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. # 37
1. Graphite is a good conductor of heat and electricity while diamond is not because :
(A) graphite has ionic bonds and diamond has covalent bonds.
(B) graphite has covalent bonds and diamond has ionic bonds.
(C*) graphite has delocalized electrons whereas diamond has not.
(D) graphite has sp3
hybridized carbon atoms and diamond has sp2
hybridized carbon atoms.
2. BF3
+ F–
 BF4
–
What is the hybridiation state of B in BF3
and BF4
–
?
(A*) sp2
, sp3
(B) sp3
, sp3
(C) sp2
, dsp2
(D) sp2
d, sp2
3. Which is the hybridization on the centre atom of SiO2
.
(A) sp (B) sp2
(C*) sp3
(D) sp3
d
4. Which of the following represents a pyrosilicate structure :
— Oxygen
— Silicon
(A) (B) (C*) (D)
5. The fundamental unit found in silicate is :
(A) SiO2
(B*) SiO4
4–
(C) SiO3
(D) Si2
O5
2–
6. Which of the following statements regarding graphite is not correct ?
(A) Graphite is a good conductor of electricity
(B) Graphite is less dense than diamond
(C*) The bond length of carbon-carbon bond is 154 pm (equal to C – C single bond length).
(D) Graphite is thermodynamically more stable than diamond.
7.* On the basis of structure of graphite, which of the following is/are true for it :
(A*) It is a diamagnetic substance
(B*) It behaves like metallic as well as semiconductor
(C*) It is less dense than diamond
(D) It reacts with F2 to form product which is more electrical conducting than graphite
CHEMISTRY
DAILY PRACTICE PROBLEMS
D P P
COURSE NAME : UDAY (UB) DATE : 07.10.2013 to 12.10.2013 DPP NO. 37 & 38
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
8. Consider the structures of the following two molecules
X : F2
C = C = CF2
Y : F2
B – C  C – BF2
In which of these two it is impossible for all the four F atoms to lie in the same plane :
(A*) X (B) Y (C) both (D) none
9. VESPR theory does not state :
(A) the order of repulsion between different pair of electrons is p – p > p – bp > bp – bp
(p = lone pair electrons, bp = bond pair electrons.)
(B) as the number of lone pair of electrons on central atom increase, the deviation in BA
from normal BA (Bond-Angle) also increase.
(C*) the number of lone pair on O in H2O is 1 while on N in NH3 is 2.
(D) the structures of Xenon-fluorides and Xenon-oxyfluorides could be explained on the
basis of VESPR theory.
10. Which of the following is V-shaped :
(A*) –2
3S (B) –
3 (C) –
3N (D) none of these
DPP No. # 38
1. Among the following, the pair in which the two species are not isostructural is :
(A) IO3
– and XeO3 (B) BH4
– and NH4
+ (C) PF6
– and SF6 (D*) SiF4 and SF4
2. Match the species in column () with that geometry in column ()
Column- Column-
(P) BH4
–
(1) 2 bond pair and 3 lone pair
(Q) Cl2
+
(2) 4 bond pair and no lone pair
(R) Cl2
–
(3) 3 bond pair and 1 lone pair
(S) Cl4
–
(4) 2 bond pair and 2 lone pair
(5) 4 bond pair and 2 lone pair
(A) P = 2; Q = 4; R = 3; S = 1 (B*) P = 2; Q = 4; R = 1; S = 5
(C) P = 2; Q = 1; R = 5; S = 4 (D) P = 2; Q = 1; R = 3; S = 4
3.* Which is true about VSEPR theory :
(A*) Lone pair-lone pair repulsion is maximum.
(B*) Lone pair and double bond occupy equitorial position in trigonal bipyramidal structure.
(C*) More electronegative atoms occupies axial position in trigonal bipyramidal structure.
(D) Bigger atoms occupy axial positions in trigonal bipyramidal structure.
4.* Choose the correct statement(s) about B3
N3
H6
.
(A*) Molecule has polar bonds but molecule is non polar.
(B*) B3
N3
H6
is isoelectronic and isostructural with benzene.
(C*) On dichloro substitution reaction it can form four isomers of dichloro derivative.
(D*) Hybridisation of all boron and nitrogen is sp2
.
5. Match the isostructure pairs :
Hint : Having same hybridization and same Number of lone pair
(a) SF4
(i) IF6
+
(b) PCl5
(ii) ClF4
+
(c) ICl3
(iii) SnCl5
–
(d) I3
–
(iv) ClF3
(e) ICl4
–
(v) ClF2
–
(f) PCl6
–
(vi) XeF4

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. STATEMENT-1 : Acidic strength of oxides of p–block elements decreases on moving down the group in
periodic table (assuming same oxidation state)
STATEMENT-2 : By increase in oxidation state of central element in oxides, acidic character increases.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B*) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
7. How many (P–O) bonds are there in P4
O6
and P4
O10
(Co-ordinate bonds may also be included).
8. How many bond pairs of electrons are around N in NO3
–
?
9. Arrange the following properties as indicated.
(a) Increasing order of number of lone pairs of electrons on central atoms.
XeF5
–, SF4, ICl2
–, SiF4
(b) Increasing order of percentage p-character of the hybrid orbitals of the central atoms.
ClO2
–, CS2, SnCl2
(c) Increasing order of bond polarity.
PH3 , AsH3, SbH3, NH3
ANSWER KEY
DPP No. # 37
1. (C) 2. (A) 3. (C) 4. (C) 5. (B) 6. (C)
7.* (A*BC) 8. (A) 9. (C) 10. (A)
DPP No. # 38
1. (D) 2. (B) 3.* (ABC) 4.* (ABCD)
5. (a-ii) (b-iii) (c- iv) (d-v) (e-vi) (f-i). 6. (B) 7. 12 and 16 8. 4
9. SiF4 < SF4 < XeF5
– < ICl2
–
0 1 2 3 no. of lone pairs
(b) CS2 < SnCl2 < ClO2
–
sp sp2 sp3
50% 66% 75%
(c) PH3 < ASH3 < SbH3 < NH3
En (2.1 – 2.1) (2.0–2.1) (1.9 – 2.1) (3.0 – 2.1)