4 TOPIC 3 MOLECULAR SHAPE AND BONDING
Topic 3
Molecular Shape and Chemical
Bonding
Content
3.2.4. Bond order, bond length and bond energy .............................................. 60
3.3.3. Further examples of Lewis structure / VSEPR ........................................ 70
M OLECULAR O RBITAL (MO) T HEORY A BRIEF INTRODUCTION .............................. 81
Learning objectives
Basic Concepts of Chemical Bonding
– chemical formulae ; ionic bonding ; covalent bonding
– octet rule, valency, bond polarity, Lewis structures, resonance, hypervalency, bond order, length and dissociation energy ; molecular geometry – VSEPR Theory ; dipole moments ; valence bond theory
– hybrid orbitals ; molecular orbital theory.
TOPIC 3. Molecular Shape and Chemical Bonding 51
3.1. Covalent Bonding
A covalent bond is a chemical bond that involves the sharing of an electron pair between neighbouring atoms.
Covalent bonding is energetically favoured. The bond energy is defined as the energy that must be supplied to separate the bonded atoms to infinity.
The bond length is the distance between the two nuclei; it is the internuclear distance for which the bond energy is minimal.
Covalent bonds are affected by the electronegativity of the connected atoms.
For two atoms with equal electronegativity, the covalent bond is non-polar.
E.g.: The hydrogen molecule: H-H; bond energy = 435 KJ mol
-1
; bond distance = 74 pm
1 e
-
2 e
-
H H
H H H
Figure.
Non-polar covalent bonding formed by the interaction of two shared bonding electrons.
In a heteroatomic bond, the more electronegative atom tends to attract the electron pair of the bond. The covalent bond is polar with a negative partial charge over the more electronegative atom and a positive partial charge over the less electronegative.
Example: The hydrochloride molecule: H-Cl unpaired electron lone pair bonding pair
1
e-
+ -
1+
+
H Cl
-
H Cl
H
+
Cl
-
Figure.
Polar covalent bonding formed by the interaction of two shared bonding electrons. The bonding pair is not shared equally but is more attracted by the more electronegative atom (here Cl) which polarises the bond.
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TOPIC 3. Molecular Shape and Chemical Bonding
3.1.1. The octet rule
52
The octet rule states that atoms tend to be more stable when their electronic configuration is similar to that of a noble gas. As a result, atoms tend to gain, lose or share electrons so that their outer shell is populated by 8 electrons. (8 = octet).
This rule applies to the main group elements of low atomic number (<20), including carbon, nitrogen, oxygen, halogens, sodium, potassium, magnesium. The number 8 correspond to a full n-shell: 2 electrons in ns-orbital plus 6 electrons in np-orbitals.
On ammonia (NH
3
) the nitrogen (which has 5 valence electrons) is bond to three hydrogen atoms: each hydrogen brings one electron to the molecule - so the total number of electron in the nitrogen's outer shell is 8 (5(N) + 3(H)).
On water molecule, the oxygen (6 valence electrons) is bond to two hydrogens. Each hydrogen brings one electron to the oxygen's outershell, which is then populated by 8 electrons. (6(O) + 2(H))
3.1.2. Valency
The valency or valence number of an atom is the number of chemical bonds that an atom may form to satisfy the octet rule. It is equal to the number of unpaired electron (upe). The lone pair electrons are not counted as bonding, therefore the valency is the number of valence electrons, less the lone pair electrons (lp).
The valency of an atom in a molecule is equal to the number of bond to the atom (single bond count as one bond, double bond two and triple bond three).
Example:
Nitrogen , has 5 valence electrons. Two of them are organised in one lone pair and three are available for bonding: the valency is 3. lp upe N upe upe
Nitrogen has one lone pair (lp) and three unpaired electrons (upe)
Chlorine , has 7 valence electrons. Six electrons are organised in three lone pairs and one electron is available for bonding: the valency is 1. lp lp
Cl lp upe
Chlorine has three lone pairs (lp) and one unpaired electron (upe)
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TOPIC 3. Molecular Shape and Chemical Bonding 53
3.1.3. Co-ordinate covalent bonds
This is also known as a dipolar bond ; a dative covalent bond or a co-ordinate bond and is a covalent bond in which the two electrons of the bond are provided by the same atom.
Example : Adduct of ammonia and boron trifluoride. The two electrons of the B-N bond are provided by the nitrogen: as a result a positive charge appears on the nitrogen and a negative charge on the boron.
Ground state Boron: [He] 2s
2
2p
1 eo lp upe empty orbital (eo)
B upe upe empty orbital
F B
F
F
Boron trifluoride
H N H
H ammonia
F
F
B
F
H
N
H
H F
F
B
F
H
N
H
H
Excited state Boron: [He] 2s
1
2p
2 Adduct of ammonia and boron trifluoride
3.1.4. Multiple bonds
A double bond is a chemical bond involving four bonding electrons shared between two atoms.
Example: Ethylene C
2
H
4
. Each carbon is surrounded by 8 valence electrons. The bonding between the two carbons involves four valence electrons making a double bond.
H H H H
C C C C
H H H H
Figure. Double bond of ethylene.
A triple bond involves six bonding electrons shared by two atoms.
Example: Acetylene C
2
H
2
. Each carbon is surrounded by 8 valence electrons. The bonding between the two carbons involves six valence electrons: a triple bond.
H C C H H C C H
Figure. Triple bond of acetylene.
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TOPIC 3. Molecular Shape and Chemical Bonding 54
The valency of carbon is 4 in ethylene and acetylene:
In ethylene, each carbon has four bonds: two single bonds with hydrogen plus one double bond with carbon.
In acetylene, each carbon has four bonds: one single bond with hydrogen plus one triple bond with carbon.
3.1.5. Bond polarity
When one of the two atoms involved in a covalent bond attracts the bonding electron pair more than the other atom, partial charges appear and the bond is polar. A negative partial charge ( -) appears over the atom attracting the electronic density and a positive partial charge ( +) appears over the other one.
The ability to attract the electronic density is evaluated using the empirical value of electronegativity . The higher the value the greater the pull on electrons and the higher the difference in electronegativity between the two atoms bonded the higher the polarity.
The polarity scale spreads continuously from non polar (the two atoms share the electron density equally -have equal ) to ionic (one atom takes the totality of the electron density leaving nothing to the other atom - the difference in between the two atoms is well over
0.5). Between these two extremes the covalent bond is essentially non polar (difference less than 0.5) or polar (difference over 0.5).
The polarity of a bond is measured by the values of the partial charge ( + and -) created. is the dipole moment measured in Debye (D)
( 1D = 3.336 x 10
-30
Cm)
= q x e x r (C m) q is the partial charge on the atoms e is the charge of an electron in C (coulomb) r is the bond length in m (metre
Example. HF has a dipole moment of 1.83 D and a bond length of 92 pm (92x10
-12 m)
Calculate the value of q q = / (e x r) = 0.41
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TOPIC 3. Molecular Shape and Chemical Bonding 55
3.2. Lewis Structure of Molecules
Lewis electron dot diagrams show the valence electrons arround the atoms of a molecule and identify their nature as lone pair (lp), bonding pair (bp) or unpaired electrons (upe).
The molecular shape is dictated by the position of the electrons around the central atoms.
Building the Lewis structure of molecules in 5 steps: Methanol, CH
3
OH
1. Start from the molecular formula. Determine for each atom the total number of valence electrons. When dealing with an anion, add one electron per charge. For a cation, remove one electron per charge.
2. Place the element of lowest valence around the element of highest valence. Then assign the bonds (single double or triple) to satisfy the valency of the atoms of lowest valency.
3. Calculate the number of valence electrons around the central atom, adding the shared electrons from the peripheral atoms.
4. Assign the shared electrons to bond pairs (bp).
5. Assign the residual valence electrons to lone pairs (lp).
3.2.1. Charged and delocalized species
To draw the Lewis structure of negatively charged species, the general method of construction is followed, adding one extra valence electron per negative charge.
Example. Tetrahydroborate ion: BH
4
-
. eo
H H H H
B H H
H B H H B H
Boron Electron Four hydrogens
H H
Tetrahydroborate
Note: Boron is surrounded by 8 electrons. The octet rule is satisfied.
The Lewis structure of positively charged species is constructed following the general method, removing one valence electron per positive charge.
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Example. Ammonium ion: NH
4
+
.
56
H H
H
H
H
H
H N H H N H
N
H H
Nitrogen Four hydrogens, of which one had its electron removed (proton with positive charge)
Ammonium ion
Note: Nitrogen is surrounded by 8 electrons. The octet rule is satisfied.
When dealing with delocalised bonding: more than one Lewis structure can be drawn. This happen when a bond pair does not have a definite position between two atoms only but over three or more atoms. Each possible Lewis structure is called a resonance structure; the molecular skeleton does not change but the arrangements of the electrons change.
Example.
Nitrate: [NO
3
]
-
. Three equivalent Lewis structures (resonance structures) are possible, because one bond pair is delocalized over four atoms. The nitrogen is positively charged and two negative charges are delocalized between the three oxygens. Each oxygen has a charge of -2/3.
N O O O
Nitrogen Three oxygens One electron
O
N O
O
O
O
N O
O
O
N
O
O
O
O
-2/3
O
N O N O N O or
N O -2/3
O O
O O
-2/3
Nitrate ion: three equivalent Lewis structure are possible
Note.
The phenomenon of resonance is symbolized by double arrow between the resonance structures.
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TOPIC 3. Molecular Shape and Chemical Bonding 57
3.2.2. Exceptions to octet rule
Hypervalence. (more than 8 valence electrons)
A hypervalent molecule contains one or more main group elements formally bearing more than eight electrons in their valence shells.
It can happen when: o The central atom is a main group element with a valence shell n ≥ 3. * o Central atom must use more than 4 valence orbitals, o Peripheral atoms have large
(O, F, Cl, …)
Example of hypervalent molecules: Phosphorus pentafluoride (PF
5
), sulfur hexafluoride
(SF
6
), SF
4
,Chlorine trifluoride (CIF
3
), Triiodite (I
3
-
), SOCl
2
, XeOF
4
, … ClF
5
, XeF
6
(*) Note. The central atom must use more than 4 valence orbitals, so hypervalence is naturally limited to valence shell, n ≥ 3. upe
F [He] 2s
2
2p
5 s p d
Ground state P [Ne] 3s
2
3p
3
3d
0 s p d
P in PF
5
[Ne] 3s
2
3p
3
3d
0
F
F
F
P F
F
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TOPIC 3. Molecular Shape and Chemical Bonding 58
Figure.
Hypervalent phosphorus pentafluoride (PF
5
). The phosphorus undergoes excitation to promote one electron from 3s orbital to one empty 3d orbital, providing five unpaired electrons and allowing the formation of PF
5
molecule.
5s
5p
5d
I [Kr] 4d
10
5s
2
5p
5
5s
5p
5d ground state of I
-
[Kr] 4d
10
5s
2
5p
6
5s
5p
5d
I
-
in I
3
-
[Kr] 4d
10
5s
2
5p
5
5d
1
Lewis structure of triiodide
I I I
Figure. Hypervalent triiodide (I
3
-
). The central atom is ion iodide; its extra electron is promoted to the empty 5d orbital – allowing the formation of two single bounds. The central atom has therefore three lone pairs.
Less than 8 valence electrons. The case of boron trifluoride.
Looking at the atomic orbitals, the central atom of boron trifluoride has only 6 valence electrons, hence does not follow the octet rule.
2s
2p
B [He] 2s
2
2p
1
F
B
2s
2p
F F
B in BF
3
[He] 2s
1
2p
2
Lewis structure of BF
3
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TOPIC 3. Molecular Shape and Chemical Bonding 59
However, three resonance structures can be drawn in which one fluoride, already bonded to the boron by a single bond, provides co-ordinate pibond to the central atom’s empty porbital. In these three resonance structures, the octet rule is respected (but fluorine has to bear a positive charge).
F F F
B B B
F F F F F F
Resonance structures of BF
3
Which Lewis structure best describe BF
3
?
If several Lewis structure are possible; choose the one with the lowest charges
The zwitterion description shows a positive charge on fluorine (not best).
3.2.3. Bridging atoms
Bridging chloride.
If boron is stable with 6 valence electrons as in BF
3
, aluminium although in the same group is not so stable in AlCl
3
.
Al [Ne] 3s
2
3p
1
Cl high temperature, gas phase
Al Cl
Cl lower temperature, gas phase and liquid phase
Cl
Al
Cl
Al
Cl
Cl
Al
Cl
Al
Cl
Cl
Cl
Cl
Cl
Cl
Cl
At lower temperatures dimers form via Cl bridging using Cl lone pairs and the Al empty porbital, forming a co-ordinate covalent bond (2 centers and 2 electrons ).
Note.
In the solid phase, the structure is polymeric
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TOPIC 3. Molecular Shape and Chemical Bonding 60
Bridging hydrogen.
Borane, BH
3
dimerises to form diborane B
2
H
6
. Here the bridge uses bond pair: the electron pair from BH bond is shared to make a second BH bond: 3 center 2 electrons bond.
B [He] 2s
2
2p
1
2
H
H
B H
H
H
B
H
H
B
H
H
H
H
B
H
H
B
H
H
H
H
B
H
H
B
3.2.4. Bond order, bond length and bond energy
Bond order (bo) is the number of chemical bonds between two atoms.
Example :
The bond order in diatomic nitrogen (N ≡N) is 3
The bond order in hydrogen chloride (H-Cl) is 1.
Covalent radius is half the internuclear distance (the bond length) in homonuclear bond and is bond order specific.
Example:
Molecules Bond length Covalent radius
H
H
H
3
C CH
3
154 pm 77 pm
H
2
C CH
2
134 pm 67 pm
HC CH 120 pm 60 pm
Bond length (bl) between two atoms is the addition of their covalent radii.
Example:
Molecules Bond length Covalent radius
Cl Cl
H
3
C CH
3
H
3
C Cl
198 pm
154 pm
178 pm
Cl cr
= 99 pm
C cr
= 77 pm
Cl cr
+ C cr
= 176 pm
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TOPIC 3. Molecular Shape and Chemical Bonding 61
Bond energy (be) is directly correlated with bond order (bo) and inversely correlated with bond length (bl).
Example:
Molecules Bond order Bond length Bond energy (kJ
.
mol
-1
)
H
3
C CH
3
1 154 pm 346
H
2
C CH
2
2 134 pm 598
HC CH 3 120 pm 813
O
S
F
Cl
H
C
N
Br
I
Bond energy is always positive: it is the energy required to break up the bond. When a bond is made, energy is released (exothermic).
Single Bonds
H C
435
416 346
N
391 285 159
O S F Cl Br I
464 359 201 146
366 272
570
432
366
298
485
327
285
213
272
193
190
218
201
201
266
326
255
217
Multiple Bonds
159
247
249
278
242
216
208
193
175 151
C=C
C≡C
598 C=N 615 C=O
813
C≡N
866
C≡O
806
1072
N=N 400 N=O 607 S=O (in SO
2
) 532
N≡N 945 O=O 498 S=O (in SO
3
P≡P 490 S=S 425 Si=O
) 469
642
Table of single and double bonds energy (in kJ
.
mol
-1
).
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TOPIC 3. Molecular Shape and Chemical Bonding
Exercise 1: Calculate the heat of reaction H (ie enthalpie change) for the reaction:
C
2
H
2
+ 2H
2
C
2
H
6
62
H C C H
H H
H H
Bond broken
1
2
C
H
C
H be total
813 813
2 X 435 870
1683 kJ
H H
H
H
C C
H
H bond made energy released
1
C C
4 C H
-346
4 X -416 total
-346
-1664
- 2010 kJ
H = 1683 -2010 = -327 kJ
Exercise 2: Calculate the heat of reaction ( H ) for the formation of sulfur bromide (S
2
Br
2
) from the elements in their standard states and comment on its sign.
S
8
+ 4 Br
2
4 S
2
Br
2
S S
S
S
S S
S
S
4 Br
Bond broken
4 x S S
4 x Br Br be
4 x 266
4 x 193 total
1064
772
1836 kJ
Br 4
Br
S S
Br bond made energy released
8 x Br S 8 x - 217 total
- 1736
- 1736 kJ
H = 1836 -1736 = + 100 kJ (ie reaction is endothermic)
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3.3. Molecular Shape
63
The molecular shape is determined experimentally:
By electron diffraction in gas phase
By X-ray diffraction in crystalline phase
The molecular shape can be predicted from the Lewis structure of the molecule, using the
VSEPR theory (Valence Shell Electron Pair Repulsion). Valence electron pairs form groups and are arranged around the central atom and repel each other. Therefore the shape of the molecule depends primarily on the number of groups.
Example: the molecular shapes of boron trichloride and nitrogen trichloride . lp
Cl Cl
N
N
Cl
B
Cl Cl
B
Cl
Cl
Cl
Cl
Cl
Cl
Cl
BCl
3
: planar shape NCl
3
: non-planar shape
The shape of a molecule is determined by the three-dimensional arrangement of the valence electrons; whether in bonding pairs, in lone pairs or unpaired.
3.3.1. Geometry and shape
The shape of a molecule is the disposition of all the molecule’s atoms . It is determined by the arrangement of the bond pairs (bp) around central atoms.
The geometry of a molecule is the threedimensional arrangement of the molecule’s valence electrons (bond pairs, lone pairs and unpaired electrons). The molecular geometry determines many of the substance’s properties such as chemical reactivity, polarity, colour or magnetism.
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TOPIC 3. Molecular Shape and Chemical Bonding 64
Examples: sulfur hexafluoride (SF
6
), chlorine pentafluoride (ClF
5
), xenon tetrafluoride
(XeF
4
).
SF
6
ClF
5
XeF
4
F F
Shape
F
F F
S
F F
F octahedral
F
F
Cl
F
F square pyramidal
F Xe F
F square planar
F
F F
S
F F
F octahedral
F
F
F
Cl
F
F
F
F
Xe
F
F
Geometry octahedral octahedral
3.3.2. VSEPR
The Valence Shell Electron Pair Repulsion (VSEPR) theory is used to predict the shape of molecules and is based on electron-pair electrostatic repulsion. The model is based on a central atom A and a number of valence electron group (electron-pairs) which can be bonding ( X ) to a substituent or non-bonding (lone pair) ( E ). All valence electrons groups repel each other and dictate the shape of the molecule.
The basic geometry.
For simple molecules, made of a central atom A and where all electron-pairs X are bonding to substituent(s) we describe five basic geometries depending on the number of valence electron group X ( from 2 to 6X).
Example:
2X 3X 4X 5X 6X
X
A
X linear
108 o
X
X
120 o
A
X trigonal planar
X
X
X
A
109 o
X tetrahedral
X
X
X
90 o
A X
120 o
X trigonal bipyramidal
X
X
X
A
X
90 o
X
X octahedral
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TOPIC 3. Molecular Shape and Chemical Bonding 65
Predicting molecular shape with VSEPR theory.
Determine the basic geometry from the Lewis structure.
Lone pairs have greater electrostatic repulsion than bonding pairs, therefore they require more space than bonding pairs. When they are present; refine the geometry and determine the molecular shape.
Examples: with one lone pair AXnE .
1 lp 1 lp 1 lp 1 lp
AX
2
E AX
3
E AX
4
E AX
5
E
X
<120 o
A
X
X
X
A
X
<109 o bent or angular trigonal pyramid
Example: with two lone pairs AXnE
2
.
2 lp
AX
2
E
2
<90 o
X
X
<120 o
X
A
X see-saw
2 lp
AX
3
E
2
X X
A
X
X
X
<90 o square pyramidal
2 lp
AX
4
E
2
X
<<109 o
X
A bent or angular
<90 o
X
X
A
X
T-shape
X
X
A
X
X
90 o square planar
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Example: with three lone pairs AXnE
3
.
3 lp 3 lp
AX
2
E
3
AX
3
E
3
66
X
A
180 o
X
A
X
<90 o
X
X linear
The VSEPR theory in four simple rules:
T-shape
rule 1. The distances between valence electron groups X are maximized
rule 2. Lone pairs (lp) require more space than bonding pairs (bp).
rule 3. Multiple bonds require more space than single bonds
rule 4. The space required by bp decreases with increasing of peripheral atom.
The rules 1 and 2 give the basic shape of the molecule and rules 3 and 4 refine the shape.
Rule 1: The distances between valence electron groups X are maximized
- Two groups (AX
2
): linear geometry.
Note: a multiple bond counts as one group.
180 o
F Be F
2 bp, 2 single bonds AX
2
F Be F
180 o
Linear shape
O C O
4 bp, 2 double bonds AX
2
O C O
180 o
Linear shape
H C N
4 bp, 1 triple bond and 1 single bond
AX
2
H C N Linear shape
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- Three groups (AX
3
): trigonal geometry.
Note: a lone pair E counts as one group.
F
F
B F 3 bp, 3 single bonds AX
3
F
120 o
F
B
F trigonal planar geometry
(trigonal planar shape)
67
O
S O 5 bp, 2 double bonds and 1 lone pair
AX
2
E
- Four groups (AX
4
): tetrahedral geometry.
H
H C H 4 bp, 4 single bonds AX
4
H
O
S
O trigonal planar geometry
(angular shape)
H
H
H
C
109.5
o tetrahedral
H
H N H
H
4 bp, 3 single bonds AX
3
E tetrahedral geometry
H
H
N
H
(trigonal pyramidal shape)
F
O H
F
4 bp, 2 single bonds AX
2
E
2
H O tetrahedral geometry
H
H
F
(bent or angular shape)
-Five groups (AX
5
): trigonal bipyramidal (tbp) geometry.
F
F
P
F
5 bp, 5 single bonds AX
5
F
P
90 o
F trigonal bipyramidal
F
F
F
F
S
F
F
5 bp, 4 single bonds and 1 lone pair
F
AX
4
E F
S
F
F
(see-saw shape) trigonal bipyramidal
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-Six groups (AX
6
): octahedral geometry.
F
F F
P
6 bp, 6 single bonds AX
6
F
F
F
F
F
F
F
F
F
I
F
6 bp, 5 single bonds and 1 lone pair
F
P
F
68
F
F octahedral geometry
AX
5
E F
I
F octahedral geometry
F F
F
(square pyramidal shape)
O
F
Xe
F
F
7 bp, 4 single bonds,
1 double bond, and
1 lone pair
AX
5
E F
F
Xe
F
F octahedral geometry
F
O
(square pyramidal shape)
Rule 2: Lone pairs (E) require more space than bonding pairs (X).
The electrostatic repulsion caused by lone pairs is greater than the repulsion caused by bonding pairs.
This has an effect on the molecule’s angles (the lone pair(s) push the bonding pairs away).
H
H C
H N
O
H H
H H H
H
109.5
o
107 o
105 o
And has an effect on molecular shape (lone pairs have more space in equatorial position than in axial):
F
S
S
X
F
F
F
F
F
F
F
(see-saw shape) (trigonal pyramidal shape) lp axial is not favourable prefered isomer: lp equatorial
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F F equatorial to equatorial angle 102 o
F
F
S
F
Cl
F F axial to axial angle 173 o
F
Xe
F
69 lone pairs in equatorial position
Rule 3: Multiple bonds require more space than single bonds.
A double bond is made of four valence electrons, a triple bond of six; they have a greater electrostatic repulsion than a bond pair .
H H
F O C
B C C
F F H H H H
120 o
116 o
117 o
Rule 4: The space required by bp decreases with increasing of peripheral atom.
The bonding pairs are attracted by the more electronegative atom. The more electronegative the peripheral atom the greater the attraction toward the periphery and less space is required by the bonding pair around the central atom.
O O
H H
F
+
F
electronic density is between C and H
O electronic density attracted by F more space available around the central atom
O
H H F F
116 o
108 o
The rules 1 and 2 give the basic shape of the molecule and rules 3 and 4 refine the shape.
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TOPIC 3. Molecular Shape and Chemical Bonding 70
When the peripheral atoms are different the refined shapes is inevitably a distorted basic shapes.
Example of sulfur hexafluoride (SF
6
) and sulfur monochloride pentafluoride (SClF
5
).
Chloride is less electronegative than fluoride, therefore requires more space .
90 o
F 90 o
Cl >90 o
F
F
S
F
F
F
F
S
F
F
F each angle is 90 o no distorsion
F
Cl is less than F
Cl requires more space than F each angle is 90 o distorsion octahedral octahedral
3.3.3. Further examples of Lewis structure / VSEPR
Chlorine dioxide (ClO
2
+
)
First; write the Lewis structure.
6 e per O plus 7 e on Cl each O needs 8 valence electrons O distorted octahedral
Cl O remove one electron to make positive charge O Cl O
Lewis structure O Cl O
Then apply the VSEPR rules; count the number of groups and apply rules 1 to 4 to determine and refine the molecular shape.
3 groups of valence electron-pairs gives a basic trigonal planar geometry. 2 groups are double bond Cl=O and 1 is a lone pair.
AX
2
E
Cl
O O bent shape trigonal planar geometry
The molecule has a bent or angular shape and a trigonal planar geometry.
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TOPIC 3. Molecular Shape and Chemical Bonding 71
Azide ion (N
3
-
)
First; write the Lewis structure. one lone pair from central N is used to make a co-ordinate covalent bond
5 e per N
N lp
N N each N needs 8 valence electrons
N N N add one electron to make negative charge
Lewis structure N N N
Then apply the VSEPR rules; count the number of groups and apply rules 1 to 4 to determine and refine the molecular shape.
2 groups of valence electron-pairs, Linear geometry. AX
2.
N N N
The molecule has a linear shape and a linear geometry.
Note: the charges on the nitrogens; the central atom forms a co-ordinate covalent bond with one peripheral nitrogen atom and therefore gets a positive charge (for the same reason the peripheral nitrogen gets a negative charge). The electron is added to the second peripheral nitrogen which comes with a negative charge .
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Nitric acid (HNO
3
)
First; write the Lewis structure.
5 valence electrons on N, 1 on H and 6 per O lone pair of N forms a co-ordinate bond with one O
H
O N O
72
Lewis structures:
H
O N O
H
O N
O
O
O
O
Then apply the VSEPR rules; count the number of groups and apply rules 1 to 4 to determine and refine the molecular shape.
3 groups of valence electron-pairs, Trigonal planar geometry. AX
3.
1 group is a single bonding pair (N-O) and 2 groups are partial double bond (the structure is delocalized). These two groups require more space than the first one: therefore the molecule has a distorted trigonal planar shape.
-1/2
O
HO N
O
-1/2
3.3.4. Inert pair effects
The inert pair effect is a tendency of the s valence electrons to remain as a lone pair in compounds of post-transition metals.
Selenium hexachloride (SeCl
6
2-
) s p d
Se [Ar] 4s
2
4p
4 s p d
Ground state Se
2-
[Ar] 4s
2
4p
6 s p d
Excited state Se
2-
[Ar] 4s
2
4p
3
4d
3
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We expect the 8 valence electrons to form seven valence electron groups: six bonding pairs with Cl and one lone pair; a AX
6
E geometry (pentagonal bipyramidal geometry). pentagonal bipyramidal
We do not obverse a AX
6
E geometry but a AX
6
octahedral geometry.
2
Cl
Cl Cl
Se
Cl Cl
Cl octahedral geometry
The valence electrons of the 4s-orbital are inert, they do not affect the shape of the molecule: they are stereochemically inactive.
For post-transition elements, such as Se, the valence s-electrons are relatively deeply buried and are not really valence. So if s-electrons are not part of the valence, only 12 valence electrons are involved in forming six groups and we observe a AX
6 octahedral geometry.
Note.
This is related to the tendency of post-transition elements to form two oxidation states: a normal oxidation state (using all valence s- and p-electrons and a lower oxidation state
(leaving the s-electrons).
This phenomenon is known as the thermodynamic inert pair effect .
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Example:
Tin Sn
Sn (IV)
[Kr] 4d
10
5s
2
5p
2
[Kr] 4d
10
Lead Pb
[Xe] 4f
14
5d
10
6s
2
6p
2
Pb (IV)
[Xe] 4f
14
5d
10
Sn (II)
[Kr] 4d
10
5s
2
Pb (II)
[Xe] 4f
14
5d
10
6s
2
74
Indium In
[Kr] 4d
10
5s
2
5p
1
In (III)
[Kr] 4d
10
In (I)
[Kr] 4d
10
5s
2
Thallium Tl
[Xe] 4f
14
5d
10
6s
2
6p
1
Tl (III)
[Xe] 4f
14
5d
10
Tl (I)
[Xe] 4f
14
5d
10
6s
2
3.3.5. Shape and Molecular Dipoles
The shape of molecule has an effect on the molecular magnetism. When two bonded atoms have different electronegativity the bond is polarized. In molecules where there is more than one polarized bond; the dipoles resulting from polar bonds can either cancel each other
(and the molecule is not polar) or contribute to create a molecular dipole and the molecule is polar, depending on the shape of the molecule.
Bond dipole moment (expressed in Debye) is used to measure the polarity of a chemical bond, where d is the bond length and the value of the partial charge .
= d
It is a vector, parallel to the bond axis.
It is pointing from negative to positive charge.
In polyatomic molecules the total molecular dipole moment is the vector sum of individual bond dipole moment and is depending on the molecular shape.
The measure of the molecular dipole gives information on molecular shape.
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TOPIC 3. Molecular Shape and Chemical Bonding 75
Example: The molecular dipole moment of water H
2
O is 1.85 Debye, of carbon dioxide is 0
Debye.
-
O net dipole = 1.85 D molecule is bent
H
+
H
+
O C
+
O net dipole = 0 D molecule is linear
Example 2 : Two isomers of 1,2-Dichloroethylene are isolated, one (a) is polar with a molecular dipole moment of 1.90 D and the second (b) is non-polar with a molecular dipole moment of 0 D. Determine which is the cis and trans-isomers from the value of the molecular dipole moment.
-
Cl Cl
+ +
C C net dipole = 1.90 D (a) is cis-isomer
H
Cl
-
H
+ +
C C
H
H Cl
net dipole = 0 D (b) is trans-isomer
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TOPIC 3. Molecular Shape and Chemical Bonding 76
3.4. Valence Bond (VB) Theory
The valence bond (VB) theory is a complement to the molecular orbital (MO) theory and was developed to explain chemical bonding.
It focuses on localized bonding (when MO theory describes orbitals covering the whole molecule).
Is not suited to describing delocalization.
Is a good qualitative model
3.4.1. Basis
According to the theory, a covalent bond is formed between two atoms by the overlap of two atomic orbitals of the same phase. Each atomic orbital containing one unpaired electron.
Example.
Two s-orbitals of the same phase form the single bond in H
2
.
AO AO orbital overlap
H (1s) H (1s) H
2
(Sigma s)
AO
AO
Example. One s-orbital from H and one p-orbital from Cl form HCl single bond. orbital overlap
HCl (Sigma sp)
H (1s)
Cl (3p)
Note.
The electrons have opposite spin in the overlapped orbital.
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3.4.2. Hybrid Orbitals
Atomic hybrid orbitals are the result of mixing atomic orbitals (s, p, d, etc) from the same atom and have their own shape and energy. Hybrid orbitals are very useful in explaining molecular geometry and bonding properties.
Example.
Methane CH
4
is a tetrahedral molecule in which each bond is similar and identical in energy.
Looking at the atomic orbital of carbon:
Ground state of C [He] 2s
2
2p
2
Energy
2s
2p 2p
2s
In order to create four bonding pairs, one electron from the 2s-orbital is promoted to an empty 2p orbital. The result is an excited state where all four valence electrons (upe) from C are available for bonding with the upe of the four hydrogen atom .
Excited state of C [He] 2s
1
2p
3 Energy
2s
2p 2p
2s
In the excited state, four upe are available for bonding, but are from two different energy levels. To reflect the fact that all bonds formed are of identical energy one s-orbital is mixed with all three p-orbitals resulting in four hybrid sp
3
orbitals (of identical energy level). sp
3
hybridisation
Energy sp
3 sp
3
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The shape of the hybrid sp
3
atomic orbitals
One s-AO is mixed with three p-AO
Each of the four hybrid sp
3
orbitals retains 25% of the s-AO character and 75 % of the p-AO.
The energy level of the new hybrid orbitals is intermediate, between the s and p AO’s energy levels.
25% 75% s p sp
3
Bonding hybrid orbital:
Four covalent bonds are formed between the four sp
3
hybrid orbitals of carbon and by the overlap of four s-atomic orbitals of the four hydrogen atoms making four sigma ( ) bonds.
C + 4 H
The hybridization reflects the geometry; sp
CH
4
3
hybridisation of the central carbon is consistent with the observed tetrahedral geometry on CH
4
.
For other geometries, other hybridisation schemes are used .
Example:
H
H sp
H
H C C H C C sp
2
C sp
3
H H
H
H
H linear trigonal tetrahedral
F
F
F
P dsp
3
F
F
F
F
S
F
F d
2 sp
3
F F trigonal bipyramidal octahedral
Valence bond theory allows a simple view of chemical bonding once the geometry is known.
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3.4.3. VB construction
To construct the valence bonds, first the Lewis structure and then the geometry (using
VSEPR) must be determined.
Example. BF
3
B [He] 2s
2
2p
1 sp
2
hybridised B
79
F
2p
2p F
F sp
2
2s two p-orbitals are mixed with s-orbitals, making three sp
2
hybrids
Example. H
2
O one p-orbital is not hybridised and is perpendicular to the molecular plane
AO
O [He] 2s
2
2p
4 sp
3
hybridised O
2p
2s lp lp sp
3
H
H three p-orbitals are mixed with s-orbital, making four sp
3
hybrids.
two sp
3 orbitals overlap a H (1s) orbital and two are lone pairs.
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Example. SF
6
AO
S [Ne] 3s
2
3p
4 d
2 sp
3
hybridised S
F
F
F
F
F
F
80
3d 3d
3p d
2 sp
3
3s six d
2 sp
3
orbitals result from mixing two d threep and one s AO.
Example : C
2
H
4
. Representation of multiple-bonds with VB theory
Three sp
2
orbitals result from mixing one s and two p-orbitals. These three new orbitals overlap with two hydrogen (1s) orbital and one (sp
2
) orbital from the second carbon; forming three single bonds (sigma bonds). ground state of C [He] 2s
2
2p
2 hybridisation sp
2 p p
Energy
H H
C C
H H
2p p sp
2
2s
The sp
2
hybridisation leaves one p-orbital unchanged on each carbon. Each p-orbitals are perpendicular to the molecular plan and parallels to each other: they overlap and form a bond . hybridisation sp
2
+ p bond
H H
C C
H H p sp
2
A double bond is made of one sigma ( ) bond and one pi ( bond.
Note. The bond has two opposite phases, one above and one bellow the molecular plan.
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3.5. Molecular Orbital (MO) Theory- a brief introduction
81
The molecular orbital theory is a method for determining molecular structures in which electrons are not assigned to individual bonds between atoms, but are delocalised on the whole molecule.
Atomic orbitals have phase (+ or -).
Bonding molecular orbitals (BMO) are formed when two AO of the same phase combine, they are more stable than the two AO.
Antibonding molecular orbitals (ABMO) are formed when two AO of opposite phase combine, they are less stable than the two AO.
In AMO, the electronic density between the two atoms is zero.
Molecule of H
2
, constructed from two H, called H
A
and H
B
Energy
ABMO
A
-
B
A B
BMO
A
+
B
The bond order of a pi orbital is:
Bo = (number of electron pairs in BMO) – (number of ep in ABMO) = 1
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