CHEMICAL BONDING GRAND TEST - 3
READ THE INSTRUCTIONS CAREFULLY
1.
2.
3.
The test is of 1 hour 30 minutes duration.
The maximum marks are 323.
This test consists of 78 questions.
SECTION – A (Single Correct Choice Type)
This Section contains 40 multiple choice questions. Each question has four choices A), B), C) and D)
out of which ONLY ONE is correct. (Mark only one choice)
(40 × 4 = 160 Marks)
1.
Which of the following is sp hybridised
2.
a. NO2
b. XeF2
c. BeCl2(g)
C
In which of the following reaction hybridisation will be changed

a. BH3 BH4
d. BeCl2(s)

b. N H3  NH 4
c. SO2  SO3
A
d. H2O  H3O
+
3.
Which of the following compound have only ionic bond
d.
4.
a. NaCl; AlF3; AlCl3
b. Na2O2; BF3; KOH
c. NaCl; BaCl2; Na2O
NaCN
C
Which of the following compound will have ionic, covalent and H-bond
a. KCN
b. NH4Cl
c. KHF2
C
Which of the following is correct order of Hybridisation in given molecule
HC  C CH 2  CH  C H2
d. AlCl36H2O
5.
5
6.
4
3
2
NaOH;
1
a. sp2, sp2, sp3, sp, sp
b. sp; sp2; sp3; sp2; sp2
2
3
2
c.sp , sp ; sp; sp; sp
d. sp3; sp2; sp; sp; sp2
A
In which of the following compound central atom is sp3d3 hybridised
a. XeF2O2
C
b. IF5
d. I3+
c. XeF6
CH 3
|
7.
How many sp3 hybridised orbitals are present in given compound CH 3  C  CH 3
|
CH 3
a. 4
b. 5
Dr. Sangeeta Khanna Ph.D
c. 20
1
d. 15
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HCN;
Dr. Sangeeta Khanna Ph.D
C
Sol. Five carbon, each having four sp3 orbital.
8. Which of the following is the correct order of p - p back bonding
9.
a. BF3 > BI3 > BBr3 > BCl3
b. BF3 > BCl3 > BBr3 > BI3
c. BI3 > BBr3 > BCl3 > BF3
d. BF3 = BCl3 = BBr3 = BI3
B
Which of the following compound will have linearly placed carbon atoms
a. HC  CH
b. CH3 – CH2 – CH3
c. CH3 = CH – CH3
A
10. Which of the following is correct geometry & Hybridisation of XeF4.
a. sp3d; tetrahedral
c. octahedral sp3d2
D
11. The incorrect order of bond angle:
d.
b. seesaw, sp3d
d. square planer sp3d2
b. NO2+ > NO3– > NO2–
d. PH3 > AsH3 > SbH3
a. CO2 > CO32– > CF2Cl2
c. XeF2 > XeF4 > XeO4
C
Sol. XeF2  XeF4  XeO 4 | CO 2  CO 3-2  CF2Cl 2 | NO 2  NO 3  NO 2
(180)
(90)
(109)
(180)
(120)
(109)
(180)
(120)
(118)
12. Correct order of Acidic strength
a. H – I > H – Br > H – Cl > H – F
c. H – F > H – Br > H – I > H – Cl
A
13. Which of the following is most covalent compound
b. H – Br > H – Cl > H – F > H – I
d. H – Cl > H – I > H – F > H – Br
a. AlCl3
b. PCl5
c. SiCl4
B
Sol. Phosphorus in +5 state has more polarising power.
14. Which of the following is correct increasing order of Bond angle
b. BF3  BF4  B2H6
a. OF2 < SO2 < CO2
c. CO2 < CO 32 < CH4
A
O
Sol.
F
F
d. ClO 4  ClO 3  ClO 2
S
<
d. CCl4
O
<
O=C=O
O
15. The nature of Pi-Bonds in XeO3 is
a. O(p)  Xe(p)
b. O(p)  Xe(d)
c. O(d)  Xe(d)
d. O(d)  Xe(p)
B
Sol. In case of XeO3 there is d-p bonding because d orbital of Xe atom combine with p orbital of
oxygen atom to form  bond.
16. The angular shape of ozone molecule (O3) consists of:
a. 1 sigma and 2  bonds
c. 1 sigma and 1  bond
Dr. Sangeeta Khanna Ph.D
b. 2 sigma and 2  bonds
d. 2 sigma and 1  bond
2
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D
Sol. Ozone molecule is:
O 1
1
O
1
O
2 sigma and 1 pi bonds
17. Which of the following has a bond formed by the overlap of sp – sp3 hybrid orbitals?
a. CH3 – C  C – H
c. CH2 = CH – CH = CH2
A
sp3
b. CH3 – CH = CH – CH3
d. HC  CH
sp
Sol. CH 3  C  C  H
18. In BrF3 molecule, the lone pairs occupy equatorial positions to minimize:
a. lone pair – lone pair repulsion only
b. Lone pair – bond pair repulsion only
c. bond pair – bond pair repulsion only
d. lone pair – lone pair repulsion and lone pair – bond pair repulsion
D
Sol. In sp3d or dsp3 hybridisation lone pairs occupy equatorial positions to
minimize lone pair-lone pair and lone pair – bond pair repulsion.
F
Br
F
F
BrF3
19. Which of the following has a geometry different from the other three species which have the
same geometry?
a. BF4
b. SO 24 
d. PH 4
c. XeF4
C
Sol. XeF4 is square planer rest are tetrahedral
20. A molecule XY4 contains four , two  bonds and one lone pair of electron in valence shell of X. the
arrangement of lone pair as well as bond pairs is:
a. square pyramidal
b. Trigonal bipyramidal
c. trigonal planar
d. seesaw
B
Sol. The valence shell of x has five pairs of electrons which have trigonal bipyramidal arrangement 4 and
one lone pair therefore it has seesaw geometry.  pair of electrons are not considered for deciding
geometry.
21. In which of the following sets the central atom of each member involves sp3 hybridisation?
a. IO 4 , ICl 4- , IF 4
b. XeO3, XeO4, XeF4
c. SO3, SO 32  , SO 24 -d. PCl 4 , BF4-- , ClO -4D
Sol. Hybridisation is calculated by steric number rule.
Steric no. = no. of atoms attached to central atom + no. of lone pairs present on central atom.
Steric no. for IO 4 = 4 + 0 = 4
sp3;
Steric no. for ICl 4 = 4 + 2 = 6
sp3d2
Steric no. for IF 4 = 4 + 1 = 5
sp3d;
Steric no. for SO3 = 3 + 0 = 3
sp2
Steric no. for SO42– = 4 + 0 = 4
sp3;
Steric no. for SO 32  = 3 + 1 = 4
sp3
Steric no. for PCl4+ = 4 + 0 = 4
Steric no. for XeO3 = 3 + 1 = 4
Steric no. for XeF4 = 4 + 2 = 6
sp3;
sp3;
sp3d2;
Dr. Sangeeta Khanna Ph.D
Steric no. for BF4 = 4 + 0 = 4
Steric no. for XeO4 = 4 + 0 = 4
Steric no. for ClO 4 = 4 + 0 = 4
3
sp3
sp3
sp3
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22. In the metallic crystal:
a. the valence electrons remain within the field of influence of their own kernels
b. the valence electrons constitute a sea of mobile electrons
c. the valence electrons are localized between the two kernels
d. both kernels as well as electrons move rapidly.
B
23. In a crystal, the atoms are located at the positions of:
a. Maximum potential energy
b. Minimum potential energy
c. Zero potential energy
d. Infinite potential energy
B
Sol. Minimum potential energy
24. The BF3 is a planar molecule where as NF3 is pyramidal because:
a. B – F bond is more polar than N – F bond.
b. boron atom is bigger than nitrogen atom
c. nitrogen is more electronegative than boron
d. BF3 has no lone pair but NF3 has a lone pair of electron
D
F
2
B
Sol.
F
sp
3
N
F
sp
F
F
F
Trigonal pyramidal
Trigonal planar
25. Which of the following is correct structure of PF3Cl2
Cl
F
F
a.
P
b.
F
Cl
F
Cl
F
Cl
P
c.
Cl
P
Cl
F
Cl
F F
B F Cl
F
26. How many orbitals at 120° angle will be present in Benzene
a. 6
b. 12
C
Sol. 6 carbon; each having three sp2 hybrid orbital.
d.
F
P
c. 18
Cl
d. 20
27. Which of the following is the correct order of stability of the resonating structure of vinyl chloride


 Θ
 Θ

CH 2  CH  Cl   CH 2  CH  Cl   C H2  CH  Cl 


I

II
III
a. I > II > III
b. I > III > II
c. III > II > I
d. II > I > III
B
Sol. (II) structure is least the stable due to violation from octet
28. What will be the formal charge on nitrogen & Hybridisation of Boron in the complex formed between
NH3 and BF3
a. zero, sp3
b. zero, sp2
Dr. Sangeeta Khanna Ph.D
c. +1, sp3
4
d. +1, sp2
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C
1
Sol. H3N  BF 3 complex
sp3
1
sp3
29. Which drawing represents a * antibonding molecular orbital for a homo-nuclear diatomic molecule?
a.
b.
c.
C
30. Which of the following will have maximum no. of -bond
a. C2H2
b. PCl3F2
c.
d.
NC \
/ CN
C

C
\ CN
NC /
CH3
d.
\ C  C /H
\CH3
H/
C
31. Which of the following is most electron deficient:
a. BF3
b. SO2
c. BeCl2
C
32. Which of the following has minimum single bond length:
d. Al2Cl6
a. B
b. I
c. C
d. Si
C
Sol. Smallest atom is carbon
33. In which of the following d-orbitals are not used by central atom in hybridisation?
a. PF5
b. PCl5
c. XeO4
d. AlCl3(s)
C
Sol. PF5 & PCl5 = sp3d; AlCl3(s) = sp3d2
34. Which of the following molecular geometry is not produced by sp3d hybridisation?
a. Triangular planar
b. Sea-saw
c. “T” shape
A
Sol. Triangular planar structure is not observed in sp3d - hybridisation
35. I 3 and I 3 have
a. same geometry
c. same bond angle
D
d. Linear
b. same no. of lone pair(s)
d. different hybridisation
+
Sol.
+
I3
3
I
= sp
I
–
3
I3 = sp d
,
[I – I – I]
–
I
36. Molecule which does not contain any F – X – F bond angle which is less than 90°: (X = central atom)
a. IF7
b. BrF3
c. PF5
C
Sol. In PF5 there is no bond angle which is less than 90°.
d. SF4
F
F
90°
F
P
120°
F
F
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37. Which of the following structure is correctly drawn according to fundamental idea of VSEPR theory?
F
P
a.
( = greater than 109°28)
F

b.
( = less than 90°)
S
F
F
F
F
F
Cl
Cl
F
I–
c.
Cl
d.
( = 90°)
F

Si
F
Cl
( = 90°)
F
C
Sol. (A)  = less than 109°28
(B) Incorrect position of L.P.
(C) Correct square planar structure with  = 90°
(D)   90° in SiF4 because it is tetrahedral.
38. In which of the following pairs of molecules/ions, both the species are not likely to exist?
a. H22, He 2
A
Sol. He2 (B.O. = Zero),
b. H2 , He 22 
c. H2 , He 22-
d. H2 , He 22-
H22  (B.O. = Zero)
He2 = 1s2, 1s2,
H22  No electron in ion
39. Select correct statement about CH4.
a. All the hybrid orbitals of carbon are equivalent
b. All the hybrid orbitals of carbon have 25% 2s property
c. All the hybrid orbitals of carbon are projected at the four vertex of a regular tetrahedron.
d. All are correct statement
D
Sol. CH4 sp3, tetrahedral.
40. Select correct about change in C – C - H bond angle:
C2H6
C2H6
Decrease
c.
C2H2
Increase
Decrease
b.
Decrease
C2H4
Decrease
C2H2
C2H6
Decrease
C2H2
Increase
C2H4
C2H6
Increase
c.
Decrease
Increase
d.
C2H4
C2H2
Decrease
Increase
C2H4
C
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H
Sol.
sp3
sp3
C
C
H
H
H
H
H
sp2
sp2
C
C
H
sp
sp
HCCH
H
H
H
109°
120°
180°
SECTION – B (Assertion and Reason) Negative Marking (-1)
This Section contains 10 multiple choice questions. Each question has four choices A), B), C) and
D) out of which ONLY ONE is correct.
10 × 4 = 40 Marks
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) If assertion is true but reason is false.
(d) If both assertion and reason are false.
1.
Assertion (A): NH 2 has a square planar shape.
Reason (R): Species which involve sp3d hybridization have square pyramidal structure.
a. (a)
D
b. (b)
c. (c)
d. (d)

3
Sol. N HΘ
2 is bent & sp
2. Assertion (A): Magnetic moment of super oxide ion is more than that of peroxide ion.
Reason (R): Peroxide ion has no unpaired electrons but superoxide ion has unpaired electrons.
3.
a. (a)
b. (b)
c. (c)
d. (d)
A
Assertion (A):All the P – F bonds in PF5 do not have equal bond lengths
Reason (R): The hybridization of P-atom involves not only s and p-orbitals but also d-orbitals
a. (a)
b (b)
c. (c)
d. (d)
B
Sol. Correct explanation: The repulsions on the axial bond pairs are more than on the equatorial bond
pairs.
4. Assertion (A): NO 3 ion has same geometry as NH3.
Reason (R): The hybrid state of N in NO 3 ion and NH3 are sp3.
a. (a)
D
b. (b)
c. (c)
d. (d)
Sol. NO 3 = triangular but NH3 = pyramidal (sp3)
5. Assertion (A): CO2 is planar but CH2 = C = CH2 is non-planar
Reason (R): In CO2 all the  electrons are in same plane but in allene all -electrons are not in same
plane.
a. (a)
C
b. (b)
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c. (c)
7
d. (d)
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6.
7.
8.
9.
Assertion (A): All carbon-carbon bond length in benzene are identical.
Reason (R): Carbon-carbon bond length is in between single & double bond.
a. (a)
b. (b)
c. (c)
B
Assertion (A): B.pt. of HF is highest in its group hydride.
Reason (R): It is due to presence of intermolecular H-Bond.
d. (d)
a. (a)
b. (b)
c. (c)
A
Assertion (A): Al2Cl6 is an electron deficient compound.
Reason (R): It has 3C – 2e bond.
d. (d)
a. (a)
b. (b)
c. (c)
d. (d)
D
Assertion (A): s-p mixing is observed in N2 but not in O2.
Reason (R): Energy difference between 2s-2p is much higher in oxygen atom than in nitrogen atom.
a. (a)
b. (b)
c. (c)
d. (d)
A
10. Assertion (A): PF5 and IF5 are isostructural molecule.
Reason (R): Both contain same number of bond pair & lone pair around central atom.
a. (a)
D
b. (b)
c. (c)
d. (d)
SECTION – C (Paragraph Type) Negative Marking (-1)
This Section contain 1 paragraph. Each of these questions has four choices A), B), C) and D) out of
which ONLY ONE is correct.
6 × 4 = 24 Marks
Comprehension 1
The distribution of electrons among various molecular orbitals is called the electronic configuration of
the molecule which provides us the following very important informations about the molecule.
(A) Stability of molecule. The molecule is stable if number of bonding molecular orbital electrons
(Nb) is greater than the number of antibonding molecular orbital electrons (Na).
1
(B) Bond order: Bond order = (Nb – Na)
2
A positive bond order means a stable molecule while a negative or zero bond order means an
unstable molecule.
(C) Nature of the bond: Bond order 1, 2 and 3 corresponds to single, double and triple bonds
respectively.
(D) Bond length: Bond length decreases as bond order increases.
(E) Magnetic nature: Molecular orbitals in a molecule are doubly occupied, the substance is
diamagnetic and if one or more molecular orbitals are singly occupied, it is paramagnetic.
1.
Which of the following statements is incorrect?

a. Among O 2 , O2 and O 2 the bond length decreases as O 2  O 2  O 2
b. He2 molecule does not exist as the bonding and anti-bonding orbitals cancel each other.
c. C2, O 22  and Li2 are paramagnetic
d. In N2 molecule, the energy of  2pz is more than  2px and  2py
C
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Sol. (A) Bond order 
1
bond length
Bond order O2+ = 2.5, O2 = 2
O 2 = 1.5

so correct order of bond length is O 2  O 2  O 2 .
22
(B) Bond order for He2 =
= 0. So He2 does not exist.
2
(C) In all these molecules all electrons are paired in molecular orbitals. So they are all diamagnetic.
2.
The following molecules/species have been arranged in the Bond orders. Identify the correct order.
(I) O2;
(II) O 2 ;
a. III < II < I < IV
A
Sol.
O2
3.
(III) O 22  ;
(IV) O 2
b. IV < III < II < I
O 2
O 22 
c. III < II < IV < I
d. II < III < I < IV
O 2
Bond Order 2
1.5
1
2.5
N2 has greater dissociation energy than N 2 , where as O2 has a lower dissociation energy than O 2
because:
a. bond order is reduced when O2 is ionized to O 2 & bond order is increased when N2 is ionized to
N 2
b. bond order is increased when O2 is ionized to O 2 & bond order is decreased when N2 is ionized to N 2
c. bond order is decreased when O2 is ionized to O 2 & bond order is decreased when N2 is ionized to
N2
d. bond dissociation energy does not depend on bond order
B
Sol. Species
N2
N 2
O2
O2+
Bond Order
3
2.5
2
2.5
4. Match the orbital overlap figures shown in List-I with the description given in List-II and select the
correct answer using the code given below the lists.
List – I
List – II
(1)
p – d  antibonding
(2)
d – d  bonding
(P)
(Q)
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(R)
(3)
p – d  bonding
(4)
d – d  antibonding
(S)
Select correct code for your answer.
5.
P
Q
R
S
P
Q
a. 2
1
3
4
b. 4
3
c. 2
3
1
4
d. 4
1
C
The correct order of decreasing bond length for N2, N 2 , NO and NO+ is
a. N 2 = NO > N2 > NO+
b. NO+ = N2 > N 2 = NO
c. N 2 = NO > NO+ > N2
d. N 2 = NO > N2 = NO+
R
1
3
S
2
2
D
Sol. As the bond order decreases bond length decreases.
b.o. for N2 = 3, b.o. for NO+ = 3
5
5
b.o. for N 2 = , b.o. for NO =
2
2
6. In a homonuclear molecule which of the following set of orbitals are degenerate?
a. 2s and 1s
b. 2p x and 2p y
c. 2p x and 2p z
d. 2p z and   2p x
B
Sol. Degenerate orbitals are orbitals with same energy
SECTION – D (More Than One Type)
This Section contains 9 multiple choice questions. Each question has four choices A), B), C) and
D) out of which More than one may be correct.
9 × 5 = 45 Marks
1.
Which of the following statement is not true for given resonating structures of N2O

(I)

N  N  O 



(II)  N  N  O


(III) O  N  N
a. Structure I is most stable structure due to –ve formal charge on oxygen atom
b. Structure III is most stable due to maximum number of covalent bonds
c. Structure II is most stable due to –ve charge and +ve charge on both nitrogen atoms
d. Structure I is most stable due to zero formal charge
B,C,D
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2.
Which of the following is true?
a. The bond order of C2 is 2.
b. There are 12 electrons in C2.
c. Electronic configuration of C2 is KK (2s)1(*2s)2 (2p 1x = 2p 2y )
3.
d. Li2 molecules is diamagnetic
A, B, D
Which of the following is true statement?
a. Maximum number of H-bond in H2O are two
b. Sharing of two pairs of electrons forms a double bond between atoms
c. Element “X” of Z = 37 and element Y of Z = 53 will form X+ Y–
d. KO2 is paramagnetic
B, C, D
Sol. Maximum H-bond in H2O = 4
x = Rb & y = Br
KO2 is paramagnetic due to superoxide ion
4. Which of the molecules is/are planar?
d. CO2
5.
a. Benzene
b. CH2 = C = CH2
c. SO3
A,C,D
Which of the following molecule have 3C – 2e– bond
d. Be2H4
6.
a. BH3
b. B2H6
c. BeH2
B,D
Which of the following molecules have d - p bonding
d. OF2
7.
a. SO2
b. SO3
A,B,C
Which of the following statement is correct
8.
a. Metals are malleable not Brittle
b. Strength of metallic bond is inversely proportional to size
c. Conductivity of metal increases with increase in temperature
d. Metallic bond is a kind of attractive force between metal ion and electron
A,B,D
Which of the following is not true for molecular orbital theory?
9.
a. Bonding molecular orbitals are formed by addition by wave function of two atoms.
b. Energy of Bonding molecular orbital is more than parent atomic obital
c. Electron density increases between two nuclei during formation of bonding molecular orbital
d. Both the bond in C2 molecules are  & 
B,D
Select correct statement about B2H6 & C2H6
c. XeO4
a. B2H6 has total 12 valence electron but C2H6 has total 14 valence electron.
b. Both have four identical M – H bond
c. Both the compound have sp3 hybridised atom
d. Free rotation around central atom in ethane
A,C,D
Dr. Sangeeta Khanna Ph.D
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Dr. Sangeeta Khanna Ph.D
SECTION – E (Matrix Type)
This Section contains 3 questions. Each question has four choices (A, B, C and D) given in Column I
and five statements (p, q, r, and s) in Column II.
3 × 8 = 24 Marks
1.
Match Column – I with Column – II (One or More than One Match)
Column– I
(A)
(B)
(C)
(D)
Column–II
CH 3 , BF3 , NH 3
Θ



C
 H3 , BF4 , NH 4
CH3 , BF3, BH3
(P)
Same hybridisation
(Q)
Same bond angle
(R)
All are planar
XeF2, BeF2, CO2
(S) All are non-planar
(T) Have no matching in any of the above
Sol. (A) – T; (B) – P, S; (C) – P, Q, R; (D) – Q, R

CH 3 = sp3, pyramidal
BF3 = sp2, triangular
NH 4 = sp3, tetrahedral

2.
C H3 = sp2, triangular
BH3 = sp2-triangualr
Match Column – I with Column – II (One or More than One Match)
Column– I
(LCAO of atomic orbitals on x-axis)
(A)
+
-
+
+
-
+
+
Resulting orbital is formed by addition of
wave function of two atomic orbitals
-
(Q)
Resulting molecular orbital have
intermolecular nodal plane/node
(R)
Resulting molecular orbital is
occupied molecular orbital of N2
(S)
Resulting molecular orbitals have higher
energy than participating atomic orbitals
one
2px
2px
(C)
(P)
1s
1s
(B)
Column–II
+
+
+
-
highest
2px
2s
(D)
-
+
+
-
+
2pz
2pz
Sol. (A) – Q, S; (B) – P, R; (C) – P; (D) – S
A = destructive interference to give antibonding 1s
B = additive overlap give 2px
C = additive overlap
Dr. Sangeeta Khanna Ph.D
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Dr. Sangeeta Khanna Ph.D
3.
D = negative overlap & have two nodal plane
Match Column – I with Column – II (One or More than One Match)
Column– I
Column–II
Highest occupied orbital is 2pz
(P)
B2  B 2
(B)
Paramagnetic (either reactant or product or both)
(Q)
O2  O2
(C)
Diamagnetic (only reactant or only product)
(R)
O22   O2
(D)
Bond order increases
(S)
C2  C22 
(A)
Sol. (A) – S; (B) – P, Q, R; (C) – R; (D) – P, Q, R, S
Paramagnetic = B2, B2–, O2, O2+, O2–;
Diamagnetic = O2–2, C2, C2–2
Bond order = B2  B2
(1)
1.5
= O2  O2
=
=
2
O22
1
2.5
 O2
1.5
C2  C22
2
3
SECTION – F (Integer Type)
This Section contains 10 questions. The answer to each question is a integer ranging from 0 to 10.
The correct digit below the question number in the ORS is to be bubbled.
10 × 3 = 30 Marks
What is the ratio of  & -bond in ethene (C2H4) molecule
H 5
H\
Sol. 5;
C  C/\ 
/
H 1
H
2. How many orbitals of central atom are involved in Hybridisation of XeF6
Sol. 7; (sp3d3)
3. How many of the following orders are correct as indicated.
(a) IF7 > BrF5 > ICl3 [No. of Hybrid orbitals involved]
(b) PCl3 > PF3 > PI3 [Covalent character]
(c) NO > NO– > NO+ [Bond order]
(d) H2O > H2Se > H2Te [Boiling point]
CH 3
CH 3
1.
|
|
(e) CH3 – CH2 – CH2 – CH2 – CH3 > CH 3  C H  CH 2CH 3  CH 3  C  CH 3 [Boiling point]
|
CH 3
(f) HF > HI > HBr [Bond polarity]
(g) O22  O2  O2 [Bond length]
Sol. 3;
(A) = sp3d3, sp3d2, sp3d
(C) = NO+ > NO > NO–
(B) = PI3 > PCl3 > PF3
(D) = H2O > H2Te > H2Se
(F) = HF > HBr > HI

4. The bond order (No. of bonds) of NO molecule is:
Sol. 3; [N  N]+
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Dr. Sangeeta Khanna Ph.D
5.
How many of the following molecule does not follow octet Rule
BeCl2; BF3;
SF6;
SF2;
NO;
ClO2;
O3;
PF5,
NO+;
Sol. 6; BeCl2; BF3;
NO;
ClO2;
PF5;
SF6
6. How many of the following have sp3 hybridised central atom
CCl4;
CH4;
H2O;
PH3;
SO4–2 ; SO3;
CO2;
NCl3;
H2S;
2
Sol. 7 SO3 = sp ; CO2 = sp;
SF4 = sp3d
7. What is the co-ordination number of Aluminium in [Al(H2O)6]Cl3
Sol. 6
8. How many H-bonds are present in Ice around each water molecule.
Sol.4
9. How many of the following statements are correct:
(i) Intramolecular H-bonds are stronger than intermolecular H-bond
(ii) Hydrogen Bonds are non-directional bonds
(iii) Strength of metallic bond is directly proportional to size of metal ion
(iv) KHCl2 molecule have one H-bond
NO2
CH4;
CO2
SF4
NO2
OH
(v) Boiling point of
is less than
OH
(vi) Boiling point of CH3CH3OH is higher than CH3 – O – CH3
(vii) Order of Boiling point is HF > HCl > HBr > HI
(viii) Boiling point of H2O is higher than HF
Sol. 4; I, V, VI, VIII
10. How many of the following molecule have unpaired electron in Antibonding M.O.
N2; N2 ; N2 , O2; O 2 ; B2; C2; O 2 , O22
Sol. 4
N2 ; O2; O2 ; O-2
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