JH SIR DPP

1. PAGE NO. # 1
ETOOS ACADEMY Pvt. Ltd
F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor,
BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
* Marked Questions are having more than one correct option.
DPP No. # 35
1. The strength of bonds by ps,pp,ss  overlap is in the order :
(A) pppsss  (B) psppss  (C) ppssps  (D) pssspp 
2. Which of the following atomic orbital overlappings are not allowed
(i) (ii) (iii)
(iv) (v)
(A) All (B) (i) (ii) (iii) (C) (i) (iii) (v) (D) (ii) only
3. Which of the following overlaps is incorrect [assuming z-axis to be the internuclear axis]
(a) 2 py + 2 py   2 py (b) 2 pz + 2 pz   2pz
(c) 2 px + 2 px   2 px (d) 1 s + 2 py  (1 s-2 py )
(A) ‘a’ & ‘b’ (B) ‘b’ & ‘d’ (C) only ‘d’ (D) None of these
4. Which of the following configuration corresponds to a pseudo-inert gas configuration ?
(A) ns2
np6
(B) ns2
np6
nd10
(C) (n – 1)d10
ns2
np6
(D) None of these.
5. An ion without pseudo-inert gas configuration is :
(A) Ag+
(B) Cd2+
(C) Zn2+
(D) Fe3+
6. Arrange the solubility of NaCl in decreasing order in the following solvents :
CH3
COCH3
(dielectric constant = 12) ; CH3
CH2
OH (dielectric constant = 35)
H2
O (dielectric constant = 81)
6.* Indicate the wrong statement :
(A) A sigma bond has no free rotation along its axis
(B) p-orbitals always have only sidewise overlapping
(C) s-orbitals never form  - bonds
(D) There can be more than one sigma bond between two atoms
8.* Hypervalent compound is(are) :
(A) SO3
2
(B) PO4
3
(C) SO4
2
(D) CIO4

9.* Answer the following :
(i) What is the valency of carbon in C2H4 and C2H2 ?
(ii) What types of bonds and how many of each are present in NH4
+ ?
10. Find  and  bonds in the following molecules :
CH3
– CH3
, CH2
= CH2
, CH  CH , CH2
= CHCOOH, C2
(CN)4
.
PHYSICAL CHEMISTRY
DAILY PRACTICE PROBLEMS
D P P
COURSE NAME : UDAY (UB) DATE : 30.09.2013 to 05.10.2013 DPP NO. 35 & 36
TARGET
JEE (ADVANCED) : 2015

2. PAGE NO. # 2
ETOOS ACADEMY Pvt. Ltd
F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor,
BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
DPP No. # 36
1. Predict the hybridisation on the central atom of following molecules :
1. BeH2 ....................... 2. BeF2 .......................
3. CO2 ....................... 4. CO .......................
5. BH3 ....................... 6. BF3 .......................
7. CH2=CH2 ....................... 8. CH4 .......................
9. HNO3 ....................... 10. HNO2 .......................
11. SO2 ....................... 12. SO3 .......................
13. HCO3
– ....................... 14. CO3
–2 .......................
15. CO3
–2 ....................... 16. SnCl2 .......................
17. AlCl3 ....................... 18. CH4 .......................
19. NH4
+ ....................... 20. BF4
– .......................
21. AlF4
– ....................... 22. NF3 .......................
23. PF3 ....................... 24. AsCl3 .......................
25. NH3 ....................... 26. :CH3 .......................
27. H2O ....................... 28. OF2 .......................
29. H2SO4 30. PCl5 .......................
31. SbCl5 ....................... 32. SF6 .......................
33. SeF6 ....................... 34. PF6
– .......................
35. IF7 ....................... 36. ICl2
– .......................
37. ICl5 ....................... 38. XeF4 .......................
39. ICl4
– ....................... 40. XeF6 .......................
2. 5PCl exists but 5NCl does not because :
(A) Nitrogen has no vacant 2d-orbitals (B) 5NCl is unstable
(C) Nitrogen atom is much smaller than P (D) Nitrogen is highly inert
3. The correct representation of lewis dot structure of HNO3 is :
(A) H – N = O
||
O
||
O
(B) H – O – N = O
||
O
(C) H – O – N = O
O
(D) H – N O
O
O
3. Consider the following statements
In HCCCHCH2 
 v
1. There are 6  and ‘3’  and 2. Carbon  &  are sp2 hybridised
3. Carbon  & V are sp hybridised,
The above statements 1, 2, 3 respectively are (T = True, F = False)
(A) T T T (B) F T T (C) F T F (D) T F T
4. Shape of NH3
is very similar to :
(A) CH4
(B) CH3
–
(C) BH3
(D) 
3CH
5. Two types of carbon-carbon covalent bond lengths are present in :
(A) diamond (B) graphite (C) C60
(D) benzene

3. PAGE NO. # 3
ETOOS ACADEMY Pvt. Ltd
F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor,
BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
6. Diamond is a hard substance because :
(A) it has ionic bond.
(B) it has planar arrangement of carbon atoms.
(C) it has sp3
hybridized carbon atoms which are arranged tetrahedrally in a cross net-work structure.
(D) it has sp2
hybridization carbon atoms arranged in a planar geometry.
Answer
DPP No. # 35
1. (A) 2. (B) 3. (C) 4. (B) 5. (D)
6. Solubiltiy of NaCl lies in following sequence
H2
O > CH3
CH2
OH > CH3
COCH3
Greater is the dielectric constant of solvent more is the solubility of an ionic compoind in it.
6.* (A, B, D) 8.* (A, B, C, D)
9.* (i) four (ii) three covalent and one coordinate.
10. CH3
– CH3
: 7
CH2
= CH2
: 5 and 1
CH  CH : 3 and 2
CH2
= CHCOOH : 8 and 2
C2
(CN)4
: 9 and 9
DPP No. # 36
1. 1. BeH2 sp 2. BeF2 sp
3. CO2 sp 4. CO sp, sp
5. BH3 sp2 6. BF3 sp2
7. CH2=CH2 sp2 8. CH4 sp3
9. HNO3 sp2 10. HNO2 sp2
11. SO2 sp2 12. SO3 sp2
13. HCO3
– sp2 14. CO3
–2 sp2
15. CO3
–2 sp2 16. SnCl2 sp2
17. AlCl3 sp2 18. CH4 sp3
19. NH4
+ sp3 20. BF4
– sp3
21. AlF4
– sp3 22. NF3 sp3
23. PF3 sp3 24. AsCl3 sp3
25. NH3 sp3 26. :CH3 sp3
27. H2O sp3 28. OF2 sp3
29. H2SO4 sp3 30. PCl5 sp3d
31. SbCl5 sp3d 32. SF6 sp3d2
33. SeF6 sp3d2 34. PF6
– sp3d2
35. IF7 sp3d3 36. ICl2
– sp3d
37. ICl5 sp3d2 38. XeF4 sp3d2
39. ICl4
– sp3d2 40. XeF6 sp3d3
2. (A) 3. (C) 3. (B) 4. (B) 5. (C) 6. (C)