Inorganic Chemistry

M.S.C. / First Semester

(4) Lecturer 2021-2022 Pro .Dr. Mohammed Hamid
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HYBRIDIZATION:-
The valence bond theory (overlapping concept) explains satisfactorily the formation of various
molecules but it fails to account the geometry and shapes of various molecules. It does not give
explanation why BeCl2 is linear, BF3 is planar. CH4 is tetrahedral, NH3 is pyramidal and water is V-
shaped molecule. In order to explain these cases, the valence bond theory has been supplemented
by the concept of hybridization. This is a hypothetical concept and has been introduced by Pauling
and Slater. According to this concept any number of atomic orbitals of an atom which differ in
energy slightly may mix with each other to form new orbitals called hybrid orbitals. The process
of mixing or amalgamation of atomic orbitals of nearly same energy to produce a set of entirely new
orbitals of equivalent energy is known as hybridization. The following are the rules of
hybridization:
(i) Only orbitals (atomic) of nearly same energy belonging to same atom or ion can take part in
hybridization.
(ii) Number of the hybrid orbitals formed is always equal to number of atomic orbitals which have
taken part in the process of hybridization.
(iii) Most of the hybrid orbitals are similar but they are not necessarily identical in shape. They may
differ from one another in orientation in space.
(iv) Actually the orbitals which undergo hybridization and not the electrons. For example, for
orbitals of nitrogen atom (2s2 2px1 2py12pz1) belonging to valency shell when hybridize, form four
hybrid orbitals, one of which has two electrons (as before) and other three have one electron each.
(v) The electron waves in hybrid orbitals repel each other and this tend to the farthest apart.
(vi) Hybrid orbitals form only sigma bonds.
(vii) Depending on the number and nature of orbitals undergoing hybridization, various types of
hybrid orbitals directing towards the corners of specified geometrical figures come into existence.
The molecule has a regular geometry if all the hybrid orbitals after overlapping contain shared pair
of electrons, i.e., there are no orbitals containing lone pairs in the valency shell. If, however, the
central atom is surrounded by one or more orbitals containing lone pairs of electrons in the valency
shell, the geometry of the molecule is distorted to some extent. Thus, the presence of one or more
orbitals with lone pairs affect the bond angle to some extent due to repulsion between lone pair
(pairs) with bonded (pairs). This type of observation has been made, for example, in the molecules
of ammonia and water. The following table shows the type of hybridization and the geometry of the
molecules containing only bond pairs of electrons.
Some Typical Cases of Hybridization
(i) BeF2 molecule : Beryllium atom has the configuration 1s2, 2s2. Since there are no unpaired
electrons in the valency shell, it cannot form any covalent bond. Thus, 2s-orbital is first unpaired
and an electron is shifted to 2p-orbital.

Now, there is hybridization between one s- and one p-orbital. Two orbitals (hybrid) of same shape
and energy come into existence. These overlap with p-orbital (singly occupied) each of the two
fluorine atoms forming two sigma bonds. The molecule formed is linear with a bond angle 180°.

Formation of BeF2 Molecule

Hybridization of BF3 (BoronTrifluoride)
BF3 has a boron atom with three outer-shell electrons in its ground state and three fluorine atoms
containing seven outer electrons. Further, if we observe closely, one boron electron is unpaired in
the ground state. During the formation of this compound, the 2s orbital and two 2p orbitals
hybridize. Only one of the empty p-orbital is left behind as the lone pair. In short, Boron needs 3
hybridized orbitals to make bonds with 3 atoms of F where the 2p z orbitals get overlapped with
these hybridized sp2 orbitals and bonds are formed.

Important Points to Remember

• The three hybridized sp2 orbitals are usually arranged in a triangular shape.
2
• BF3 molecule is formed by bonding between three sp orbitals of B and p of 3 F atoms.
• All the bonds in BF3 are sigma bonds.

Hybridization of CH4 (Methane)

Hybridization of CH4 (Methane)
In order to understand the hybridization of CH4 (methane), we have to take a look at the atomic
orbitals which are of different shape and energy that take part in the process. The type of
hybridization involved with CH4 is sp3. We will discuss in detail how this hybridization occurs
below.
In order to explain this observation, valence bond theory relies on a concept called orbital
hybridization. In this picture, the four valence orbitals of the carbon (one 2s and three 2p orbitals)
combine mathematically (remember: orbitals are described by equations) to form four
equivalent hybrid orbitals, which are named sp3 orbitals because they are formed from mixing
one s and three p orbitals. In the new electron configuration, each of the four valence electrons on
the carbon occupies a single sp3 orbital creating four unpaired electrons.
The shape of an sp3 hybridized orbital is a combination of s and p atomic orbitals.

Each sp3-hybridized orbital bears an electron, and electrons repel each other. To minimize the
repulsion between electrons, the four sp3-hybridized orbitals arrange themselves around the carbon
nucleus so that they are as far away as possible from each other, resulting in the tetrahedral
arrangement predicted by VSPER. The carbon atom in methane is called an “sp3-hybridized carbon
atom.” The larger lobes of the sp3 hybrids are directed towards the four corners of a tetrahedron,
meaning that the angle between any two orbitals is 109.5 o.

Bonding in Methane

Each C-H bond in methane, then, can be described as an overlap between a half-filled 1s orbital in
four hydrogen atoms and the larger lobe of one of the four half-filled sp3 hybrid orbitals form a four
equivalent sigma (σ) bond. This orbital overlap is often described using the notation: sp 3(C)-1s(H).
The formation of sp3 hybrid orbitals successfully explains the tetrahedral structure of methane and
the equivalency of the the four C-H bonds.
What remains is an explanation of why the sp3 hybrid orbitals form. When the s and 3 p orbitals in
carbon hybridize the resulting sp3 hybrid orbital is unsymmetrical with one lobe larger than the
other. This means the larger lobe can overlap more effectively with orbitals from other bonds
making them stronger. Hybridizing allows for the carbon to form stronger bonds than it would with
unhybridized s or p orbitals.

The four carbon-hydrogen bonds in methane are equivalent and all have a bond length of 109 pm
(1.09 x 10-10 m), bond strength of of 429 kJ/mol. All of the H-C-H bond angles are 109.5o.
Hybridization of PCl5
Atomic number of Phosphorous is 15, The Ground state configuration is

& the excited state configuration is

The 5 electrons present in valence shell form bond pairs with the electrons of five chlorine atoms.
Phosphorus atom is sp3d hybridized in the excited state.

PCl5

The geometry of PCl5 is trigonal bipyramidal .The P atom lies in the centre of an equatorial triangle
& three P-Cl bonds (equatorial bonds) are directed towards its three corners with 120° bond angle.
The remaining two P-Cl bonds (axial bonds) lie above & below the plane of the triangle at bond
angle 90° .

The axial bonds are longer than equatorial bonds because the axial Cl atoms suffer from more
repulsion then the equatorial Cl atoms,as a result the axial Cl atoms tries to reside far away from
the equatorial Cl atoms, & hence axial bond are longer than equatorial bonds.
Hybridization of SF6
The outer electronic configuration of S atom is 3s2,3p4 . It has also vacant 3d orbital . In excited
state by absorbing energy one 3s and one 3p electron shifted to 3dxy and 3dyz orbital . As a result,
the number of unpaired electron is six . Now one S , three P and two ‘d’ orbital mixed and form six
equivalent new orbital ,each are called sp3d2 hybridized orbital. so in SF6, the hybridization of ‘S’
is sp3d2 .
Geometry of molecules contaning lons paris of electrons

When we travel, we often take a lot more stuff than we need. Trying to fit it all in a suitcase can be
a real challenge. We may have to repack or just squeeze it all in. Atoms often have to rearrange
where the electrons are in order to create a more stable structure.

Central Atom with One or More Lone Pairs

The molecular geometries of molecules change when the central atom has one or more lone pairs
of electrons. The total number of electron pairs, both bonding pairs and lone pairs, leads to what is
called the electron domain geometry. When one or more of the bonding pairs of electrons is
replaced with a lone pair, the molecular geometry (actual shape) of the molecule is altered. In
keeping with the A and B symbols established in the previous section, we will use E to represent a
lone pair on the central atom (A). A subscript will be used when there is more than one lone pair.
Lone pairs one th surrounding atoms (B) do not affect the geometry.

AB3E: Ammonia, NH3

The ammonia molecule contains three single bonds and one lone pair one central nitrogen atom (see
figure below).

Figure : Lone pair electrons in ammonia.

The domain geometry for a molecule with four electron pairs is tetrahedral, as was seen
with CH4CH4. In the ammonia molecule, one of the electron pairs is a lone pair rather than a
bonding pair. The molecular geometry of NH3 is called trigonal pyramidal (see figure below).

Figure : Ammonia molecule.

Recall that the bond angle in the tetrahedral CH4 molecule is 109.5o. Again, the replacement of one
of the bonded electron pairs with a lone pair compresses the angle slightly. The H−N−H angle is
approximately 107o.

AB2E2: Water, H2O

A water molecule consists of two bonding pairs and two lone pairs (see figure below).

Figure : Lone pair electrons on water.

As for methane and ammonia, the domain geometry for a molecule with four electron pairs is
tetrahedral. In the water molecule, two of the electron pairs are lone pairs rather than bonding pairs.
The molecular geometry of the water molecule is bent. The H−O−H bond angle is 104.5o, which is
smaller than the bond angle in NH3NH3 (see figure below).

Figure : Water molecule.

AB4E: Sulfur Tetrafluoride, SF4

The Lewis structure for SF4 contains four single bonds and a lone pair on the sulfur atom (see figure
below).
 Figure : Lone pair electrons in SF4.

The sulfur atom has five electron groups around it, which corresponds to the trigonal bipyramidal
domain geometry, as in PCl5PCl5 (see figure below). Recall that the trigonal bipyramidal geometry
has three equatorial atoms and two axial atoms attached to the central atom. Because of the greater
repulsion of a lone pair, it is one of the equatorial atoms that are replaced by a lone pair. The
geometry of the molecule is called a distorted tetrahedron or seesaw.

Figure : Ball and stick model for SF4

Table Geometries in Which the Central Atom Has One or More Lone Pairs
Number
Total Number Number
of Molecular
of Electron of Lone Electron Domain Geometry Examples
Bonding Geometry
Pairs Pairs
Pairs
3 2 1 Trigonal Planar Bent O3
Trigonal
4 3 1 Tetrahedral NH3
Pyramidal
4 2 2 Tetrahedral Bent H2O
Distorted
5 4 1 Trigonal Bipyramidal Tetrahedron SF4
(Seesaw)
5 3 2 Trigonal Bipyramidal T-shaped ClF3
5 2 3 Trigonal Bipyramidal Linear I3−
Square
6 5 1 Octahedral BrF5
Pyramidal

6 4 2 Octahedral Square Planar XeF4

• The presence of lone pair electrons influences the three-dimensional shape of the molecule.
Geometry of molecules involving sigma and pi bonds:-
Our minds can handle two electrons interacting with one another in a sphere of space. But then we
start putting in double bonds and triple bonds. The way we draw these bonds suggests we are
squeezing more electrons into the same space, and that doesn't work. Electrons don't like to be
pushed together (especially since they all have negative charges that repel one another). So we need
a more complex picture that works for all these electrons.
The hybridization model helps explain molecules with double or triple bonds (see figure below).
Ethene (C2H4) contains a double covalent bond between the two carbon atoms and single bonds
between the carbon atoms and the hydrogen atoms. The entire molecule is planar.

Figure : Geometry of ethene molecule.

As can be seen in the figure below, the electron domain geometry around each carbon independently
is trigonal planar. This corresponds to sp2 hybridization. Previously, we saw carbon
undergo sp3 hybridization in a CH4 molecule, so the electron promotion is the same for ethene, but
the hybridization occurs only between the single S orbital and two of the three p orbitals. Thus
generates a set of three sp2 hybrids along with an unhybridized 2pz orbital. Each contains one
electron and so is capable of forming a covalent bond.

Figure : Hybridization in ethene.

The three sp2 hybrid orbitals lie in one plane, while the unhybridized 2pz2pz orbital is oriented
perpendicular to that plane. The bonding in C2H4 is explained as follows. One of the
three sp2 hybrids forms a bond by overlapping with the identical hybrid orbital on the other carbon
atom. The remaining two hybrid orbitals form bonds by overlapping with the 1s orbital of a
hydrogen atom. Finally, the 2pz orbitals on each carbon atom form another bond by overlapping
with one another sideways.
It is necessary to distinguish between the two types of covalent bonds in a C2H4C2H4 molecule.
A sigma bond (σ bond) is a bond formed by the overlap of orbitals in an end-to-end fashion, with
the electron density concentrated between the nuclei of the bonding atoms. A pi bond (π bond) is
a bond formed by the overlap of orbitals in a side-by-side fashion with the electron density
concentrated above and below the plane of the nuclei of the bonding atoms. The figure below shows
the two types of bonding in C2H4. The sp2 hybrid orbitals are purple and the pzpz orbital is blue.
Three sigma bonds are formed from each carbon atom for a total of six sigma bonds total in the
molecule. The pi bond is the "second" bond of the double bonds between the carbon atoms and is
shown as an elongated green lobe that extends both above and below the plane of the molecule. This
plane contains the six atoms and all of the sigma bonds.

Figure : Sigma and pi bonds.

In a conventional Lewis electron-dot structure, a double bond is shown as a double dash between
the atoms as in C=C. It is important to realize, however, that the two bonds are different: one is a
sigma bond, while the other is a pi bond.
Ethyne (C2H2) is a linear molecule with a triple bond between the two carbon atoms (see figure
below). The hybridization is therefore sp.

Figure : Ethyne structure.

The promotion of an electron in the carbon atom occurs in the same way. However, the hybridization
now involves only the 2s orbital and the 2px orbital, leaving the 2py2py and the 2pz2pz orbitals
unhybridized.

Figure : Hybridization in ethyne.

The sp hybrid orbitals form a sigma bond between each other as well as sigma bonds to the hydrogen
atoms. Both the py and the pz orbitals on each carbon atom form pi bonds between each other. As
with ethene, these side-to-side overlaps are above and below the plane of the molecule. The
orientation of the two pi bonds is that they are perpendicular to one another (see figure below). One
pi bond is above and below the line of the molecule as shown, while the other is in front of and
behind the page.

Figure : The C2H2 molecule contains a triple bond between the two carbon atoms, one of
which is a sigma bond, and two of which are pi bonds.

In general, single bonds between atoms are always sigma bonds. Double bonds are comprised of
one sigma and one pi bond. Triple bonds are comprised of one sigma bond and two pi bonds.

Resonance:-
It is generally observed that a single valence bond structure of a molecule cannot correctly account
for the properties of the molecule. In such cases, the concept of resonance is introduced. According
to this concept if two or more alternate valence bond structures can be written for a molecule,
the actual structure is said to be a resonance or mesomeric hybrid of all these alternate
structures. For example, carbon dioxide molecule can be represented by the following three
structures:

The calculated values of bond distances between carbon and oxygen in C=0 and C≡0 are 1.22 Å0
and 1.10 A0, respectively but the observed bond distance between carbon and oxygen in carbon
dioxide is 1.15 A0. Thus, none of the above structures correctly explains the observed bond length.
It is, thus, said that a hybrid form of these structures can provide the exact explanation. The various
structures of which the molecule is a resonance hybrid are known as canonical forms or mesomeric
forms. Actually resonance hybrid does not oscillate between the canonical forms of a mixture
of these forms but it is a definite form and has definite structure which cannot be written on
paper.

Rules for Selecting Canonical Forms or Mesomeric Forms :

(i) The relative position of all the atoms in each of the canonical forms must be the same. They
should differ only in the position of electrons,
(ii) The number of unpaired and paired electrons in each of the canonical forms must be same.
(iii) The contributing structures should not differ much in energy.
 (iv) The contributing structures should be such that negative charge resides on more electronegative
and positive charge on the electropositive. Like charges should not reside on atoms close together
in the canonical forms.
Formal charges on the atoms in the molecule help us in choosing the most appropriate resonance
structure. For example, nitrous oxide molecule is represented by two resonance electron dot
structures, one of which has a negative formal charge on the oxygen atom and the other of which
has a negative charge on the terminal nitrogen atom.

Since oxygen is a more electronegative element than nitrogen, the structure that places a negative
formal charge on oxygen is probably lower in energy than the structure that has a negative formal
charge on nitrogen. Thus, the actual structure of N2O is :structure 1

(v) As a result of resonance, the bond order changes in many molecules or ions.

Some Examples Showing Resonance

PREDICTION OF GEOMETRY (SHAPE) OF COVALENT MOLECULES

Geometry of a molecule or ion can be predicted if the state of hybridization at central atom is
known. The type of hybridization can be known by the following methods:

First Method: The state of hybridization at central atom in a molecule can be known by counting
the number of orbitals involved in co-axial overlappings and the number of orbitals with lone pair
of electrons, ie., by counting the number of σ-bonds and the number of lone pair of electrons at
central atom. Adding the two, if the total is 4, the hybridization is sp3 If this total is 3, the
hybridization is sp2 and if this total is 2, the hybridization is sp.

A few examples are being given below:

(a) Beryllium chloride (BeCl2)

No. of σ-bonds at Be atom = 2

No. of lone pair of electrons at Be atom = 0

Total = 2 + 0 = 2 Hybridization is sp, ie., BeCl2 is linear and bond angle is 180°.

(b) Boron trichloride (BCI3)

No. of σ-bonds at B atom = 3

No. of lone pair of electrons at B atom = 0

Total = 3 Hybridization is sp2, i.e., BCl3 is trigonal planar and bond angles are 120° each.

(c) Carbon tetrachloride (CCl4)

No. of σ-bonds at C atom = 4

No. of lone pair of electrons at C atom = 0

Total = 4 Hybridization is sp3, i.e., CCl, is tetrahedral and bond angles are 109°28' each.

(d) Ammonia (NH3)

No. of σ-bonds at N atom = 3

No. of lone pair of electrons at N atom =1

Total = 3 + 1 = 4 Hybridization is sp3 Shape is trigonal pyramidal ( not tetrahedral ) and bond
angles are not 109°28' but 106°45'

Second Method: Number of electron pairs in the valency shell of the central atom can be
determined by applying the following formula:
For neutral molecules:

No. of electron pairs = No. of atoms bonded to it + 1/l2(Group number of central atom -
Valency of the central atom]

No. of electron pairs at the central atom in CH4 = 4 + 1/2(4-4) = 4 (sp3 hybridization)

No. of electron pairs at the central atom in PCl5 = 5 + 1/2(5-5) = 5 (sp3d hybridization)

No. of electron pairs at the central atom in SF6 = 6 + 1/2 (6-6) = 6 (sp3d² hybridization)

No. of electron pairs at the central atom in H2O = 2 +1/2(6 – 2) = 4 (sp³ hybridization)

For ions :

No. of electron pairs = No. of atoms bonded to it +1/2 [Group number of central atom - Valency of
+
the central atom No. of electron]
−
No. of electron pairs in NH4+ = 4+1/2 (5 – 4 – 1) = 4 (sp³ hybridization)

No. of electron pairs in SO42- = 4 + 1/2(6-8 +2) = 4 (sp³ hybridization)

No. of electron pairs in BF4- = 4+(3 - 4 + 1) = 4 (sp³ hybridization)

Third Method : The number of orbitals involved in hybridization can be determined by the
application of the following formula :

H= + 1/2[V + M –C +A ]

where H = number of orbitals involved in hybridization

V = valence electrons of central atom

M = number of monovalent atoms linked with central atom

C = charge on the cation

A= charge on the anion

From the value of H, the type of hybridization can be predicted.

Value of H 2 3 4 5 6 7

Hybridization sp sp2 sp3 sp3d sp3d2 sp3d3

Type 1. Neutral molecules :

(i) CO2 : V = 4, M = 0, C = 0, A = 0
H=1/2 [4 + 0 -0 + 0] = 2, sp hybridization

(ii) SO2 or SO3 V = 6, M = 0, C = 0, A = 0

H= 1/2[6 + 0 – 0 + 0] = 3, sp² hybridization

(iii) BCl3 : V = 3, M = 3, C = 0, A = 0

H = 1/2 [3 + 3-0 + 0] = 3, sp2 hybridization

(iv) SiCl4 : V = 4, M = 4, C = 0, A = 0

H = 1/2 [4 + 4-0+ 0] = 4, sp3 hybridization

Type 2. When the species is a cation

(i) NH4+ : V= 5 , M= 4 , C= 1 , A= 0

H= 1/2 (5+4 – 1+0 ) = 4 , sp3 hybridization

(ii) CH3+ : V= 4 , M= 3 , C= 1 , A= 0

H= 1/2 (4+3 – 1+0 ) = 3 , sp2 hybridization

(iii) NO2+ : V= 5 , M= 0 , C= 1 , A= 0

H= 1/2 (5+0 – 1+0 ) = 2 , sp hybridization

Type 3. When the species is an anion

(i) CO32- : V = 4, M = 0, C = 0, A = 2

H=1/2 [4 + 0 -0 + 2] = 3, sp2 hybridization

(ii) SO42- : V = 6, M = 0, C = 0, A = 2

H= 1/2[6 + 0 – 0 + 2] = 4, sp3 hybridization

(iii) ICl4-: V = 7, M = 4, C = 0, A = 1

H= 1/2[7 + 4 – 0 + 1] = 6, sp3d2 hybridization