CHARACTERISTICS
OF
COVALENT BOND
1. Bond
polarity
Uneven sharing of electrons
produces partial negative charges on
the atom having greater than half
share, leaving partial positive charge
on the atom having less than half
share. There will be surplus of
negative charge around one nucleus
and deficiency around the other.
Bond polarity depends on electronegativity
difference. The greater the electronegativity
difference, the greater the unevenness of
sharing, the greater share is acquired by the
initially more electronegative atom, thus
electron density will be displaced in the
direction of the more electron attracting atom.
The greater the difference between the electron
attracting power of the two atoms, the greater
is the ionic character of the covalent bonds.
Electronegativity:
F > O > Cl, N > Br > C, H
Question: The electronegativity difference of
C - I bond is zero but the bond is polarizable .
Why?
• In polar environment this covalent bond can
acquire an appreciable degree of ionic character.
This is due to the polarizability of the outermost
electrons of iodine as a consequence of the big
size of iodine atom. The distance between the
nucleus and the outermost shell is great so that
the outermost electrons are not as strongly
drawn towards the nucleus as are the electrons
of small atoms.
• Formal positive charge effectively increases
the electronegativity, thus increases polarity.
Calculation of Formal charge:
FC= group # - ½ ( # of shared electrons) – # of
unshared electrons
• Example: 1. N in ammonia
H
H
N
H
FC= 5- 3- 2 = 0
•
H
N
H
H
Example: 2. N in ammonium H
FC = 5 – 4 = +1 more polar compared to
ammonia
• The formal positive charge effectively
increases the electronegativity of nitrogen.
There will be a tendency to pull the electron
closer to nitrogen resulting in increased partial
positive charges on each of the four hydrogen
atoms
a. Determine which C – O is more polar and explain.
H
H
C
O
H
H
H
( FC = 0)
or
H
H
C
O
H
H
+
(FC = +1)
The excess positive charge increases its
electron attracting power so that there will
be a greater displacement of electron
density towards oxygen.
b. Which C – H is more polar?
Cl
Cl
C
Cl
H
H
or
H
C
H
H
The C-H bond of chloroform is more polar than
the C-H bond in methane because in chloroform
adjacent atoms are chlorine atoms which possess
greater
electron-attracting
capacity
than
hydrogen atoms in methane. The 3 chlorine will
reduce the electron density around C, increasing
the partial positive charge on C which in turn
will attract electron from hydrogen increasing
the partial positive charge on this atom.
c. Which C – H is more polar?
HC
CH
or
H2C
CH2
The C-H bond in acetylene is more polar than C-H
bond in ethylene because C in acetylene is an sp (
½ s and ½ p character) while C in ethylene is an
sp2 ( 1/3 s and 2/3 p). The s character of hybrid
orbital is a measure of the electron-attracting
capacity of the atom. The electron-attracting
capacity of C in acetylene is greater than in that of
ethylene. The electron attracting power as a result
of s character of the hybrid orbital, responsible for
the polarity of C-H bond in acetylene is also known
as
ORBITAL
ELECTRONEGATIVITY.
d. Which C – H is more polar?
H
H
C C
H
or
H
H C
H
H
H
C H
H
The C-H bond in ethylene is more polar than the C-H
bond in ethane because of orbital electronegativity.
Importance of bond Polarity:
1. It contributes to the characteristic of bond that is it
influenced bond length and energy.
2. It influences molecular polarity, thus in the end
determines the physical properties like melting point,
boiling point, solubility.
3. It influences chemical reactivity.
2. BOND LENGTH
• The bond length is the distance between those two
atoms. The greater the number of electrons between
two atoms, the closer the atoms can be brought
towards one another, and the shorter the bond. This
is measured in Angstrom units which 1A is equal to
10-8 cm. It has been shown that bond length is
correlated with bond polarity, hybridization of orbital
and
delocalization
of
electrons.
When two atoms approach each other, their
interaction is influenced by:
a. repulsion between two electron, clouds
b. repulsion between the two nuclei
c. attraction between the nucleus of each, and
the
electrons
of
the
other.
Factors influencing bond length
a. Bond Polarity
Increasing polarity decreasing bond length
Type of Single Bonds
Bond Length
C-C
1.538
C-N
1.471
C–O
1.430
C–F
1.380
b. Hybridization
Increasing s ( near the nucleus) character of hybrid orbital
decreasing bond length or orbital electronegativity leads to
reduction of bond distance.
Ethane
Propylene
Methylacetylene
Ethylene
Acetylene
CH3-CH3
CH3-CH=CH2
H3C
C
CH2=CH2
HC
CH
CH
sp3 – sp3
sp3 – sp2
sp3-sp
sp2-sp2; p-p
sp-sp; p-p; p-p
1.538
1.501
1.459
1.339
1.207
c. atomic size
small atoms will form shorter bond.
Example: H-H < CH4-CH4
d. bond order
decreasing bond order increasing bond
length.
Triple bond< double bond< single bond
sp
sp2
sp3
• e. π and σ electron delocalization
illustrated by system containing atoms in the trigonal state
of hybridization.
Example:
a. CH2 = CH - Cl
b. O = C
c. CH3 – CH = CH2
3. BOND ENERGY or
BOND STRENGTH
energy required to break the bond and at the
same time the energy release when the bond
is formed.
This is express in terms of kilo calories per
mole
Bond energy is variable depending on length,
the shorter the bond the stronger the bond.
Bond strengthening are attributed to:
a. orbital hybridization – when the s character of the bonding
orbitals is high, there will be closer interaction between the
bonding electrons and the nuclei thus covalent bond is stronger.
sp3 – sp3
sp3- sp2
sp3 – sp
sp2 – sp2
sp2 – sp
sp - sp
Bond energy increases bond length decreases
b. bond polarity - when bond polarity increases
bond energy also increases
-C–C
=C–C=
-C=C= C – C=
- C = C-
sp3 –sp3
sp2 – sp2
sp2 – sp2
sp – sp
sp-sp
0
0
1
0
2
increasing pi bond increasing
bond energy
c. pi bonding
- C – C sp3 –sp3
= C – C = sp2 – sp2
- C = C - sp2 – sp2
= C – C= sp – sp
-C = C- sp-sp
0
0
1
0
2
increasing pi bond
increasing bond
energy
d. Reduction of bond strength is observed
when there is loss of overlap of hybrid
orbitals. The loss of overlap is a consequence
of forced bending of orbitals. A vivid example
of loss of overlap as a result of forced bending
of sp3 orbitals is seen in cyclopropane. Forced
bending results in angular strain due to
compression of the tetrahedral bond angle.
4. BOND order
• The bond order is equal to the number of
bonds between two atoms.
The BO is an indication of the bond length,
the greater the bond order, the shorter the
bond and the greater the bond strength.
5. BOND ANGLE
a. governed by hybridization of central atom
sp3 - 109.5o
sp2 – 120 o
sp - 180 o
b. influenced by the presence of lone pairs
CH4
sp3
109o 28’
NH3
sp3
106o 47’
H2O
sp3
104o 31’
As the number of lone pair increases, the bond angle
decreases.