chemical bonding part 1
TRANSCRIPT
29/11/2010
1
CHEMICAL BONDING
TOPIC TWO
1mov29/11/2010
CHEMICAL BOND• This is the force of attraction that binds two or more atoms
together.
• Types of chemical bonds1) Ionic
2) Covalent
3) Metallic
4) Coordinate
5) Van-derwaal’s
6) Hydrogen
29/11/2010 mov 2
29/11/2010
2
Bond formation• Why do atoms form chemical bonds ?�so that the system can achieve the lowest possible
potential energy• Example covalent bonding in H2
H • • H
3mov29/11/2010
Ionic bond• This is the electrostatic attraction between oppositely
charged ions. It involves complete transfer of one or more electrons from a highly electropositive to a highly electronegative element.
• The atom that gives out electron becomes positively charged, whereas the one that gains electrons becomes negatively charged
Cation anion
• Formed between elements with the biggest difference in electronegativity
Example: 4mov29/11/2010
29/11/2010
3
Coulomb’s Law• Especially prevalent in compounds formed between group
1A and 2A elements with group 6A and 7A elements.
• Ionic bond formation is influenced by the of electrostatic force of attraction (F), which is directly proportional to product of charge and inversely proportional to distance between the ions
F = k Q1xQ2/r2
k = (2.31 x 10-19 J nm)
� therefore, strong lattices are favoured when the ions have a high charge to size ratio
5mov29/11/2010
Conditions for formation of ionic bond
1) Low ionisation potential of the metal –the lower the ionisation potential the greater the ease to form cation
2) High electron affinity of the non-metal-the higher the electron affinity, the greater the ease to firm anion
3) High lattice energy – the greater the lattice energy the stronger the bond formed
29/11/2010 mov 6
29/11/2010
4
Characteristics of ionic compounds1) They have high melting and boiling points due to strong
electrostatic force of attraction between the ions in the solid
2) They are non-volatile due to high melting and boiling points
3) Electrical conductivity:1) They are poor conductors of electricity in solid state due to strong
electrostatic force of attraction-hence ions are immobile
2) Good conductors of electricity in aqueous state-ions are mobile
3) Good conductors of electricity in molten or fused state
4) Soluble in polar solvents but in soluble in organic solvents
29/11/2010 mov 7
Formation of Ionic Compounds• Electron affinity : is the energy change that occurs when
an electron is accepted by an atom in the gaseous state (kJ/mol)
X(g) + e- X-(g)
• The higher the electron affinity, the greater the tendency of the atom to accept an electron
• Chlorine has the greatest electron affinity of any element:Cl(g) + e- Cl-(g) ∆H = -349 kJ
8mov29/11/2010
29/11/2010
5
Formation of Ionic Compounds
• Lattice energy: t his is the energy required to completely separate one mole of a solid ionic compound into gaseous ions.
• or the energy released when an ionic solid forms from its ions
• It has a negative sign (-)
• Calculation of lattice Energy is based on two metho ds1) Born-lande Equation (Theoretical method)
2) The Born-Haber Cycle (Experimental method)
9mov29/11/2010
Born-lande Equation (Theoretical method)
• It is based on the coulombic interaction within the ionic crystal:1) Electrostatic force of attraction between oppositely charged ions
(cation and anion)
2) Repulsive interaction due to interpenetration of electron charge clouds
• Based on these two interactions, Born-Lande theoretically derived the equation for calculating lattice energy.
• The equation is called Born-Lande Equation and is given by:
• U = -NoAZ+Z-e2 (1-1/n)ro
10mov29/11/2010
29/11/2010
6
Born-lande Equation Cont.• Where,
No = Avogadro’s number = 6.022x1023 atom-1
A = Madlung constantZ+ = Charge on cationZ- = charge on anione- = Electronic chargero = Sum of radii of cation and anionn = Born exponent >1
• Significance of Born-Lande Equation
1) From U α Z+Z-, the higher the charge on cation and anion, the greater the magnitude of lattice energy e.g. U (LiF) < U(CaF2) < U(MgS)
11mov29/11/2010
Born-lande Equation Cont.
2) U α 1/ro
� The smaller the size of ions, the higher the lattice energy e.g.
U (NaF) > U(NaCl) > U(NaBr) > U(NaI)
Exercise:
1) Zinc oxide, ZnO, is a very effective sun screen. How would the lattice energy of ZnO compare
with that of NaCl
2) The precious gem ruby is aluminium oxide, Al2O3, containing traces of Cr3+. The compound
Al2Se3 is used in the fabrication of some semiconductor devices. Which of the two has a
larger lattice energy?
12mov29/11/2010
29/11/2010
7
Born-Haber Cycle• Direct experimental determination of lattice energy is difficult,
hence determined indirectly by a cyclic process called Born-Haber Cycle.
• Born-Haber cycle relates lattice energies of ionic compounds with other thermodynamic data such as sublimation energy (SE), ionization energies (IE), electron affinities (Ea), dissociation energy (DE) and other atomic and molecular properties
• The cycle is based on Hess’s law that total amount of heat evolved or absorbed in a chemical reaction is const ant whether the reaction is carried out in a single ste p or multiple steps
13mov29/11/2010
Born-Haber Cycle
14mov29/11/2010
e.g. ∆Hf = ∆Hs + IP + ½ ∆HD + Ea + U
∆Hs
IP
½ ∆HD
Ea
U
∆Hf
29/11/2010
8
Ionic Bond Formation• requires Coulombic attractive energy (lattice energy) to be
sufficiently large to overcome ionization energy of the element that forms the cation.
• balance between energy input (ionization energies) and stability gained from formation of the solid.
• The main impetus for the formation of an ionic compound rather than a covalent compound results from the strong mutual attraction among the ions
15mov29/11/2010
Ionic Bond Formation cont.
• Bonding in MgCl2 is ionic; Ionization energies ( I.E.)
Mg(g) Mg2+(g) + 2ebonding
∆H = I.E.1 +I.E.2= 753 +1435 = +2180 kJ/mol
• bonds in AlCl3 are polar covalentAl(g) Al3+
(g) + 3e-
∆H = I.E.1 + I.E. 2 + I.E. 3= 580 +1815 + 2740 = +4125 kJ/mol
16mov29/11/2010
29/11/2010
9
Ionic Bond Formation cont.
• Bonds in AlCl3 are polar covalent• energy input (ionization energies) out weighs stability
gained from formation of an ionic solid
Al (g) Al3+(g) + 3e-
∆H = I.E.1 + I.E. 2 + I.E.3= 580 +1815 + 2740 = +4125 kJ/mol
17mov29/11/2010
Application of Born-Haber Cycle
1) To determine lattice energy of unknown crystals
eg. Calculate the lattice energy for formation of NaCl crystal based on the
data below
1) ∆Hf (NaCl) = -99 Kcal/mol2) ∆Hs (Na) = 26 Kcal/g atom3) IP (Na) = 117 Kcal/atom4) ∆HD (Cl2) = 54 Kcal/mol5) Ea (Cl) = -84 Kcal/mol
(Ans = -185 Kcal/mol)
29/11/2010 mov 18
29/11/2010
10
Application of Born-Haber Cycle cont.
2) To determine electron affinity of elements which are difficult to determine by other methods
eg. Determine electron affinity of iodine given the following thermodynamic data:
1) ∆Hf (Nal) = -68.8 Kcal/mol2) ∆Hs (Na) = 25.9 Kcal/atom3) IP (Na) = 118.4 Kcal/atom4) ∆HD (l2) = 25.5 Kcal/mol5) U (Nal) = -165.4 Kcal/mol
(Ans = -73.2 Kcal/atom)
29/11/2010 mov 19
Application of Born-Haber Cycle cont.
• Explain why group IIA oxides and chlorides are stable in higher oxidation state
despite the fact that the formation of Mg2+, Ca2+ etc require more energy than
Mg+ and Ca+
Ans. the higher lattice energy involved compensates for the energy required for formation of Mg2+, Ca2+ making MgCl2, CaCl2, Al2O3 more stable.
29/11/2010 mov 20
29/11/2010
11
Valence Bond Theory
• It is based on linear combination of atomic orbitals.• Main features:
1) A covalent bond is formed by the overlap of half filled atomic orbitals of different atoms
2) Overlapping atomic orbitals must have electrons with opposite spins
3) The bonded electron pair is localized between the two linked atoms
4) The stability of covalent bond is due to exchange of valence electrons between participating atoms, which lowers the potential energy of the bonded atoms
5) Each atom of the covalent compound tends to acquire a noble gas configuration by sharing electrons
• s 21mov29/11/2010
22
Bonding Types• Two types of bonds result
from orbital overlap:
• sigma σ bonds • from head-on overlap• lie along the bond axis • account for the first bond • Can freely rotate around
bond
29/11/2010
29/11/2010
12
• pi π bonds • from lateral overlap by
adjacent p or d orbitals• pi bonds are
perpendicular to bond axis
• account for the second and third bonds in a multiple bond
• Cannot undergo rotation around bond
Bonding Types
23mov29/11/2010
• One pair of electrons can occupy this overlapping area
• Electron density is maximizes in overlapped region
• Example : H2 bonds form because atomic valence orbitals overlap
• Each hydrogen contributed 1s orbital
1s 1s
24mov29/11/2010
Valence Bond Theory cont.
29/11/2010
13
25
Valence Bond Theory
• HF involves overlaps between the s orbital of H and the 2p orbital of F
1s 2s 2p
mov29/11/2010
26
VB Theory And H 2S
• Assume that the unpaired e- in S and H are free to form a paired bond
• We may assume that the H-S bond forms between an s and a p orbital
mov29/11/2010
29/11/2010
14
27
According to Valence Bond Theory:Which orbitals overlap in the formation of NH3?• Ground state of nitrogen
2s ↑↓ 2p _↑ ↑ _↑__
mov
VB Theory and NH 3 cont.
28
Difficulties With VB Theory• Most experimental bond angles do not support those
predicted by mere atomic orbital overlap
• For example: C 1s22s22p2 and H 1s1
• Experimental bond angles in methane are 109.5° and all are the same
• p orbitals are 90° apart, and not all valence e- in C are in the p orbitals
• How can multiple bonds form?
mov29/11/2010
29/11/2010
15
29
Hybridization• The mixing of atomic orbitals to allow formation of bonds
that have realistic bond angles
• The new shapes that result are called “hybrid orbitals”
• The number of hybrid orbitals required = the number of bonding domains + the number of non-bonding domains on the atom
mov29/11/2010
Hybrid between s and p OrbitalsTwo sp Orbitals in Linear Arrangement Formed by
Hybridization of a single s and a single p Orbital
30mov29/11/2010
29/11/2010
16
31
Hybrid orbitals• Naming of the hybrid orbital is based on the combination
of the orbitals used to form the new hybrid
• The name shows what type of atomic orbitals, and how many of each were used in formation of hybrid orbitals e.g. sp, sp2, sp3 etc.
• One atomic orbital is used for every hybrid formed (orbitals are conserved)
mov29/11/2010
32
Hybrids From s & p Atomic Orbitals explain VSEPR Geometry
Hybrid Atomic Orbitals Used
Electron Geometry
sp3 s + px + py + pz Tetrahedral, bond angles 109.5˚
mov
29/11/2010
17
Hybrids From s & p Atomic Orbitals explain VSEPR Geometry
Hybrid Atomic Orbitals Used
Electron Geometry
sp2 s + px + py Trigonal planar, bond angles 120 ˚
33mov29/11/2010
Hybrids From s & p Atomic Orbitals explain VSEPR Geometry
Hybrid Atomic Orbitals Used
Electron Geometry
sp s + px Linear,
bond angles 180 ˚
34mov29/11/2010
29/11/2010
18
Hybrid Orbitals in BeH 2
• Ground state of Be2s↑↓ 2p _ ___ [He]2s2
No half filled orbitals available for bonding
• For hybrid sp hybrid orbitals sp ↑ ↑ 2p ___
• Now bond can form between 1s orbital of hydrogen and sp hybrid orbital or berylllium
sp ↑↓ ↑↓ 2p ___
35mov29/11/2010
36mov29/11/2010
29/11/2010
19
37
Consider the CH 4 molecule
• Ground state for carbon 2s ↑ ↓ 2p ↑ _ ↑ ___ [He]2s2 2p2
• Form sp3 hybrid orbitals↑ ↑_ ↑ ↑ .
• Each sp3 hybrid can now overlap with 1s orbital of hydrogen
↑↓ ↑↓_ ↓↑ ↑↓ .
mov29/11/2010
38
Bonding in CH 4
• The 4 hybrid orbitals are evenly distributed around the C
• The H s-orbitals overlap the sp3 hybrid orbitals to form the bonds.
mov29/11/2010
29/11/2010
20
39
Bonding in NH 3
• The 4 hybrid orbitals are evenly distributed around the N
• The H s-orbitals overlap the sp3 hybrid orbitals to form three bonds bonds
• The remaining lone pair occupies the last hybrid orbital
mov29/11/2010
Hybridization for form sp 2 orbitals
40mov29/11/2010
29/11/2010
21
Ethene and Double BondsSigma Bonds σ : Direct overlap of
orbitals between the two nuclei
Direct overlap of atomic orbitals is not
affected by rotation around that bond
C--C
41mov29/11/2010
C=C double bond consists of one sigma
σ σ σ σ bond and one pi ππππ bond
42mov29/11/2010
29/11/2010
22
Orbitals used for bonding in Ethene
Sigma bond: sp2 overlaps sp2
Pi bond: unhybridized p orbital overlaps
unhybridized p orbital
43mov29/11/2010
Formation of sp Hybrid Orbitals
44mov29/11/2010
29/11/2010
23
Hybrid Orbitals is CO 2 molecule
First bond is sigma bond
Second bond is pi bond
45mov29/11/2010
CO2 Molecule
• Unhybridized p orbitals are used to form pi bonds between
carbon and oxygen
• sp hybrid orbitals of carbon overlap sp2 hybrid orbitals from
oxygen to form sigma bonds
46mov29/11/2010
29/11/2010
24
• Triple bond in N2 molecule consists of one sigma bond and 2 pi bonds
• Sigma bond stems from sp hybrid overlap• Pi bonds come from unhybridized p orbital overlap
47mov29/11/2010
48
Expanded Octet Hybridization
• Can be predicted from the geometry as well
• In these situations, d orbitals are be needed to provide room for the extra electrons
• One d orbital is added for each pair of electrons in excess of the standard octet
mov29/11/2010
29/11/2010
25
Expanded Octet hybridization
• dsp3 hybridization gives rise to trigonal bipyramid geometry of PCl5
49mov29/11/2010
d2sp3 hybridization gives rise to octahedral geometry
50mov29/11/2010
29/11/2010
26
Consider SF 6• Ground state for sulfur
3s ↑ ↓ 3p ↑ ↓ ↑ ↑_ 3d _ _ _ _ _
• Six hybrid orbitals neededsp3d2 ↑ ↑ ↑ ↑ ↑ ↑ . 3d _ _ _
• Each sp3d2 hybrid can now overlap with 2p orbital of fluorine↑↓ ↑↓_ ↓↑ ↑↓ . ↓↑ ↓↑
51mov29/11/2010
Bonding is XeF 4
• Placing lone pairs at axial positions lets them be as far as possible from one another
• Square planar geometery
52mov29/11/2010
29/11/2010
27
• Hybrid orbital can also hold nonbonding electrons
• Usually results in polar molecules
53mov29/11/2010
Consider the SF 4 molelcule
• Four bonding + 1 nonbonding pairs around sulphur• Five hybrid orbitals needed
sp3d ↑↓ ↑ ↑ ↑ ↑ . 3d _ _ _ _
• Four half filled orbitals available to overlap with 2p orbital of fluorine
sp3d ↑↓ ↓↑ ↓ ↑ ↓↑ ↓ ↑ . 3d _ _ _ _
Lewis Structure
S F
F
F
F
:
54mov29/11/2010
29/11/2010
28
Geometry of SF 4
• sp3d requires trigonal bipyramid geometry
• Nonbonding pair goes on equatorial position
• Distorted tetrahedron geometry
F
F
F
F
:
S
55mov29/11/2010
Bonding in Ethene C 2H4
• Carbon forms sp2 hybrid orbitals, and one unhybridized p orbital
56mov29/11/2010
29/11/2010
29
57mov29/11/2010
58
Sigma and Pi Bonding O
H HC
mov29/11/2010
29/11/2010
30
59
H−C≡C −H
• Each C has a triple bond and a single bond
• Requires 2 hybrid orbitals, sp
• unhybridized p orbitals used to form the pi bond
mov29/11/2010
Types of bonds in Acetylene
60mov29/11/2010
29/11/2010
31
Summary of Multiple Bonds
• Molecular skeleton held together by σ bonds. First bond between two atoms always σ.
• Hybrid orbitals are used to form σ bonds, and to hold nonbonding electrons• Number of hybrid orbitals needed = # atoms bonded +
# of nonbonding pairs
• π bonds are formed using non-hybridized p or d orbitals
• Double bond is one σ and one π bond
• Triple bond consists of one σ and two π bonds61mov29/11/2010
62
Molecular Orbital Theory• Modification of VB theory, considers that the orbitals
may exhibit interference.
• Waves may interfere constructively or destructively
• Bonding orbitals stabilize, antibonding destabilize.
mov29/11/2010
29/11/2010
32
Molecular Orbital Theory concept1) Atomic orbitals having same energy and symmetry combine
to form molecular orbitals by linear combination of atomic orbitals (LCAO)
• If ΨA and ΨB are the wave functions of atoms A and B, then by LCAO, Ψ = ΨA ± ΨB
2) Molecular orbital formed by addition of two atomic orbital wave functions is called Bonding Molecular Orbital i.e.
Ψb = ΨA + ΨBProbability of finding electron in BMO is given by
Ψb2 = ΨA2 + ΨB2 + 2 ΨAΨB> ΨA2 + ΨB2
� The energy of bonding molecular orbital is less than the energy of individual atomic orbitals
29/11/2010 mov 63
Molecular Orbital Theory concept
3) Molecular orbital formed by subtraction of two atomic orbital wave functions is called Anti-Bonding Molecular Orbital i.e.
Ψa = ΨA + ΨB
Probability of finding electron in ABMO is given byΨa2 = ΨA2 + ΨB2 - 2 ΨAΨB< ΨA2 + ΨB2
� The energy of anti-bonding molecular orbital is higher than the energy of individual atomic orbitals
29/11/2010 mov 64
29/11/2010
33
Molecular Orbital Theory concept4) Molecular Orbitals that do not participate in bonding are
called Non Bonding molecular orbitals. They occur at same energy as individual atomic orbitals
5) Atomic orbitals are monocentric, whereas molecular orbitals are polycentric
6) The Stability of the bond is expressed in terms of bond order- defined as one half the difference between the number of electrons in bonding molecular orbitals (Nb) and atibonding molecular orbitals (Na) i.e.
Bond Order = (Nb – Na)/229/11/2010 mov 65
Molecular Orbital Theory concept
• Bond order my be zero, fraction or positive but not negative
• The greater the bond order, the stronger the bond formed, hence the greater the stability of the molecule
� BO α Bond Strength α Stability α -1/bond length α1/reactivity
29/11/2010 mov 66
29/11/2010
34
Molecular orbital Energy Diagram• This is a potential energy diagram showing the atomic
orbitals combining and molecular orbitals formed.
• Electrons fill molecular orbitals in order of increasing energy levels
• Sequence of molecular orbital energies is based on light (H to N) and heavy (≥ O )molecules
• Light molecules : σ1s< σ*1s< σ2s< σ*2s< π2px = π2py<σ2pz< π*2px = π*2py<σ*2pz
• Heavy Molecules: σ1s< σ*1s< σ2s< σ*2s< σ2pz< π2px = π2py< π*2px = π*2py<σ*2pz
29/11/2010 mov 67
Differences between bonding and anti-bonding molecular orbitals
BMO ABMO
1. BMO is formed by the addition of
overlapping atomic orbitals i.e.
Ψmo= ΨA+ΨB
1. ABMO is formed by the subtraction of
overlapping atomic orbitals
Ψ*mo= ΨA-ΨB
2. It has greater electron density in the
region between the two nuclei of bonded
atoms
2. It has lower electron density in the
region between the two nuclei of the
atoms
3. Electrons in BMO contribute to
attraction between the two atoms
3. Electrons in ABMO contribute to
repulsion between the two atoms
4. It possesses lower energy than
associated atomic orbitals
4. It possesses higher energy than
associated atomic orbitals
29/11/2010 mov 68
29/11/2010
35
MO diagram for H 2
• Show atomic energy level diagram for each atom
• Show molecular orbitals (bonding and antibonding*)
• 1 MO for each Atomic orbital.
• Show electron occupancy of the orbitals.
69mov29/11/2010
70
Filling MO diagrams
1. Electrons fill the lowest-energy orbitals that are available.
2. No more than two electrons, with spins paired, can occupy any orbital.
3. Electrons spread out as much as possible, with spins unpaired, over orbitals that have the same energy.
bondelectrons/ 2
)e gantibondin #()e bonding (#Order Bond
-- −=
mov29/11/2010
29/11/2010
36
H2 vs He2
12
02Order Bond H2 =−=
02
22Order Bond He2 =−=
71mov29/11/2010
Molecular Orbitals Using p Orbitals
Two boron atoms have one set of p orbitals that
can directly overlap to for sigma bond.
Two parallel p orbitals can form pi bonds72mov29/11/2010
29/11/2010
37
• Two px orbitals overlap for form sigma bonding and antibonding molecular orbitals
73mov29/11/2010
• Two p orbitals overlap to form pi bonding and anti-bonding orbitals
• Can happen both to pypair and to pz pair, resulting in two bonding and two anti-bonding orbitals
74mov29/11/2010
29/11/2010
38
Molecular Orbital Diagram for B 2
12
2-4
2gantibondin Total - Bonding Total
Order Bond
==
=
B2
should be a stable
molecule
75mov29/11/2010
76
Diatomic MO diagrams differ by group
A) I - V B) VI-VIIIASecond period used s and
p orbitals
mov29/11/2010
29/11/2010
39
Molecular Orbitals Explains Paramagnetic O 2
• Paramagnetic; weakly attracted to magnetic field• Usually a result of unpaired electron
• Simple Lewis structure has no unpaired electrons
• However, MO treatment shows two unpaired electrons in π* orbitals
77mov29/11/2010
Molecular Orbital Diagrams for B 2 to F2
78mov29/11/2010
29/11/2010
40
MO Diagram for Group I-V
2s2s
σ2s
σ∗2s
2p 2p
π2p
π∗2p
σ∗2p
σ2p
Draw the MO
diagram for N2
79mov29/11/2010
MO Diagram for Group VI-VIII
2s2s
σ2s
σ∗2s
2p 2pπ2p
π∗2p
σ∗2p
σ2p
Draw the MO
diagram for O2
80mov29/11/2010
29/11/2010
41
2s
2s
2p 2p
Draw the MO
diagram for NO
nitrogen oxygen
MO also works for heteronuclear diatomics
81mov29/11/2010
2s2s
σ2s
σ∗2s
2p
2p
π2p
π∗2p
σ∗2p
σ2p
nitrogen oxygen
MO’s and Free Radicals NOFree radicals are molecules with an unpaired electron
82mov29/11/2010
29/11/2010
42
Resonance Structures for O 3 and NO3
83mov29/11/2010
Figure (a) benzene molecule (b) two resonance structures for benzene molecule
84mov29/11/2010
29/11/2010
43
85
Delocalized Electrons• Lewis structures use resonance to explain that the
actual molecule appears to have several equivalent bonds, rather than different possible structures
• MO theory shows the electrons being delocalized in the structure
mov29/11/2010
Calculation of Hybridization in a molecule• Steps:1) Calculate the total number of valence electrons2) Calculate the number of duplet or octate
Duplet = Total valence electrons/2Octet = Total valence electrons/8
3) Calculate the number of lone pairs of electronsLone pairs = (Total valence electrons – 2xNumber of
duplets)/2Lone pairs = (Total valence electrons – 8xNumber of
duplets)/84) Calculate the number of orbitals used = No. Of duplets +
No. Of lone pairs or electrons5) If there are no lone pairs of electrons, the structure and
shape are ideal, else distortion in shape occurs29/11/2010 mov 86
29/11/2010
44
Calculation of Hybridization in a molecule
1: Calculate the type of hybridisation, geometry an d shapes of the following molecules: H 2O, SO4
2- ionsH2Oi) Total No. Electrons = 1x2+6 = 8
ii) Required number of orbitals = 2
iii) Required electrons for duplets = 2x2 = 4
iv) No. Of lone pair electrons = (i)-(iii)/2 = 8-4/2 = 2
v) Number of orbitals = (ii) + (iv) = 2+2 = 4
• Thus, oxygen exhibits SP3 hybridisation and its structure is tetrahedral with distortion bond angle due to presence of lone pairs
29/11/2010 mov 87
Calculation of Hybridization in a molecule
SO42-
i) Total No. Electrons = 6+6x4+2 = 32
ii) Required number of orbitals = 4iii) Required electrons for octet = 4x8 = 32iv) No. Of lone pair electrons = ((i)-(iii))/2 = (32-32)/2 = 0v) Number of orbitals = (ii) + (iv) = 4+0 = 4
• Thus, SO42- involves SP3 hybridisation and its structure is a regular tetrahedron
29/11/2010 mov 88
29/11/2010
45
Calculating the percentage s- and p-character
• It is based on the equation: Cosθ = s/s-1 = p-1/s, where θ = bond angle
Example: Calculate the percentage s-orbital and p-orbital character of a hybrid
orbital if the bond angle between the hybrid orbital is 105o
Cosθ = s/s-1 = p-1/s, where θ is the bond angles/s-1 = cos 105o = cos (90 + 15) = -sin 15o = -0.2588s = 0.2588/1.2588 = 0.2056
similarly, p-1/p = cos 105o = -0.2588� p = 0.7944
29/11/2010 mov 89
Calculating the number σ and πbonds
• Steps:1) The number of π electrons in a noncyclic molecule which
does not contain hydrogen atom is given by 6n+2-v, where n = total number of atoms of molecules and v is the sum of the valence electrons in the molecule
2) The number of π electrons in a noncyclic molecule containing hydrogen atoms is given by 6n-6q+2-v, where q = the number of hydrogen atoms in the molecule
3) If the molecule is of Ax n type, and none of the them is a hydrogen atom, then,
Number of σ bonds = Total No. Valence electrons/84) If the molecule contains a hydrogen atom, then,
Number of σ bonds = total no. Valence electrons/229/11/2010 mov 90
29/11/2010
46
Calculating the number σ and π bonds• Example: Calculate the number of π and σ bonds in the following molecules:
CO2, CO32-, H2O, NH3, C2H2, C2H4
CO2No. Of π electrons = 6x3+2-(4+2x6) = 4
No. π bonds = 4/2 = 2
No. σ bonds = (4+2x6)/8 = 2
C2H4
No. Of π electrons = 6x6-6x4+2-(2x4+4) = 2No. π bonds = 2/2 = 1No. σ bonds = (2x4+4x1)/2 = 6
29/11/2010 mov 91
Calculating the number σ and π bonds
• Calculate the number of π and σ bonds in the following molecules: CO2, CO32-,
H2O, NH3, C2H2, C2H4
H2O
No. Of π electrons = 6x3+6x2+2-8 = 0No. π bonds = 0No. σ bonds = (2+6)/2 = 4
CO32-
No. Of π electrons = 6x4+2-(4+3x6+2) = 2No. π bonds = 2/2 = 1No. σ bonds = (4+3x6+2)/8 = 3
29/11/2010 mov 92