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BRINGiiT on – Study Pack By ASKIITIANS.COM – powered by IITians
SUBJECT – CHEMISTRY
TOPIC – CHEMICAL BONDING
COURSE CODE – AISM-09/C/CMB
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Content :- Chemical Bonding
Introduction………………………………………………………………………………3 Electrovalency……………………………………………………………………………4 Energy change during the formation of ionic bond……………………6 Born Haber Cycle……………………………………………………………………….8 Covalency…………………………………………………………………………………12 Co-ordinate Covalency…………………………………………………………….15 Hybridization……………………………………………………………………………19 Maximum Covalency……………………………………………………………….34 VSEPR theory…………………………………………………………………………..35 Resonance……………………………………………………………………………….41 FAJAN’s Rule……………………………………………………………………………45 Dipole Moment……………………………………………………………………….49 Bond Characteristics……………………………………………………………….56 Hydrogen Bonding………………………………………………………………….59 Intermolecular Forces…………………………………………………………….62 Molecular Orbital Theory………………………………………………………..65 Miscellaneous Exercises………………………………………………………….72 Solved Problems…………………………………………………………………….77
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INTRODUCTION
A molecule is formed if it is more stable and has lower energy than the
individual atoms. Normally only electrons in the outermost shell of an atom
are involved in bond formation and in this process each atom attains a stable
electronic configuration of inert gas. Atoms may attain stable electronic
configuration in three different ways by loosing or gaining electrons by
sharing electrons. The attractive forces which hold various constituents
(atoms, ions etc) together in different chemical species are called chemical
bonds. Elements may be divided into three classes.
Electropositive elements, whose atoms give up one or more electrons
easily, they have low ionization potentials.
Electronegative elements, which can gain electrons. They have higher
value of electronegativity.
Elements which have little tendency to loose or gain electrons.
Three different types of bond may be formed depending on electropositive or
electronegative character of atoms involved.
Electropositive element + Electronegative element = Ionic bond
(electrovalent bond)
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Electronegative element + Electronegative element = Covalent bond
or less electro positive + Electronegative element = Covalent bond
Electropositive + Electropositive element = Metallic bond.
ELECTROVALENCY
This type of valency involves transfer of electrons from one atom to another,
whereby each atom may attain octet in their outermost shell. The resulting
ions that are formed by gain or loss of electrons are held together by
electrostatic force of attraction due to opposite nature of their charges. The
reaction between potassium and chlorine to form potassium chloride is an
example of this type of valency.
K Clxx
xx
xx
x K Clxx
xx
xxx or K Cl
Here potassium has one electron excess of it‘s octet and chlorine has one
deficit of octet. So potassium donates it‘s electron to chlorine forming an
ionic bond.
Ca++
O2–
(Ionic bond)
xx
CaO
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Here the oxygen accepts two electrons from calcium atom. It may be noted
that ionic bond is not a true bond as there is no proper overlap of orbitals.
Criteria for Ionic Bond:
One of the species must have electrons in excess of octet while the other
should be deficit of octet. Does this mean that all substance having surplus
electron and species having deficient electron would form ionic bond? The
answer is obviously no. Now you should ask why? The reasoning is that in an
ionic bond one of the species is cation and the other is anion. To form a
cation from a neutral atom energy must be supplied to remove the electron
and that energy is called ionization energy. Now it is obvious that lower the
ionization energy of the element the easier it is to remove the electron. To
form the anion, an electron adds up to a neutral atom and in this process
energy is released. This process is called electron affinity.
So for an ionic bond one of the species must have low ionization energy and
the other should have high electron affinity. Low ionization energy is mainly
exhibited by the alkali and alkaline earth metals and high electron affinity by
the halogen and chalcogens. Therefore this group of elements are
predominant in the field of ionic bonding.
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Energy Change During the Formation of Ionic Bond
The formation of ionic bond can be consider to proceed in three steps
(a) Formation of gaseous cations
+ -A(g)+I.E. A (g)+ e
The energy required for this step is called ionization energy (I.E)
(b) Formation of gaseous anions
- -X(g)+ e X (g)+E.A
The energy released from this step is called electron affinity (E.A.)
(c) Packing of ions of opposite charges to form ionic solid
+ -A (g)+ X (g) AX(s)+ energy
The energy released in this step is called lattice energy.
Now for stable ionic bonding the total energy released should be more than
the energy required.
From the above discussion we can develop the factors which favour
formation of ionic bond and also determine its strength. These factors have
been discussed below:
(a) Ionization energy: In the formation of ionic bond a metal atom loses
electron to form cation. This process required energy equal to the
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ionization energy. Lesser the value of ionization energy, greater is the
tendency of the atom to form cation. For example, alkali metals form
cations quite easily because of the low values of ionization energies.
(b) Electron affinity: Electron affinity is the energy released when
gaseous atom accepts electron to form a negative ion. Thus, the value
of electron affinity gives the tendency of an atom to form anion. Now
greater the value of electron affinity more is the tendency of an atom
to form anion. For example, halogens having highest electron affinities
within their respective periods to form ionic compounds with metals
very easily.
(c) Lattice energy: Once the gaseous ions are formed, the ions of
opposite charges come close together and pack up three dimensionally
in a definite geometric pattern to form ionic crystal.
Since the packing of ions of opposite charges takes place as a result of
attractive force between them, the process is accompanied with the
release of energy referred to as lattice energy. Lattice energy may be
defined as the amount of energy released when one mole of ionic solid
is formed by the close packing of gaseous ion.
In short, the conditions for the stable ionic bonding are:
(a) I.E. of cation forming atom should be low:
(b) E.A. of anion – forming atom should be high;
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(c) Lattice energy should be high.
Born Haber Cycle
Determination of lattice energy
The direct calculation of lattice enthalpy is quite difficult because the
required data is often not available. Therefore lattice enthalpy is determined
indirectly by the use of the Born – Haber cycle. The cycle uses ionization
enthalpies, electron gain enthalpies and other data for the calculation of
lattice enthalpies. The procedure is based on the Hess‘s law, which states
that the enthalpy of a reaction is the same, whether it takes place in a single
step or in more than one step. In order to understand it let us consider the
energy changes during the formation of sodium chloride from metallic
sodium and chlorine gas. The net energy change during the process is
represented by Hf.
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Example 1:
Calculate the lattice enthalpy of2MgBr . Given that
Enthalpy of formation of 2MgBr = -524 kJ -1mol
Some of first & second ionization enthalpy (IE1 + IE2 ) = 148 kJ mol 1
Sublimation energy of Mg = +2187 kJ -1mol
Vaporization energy of 2Br (I) = +31kJ -1mol
Dissociation energy of 2Br (g) = +193kJ -1mol
Electron gain enthalpy of Br(g) = -331 kJ -1mol
Solution:
f vapH S I.E H D 2 E.A. U
2 f
1Na(s) Cl (g) NaCl(s) ;Energychange ( H )
2
S1
D2
Na(g) Cl(g)
IEEA
Na (g) Cl (g)
e
2
S Sublimation of sodium
D Dissociation energyof Cl
IE IE of sodium(IE)
EA EA of chlorine(EA)
U Lattice energy of NaCl
U
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Or f vapU H S I.E H D 2 E.A.
Or U = 524 [2187 148 + 31 + 193 + 2 ( 331)]
= 524 1897= 12421kJ mol
Characteristics of ionic compounds:
The following are some of the general properties shown by these compounds
(i) Crystalline nature: These compounds are usually crystalline in nature
with constituent units as ions. Force of attraction between the ions is
non-directional and extends in all directions. Each ion is surrounded by a
number of oppositely charged ions and this number is called co-
ordination number. Hence they form three dimensional solid aggregates.
Since electrostatic forces of attraction act in all directions, therefore, the
ionic compounds do not posses directional characteristic and hence do
not show stereoisomerism.
(ii) Due to strong electrostatic attraction between these ions, the ionic
compounds have high melting and boiling points.
(iii) In solid state the ions are strongly attracted and hence are not free to
move. Therefore, in solid state, ionic compounds do not conduct
electricity. However, in fused state or in aqueous solution, the ions are
free to move and hence conduct electricity.
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(iv) Solubility: Ionic compounds are fairly soluble in polar solvents and
insoluble in non-polar solvents. This is because the polar solvents have
high values of dielectric constant which defined as the capacity of the
solvent to weaken the force of attraction between the electrical charges
immersed in that solvent. This is why water, having high value of
dielectric constant, is one of the best solvents.
The solubility in polar solvents like water can also be explained by the dipole
nature of water where the oxygen of water is the negative and hydrogen being
positive, water molecules pull the ions of the ionic compound from the crystal
lattice. These ions are then surrounded by water dipoles with the oppositely
charged ends directed towards them. These solvated ions lead an independent
existence and are thus dissolved in water. The electrovalent compound
dissolves in the solvent if the value of the salvation energy is higher than the
lattice energy of that compounds.
AB Lattice energy A B
These ions are surrounded by solvent molecules. This process is exothermic
and is called solvation.
x
A x solv. A solv. energy
y
B y solv. B solv. energy
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The value of solvation energy depend on the relative size of the ions. Smaller
the ions more is the solvation. The non-polar solvents do not solvate ions and
thus do not release energy due to which they do not dissolve ionic compounds.
(v) Ionic reactions: Ionic compound furnish ions in solutions. Chemical
reactions are due to the presence of these ions. For example
2
2 4 4Na SO 2Na SO
2
2BaCl Ba 2Cl
COVALENCY
This type of valency involves sharing of electrons between the concerned
atoms to attain the octet configuration with the sharing pair being
contributed by both species equally. The atoms are then held by this
common pair of electrons acting as a bond, known as covalent bond. If two
atoms share more than one pair then multiple bonds are formed. Some
examples of covalent bonds are
Sigma and Pi Bonding:
Cl Cl – Cl N N
Cl
N x x x
x x x
N x x
x x
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When two hydrogen atoms form a bond, their atomic orbitals overlap to
produce a greater density of electron cloud along the line connecting the two
nuclei. In the simplified representations of the formation of H2O and NH3
molecules, the O—H and N—H bonds are also formed in a similar manner,
the bonding electron cloud having its maximum density on the lines
connecting the two nuclei. Such bonds are called sigma bonds ( -bond).
A covalent bond established between two atoms having the maximum
density of the electron cloud along the line connecting the centre of the
bonded atoms is called a -bond. A -bond is thus said to possess a
cylindrical symmetry along the internuclear axis.
Let us now consider the combination of two nitrogen atoms. Of the three
singly occupied p-orbitals in each, only one p-orbital from each nitrogen
(say, the px may undergo ―head –on‖ overlap to form a -bond. The other
two p-orbitals on each can no longer enter into a direct overlap. But each
p-orbital may undergo lateral overlap with the corresponding p-orbital on the
neighbour atom. Thus we have two additional overlaps, one by the two py
orbitals, and the other by the two pz orbitals. These overlaps are different
from the type of overlap in a -bond. For each set of p-orbitals, the overlap
results in accumulation of charge cloud on two sides of the internuclear axis.
The bonding electron cloud does no more posses an axial symmetry as with
the -bond; instead, it possess a plane of symmetry. For the overlap of the
pz atomic orbital, the xy plane provides this plane of symmetry; for the
overlap of the py atomic orbitals, the zx plane serves the purpose. Bonds
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arising out of such orientation of the bonding electron cloud are designated
as -bonds. The bond formed by lateral overlap of two atomic orbitals having
maximum overlapping on both sides of the line connecting the centres of the
atoms is called a -bond. A -bond possess a plane of symmetry, often
referred to as the nodal plane.
-Bond
(a) s-s
overlapping
s s
(b) s-p
overlapping
+s p
(c) p-p
overlapping +
p p
- Bond:
This type of bond is formed by the sidewise or lateral overlapping of two half
filled atomic orbitals.
p p
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CO-ORDINATE COVALENCY
A covalent bond results from the sharing of pair of electrons between two
atoms where each atom contributes one electron to the bond. It is also
possible to have an electron pair bond where both electrons originate from
one atom and none from the other. Such bonds are called coordinate bond
or dative bonds. Since in coordinate bonds two electrons are shared by two
atoms, they differ from normal covalent-bond only in the way they are
formed and once formed they are identical to normal covalent –bond.
It is represented as [ ]
Atom/ion/molecule donating electron pair is called Donor or Lewis base.
Atom / ion / molecule accepting electron pair is called Acceptor or Lewis
acid [ ] points donor to acceptor
NH4+, NH3 has three (N – H) bond & one lone pair on N – atom. In NH4
+
formation this lone pair is donated to H+ (having no electron)
NH3 + H+ NH4+
Lewis base Lewis acid
NH
H
H
H+
NH
H
H
or H
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Properties of the coordinate compounds are intermediates of ionic and
covalent compounds.
Comparison of ionic, covalent & coordinate compounds
Property Ionic Covalent Coordinate
1. binding force Between ions strong
(coulombic)
Between molecules
smaller (Vander
Waal‘s)
in between
2. mp/bp High less than ionic in between
3. condition conductor of
electricity in fused
state & in aqueous
solution
bad conductor Greater than
covalent
4. solubility in
polar solvent
(H2O)
High Less in between
5. Solubility in non Low High in between
O S
O
O
SO3
P
ClCl
ClCl
Cl ClPCl 6
Sb
FF
FF
F FSbF6
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polar solvent
(ether)
6. Physical state generally solid liquid & gaseous solid, liquid
gas
Example 2:
Which of the following statement is/are not true for -bond.
1. It is formed by the overlapping of s s or s p orbitals
2. It is weaker than pi bond
3. It is formed when bond exists already.
4. It is resulted from partial overlapping of orbitals.
(A) 1, 2, 3, 4 (B) 2, 3 and 4
(C) 2 and 4 (D) 1, 2 and 4
Solution:
(B)
Example 3:
The types of bond present in ZnSO4.7H2O are only
(A) Electrovalent and covalent
(B) Electrovalent and co-ordinate
(C) Electrovalent, Covalent and co- ordinate
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(D) None of these
Solution:
(C)
Example 4:
Classify the following bonds as ionic, polar covalent or covalent and
give your answer
(a) Si Si bond in Cl3 SiSiCl3 (b) SiCl bond in Cl3SiSiCl3
(c) CaF bond in CaF2 (d) NH bond in NH3
Solution:
(a) Covalent due to identical electronegativity
(b) One electron pair is shared between Si & Cl and thus, covalent
bond is expected but electron negativity of Cl is greater than
that of Si & some polarity develops giving polar – covalent
nature
(c) Ionic since Ca completes its octet by transfer of two outershell
electrons thus, completing their octets
Ca [Ar]4s2, F[He)2s22p5
(d) Polar covalent, explanation as in (b)
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Example 5:
Arrange the bonds in order of increasing ionic character in the
molecules:
LiF, K2O, N2, SO2 and ClF3
Solution:
N2 < ClF3 < SO2 < K2O < LiF
Example 6:
Arrange the following in order of increasing ionic character:
C H, F H, Br H, Na I, K F and Li Cl
Solution:
C H < Br H < F H < Na I < Li Cl < K F
HYBRIDIZATION
The tetravalency shown by carbon is actually due to excited state of carbon
which is responsible for carbon bonding capacity.
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C excited state
1s 2sx2P y2P z2P
If the bond formed is by overlapping then all the bonds will not be
equivalent so a new concept known as hybridization is introduced which can
explain the equivalent character of bonds.
s and p orbital belonging to the same atom having slightly different energies
mix together to produce same number of new set of orbital called as hybrid
orbital and the phenomenon is called as hybridization.
Important characteristics of hybridization
(i) The number of hybridized orbital is equal to number of orbitals that
get hybridized.
(ii) The hybrid orbitals are always equivalent in energy and shape.
(iii) The hybrid orbitals form more stable bond than the pure atom orbital.
(iv) The hybrid orbitals are directed in space in same preferred direction to
have some stable arrangement and giving suitable geometry to the
molecule.
Depending upon the different combination of s and p orbitals, these types of
hybridization are known.
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(i) sp3 hybridization: In this case, one s and three p orbitals hybridise
to form four sp3 hybrid orbitals. These four sp3 hybrid orbitals are
oriented in a tetrahedral arrangement.
For example in methane CH4
109028'
2s
2px
2py2pz
(ii) sp2 hybridization: In this case one s and two p orbitals mix together
to form three sp2 hybrid orbitals and are oriented in a trigonal planar
geometry.
1200
2s2px 2py
The remaining p orbital if required form side ways overlapping with the
other unhybridized p orbital of other C atom and leads to formation of
bond as in H2C = CH2
(iii) sp hybridization: In this case, one s and one p orbital mix together
to form two sp hybrid orbitals and are oriented in a linear shape.
2s 2ps
1800
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The remaining two unhybridised p orbitals overlap with another
unhybridised p orbital leading to the formation of triple bond as in
HC CH.
Shape Hybridisation
Linear sp
Trigonal planar sp2
Tetrahedral sp3
Trigonal bipyramidal sp3d
Octahedral sp3d2
Pentagonal bipyramidal sp3d3
Example 7:
Which of the following molecule has trigonal planer geometry?
(A) CO2 (B) PCl5
(C) BF3 (D) H2O
Solution:
BF3 has trigonal planer geometry (sp2 - hybridized Boron).
Hence (A) is correct.
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Rule for determination of total number of hybrid orbitals
Detect the central atom along with the peripheral atoms.
Count the valence electrons of the central atom and the peripheral
atoms.
Divide the above value by 8. Then the quotient gives the number of
bonds and the remainder gives the non-bonded electrons. So number
of lone pair =non bonded electrons
2.
The number of bonds and the lone pair gives the total number of
hybrid orbitals.
An Example Will Make This Method Clear
SF4 Central atom S, Peripheral atom F
total number of valence electrons = 6 + (4 7) = 34
Now 34/8 = 4 2
8
Number of hybrid orbitals = 4 bonds + 1 lone pair)
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So, five hybrid orbitals are necessary and hybridization mode is sp3d and it
is trigonal bipyramidal (TBP).
Note:
Whenever there are lone pairs in TBP geometry they should be placed
in equatorial position so that repulsion is minimum.
1. NCl3 Total valence electrons = 26
Requirement = 3 bonds + 1 lone pair
Hybridization = sp3
Shape = pyramidal
2. BBr3 Total valence electron = 24
Requirement = 3 bonds
Hybridization = sp2
Shape = planar trigonal
3. SiCl4 Total valence electrons = 32
Requirement = 4 bonds
Hybridization = sp3
Shape = Tetrahedral
F
F
F
S
F
S
F
F
F
F
(A) (B)
N
ClCl
Cl
B
Br Br
Br
Si
Cl Cl Cl
Cl
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4. CI4 Total valence electron = 32
Requirements = 4 bonds
Hybridization = sp3
Shape = Tetrahedral
5. SF6 Total valence electrons = 48
Requirement = 6 bonds
Hybridization = sp3d2
Shape = octahedral/square bipyramidal
6. BeF2 Total valence electrons : 16
Requirement : 2 bonds
Hybridization : sp
Shape : Linear
F – Be – F
7. ClF3 Total valence electrons : 28
Requirement: 3 bonds + 2 lone pairs
Hybridization : sp3d
Shape : T – shaped
We have already discussed that whenever there are lone pairs they should
be placed in equatorial positions. Now a question that may come to your
mind that though the hybridization is sp3d, so the shape should be T.B.P.
But when all the bonds are present the actual shape is TBP. But when
instead of bond there are lone pairs in TBP the actual geometry is
C
l
l l l
S
F
F
F F
F
F
Cl F
F
F
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determined by the bonds not by the lone pairs. Here in ClF3 the bond
present (2 in axial and 1 in equatorial) gives the impression of T shape.
8. PF5 Total valence electrons : 40
Requirement : 5 bonds
Hybridization : sp3d
Shape : Trigonal bipyramidal (TBP)
9. XeF4 Total valence electrons : 36
Requirement:4 bonds+ 2 lone pairs
Hybridisation : sp3d
Shape : Square planar
Now three arrangements are possible out of which A and B are same. A and
B can be inter converted by simple rotation of molecule. The basic difference
of (B) and (C) is that in (B) the lone pair is present in the anti position which
minimizes the repulsion which is not possible in structure (C) where the lone
pairs are adjacent. So in a octahedral structure the lone pairs must be
placed at the anti positions to minimize repulsion. So both structure (A) and
(B) are correct.
P
F
F
F
F
F
Xe
F
F
F
F
(A)
Xe
F
F F
F
(B)
Xe
F
F
F
(C)
F
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10. XeF2 Total valence electrons : 22
Requirements : 2 bonds + 3 lone pairs
Hybridisation: sp3d
Shape : Linear
[ l.p. are present in equatorial position and ultimate shape is due to the
bonds that are formed]
11. PF2Br3 Total valence electrons : 40
Requirements : 5 bonds
Hybridisation: sp3d
Shape : trigonal bipyramidal
Here we see that fluorine is placed in axial position whereas bromine is
placed in equatorial position. It is the more electronegative element that is
placed in axial position and less electronegative element is placed in
equatorial position. Fluorine, being more electronegative pulls away bonded
electron towards itself more than that is done by bromine atom which results
in decrease in bp – bp repulsion and hence it is placed in axial position.
In this context it can also be noted that in T.B.P. shape the bond lengths are
not same. The equatorial bonds are smaller than axial bonds. But in square
bipyramidal shape, all bond lengths are same.
Xe
F
F
P
FBr
BrF
Br
AISM-09/C/CMB
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12. 3-
4PO Total valence electrons : 32
Requirement : 4 bonds
Hybridisation: sp3
Shape: tetrahedral
Here all the structures drawn are resonating structures with O– resonating
with double bonded oxygen.
13. NO2– Total valence electron: 18
Requirement : 2 bonds + 1 lone pair
Hybridisation: sp2
Shape: angular
14. CO32– Total valence electrons: 24
Requirement = 3 bonds
Hybrdisation = sp2
Shape: planar trigonal
But C has 4 valence electrons of these 3 form bonds the rest will form a
bond.
O
P
O O O
O–
P
O O O
O
P
O O O
etc.
N
O O
C
O O
O
C
O O
O–
C
O O
O–
AISM-09/C/CMB
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In the structure one bond is a double bond and the other 2 are single. The
position of the double bonds keeps changing in the figure. Since peripheral
atoms are isovalent, so contribution of the resonanting structures are equal.
Thus it is seen that none of the bonds are actually single or double. The
actual state is
15. CO2 Total valence electrons:16
Requirement: 2 bonds
Hybridisation: sp
Shape: linear
O = C = O
16. -
4BF Total valence electrons = 32
Requirement= 4 bonds
Hybridisation: sp3
Shape: Tetrahedral
17. -
3ClO Total valence electron = 26
Requirement = 3 bond + 1 lone pair
Hybridization: sp3
Shape: pyramidal
18. XeO2F2 Total valence electrons : 34
Requirement: 4 bonds +1 lone pairs
O
O
Xe
F
F
C
O–2/3
O–2/3
O–2/3
Bond order = 3/2 = 1.5
B
FF
F
F–
Cl
OO
O-
Cl
OO
- O
Cl
O-
OO
AISM-09/C/CMB
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Hybridization : sp3d
Shape: Distorted TBP (sea-saw geometry)
19. XeO3 Total valence electrons : 26
Requirement: 3 bonds + 1 lone pair
Hybridization: sp3
Shape: Pyramidal
20. XeOF4 Total valence electrons : 42
Requirement: 5 bonds + 1 lone pair
Hybridization: sp3d2
Shape: square pyramidal.
Example 8:
Predict the hybridization for the central atom in 3POCl , 4OSF , 5OIF
Solution:
3POCl Total No. of V.E. = 5 6 21 32
48 8
So, hybridization = 3sp
4OSF = 6 6 28 40
58 8
So, hybridization of s = 3dsp
Xe
O OO
Xe
F
F
F
FO
AISM-09/C/CMB
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5OIF
6 7 35 486
8 8
So, hybridization of I = 2 3d sp
Example 9:
Out of the three molecules XeF4, SF4 and SiF4 one which has