chemical bonding (package solutions)
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Solutions of Assignment (Set-2) Chemical Bonding and Molecular Structure (Solutions)
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Section-A
Q.No. Solution
1. Answer (4)
Cl – is common hence larger the cation more will be the ionic character (Fajan’s rule). Hence, CsCl willhave highest ionic character.
2. Answer (3)
NH3
)ACMV(2
1VSEP
=2
1 (5 + 3 – 0 + 0) = 4 sp
3
3. Answer (4)
BF3 , = 0
NH3 has greater dipole moment than NF3 hence order is
BF3 < NF3 < NH3
4. Answer (4)
22O has zero unpaired electrons.
5. Answer (3)
Both S and P are sp3 hybridized in 2
4SO and 34PO respectively.
6. Answer (3)
Both 24SO and
4BF have sp3 hybridization.
7. Answer (2)
Number of Hydrogen-bonds formed by water molecule = 4
8. Answer (1)
Difference of electronegativity is maximum between H and F in H.
It will form strongest H-Bond.
4Chapter
Chemical Bonding andMolecular Structure
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Chemical Bonding and Molecular Structure (Solutions) Solutions of Assignment (Set-2)
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Q.No. Solution
9. Answer (1)
OCO
Linear molecule with two equal and opposite dipoles.
10. Answer (1)
NCK
KCN has ionic and covalent bonds.
11. Answer (1)
C
OH
O Intramolecular Hydrogen bonding
H
12. Answer (2)
4NH = 7 + 4 – 1 = 10e –
– 4BH = 5 + 4 + 1= 10e –
Both have same number of electrons and same total number of atoms, hence, isostructural.
13. Answer (2)
O2 is paramagnetic because it contains unpaired electrons. N2 does not have unpaired electrons henceit is diamagnetic.
14. Answer (1)
Cl F
F
F
sp d3
T-Shaped
I
I
I
sp d3
Linear
Xe
O OO
sp3
Pyramidal
15. Answer (2)
Due to H-bonding, boiling point and solubility in water increases
16. Answer (3)
Xe
F O
sp3d
2
Squareplanar
FO
Xe
F
sp3d
3
Distorted octahedral
F
F F
F
F
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Solutions of Assignment (Set-2) Chemical Bonding and Molecular Structure (Solutions)
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Q.No. Solution
17. Answer (1)
4sp
3sp
HNHHN33
18. Answer (1)
CN – = 6 + 7 + 1 = 14
CO = 6 + 8 = 14
Both have same number of electrons and same number of atoms.
19. Answer (1)
CO2 = 6 + 2 × 8 = 22
2NO = 7 + 8 × 2 – 1= 22
Both have same number of atoms and electrons, hence, isostructural.
20. Answer (1)
ClNHHClNH 43
When NH3 converts into 4NH number of lone pair decreases, hence, bond angle increases.
21. Answer (3)
22 OKKO
2232 OAlAlO
22
22 OBaBaO
2NO
N
O O
+
Species Bond order No. of unpaired e –
2O
2
3 1
22O 1 0
only KO2 has unpaired electrons
22. Answer (1)
Cl
Cl
Compound is not planar while
Cl
Cl
is planar.
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Chemical Bonding and Molecular Structure (Solutions) Solutions of Assignment (Set-2)
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Q.No. Solution
23. Answer (3)
3I , BP = 2 , LP = 3 Total no. of hybridized orbital = 5
(5, 2, 3) linear
XeF2, BP = 2, LP = 3 Total no. of hybridized orbital = 5(5, 2, 3) linear
SO2 , OSO
, BP = 2 , LP = 1 Total no. of hybridized orbital = 3
(3, 2, 1) sp2 bent or V shaped or angular
CHC–CCH linear
24. Answer (2)
Except BF3, all molecules have complete octet hence only BF3 can act as Lewis acid.
25. Answer (4)
All are sp2 hybridized with 120° bond angle.
26. Answer (1)
BF3 has sp2 hybridization hence 3 dipole moment vectors of equal magnitude are at 120° from each
other. Hence = 0
120°
120° 120°
P
P P
= 0
27. Answer (4)
O
Hare Isoelectronic
H H
N
H H Hand
C
H H H
28. Answer (1)
2ClO
VSEP =2
1 (V + M –C + A)
=2
1 (7 + 0 – 0 + 1)
= 42
8 sp
3
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Solutions of Assignment (Set-2) Chemical Bonding and Molecular Structure (Solutions)
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Q.No. Solution
29. Answer (2)
CIF3 is T-shaped.
30. Answer (4)
P
Fis pyramidal with one lone and three bond pairs.
F F
31. Answer (1)
CCl4 ;
Cl |C |Cl
Cl Cl
all atoms in surrounding are same and no lone pairs present on central atom
Hence = 0
32. Answer (1)
Smaller the cation and larger the Anion, higher is polarization.
33. Answer (4)
lengthbondO — OIncreasing
ONaOHKOO 222222
34. Answer (1)
Cl–P–Cl | Cl
ClCl
Since P-has 5e – in its outermost shell and is forming 5 bonds hence all the 5e – are consumed and nolone pair will remain.
35. Answer (3)
Å115.1Å128.1lengthBond COCO
36. Answer (2)
++
– Nonbonding overlap
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Chemical Bonding and Molecular Structure (Solutions) Solutions of Assignment (Set-2)
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Q.No. Solution
37. Answer (2)
2
3
NO 2.5
NO 3
3NO24
NO3
Compound Bond order
Bond length1
Bond order
38. Answer (4)
Cl
Cl
I
Cl
Cl
II
Cl
ClIII
II < I < III (increasing order of Bond angle)
1
III < I < II ( Increasing order of dipole moment ).
39. Answer (4)
OO
O O
P P
P
P
O
OO
O
O
O
Each P is attached to 4 oxygen atoms.
40. Answer (1)
P
F F F
O
greatest
More electronegative substituent
41. Answer (1)
Bond order of CO = 3 & NO – = 2.5
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Solutions of Assignment (Set-2) Chemical Bonding and Molecular Structure (Solutions)
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Q.No. Solution
42. Answer (4)
43. Answer (4)
S
O FF
Pyramidal
44. Answer (4)
Xe
F
F
O
O
Seesaw.
Section-B
Q.No. Solution
1. Answer (1, 3)
CO2, Hg2Cl2 and C2H2 all have linear geometry.
2. Answer (1, 3)
SF4 BP = 4, LP = 1 VSEP = 5 , (5, 4, 1) seesaw
O ||F–Xe–F || O
BP = 4, LP = 1 VSEP = 5 , (5, 4, 1) seesaw
3. Answer (2, 3, 4)
F do not have d orbital hence it is always monovalent but Cl, Br, I have d orbital hence can showvariable valency.
4. Answer (2, 3)
If z is intermolecular axis than Px – Px and Py – Py will overlap sidewise
5. Answer (3, 4)
)CH,NH,XeO( 444 and )NF,NH,CH( 333
are isostructural.
6. Answer (1, 3)
Hybridization of P in 4PCl is sp3 while in 6PCl is sp
3d2.
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Chemical Bonding and Molecular Structure (Solutions) Solutions of Assignment (Set-2)
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Q.No. Solution
7. Answer (2, 3, 4)
SnCl2 has lone pair on central atom hence it is not linear
Dipole moment of CH3Cl is more than CH3F because bond length of C–F is less (charge in both issame).
= q × d3I has a linear shape.
8. Answer (1, 3, 4)
In (SiH3)3N. Lone pair is delocalized hence Lone pair-Bond pair repulsion decreases, hence, bondangle increases.
9. Answer (1, 2)
SF4 and XeO2F2 both have 5 VSEP, 4 bond pair and 1 lone pair
10. Answer (2, 4)NaCl is more soluble and p-hydroxy benzoic acid is more soluble.
11. Answer (1, 3)
When CO CO+, bond order increases from 3 to 3.5 and when O2 2O bond order increases
from 2 to 2.5 because in both case electron is removed from antibonding molecular orbital.
12. Answer (1, 2, 4)
Molecular orbital No. of nodes
2pz 0
* 2pz 1
2py 1
* 2py 2
N |H
H H
P |H
H H
As |H
H H
Size N < P < As
EN N > P > As
Bond angle increases with increasing electronegativity
CH3 is electron donating and CN is electron attracting, hence, order of dipole moment will be o < m < p CH3
, CH3 is donor hence bond length decreases.
CH2=CH–CH3, Here CH3 can show hyper conjugation hence bond length will be between single anddouble bond.
CH3 –CH2 –CH=CH2.
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Solutions of Assignment (Set-2) Chemical Bonding and Molecular Structure (Solutions)
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Q.No. Solution
13. Answer (1, 2)
H2 gas molecule can not form H-bonding
HCl gas molecules have dipole interaction.
14. Answer (1, 3, 4)In option 2, N-atom has 5 bonds.
15. Answer (1, 3)
MgF2 I2 Na3Bi
Ionic Covalent Metallic
BeF2 S8 Na3Sb
Ionic Covalent Metallic
Section-C
Q.No. Solution
Comprehension-I
1. Answer (1)
2N 2.5
2N 2.5
N2 3.0
2N &
2N both have same bond order but N2 has higher number of antibonding e – . Hence correct order
is 2N <
2N < N2
2. Answer (4)
NO is paramagnetic
3. Answer (3)
F2 1
Ne2 0
C2 2
N2 3
2N 2.5
2O 2.5
Both 2N &
2O have bond order 2.5
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Chemical Bonding and Molecular Structure (Solutions) Solutions of Assignment (Set-2)
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Q.No. Solution
Comprehension-II
1. Answer (1)
Cations are same, hence, smaller the size of anion, lesser is polarization and greater will be ioniccharacter.
2. Answer (1)
Li2CO3 has smaller cation hence more covalent character hence decompose on heating.
3. Answer (3)
LiCl BeCl2
RbCl MgCl2
Anions are same hence greatest ionic character is possessed by species having largest cation andleast ionic character will be possessed by species containing smallest cation.
Comprehension-III
1. Answer (2)
– 4ICl is square planar molecule having zero dipole moment.
2. Answer (2)
CH3 — is e – donar and —CN is e – acceptor groups.
3. Answer (2)
In
OH
OH
, H atoms are not in same plane. In p-xylene both moments are cancelled.
4. Answer (1)
It should be sp hybridized and should have zero lone pair.
Comprehension-IV
1. Answer (1)
Lattice energy 1 22
q q
r
2. Answer (2)
3. Answer (4)
LiF have very high lattice energy.
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Solutions of Assignment (Set-2) Chemical Bonding and Molecular Structure (Solutions)
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Q.No. Solution
Comprehension-V
1. Answer (1)
SnCl2 is bent shape molecule.
2. Answer (4)
3. Answer (4)
2zd have largest lobes and can easily takes part in all three hybridization.
Section-D
Q.No. Solution
1. Answer (2)
Factual.
2. Answer (1)
More EN atom shall repel each other strongly, increasing the bond angle. But back-bonding is possiblein PF3 due to vacand d-orbitals in P and availability of lp on F.
3. Answer (3)
AlF3 is ionic, hence, a solid. SiF4 is covalent, hence, a gas.
4. Answer (2)
S=C=S
C, S have almost same electronegativity, hence, CS2 is non polar.
CS2 is linear (BP = 2, LP = 0 , VSEP = 2)
5. Answer (3)
In PCl5 all bond angles and bond lengths are not equal.
6. Answer (4)
O=C=O
CO2 has two polar bonds although it is non polar.
7. Answer (2)
H2O is liquid due to Hydrogen-bonding; O is more electronegative than S but for Hydrogen-bondingone atom should be highly electronegative while S is not highly electronegative.
8. Answer (2)
Boiling point of H2O is higher because it forms four hydrogen bonds per molecule.
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Chemical Bonding and Molecular Structure (Solutions) Solutions of Assignment (Set-2)
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Q.No. Solution
9. Answer (1)
Larger will be anoin more will be polarisability.
10. Answer (3)
Bond order of – 2 2O and O are 2.5 and 1.5 respectively.
11. Answer (4)
o-hydroxybenzoic acid has lower boiling point as it has intramolecular hydrogen bonding.
Section-E
Q.No. Solution
1. Answer A(s), B(p), C(q), D(r)
CH3CH3CH3CH3
N |CH3
CH3
xx
CH3CH3CH3
Si(CH )
|N
3
3
(CH ) Si3 3Si(CH )3 3
In NSi(CH3)3 , Si uses its vacant d-orbital for back bonding with lone pair electrons of central N atom,hence, has less tendency to release its electrons, hence, less basic.
Ortho nitro phenol shows intramolecular hydrogen bonding while para nitro phenol showsintermolecular hydrogen bonding hence ortho isomer is more volatile.
Na+ is smaller than Ca+2, hence, Na2CO3 has more lattice energy than CaCO3. Hence, Na2CO3 is morethermally stable.
Water has high density than ice because ice forms cage like structure and large vacancies are createdin ice.
2. Answer A(q), B(r), C(s), D(p)
ClF3 :F
|F–Cl–F
322LP3BP
XeF4 :
F |F–Xe–F |
F
422LP4BP
3I
LP = 3, BP = 2 3 + 2
H2S : H–S–H
LP = 2, BP = 2 2 + 2
I –
xx
xx xx
I
I
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Solutions of Assignment (Set-2) Chemical Bonding and Molecular Structure (Solutions)
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Q.No. Solution
3. Answer A(r), B(s), C(p), D(q)
H2Te
VSEP =2
1 (V + M – C + A)
=21 (6 + 2 – 0 + 0)
=2
8 = 4 sp
3
XeF4
VSEP =2
1 (8 + 4 – 0 + 0)
=2
12 = 6 sp
3d
2
IClF2
Cl is centre atom
VSEP =2
1 (7 + 3 – 0 + 0)
=2
10 = 5 sp
3d
O ||H–C–O–H
central atom is C
If C attached with, one double bond then sp2
4. Answer A(p), B(s), C(r), D(q, s)
Section-F
Q.No. Solution
1. Answer (6)
Fact.
2. Answer (6)
Fact.
3. Answer (4)
Hybridization is sp3d
Five hybrid orbitals and one hybrid orbital is unused carrying lone pair.
4. Answer (0)
Its shape is distorted square pyramidal.
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Chemical Bonding and Molecular Structure (Solutions) Solutions of Assignment (Set-2)
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Q.No. Solution
5. Answer (1)
Na2O2 contain 22O anion having bond order one.
6. Answer (6)
PCl6 – exist
7. Answer (6)
Due to its trigonal bipyramidal shape.
8. Answer (5)
4 between C & N and one between C —C.
9. Answer (0)
Br
FeqFeq
FeqFeq
Fa
Presence of lone pair will distort all the bond angles and none of the bond angle is exactly equal to 90º
FaBrFeq = 84°
FeqBrFeq = 89°
Section-G
Q.No. Solution
1. Answer (3)
Fact.
2. Answer (4)
Bond order of CO+
is 3.5.
3. Answer (2)
4. Answer (1)
5. Answer (2)
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Solutions of Assignment (Set-2) Chemical Bonding and Molecular Structure (Solutions)
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Section-H
Q.No. Solution
1. Answer (3)
Hybridisation compound/ion shape
(i) sp3d XeF3
+
T(ii) sp3d2 XeF5
+ Square pyramid
2. Answer (1)
In PF3 due to presence of vacant d orbital.
3. Answer (2)
PBr 5 exist on PBr 4+Br – in solid.
4. Answer (4)Fact
5. Answer (1)
In hydride down the group bond angle decreases.
6. Answer (1)
Due to network structure in water.
7. Answer (2)
Fact
8. Answer (2)
Opposite sign of wave function gives antibonding.
9. Answer (4)
Due to formation of pseudo dipole.
10. Answer (4)
Based on M.O. theory.
11. Answer (2)
In PCl5, P—Claxial bond length are large.
12. Answer (3)
Fact
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Chemical Bonding and Molecular Structure (Solutions) Solutions of Assignment (Set-2)
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Q.No. Solution
13. Answer (3)
Fact
14. Answer (2)
Hybridization of S in orthorhombic S8.
15. Bond order of H2 = 1, H2+ = 0.5
He2+ = 1.5, He2 = 0
H2 is more stable than H2+ but He2
+ is more stable than He2
16. CO > SiO (stability order) because there is no tendency of p-p bonding in Si.
17.
H
S(bond angle)
O
H>
H H
O is more electronegative than S, so it has more tendency to attract bonded electron pairs.
18. More the carbonate ion is polarized by the cation more is the chance of formation of CO2 and thereforehigher is probability of decomposition.
19. AlCl3 (anhydrous) is covalent because of very high ionization energy required for formation of Al+3. Butin hydrated condition, this excessive energy is obtained by hydration energy of smaller Al+3 ion.
20. White P has following tetrahedral structure where each phosphorus is attached to 3 other phosphorusatoms by sigma bonds :
PP
P
P
In red phosphorus, molecule forms a chain structure (polymer like) where 2 inter molecular bonds aremade by each P4 unit.
21. Silicon has lower bond energy than carbon, so less tendency of catenation is present.
22. IF7 has lower covalent character i.e., higher ionic character this strengthens the I– F interaction, hence,stabilizing the molecule.
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Solutions of Assignment (Set-2) Chemical Bonding and Molecular Structure (Solutions)
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Q.No. Solution
23. Calculated µ = 4.8 × 10 –10 esu × 1.62 × 10 –8 cm = 7.776 D
Percentage ionic character 100776.7
39.0 = 5%
Percentage covalent character = 100 – 5 = 95%.
24. To avoid strong repulsion of lone pair and bond pair.