Chemical Bonding (Package Solutions)
May 8, 2017 | Author: DevarshWali | Category: N/A
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Solutions of Assignment (Set-2)
Chemical Bonding and Molecular Structure (Solutions)
4
Chemical Bonding and Molecular Structure
Chapter
Section-A Q.No. 1.
Solution Answer (4) Cl– is common hence larger the cation more will be the ionic character (Fajan’s rule). Hence, CsCl will have highest ionic character.
2.
Answer (3) NH3 VSEP
1 (V M C A) 2
=
3.
1 (5 + 3 – 0 + 0) = 4 sp3 2
Answer (4) BF3 , = 0 NH3 has greater dipole moment than NF3 hence order is BF3 < NF3 < NH3
4.
Answer (4) O 22 has zero unpaired electrons.
5.
Answer (3) Both S and P are sp3 hybridized in SO 24 and PO 34 respectively.
6.
Answer (3) Both SO 24 and BF4 have sp hybridization. 3
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.
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Chemical Bonding and Molecular Structure (Solutions)
Q.No. 9.
Solutions of Assignment (Set-2)
Solution Answer (1)
O C O
Linear molecule with two equal and opposite dipoles. 10.
Answer (1) K C N
KCN has ionic and covalent bonds. 11.
Answer (1) H
C O
12.
O Intramolecular H Hydrogen bonding
Answer (2) NH4 = 7 + 4 – 1 = 10e
–
– BH 4– = 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 hence it is diamagnetic.
14.
Answer (1)
F
I
Cl
F
Xe
I O
F
I
3
sp d T-Shaped
15.
3
O
O
3
sp d Linear
sp Pyramidal
Answer (2) Due to H-bonding, boiling point and solubility in water increases
16.
Answer (3)
O
F
F Xe
F
O 3 2
sp d Squareplanar
F
Xe
F
F3
F F 3
sp d Distorted octahedral
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Solutions of Assignment (Set-2)
Chemical Bonding and Molecular Structure (Solutions)
Q.No. 17.
Solution Answer (1)
N H3 H N H4
sp 3
18.
sp 3
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 NO 2 = 7 + 8 × 2 – 1= 22 Both have same number of atoms and electrons, hence, isostructural.
20.
Answer (1) NH3 HCl NH4 Cl
When NH3 converts into NH4 number of lone pair decreases, hence, bond angle increases.
21.
Answer (3) KO 2 K O 2 AlO 2 Al 3 O 22 BaO 2 Ba 2 O 22
+
N
NO 2
O
O
Species
Bond order
No. of unpaired e–
O 2
3 2
1
O 22
1
0
only KO2 has unpaired electrons 22.
Answer (1) Cl
Cl
Compound is not planar while Cl
is planar. Cl
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Chemical Bonding and Molecular Structure (Solutions)
Solutions of Assignment (Set-2)
Q.No. 23.
Solution Answer (3) I3
,
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 , O S O , 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
P 120° 120° P
27.
P
=0
Answer (4) O H
28.
120°
N H
H H
H H
C and
H
H H
are Isoelectronic
Answer (1) ClO 2
VSEP =
1 (V + M –C + A) 2 =
1 (7 + 0 – 0 + 1) 2
=
8 4 sp3 2
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Solutions of Assignment (Set-2)
Chemical Bonding and Molecular Structure (Solutions)
Q.No. 29.
Solution Answer (2) CIF3 is T-shaped.
30.
Answer (4)
P F
31.
F
is pyramidal with one lone and three bond pairs.
F
Answer (1)
CCl4 ;
Cl
Cl | C | 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) O 2 KO 2 H2O 2 Na 2O 2 Increasing O — O bond length
34.
Answer (1)
Cl
Cl
Cl–P–Cl | Cl Since P-has 5e– in its outermost shell and is forming 5 bonds hence all the 5e– are consumed and no lone pair will remain.
35.
Answer (3) CO Bond length 1.128 Å
36.
CO 1.115 Å
Answer (2)
+ +
Nonbonding overlap
–
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Chemical Bonding and Molecular Structure (Solutions)
Q.No. 37.
Solution Answer (2) Compound NO
Bond order 2.5
NO
3 3 2 4 3
NO2 NO3
Bond length
38.
Solutions of Assignment (Set-2)
1 Bond order
Answer (4)
Cl Cl
Cl Cl Cl
Cl III
II
I
II < I < III (increasing order of Bond angle)
1
III < I < II ( Increasing order of dipole moment ). 39.
Answer (4)
O O O
P O
P
O O
P
O P
O
O
O Each P is attached to 4 oxygen atoms. 40.
Answer (1)
O P
greatest
F
F F More electronegative substituent 41.
Answer (1) Bond order of CO = 3 & NO– = 2.5
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Solutions of Assignment (Set-2)
Q.No.
Chemical Bonding and Molecular Structure (Solutions)
Solution
42.
Answer (4)
43.
Answer (4)
Pyramidal S O
F F
44.
Answer (4)
F
O
Xe F
Seesaw.
O
Section-B Q.No.
1.
Solution
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 BP = 4, LP = 1 VSEP = 5 , (5, 4, 1) seesaw || O 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 show variable valency.
4.
Answer (2, 3) If z is intermolecular axis than Px – Px and Py – Py will overlap sidewise
5.
Answer (3, 4) ( XeO 4 , NH4 , CH4 ) and (CH3 , NH3 , NF3 ) are isostructural.
6.
Answer (1, 3) Hybridization of P in PCl4 is sp3 while in PCl6 is sp3d2.
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Chemical Bonding and Molecular Structure (Solutions)
Solutions of Assignment (Set-2)
Q.No.
7.
Solution
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 is same). =q×d I3 has a linear shape.
8.
Answer (1, 3, 4) In (SiH3)3N. Lone pair is delocalized hence Lone pair-Bond pair repulsion decreases, hence, bond angle 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 O 2 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
H
Size
N < P < As
EN
N > P > As
P | H
H
H
As | H
H
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 and double bond. CH3–CH2–CH=CH2.
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Solutions of Assignment (Set-2)
Chemical Bonding and Molecular Structure (Solutions)
Q.No.
13.
Solution
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) I2
MgF2 Ionic
Covalent S8
BeF2 Ionic
Covalent
Na3Bi Metallic Na3Sb Metallic
Section-C Q.No.
Solution Comprehension-I
1.
Answer (1) N2 2.5 N2 2.5
N2 3.0 N2 & N2 both have same bond order but N2 has higher number of antibonding e–. Hence correct order
is N2 < N2 < N2 2.
Answer (4) NOis paramagnetic
3.
Answer (3) F2
1
Ne2
0
C2
2
N2
3
N2
2.5
O 2
2.5
Both N2 & O 2 have bond order 2.5
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Chemical Bonding and Molecular Structure (Solutions)
Q.No.
Solutions of Assignment (Set-2)
Solution Comprehension-II
1.
Answer (1) Cations are same, hence, smaller the size of anion, lesser is polarization and greater will be ionic character.
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 and least ionic character will be possessed by species containing smallest cation. Comprehension-III
1.
Answer (2) ICl4– is square planar molecule having zero dipole moment.
2.
Answer (2) CH3— is e– donar and —CN is e– acceptor groups.
3.
Answer (2) OH
, H atoms are not in same plane. In p-xylene both moments are cancelled.
In OH
4.
Answer (1) It should be sp hybridized and should have zero lone pair. Comprehension-IV
1.
Answer (1) Lattice energy
2.
Answer (2)
3.
Answer (4)
q1 q2 r2
LiF have very high lattice energy.
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Solutions of Assignment (Set-2)
Chemical Bonding and Molecular Structure (Solutions)
Q.No.
Solution Comprehension-V
1.
Answer (1) SnCl2 is bent shape molecule.
2.
Answer (4)
3.
Answer (4) d
z2
have largest lobes and can easily takes part in all three hybridization.
Section-D Q.No.
1.
Solution
Answer (2) Factual.
2.
Answer (1) More EN atom shall repel each other strongly, increasing the bond angle. But back-bonding is possible in 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-bonding one 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)
Q.No.
9.
Solution
Answer (1) Larger will be anoin more will be polarisability.
10.
Answer (3) Bond order of O2 and O2– 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.
1.
Solution
Answer A(s), B(p), C(q), D(r) xx
CH3
N | CH3
Si(CH3)3 | N
CH3
(CH3)3Si
Si(CH3)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 shows intermolecular hydrogen bonding hence ortho isomer is more volatile. +
+2
Na is smaller than Ca , hence, Na2CO3 has more lattice energy than CaCO3. Hence, Na2CO3 is more thermally stable. Water has high density than ice because ice forms cage like structure and large vacancies are created in ice. 2.
Answer A(q), B(r), C(s), D(p) ClF3 :
F | F–Cl–F
BP 3 23 LP 2
XeF4 :
F | F–Xe–F | F
I
BP 4 24 LP 2
xx –
I3
xx
LP = 3, BP = 2 3 + 2
I
xx
I
H2S : H–S–H LP = 2, BP = 2 2 + 2
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Solutions of Assignment (Set-2)
Chemical Bonding and Molecular Structure (Solutions)
Q.No.
3.
Solution
Answer A(r), B(s), C(p), D(q) H2Te VSEP =
1 (V + M – C + A) 2
=
1 (6 + 2 – 0 + 0) 2
=
8 = 4 sp3 2
XeF4 1 (8 + 4 – 0 + 0) 2
VSEP = =
12 = 6 sp3d2 2
IClF2 Cl is centre atom 1 (7 + 3 – 0 + 0) 2
VSEP = =
10 = 5 sp3d 2
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.
1.
Solution
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)
Q.No.
5.
Solution
Answer (1) Na2O2 contain O22 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)
Feq Feq
Fa Br
Feq Feq
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.
1.
Solution
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)
Section-H Q.No.
1.
Solution
Answer (3) Hybridisation compound/ion shape
2.
(i) sp3d
XeF3+
T
(ii) sp3d2
XeF5+
Square pyramid
Answer (1) In PF3 due to presence of vacant d orbital.
3.
Answer (2) PBr5 exist on PBr4+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)
Q.No.
13.
Solutions of Assignment (Set-2)
Solution
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.
O
S
(bond angle)
> H
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 therefore higher is probability of decomposition.
19.
AlCl3 (anhydrous) is covalent because of very high ionization energy required for formation of Al+3. But in 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 phosphorus atoms by sigma bonds :
P P
P P
In red phosphorus, molecule forms a chain structure (polymer like) where 2 inter molecular bonds are made 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)
Q.No.
23.
Solution
Calculated µ = 4.8 × 10–10 esu × 1.62 × 10–8 cm = 7.776 D Percentage ionic character
0.39 100 = 5% 7.776
Percentage covalent character = 100 – 5 = 95%.
24.
To avoid strong repulsion of lone pair and bond pair.
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