The Solid State Practice Questions - DPP for JEE

 Page 1


1. (b)
= 4
? It has fcc unit cell.
2. (a) Wurtzite has face centred cubic structure in which each Zn
2+
 ion is
attached to four S
2– 
ions and each S
2–
 ion remains in contact with
four Zn
2+
 ions.  Hence coordination number of each ion is 4.
3. (c) The increase in pressure results in decrease in size of ions (more in
case of anion than cation), the  increases and the
coordination number also increase.
4. (a) ratio is  Hence LiF has NaCl structure with CN
= 6.
5. (a) Diamond is like ZnS. In diamond cubic unit cell, there are eight
corner atoms, six face centered atoms and four more atoms inside
the structure.
Number of atoms present in a diamond cubic cell
=  = 8
(corners) (face centered) (inside body)
6. (a) Fraction of unoccupied sites in NaCl crystal
 = 5.96 × 10
–3
7. (a) MnO
2
Page 2


1. (b)
= 4
? It has fcc unit cell.
2. (a) Wurtzite has face centred cubic structure in which each Zn
2+
 ion is
attached to four S
2– 
ions and each S
2–
 ion remains in contact with
four Zn
2+
 ions.  Hence coordination number of each ion is 4.
3. (c) The increase in pressure results in decrease in size of ions (more in
case of anion than cation), the  increases and the
coordination number also increase.
4. (a) ratio is  Hence LiF has NaCl structure with CN
= 6.
5. (a) Diamond is like ZnS. In diamond cubic unit cell, there are eight
corner atoms, six face centered atoms and four more atoms inside
the structure.
Number of atoms present in a diamond cubic cell
=  = 8
(corners) (face centered) (inside body)
6. (a) Fraction of unoccupied sites in NaCl crystal
 = 5.96 × 10
–3
7. (a) MnO
2
8. (c) I
2
 (s)
9. (a)
10. (b) It is the property of liquid crystal.
11. (d) Calculate the  ratio to get the limiting ratio value and consult
the table. All are correct.
12. (a) = 
Hence 
13. (c) For cubic geometry the limiting ratio is
 i.e., (c)
14. (a) Statement (a) is correct.
15. (a) For bcc lattice body diagonal = .
The distance between the two oppositely charged ions  = 
16. (d) Number of formulas in cube shaped crystals 
since in NaCl type of structure 4 formula units form 'a' cell
units cells  = 2.57 × 10
21
 unit cells.
17. (c) As CsCl is body-centred, 
18. (c) AB is just like NaCl. Thus twelve A
+
 are at edges and 1 within
body of fcc i.e. in octahedral voids and six B
–
 at faces and 8 at
corner.
19. (b) Among the given crystals, only silicon exists as a covalent solid. It
has diamond like structure.
20. (c) The p.f. for body centred cube = 0.68.
Page 3


1. (b)
= 4
? It has fcc unit cell.
2. (a) Wurtzite has face centred cubic structure in which each Zn
2+
 ion is
attached to four S
2– 
ions and each S
2–
 ion remains in contact with
four Zn
2+
 ions.  Hence coordination number of each ion is 4.
3. (c) The increase in pressure results in decrease in size of ions (more in
case of anion than cation), the  increases and the
coordination number also increase.
4. (a) ratio is  Hence LiF has NaCl structure with CN
= 6.
5. (a) Diamond is like ZnS. In diamond cubic unit cell, there are eight
corner atoms, six face centered atoms and four more atoms inside
the structure.
Number of atoms present in a diamond cubic cell
=  = 8
(corners) (face centered) (inside body)
6. (a) Fraction of unoccupied sites in NaCl crystal
 = 5.96 × 10
–3
7. (a) MnO
2
8. (c) I
2
 (s)
9. (a)
10. (b) It is the property of liquid crystal.
11. (d) Calculate the  ratio to get the limiting ratio value and consult
the table. All are correct.
12. (a) = 
Hence 
13. (c) For cubic geometry the limiting ratio is
 i.e., (c)
14. (a) Statement (a) is correct.
15. (a) For bcc lattice body diagonal = .
The distance between the two oppositely charged ions  = 
16. (d) Number of formulas in cube shaped crystals 
since in NaCl type of structure 4 formula units form 'a' cell
units cells  = 2.57 × 10
21
 unit cells.
17. (c) As CsCl is body-centred, 
18. (c) AB is just like NaCl. Thus twelve A
+
 are at edges and 1 within
body of fcc i.e. in octahedral voids and six B
–
 at faces and 8 at
corner.
19. (b) Among the given crystals, only silicon exists as a covalent solid. It
has diamond like structure.
20. (c) The p.f. for body centred cube = 0.68.
21. (1.46)
22. (5) To get a n – type semiconductor from silicon, it should be doped
with a substance with valency 5.
e.g., silicon is doped with phosphorus to form a n – type semiconductor.
23. (1.432) = 
2.7 =  Z = 4
Hence it is face centred cubic unit lattice.
Again 4r =  = 5.727 Å
 r = 1.432 Å
24. (4) In Na
2
O there is antifluorite structure. Here negative ions form the
ccp arrangement so that each positive ion is surrounded by 4
negative ions and each negative ion is surrounded by 8 positive
ions. So coordination no. of Na
+
 is 4.
25. (1.15) ;