic moments completely, i.e., net magnetism is zero.
Question 102.
An element crystallises in b.c.c. lattice with cell edge of 400 pm. Calculate its density if 500 g of this element contains 2.5 × 1024 atoms.
Answer:
Given : a = 400 pm = 400 × 10-10 cm
Z = 2 (for bcc) M = ? d = ?
Using formula, d = Z×Ma3×NA
∵ 2.5 × 1024 atoms of an element have mass = 500 g
∴ 6.022 × 2023atoms of an element have mass
= 240.886.4×6.022=240.8838.5408
∴ d = 6.25 g cm-3
Question 103.
An element crystallises in fee lattice with cell edge of 400 pm. Calculate its density if 250 g of this element contain 2.5 × 1024 atoms.
Answer:
Given : a = 400 pm = 400 × 10-10 cm
Z = 4 (for fee), M = ?, d = ?
Using formula, d = Z×Ma3×NA
∵ 2.5 × 1024 atoms of an element have mass = 250 g
∴ 6.022 × 1023 atoms of an element have mass
= 250×6.022×10232.5×1024
∴ M = 60.22 g
Substituting all values in formula :
d = 4×60.22(400×10−10)3×6.022×1023=240.8838.5408
∴ d = 6.25 g cm-3
Question 104.
An element crystallises in bcc lattice with cell edge of 400 pm. Calculate its density if 250 g of this element contains 2.5 × 1024 atoms.
Answer:
Given : a = 400 pm = 400 × 10-10 cm
Z = 2 (for bcc), M = ?, d = ?
Using formula, d = Z×Ma3×NA
∵ 2.5 × 1024 atoms of an element have mass = 250g
∴ 6.022 × 1023 atoms of an element have mass
= 250×6.022×10232.5×1024
∴ M = 60.22 g
Substituting all values in formula : .
d = 2×60.22(400×10−10)3×6.022×1023=120.4438.5408
∴ d = 3.125 g cm-3
Question 105.
An element exists in bcc lattice with a cell edge of 288 pm. Calculate its molar mass if its density is 7.2 g/cm3. Answer:
Given : Cell edge, a = 288 pm = 288 × 10-10 cm
Density, d = 7.2 g/cm3
For bcc formula, units per cell Z = 2, M = ?
Using formula and substituting values,