CHEMICAL BONDING

1
(No Transcript)
2
CHEMICAL BONDING SESSION III
3
Session Objectives

1. Fajans rule Covalent nature in ionic bond
2. Polarity in a covalent bond Dipole moment
3. Calculation of percentage ionic character
4. Resonance
5. Odd electron bond
6. Hydrogen bonding
7. van der Waals forces

4
Fajans rule
The nature of bond between these two ions
effect of one ion on the other.
Positive ion attracts the outermost electron of
anion and repels its nucleus, causes distortion
or polarisation of the anion.
Polarisation leads to the partial sharing of
electron cloud.
5
Factors responsible for partial covalent nature
Smaller size and increase in charge density on
cation, increases its polarising power.
6
Factors responsible for partial covalent nature
Larger anion holds its valence electronloosely
more polarisation occurs.
Decreasing order of covalent nature NaI gt NaBr
gt NaCl
7
Factors responsible for partial covalent nature
Na Ne (Noble gas configuration)
No. of protons 11 29
Size and charge are almost same.
More polarising powerof Cu increase in
effective nuclear charge
8
(No Transcript)
9
Illustrative example 1

• The decreasing ionic character of NaF,
• MgF2 and AlF3 is
• NaF gt MgF2 gt AlF3
• AlF3 gt MgF2 gt NaF
• MgF2 gt AlF3 gt NaF
• None of these

10
Solution
Increase in charge on cation increases covalent
nature. Therefore, increasing order of covalent
character Na lt Mg2 lt Al3 Therefore,
decreasing order of ionic character NaF gt MgF2 gt
AlF3 Hence, answer is (a).
11
Factors responsible for pure ionic bond

1. Large cation
2. Small anion
3. Small charge density on ions
4. Noble gas configuration

12
Polarity in covalent bond Dipole moment
No difference in electronegativity between bonded
atoms.
Pure covalent bond
Chlorine is more electronegative attracts bond
pair of electrons more towards itself.
partial charge development leads to partial
ionic character.
13
Polarity in covalent bond Dipole moment
1 D 1018 esucm 3.33 x 1030 Cm (SI
unit)
Dipole moment is vector quantity.
14
Dipole moments in diatomic molecules
No difference in electronegativity
Difference in electronegativity leads to polar
nature.
15
Dipole moments in polyatomic molecules
Dipole moment depends on the geometry of the
molecules.
16
Dipole moments in polyatomic molecules
17
Dipole moments in polyatomic molecules
18
Dipole moments in polyatomic molecules
19
(No Transcript)
20
Illustrative example 2
21
Solution
To find minimum dipole moment weneed to find
most symmetrical molecule.
The most symmetrical one is
Hence, the answer is (b).
22
Calculation of percentage ionic character
Indicates ionic character in a covalent bond.
23
(No Transcript)
24
Illustrative example 3
25
Solution
Hence, the answer is (c).
26
Resonance
It is a hypothetical phenomena which can explain
various properties of a species consists of
number of Lewis structures.
These are cannonical forms almost equal energy.
Resonance stablises molecule by lowering energy.
27
(No Transcript)
28
Illustrative example 4
29
Solution
In others no such resonance is possible.
Hence, the answer is (a).
30
Odd Electron Bonds
The bond formed by sharing odd number of
electrons results stable molecules (following
octet rule).
31
Hydrogen Bond
Weak electrostatic force of attraction between
hydrogen atom and a highly electronegative atom
(e.g., F, O, N) which must be small size.
H Cl
H Cl
No H-bond as Cl is of larger size.
H F
H F
Strong H-bond
32
Types of H-bond
(a) Intermolecular H-bond
Increase in boiling point
(b) Intramolecular H-bond
Decrease in boiling point
33
(No Transcript)
34
Illustrative example 5

• Which of the following leads to
  explosivereactions?
• HF (b) HCl (c) HBr (d) HI

Solution
(Forms very stable intermolecular H-bonding).
Requires very high energy to break the bond
between them. Hence, the answer is (a).
35
Illustrative example 6

• Which of the following having highestboiling
  point?
• HF (b) HCl (c) HBr (d) HI

Solution
Substance having intermolecular H-bonding is
highly stable and boiling point will be maximum.
Hence, the answer is (a).
36
Types of H-bond
Less solubility in H2O steam volatile
Inter molecular H-bonding with H2O, more
solubility, less volatile.
37
(No Transcript)
38
Illustrative example 7
39
Solution
The compound which forms intramolecular H-bonding
has lower boiling point and hence, steam
volatile. Hence, the answer is (b).
40
van der Waals forces
Different weak forces in covalent molecules
(polar and non-polar).
Dipole-dipole attraction
Explain high solubility of HCl gas in water.
41
van der Waals forces
(b) O O (no dipole)
Dipole-induced dipole attraction
Temporary dipole formation in O2.
42
Do you know?
Oxygen is soluble in water due to weak
dipole- induced dipole attraction.
43
(No Transcript)
44
Class Exercise - 1
If a molecule MX3 has zero dipolemoment,the
sigma bonding orbitalsused by M (atomic number lt
21) are (a) pure p (b) sp hybrid(c) sp2
hybrid (d) sp3 hybrid
Solution
Molecule of MX3 type having zero dipole
momentthen its geometry must be triangular
planar, i.e.hybridization of M will be sp2.
Hence, answer is (c).
45
Class Exercise - 2
Solution
The hydrocarbon withsymmetrical structurewill
have least dipolemoment.
Hence, answer is (b).
46
Class Exercise - 3
H2O has the boiling point of 1000Cand H2S has
boiling point of-420C.This is explained by(a)
van der Waals forces (b) molecular weight (c)
covalent bonding(d) inter-molecular H-bonding
Solution
Intermolecular H-bondingin H2O increases its
boilingpoint than H2S.
Hence, answer is (d).
47
Class Exercise - 4
NF3 BF3 are covalent compounds,but NF3 is
polar while BF3 is non-polar.The reason is
that(a) boron is a metal while nitrogen is a
gas(b) B F bonds have no dipole while N F
bonds have dipole moment(c) atomic size of
boron is smaller than that of nitrogen(d) BF3 is
planar while NF3 is pyramidal in shape
48
Solution
It has pyramidal geometry with one lone pair.
It has planar geometry with no lone pair. Hence,
dipole moment is zero.
Hence, answer is (d).
49
Class Exercise - 5
In which of the following pairs, the
firstcompound does not have a higherboiling
point than the second ?(a) Water and H2S (b)
HI and HBr(c) Argon and Krypton (d) HF and HCl
50
Solution
In given options, H2O and HF form H-bonding.
Hence, boiling point will be higher in both cases
compared to the other compound in pair.
HI is having very high molecular weight compared
to HBr leads to its high boiling point.
Only Argon and Krypton are noble gases and
Kryptonis having higher atomic weight.
Hence, answer is (c).
51
Class Exercise - 6
Following is the arrangement ofdecreasing
percentage of ioniccharacter of bonds (a) F-H
gt O-H gt C-H gt C-F(b) F-H gt C-F gt O-H gt C-H(b)
C-F gt F-H gt O-H gtC-H(d) O-H gt C-H gt C -F gt F-H
Solution
Percentage ionic character decreases with
decreasingdifference of electronegativity
between two bonded atoms. Therefore the correct
order should be H F gt C F gt O H gt C H
Hence, answer is (b).
52
Class Exercise - 7
Which of the following ions havelowest
polarising power? (a) Na (b) Ca2(c) Mg2 (d)
Al3
Solution
Cation with minimum charge will have low
polarizingpower Fajans rule.
Hence, answer is (a).
53
Class Exercise - 8
Which of the following will havemaximum covalent
character? (a) LiCl (b) KCl(c) NaCl (d) CsCl
Solution
Down the group, size of the cation increases
whichdecreases its polarizing power as well as
covalentcharacter. Hence, LiCl is maximum
covalent in nature.
Hence, answer is (a).
54
Class Exercise - 9
Phosphine has lower boiling pointthan ammonia.
Why?
Solution
NH3 is capable of forming intermolecular
H-bondingthrough nitrogen. But phosphorous is
unable to form such H-bonding. Hence, NH3 is
having higher boiling point than PH3.
55
Class Exercise - 10
Calculate the percentage of ionic character in
HCl, given the observeddipole moment is 1.03 D
and theirinternuclear distance 1.275 Ao.
Solution
If there is a complete transfer of electron from
H to Cl in HCl giving rise to completely ionic
HCl, the dipole moment would be
1D 10-18 e.s.u. cm
56
Solution
16.83
The bond in HCl is 16.83 ionic and 83.17
covalent.
57
(No Transcript)