The mixing of one s – orbital and one p – orbital of the same atom of nearly same energy to form a set of diagonally arranged two identical hybrid orbitals of equivalent energy is called sp hybridization. These hybrid orbitals are arranged in a linear manner around central atom and are at an angle of 180^o to one another.

Formation of BeF₂ Molecule:

Ground State of Beryllium Atom:

Atomic number of beryllium is 4. Its configuration in ground state is 1s², 2s², 2p⁰.

Beryllium atom in ground state:

Excited state of Beryllium Atom:

During combination with fluorine, the 2s electron pair is split up and one electron is promoted to empty 2p_x orbital. This condition is called excited state of beryllium. In excited state one electron of 2s migrates to 2p orbital forming 2 – half filled orbitals.
Beryllium atom in excited state:

Hybridization of Beryllium Atom:

One 2s orbital and one 2p orbitals of beryllium mix up forming two hybrid orbitals of equivalent energy. These two new equivalent orbitals are called sp hybrid orbitals. They are identical in all respect.

Beryllium atom in excited state:

Angle and Geometry :

Two sp hybridized orbitals formed, repel each other and These hybrid orbitals are arranged in a linear manner around central beryllium atom and are at an angle of 180^o to one another. Each sp hybrid orbital contain unpaired electron. In each sp hybrid orbital, one of the lobes is bigger because of more concentration of electron density. Only bigger lobe is involved in bond formation.

Thus in BeF₂ molecule has linear or diagonal structure with boron atom at the centre and two fluorine atoms at the either sides of beryllium. F-Be-F bond angle is 180^o.

Bond:

Two sp hybrid orbitals of boron atom having one unpaired electron each overlap separately with 1p orbitals of two fluorine atoms along the axis forming two covalent bonds (sigma bonds). The bonds between beryllium and fluorine are sp- p overlap. Thus F – Be — F bond angles are 180^o. Molecule is diagonal or linear. Both Be-F bonds in beryllium difluoride are of equal strength.

Diagram:

Type and Geometry of Beryllium difluoride Molecule:

BeF₂ Is linear molecule, while H₂O is angular molecule.

In BeF₂ molecule, beryllium undergoes sp hybridization and achieve electronic configuration 1s², 2s¹2p_x¹ in excited state. During formation of beryllium difluoride boron undergoes sp hybridization One 2s orbital and one 2p orbital of beryllium mix up forming two hybrid orbitals of equivalent energy. These two new equivalent orbitals are called sp hybrid orbitals. They are identical in all respect. Two sp hybridized orbitals formed, repel each other and they are directed diagonally opposite in space.  Angle between them is 180⁰. Two sp hybrid orbitals of boron atom having one unpaired electron each overlap separately with 1p orbitals of two fluorine atoms along the axis forming two covalent bonds (sigma bonds). Thus BeF₂ Is linear molecule.

In water, the oxygen atom is sp³ hybridized. Two hybrid orbitals have paired electrons (lone pair) and they are non – bonding orbitals.  Other two orbitals are half – filled (singly occupied) and they are bonding orbitals. These hybridized orbitals are in four directions of four corners of tetrahedron. Four sp³ hybridized orbitals formed, repel each other and they should be directed towards the four corners of a regular  tetrahedron and  Angle between them should be 109.5⁰.  The non bonding electron repel each other strongly and occupy more space than the electron pairs involved in bonding.  The force of repulsion between electron pair decreases in following order. lone pair – lone pair > lone pair- bond pair >  bond pair – bond pair. Hence the bond angle between two lone pairs i.e. H-O-H angle decreases from109.5⁰ to 104.5⁰. Thus water molecule is angular.

Formation of acetylene (C₂H ₂) Molecule:

Ground State of Carbon Atom:
Atomic number of carbon is 6. Its configuration in ground state is 1s², 2s², 2p² i.e. 1s² 2s², 2p_x¹ 2p_y¹ 2p_z⁰

Carbon atom in ground state:

Excited state of Carbon Atom:

During combination with hydrogen, the 2s electron pair is split up and one electron is promoted to empty 2p_z orbital. This condition is called excited state of carbon. In excited state one electron of 2s migrates to 2p orbital forming 4 – half filled orbitals.
Carbon atom in excited state:

Hybridization:

In acetylene there is sp hybridization. One 2s orbital and one 2p orbitals of carbon mix up forming two hybrid orbitals of equivalent energy. These two new equivalent orbitals are called sp hybrid orbitals. They are identical in all respect. Two ‘p’ orbitals of each carbon atom remains unhybridized.

Angle and Geometry:

Two sp hybridized orbitals formed, repel each other and These hybrid orbitals are arranged along x- axis in a linear manner around central carbon atom and are at an angle of 180⁰ to one another. The unhybridized p_y and p_z orbital remain perpendicular to hybrid orbitals along y-axis and z- axis ,
mutually perpendicular. Each sp hybrid orbital and unhybrid orbitals contain unpaired electron. In each sp hybrid orbital, one of the lobes is bigger because of more concentration of electron density. Only bigger lobe is involved in bond formation.

As all the four atoms in C₂H₂ molecule being in the same line, the molecule is diagonal or linear. H-C-C bond angle is 180^o.

Formation of Bonds:

1. Sigma Bond Formation :

A covalent bond formed by collinear or coaxial or in the line of internuclear axis. Overlapping of orbitals is known as sigma bond. One Sp hybrid orbital of one carbon atom overlaps with One hybrid orbital of other carbon atom by head on collision forming sigma bond. One (Sp- Sp ) overlap. Remaining one hybrid orbitals of each carbon atom overlap with ‘s’ orbital of two hydrogen atoms separately forming two sigma bonds. (2 C – H). Two (Sp– s)overlaps. Both C-H bond in Acetylene are of equal strength. Thus there are Three sigma bonds. Sigma bonds are stronger.

2) Formation of pi Bond :
The covalent bond formed by collateral or sidewise overlapping is called pi bond. The unhybridized 2 p_x and 2 p_z orbitals of each carbon atom being perpendicular to each other and to the plane of H-C-C-H axis overlap laterally with one another to form two week pi bond between two
carbon atoms by two p – p overlap. (one 2 p_y -2 p_y )and (one 2p_z-2 p_z) . Thus two (p-p) -overlaps.

In ethylene molecule there are 3 sigma bonds and 2 pi bonds. There is a triple bond between carbon and carbon consisting one sigma and two pi bonds.

Diagram:

Type and Geometry of Acetylene Molecule:

There is only one pi bond in ethylene molecule but there are two pi bonds in the acetylene molecule:

In ethylene molecule carbon atom shows sp² hybridization in excited state. The resulting three hybrid orbitals form two C-H bonds and One pi bond of sigma type. Thus each carbon atom is left with unhybridized p_z orbitals with lobes above and below the plane of hybridized orbitals. These two unhybridized orbitals overlap each other laterally and form a single pi bond between two carbon atoms.

In acetylene molecule carbon atom shows sp hybridization in excited state. The resulting two orbitals are linear and form one C-H bond and one C-C bond of sigma type. Thus each carbon is left with two unhybridized p_y and p_z orbitals, which are mutually perpendicular to H-C-C-H axis. These unhybridized orbitals overlap each other laterally and form two pi bonds between two carbon atoms.

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Mixing of one ‘s’ orbital and two ‘p’ – orbitals of nearly same energy forming set of trigonally arranged three identical orbitals of equal energy is known as sp² hybridization.

Geometry sp² Hybridization:

The hybrid orbitals are arranged around the central atom in a plane at an angle of 120° to one another. When these three orbitals overlap with appropriate orbitals of three other atoms, three bonds are formed and the resulting molecule has a trigonal planar structure. Each bond angle is 120°.

Diagram:

Formation of Boron Trifluoride Molecules:

Ground State of Boron Atom:

Atomic number of boron is 5. Its configuration in ground state is 1s², 2s², 2p¹

Boron atom in ground state:

Excited State of Boron Atom:

During combination with fluorine, the 2s electron pair is split up and one electron is promoted to empty 2p_y orbital. This condition is called excited state of boron. In excited state one electron of 2s migrates to 2p orbital forming 3 – half filled orbitals.

Boron atom in excited state:

Hybridization of Boron Atom:
One 2s orbital and two 2p orbitals of boron mix up forming three hybrid orbitals of equivalent energy. These three new equivalent orbitals are called sp² hybrid orbitals. They are identical in all respect

Boron atom in hybridized state:

Angle and Geometry:

• Three sp² hybridized orbitals formed, repel each other and they are directed towards the three corners of an equilateral triangle. The angle between them is 120°.
• Each sp² hybrid orbital contains an unpaired electron.
• In each sp² hybrid orbital, one of the lobes is bigger because of more concentration of electron density. Only bigger lobe is involved in bond formation.
• Thus BF₃ molecule has a trigonal planar structure with boron atom at the centre and three fluorine atoms at the four corners of equilateral triangle. F-B-F bond angle is 120°.

Bond Formation:

Three sp² hybrid orbitals of boron atom having one unpaired electron each overlap separately with 2p orbitals of three fluorine atoms along the axis forming three covalent bonds (sigma). Thus in BF₃ molecule has a planar structure with a boron atom at the centre and three fluorine atoms at the three corners of an equilateral triangle. F-B-F bond angle is 120°.

Bond:

• There are three sigma bonds..
• The bonds between boron and fluorine are sp²– p.
• Thus F – B — F bond angles are 120°. The molecule is trigonal planar.
• All B-F bonds in boron trifluoride are of equal strength.

Diagram:

Type and Geometry of Boron trifluoride Molecule:


The BF₃ molecule has a planar structure while NH₃ molecule has a pyramidal structure.
During formation of boron trifluoride, boron undergoes sp² hybridization and achieve the electronic configuration 1s², 2s¹2p_x¹2p_y¹. In the excited state, one 2s orbital and two 2p orbitals of boron mix up forming three hybrid orbitals of equivalent energy. These three new equivalent orbitals are called sp² hybrid orbitals. They are identical in all respect. Three sp² hybridized orbitals formed, repel each other and they are directed towards the three corners of an equilateral triangle (in one plane). Angle between them is 120^o. Three sp² hybrid orbitals of boron atom having one unpaired electron each overlap separately with 1p orbitals of three fluorine atoms along the axis forming three covalent bonds (sigma bonds). Thus boron trifluoride has triangular planar structure.

During formation of ammonia nitrogen undergoes sp³ hybridization. One 2s orbital and three 2p orbitals of mix up forming four hybrid orbitals of equivalent energy. These four new equivalent orbitals are called sp³ hybrid orbitals. They are identical in all respect and they should be directed towards the four corners of a regular tetrahedron and Angle between them should be 109.5^o. Three hybridized orbitals contain unpaired electron. The fourth hybridized orbital has lone pair of electron. The three half filled (containing unpaired electron) sp³ hybrid orbitals of nitrogen overlap axially with three half filled 1s orbitals of three hydrogen atoms separately to form three covalent N-H bonds (sigma bonds). The fourth hybrid orbital containing lone pair of electron remains non bonded. The non bonding electron repel each other strongly and occupy more space than the electron pairs involved in bonding. The force of repulsion between lone pair- bond pair is greater than bond pair – bond pair. Hence the
bond angle i.e. H-N-H angle decreases from109.5^o to 107^o.

Formation of Ethylene Molecule:

Ground State of Carbon Atom:
Atomic number of carbon is 6. Its configuration in ground state is 1s², 2s², 2p² i.e. 1s² 2s², 2p_x¹ 2p_y¹ 2p_z⁰

Carbon atom in ground state:

Excited state of Carbon Atom:

During combination with hydrogen, the 2s electron pair is split up and one electron is promoted to empty 2p_z orbital. This condition is called excited state of carbon. In excited state one electron of 2s migrates to 2p orbital forming 4 – half filled orbitals.
Carbon atom in excited state:

Hybridization of Carbon Atom:

In ethylene there is sp² hybridization. One 2s orbital and two 2p orbitals of carbon mix up forming three hybrid orbitals of equivalent energy. These three new equivalent orbitals are called sp² hybrid orbitals. They are identical in all respect. One ‘p’ orbital of each carbon atom remains unhybridized

Carbon atom in hybridized state:

Angle and Geometry:

Three sp² hybridized orbitals formed, repel each other and they are directed in a plane towards the three corners of an equilateral triangle. Angle between them is 120^o. The unhybridized p_z orbital remain perpendicular to this plane. Each sp² hybrid orbital contain unpaired electron. In each sp² hybrid orbital, one of the lobes is bigger because of more concentration of electron density. Only bigger lobe is involved in bond formation.

As all the six atoms in C₂H₆ molecule being in the same plane, the molecule is planar. H-C-H bond angle is 120^o. H-C-C bond angle is 120^o.

Bonds Formation:
1. Sigma Bond Formation :

A covalent bond formed by collinear or coaxial or in the line of internuclear axis. Overlapping of orbitals is known as sigma bond. One Sp² hybrid orbital of one carbon atom overlaps with One hybrid orbital of other carbon atom by head on collision forming sigma bond. One (Sp²– Sp² ) overlap.

Remaining two hybrid orbitals of each carbon atom overlap with ‘s’ orbital of four hydrogen atoms separately forming four sigma bonds. (C – H). Four (Sp²– s) overlaps. All C-H bond in ethylene are of equal strength.

Thus there are five sigma bonds. Sigma bonds are stronger.

2. Formation of pi Bond:

The covalent bond formed by collateral or sidewise overlapping is called pi bond. The unhybridized 2 p_z orbitals of each carbon atom being perpendicular to the plane of four hydrogen atoms and carbon atoms overlap laterally with one another to form a week pi bond between two carbon atoms by p – p overlap. One (p-p) -pi bond. This bond consists of two equal electron cloud one lying above the plane of the atom and other lying below this plane.

Hence, in ethylene molecule there are 5 sigma bonds and 1 pi bond.

Diagram :

Type and Geometry of Ethylene Molecule :

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The mixing of one s-orbital and three p- orbitals of the same atom having nearly the same energy to form four orbitals of equal in all respects and tetrahedrally arranged is known as sp3 hybridization.

Geometry of sp3 Hybridization:

sp³ hybridized orbitals repel each other and they are directed to four corners of a regular tetrahedron. The angle between them is 109.5° and the geometry of the molecule is tetrahedral (non-planar). This type of hybridization is also known as tetrahedral hybridization. The sp³ hybridization is shown pictorially in the figure.

Formation of Methane Molecule ( CH₄ ):

Step -1: Formation of the excited state of a Carbon atom:

The carbon atom in the ground state takes up some energy and goes to the excited state.  In this process, a pair of electrons in 2s orbital splits up and one of the electron from this pair is transferred to empty 2pz orbital.  Thus the excited state has four half-filled orbitals.

Step – 2: Hybridization of Orbitals:

One orbital of 2s and three orbitals of 2p mix up forming four hybrid orbitals of equivalent energy. These four new equivalent orbitals are called sp³ hybrid orbitals. They are identical in all respect.

Angle and Geometry:

Four sp³ hybridized orbitals formed, repel each other and they are directed towards the four corners of a regular tetrahedron.    The Angle between them is 109.5°. Each sp³ hybrid orbital contains one unpaired electron.

In each sp³ hybrid orbital, one of the lobes is bigger because of more concentration of electron density. Only a bigger lobe is involved in bond formation. Due to this maximum overlapping is achieved.

Four sp³ hybrid orbitals of carbon atom having one unpaired electron each overlap separately with 1s orbitals of four hydrogen atom along the axis forming four covalent bonds. Thus in CH₄ molecule has a tetrahedral structure with a carbon atom at the centre and four hydrogens at the four corners of a regular tetrahedron. H-C-H bond angle is 109.5°.

Bonds:

Four sp³ hybrid orbitals of carbon atom having one unpaired electron each overlap separately with 1s orbitals of four hydrogen atom along the axis forming four covalent bonds (sigma bonds). The bonds between carbon and hydrogen are sp³– s. Thus H – C – H bond angles are 109.5°. The molecule is tetrahedral. All C-H bonds in methane are of equal strength.

Type and Geometry of Methane Molecule:

Formation of Ammonia Molecule ( NH₃):

Need of Hybridization in Ammonia Molecule:

In nitrogen atom, there are three half-filled 2p orbitals and the valency should be 3 and it is three. These three ‘p’ orbitals are perpendicular to each other and if they form bonds, the angle between them should be 90°. But the actual angle between bonding orbitals in ammonia is 107°. To explain this difference in the angle the concept of hybridization is required.

Electron Configuration:

The Atomic number of Nitrogen is 7.  Its configuration in ground state is 1s², 2s², 2p³

Hybridization:

In the formation of ammonia one 2s orbital and three 2p orbitals of nitrogen mix up forming four hybrid orbitals of equivalent energy. These four new equivalent orbitals are called sp³ hybrid orbitals. They are identical in all respect. One hybrid orbital has paired electrons (lone pair) and it is nonbonding orbital. The other three orbitals are half-filled and they are bonding orbitals. The nonbonding pair of hybridized orbitals is called as a lone pair. These hybridized orbitals are in the directions of four corners of a regular tetrahedron.

Angle and Geometry:

Four sp³ hybridized orbitals formed, repel each other and they should be directed towards the four corners of a regular tetrahedron and the angle between them should be 109.5°. One hybrid orbital has paired electrons and it is nonbonding orbital. The other three orbitals are half-filled and they are bonding orbitals. But due to greater repulsion exerted by lone pair on bonding orbitals, the angle is reduced to 107°. Thus ammonia has distorted tetrahedral shape.

The three half-filled (containing unpaired electron) sp3 hybrid orbitals of nitrogen overlap axially with three half-filled 1s orbitals of three hydrogen atoms separately to form three covalent N-H bonds Hence geometry is tetrahedral pyramidal in which the nitrogen lies at the centre, three hydrogen atoms form the base and one pair of electrons forms the apex of the pyramid. H-N-H angle is 107°.

Bonds:

The three half-filled (containing unpaired electron) sp3 hybrid orbitals of nitrogen overlap axially with three half-filled 1s orbitals of three hydrogen atoms separately to form three covalent N-H bonds (sigma bonds). The fourth hybrid orbital containing lone pair of the electron remains non bonded. The bonds between Nitrogen and hydrogen are sp3- s. H – N – H bond angles are 107°. All N-H bonds in ammonia are of equal strength.

Type and Geometry:

Formation of Water Molecule:

Need of Hybridization:

In oxygen atom there are two half-filled 2p orbitals and valency should be 2 and it is two.  These three ‘p’ orbitals are perpendicular to each other and if they form bonds, the angle between them should be 90°.  But the actual angle between bonding orbitals in water is 104.5°.  To explain this variation the concept of hybridization is required.

Electron Configuration:

The atomic number of Oxygen is 8.  Its configuration in ground state is 1s², 2s², 2p⁴

Hybridization :

In the formation of water molecule one 2s orbital and three 2p orbitals of Oxygen mix up forming four hybrid orbitals of equivalent energy. These four new equivalent orbitals are called sp³ hybrid orbitals. They are identical in all respect. The two hybrid orbitals have paired electrons and they are non – bonding orbitals.  Other two orbitals are half-filled and they are bonding orbitals. The nonbonding pairs of hybridized orbitals are called lone pairs. These hybridized orbitals are in the directions of four corners of a regular tetrahedron.

Angle and Geometry:

Four sp3hybridised orbitals formed, repel each other and they should be directed towards the four corners of a regular tetrahedron and the angle between them should be 109.5°. But due to greater repulsion exerted by lone pair on bonding orbitals, the angle is reduced to 104.5⁰. Thus water has distorted the tetrahedral shape.

The two half-filled (containing unpaired electron) sp³ hybrid orbitals of oxygen overlap axially with two half-filled 1s orbitals of two hydrogen atoms separately to form three O-H bonds. Thus geometry is angular or V-shaped, in which the oxygen lies at the centre, two hydrogen atoms occupy two corners of the tetrahedron and lone pair of electrons occupy remaining two corners of the tetrahedron. H-O-H angle is 104.5⁰.

Bonds:

The two half-filled (containing unpaired electron) sp3 hybrid orbitals of oxygen overlap axially with two half-filled 1s orbitals of two hydrogen atoms separately to form two O-H bonds (sigma bond).  The remaining two hybrid orbitals containing a lone pair of the electron remains nonbonded.

Type and Geometry:

Science > Chemistry > Physical Chemistry > Nature of Chemical Bond > sp3 Hybridization

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In this article, we shall study the concept of hybridization of orbitals. This concept overcomes the limitations of valence bond theory.

Need for Hybridization Concept:

A simple approach based on the overlap of s and p orbitals can be applied to many molecules, but it fails to explain the formation of compounds of Beryllium (Be), Boron (B), and carbon (C). The electronic configurations of Be, B, and C in the ground state are as follows.

The atoms of Beryllium Be (Z = 4)  Electronic configuration is 1s² 2s² (With no unpaired electrons).

The atom of Boron B (Z = 5)   Electronic configuration is 1s² 2s² 2p¹ (With one unpaired electron).

The atom of Carbon C (Z = 6)   Electronic configuration is 1s² 2s² 2p² (With two unpaired electrons).

According to valence bond theory valency of an element depends on a number of unpaired electrons in the orbitals. Thus Beryllium, Boron, and Carbon should be zero-valent, monovalent and divalent respectively.  However, these elements form the compounds having valency 2, 3, and 4 respectively. Valence bond theory fails to explain this phenomenon. This is explained by hybridization.

Valence Bond Theory fails to explain the observed geometry of the molecules of water and ammonia e.g. in the formation of H₂O molecule, the H – O – H bond angle should be 90°. But the measured bond angle is 104.3° and the molecule is V-Shaped. Valence bond theory failed to explain this change. To explain The equivalence of bonds we have to use the concept of a process of mixing and recasting of atomic orbitals. For e.g. all the four C-H bonds in methane molecule are equivalent in terms of strength, energy, etc. It can be explained on the basis of hybridization.

Note: The above paragraphs give limitations of the valence bond theory.

Hybridization:

Mixing and recasting or orbitals of an atom (same atom) with nearly equal energy to form new equivalent orbitals with maximum symmetry and definite orientation in space is called hybridization.

Steps Involved in Hybridization:

Step -1: Formation of excited state:

The atom in the ground state takes up some energy and goes to the excited state.  In this process, usually, a pair of electrons in lower energy orbital is split up and one of the electron from this pair is transferred to some empty slightly higher but almost equal energy orbital.  Thus the excited state has a larger number of half-filled orbitals. This new number of half-filled orbitals decides the number of covalent bonds an atom can form.

Step = 2: Mixing and recasting of atomic orbitals:

The orbitals of nearly the same energy in an excited state now hybridize i.e. unite and redistribute themselves giving hybrid orbitals of the same energy and definite orientation in space. Mixing and recasting or orbitals of an atom(same atom) with nearly equal energy to form new equivalent orbitals with maximum symmetry and definite orientation in space is called hybridization.

One 2-s orbital and three p- orbitals mix together and recast themselves to form new four sp³ hybridized orbitals.

Step – 3: Proper Orientation of the Hybrid Orbitals in Space:

The hybrid orbitals then get arranged in space in such a way to minimize mutual repulsion. Each hybrid orbital is more concentrated on one side of the nucleus. Due to this greater overlap is achieved and a stronger bond is formed.

Thus in carbon, the four hybrid sp3 orbitals arrange themselves at four corners of a tetrahedron to minimize mutual repulsion.

Essential Condition for Hybridization:

The orbitals participating in hybridization should have nearly the same energy. Thus in the formation of methane, the 2s and 2p orbitals of carbon have nearly the same energies, so that the recasting of orbitals is possible.  But hybridization of 2s and 3p is not possible because there is much difference between their energies.

Characteristics or Rules of hybridization:

• Atomic orbitals undergoing hybridization should belong to the same atom or ion.
• Atomic orbitals participating in hybridization should have nearly the same energy. Thus 2s and 2p can hybridize, 3s and 3p can also hybridize, but 2s and 3p cannot.
• The total number of hybrid orbitals formed is equal to the number of atomic orbitals involved in the hybridization process.
• All hybrid orbitals are identical with respect to energy and directional character.
• The hybrid orbitals may differ from one other in their orientations.
• The shape of the hybrid orbitals is different from that of the original atomic orbital.
• Similar to atomic orbitals, each hybrid orbital can have a maximum of two electrons.
• The hybrid orbitals are concentrated in one particular direction to achieve greater overlapping.
• The hybrid orbitals have maximum symmetry and definite orientation in space so that the mutual force of repulsion of electrons is avoided.

Types of Hybridization and Geometry of Molecules:

The hybridization involving s and p orbitals are of the following three types:

• Tetrahedral or sp³ hybridization e.g. CH₄, NH₃, H₂O
• Trigonal or sp² e.g. BF₃, C₂H₄.
• Diagonal or sp hybridization e.g. BeF₂, C₂H₂

Their names indicate the orientation of the orbitals in space and the designation (sp², sp³, etc) indicates the number and types of atomic orbitals involved in hybridization.

Science > Chemistry > Physical Chemistry > Nature of Chemical Bond > Hybridization of Orbitals

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In previous articles, we have seen formation of ionic bonds. In this article, we shall study the concept of the formation of a covalent bond.

Covalent Bonds or Electron Pair Bonds:

A bond established between two identical or different atoms by sharing one or more pairs of electrons is known as a covalent bond. The attractive force which comes into existence due to the mutual sharing of electrons between two atoms having similar electronegativity or having a small difference in electronegativities is called a covalent bond.

No ionic bond is possible between two atoms having similar electronegativities. Hence to explain the bonding between such atoms Lewis introduced the concept of covalent bonds.

Each atom contributes one electron to form a common pair i.e. equal contribution of electrons followed by equal sharing. If one electron pair is shared, it is known as a single covalent bond.  If two electron pairs are shared, it is a double covalent bond and so on. In CI2, O2, CH4, H2O, etc. there are covalent bonds.

Examples of Covalent Bonds:

Formation of Hydrogen Molecule:

Electronic configuration of hydrogen H (Z = 1) is (1) Hydrogen has one electron in its valence shell. It tries to acquire a stable configuration (2) by sharing its one electron with another hydrogen atom.

When two hydrogen atoms approach each other, at a certain distance between the nuclei, they share their valence electrons and form a shared pair of electrons. The shared pair belongs to each atom equally. As only one pair is shared between the two hydrogen atoms, they are joined to each other by a single covalent bond.

Formation of Chlorine Molecule:

Electronic configuration of chlorine Cl (Z = 17) is (2, 8, 7). Chlorine has 7 electrons in its valence shell. It can acquire a stable configuration (2, 8, 8) of the nearest inert gas (Ar) by sharing one electron with another chlorine atom.

When two chlorine atoms approach each other, at a certain distance between the nuclei, they share their valence electrons and form a shared pair of electrons. The shared pair belongs to each atom equally. As only one pair is shared between the two chlorine atoms, they are joined to each other by a single covalent bond.

Formation of Hydrogen Chloride Molecule:

Electronic configuration of hydrogen H (Z = 1) is  (1)  Hydrogen has one electron in its valence shell. It tries to acquire a stable configuration (2) by sharing its one electron with the chlorine atom. Electronic configuration of chlorine Cl (Z = 17) is (2, 8, 7). Chlorine has 7 electrons in its valence shell. It can acquire a stable configuration (2, 8, 8) of nearest inert gas (Ar) by sharing one electron with the hydrogen atom.

When hydrogen and chlorine atoms approach each other, at a certain distance between the nuclei, they share their valence electrons and form a shared pair of electrons. The shared pair belongs to each atom equally. As only one pair is shared between the hydrogen and chlorine atoms, they are joined to each other by a single covalent bond.

Formation of Ammonia Molecule:

Electronic configuration of nitrogen N (Z = 7) is (2, 5). Nitrogen has 5 electrons in its valence shell. It can acquire stable configuration (2, 8) of nearest inert gas (Ne) by sharing one electron each with three hydrogen atom.

When three hydrogen atoms approach nitrogen atom, at a certain distance between the nuclei, it shares one valence electron each with the nitrogen atom and forms three shared pairs of electrons with three hydrogen atoms. The shared pair belongs to each atom equally. As only one pair is shared between the atoms, they are joined to each other by a single covalent bond.

Formation of Water Molecule:

Electronic configuration of oxygen O (Z = 8) is (2, 6). Oxygen has 6 electrons in its valence shell. It can acquire stable configuration (2, 8) of nearest inert gas (Ne) by sharing one electron each with two hydrogen atom.

When two hydrogen atoms approach oxygen atom, at a certain distance between the nuclei, it shares one valence electron each with two hydrogen atoms and forms two shared pairs of electrons with two hydrogen atoms. The shared pair belongs to each atom equally. As only one pair is shared between the atoms, they are joined to each other by a single covalent bond.

Formation of Phosphorous Trichloride Molecule:

Electronic configuration of phosphorous P (Z = 15) is (2, 8, 5). Phosphorous has 5 electrons in its valence shell. It can acquire stable configuration (2, 8, 8) of nearest inert gas (Ar) by sharing one electron each with three chlorine atoms. Electronic configuration of chlorine Cl (Z = 17) is (2, 8, 7). Chlorine has 7 electrons in its valence shell. It can acquire a stable configuration (2, 8, 8) of the nearest inert gas (Ar) by sharing one electron with another atom.

When three chlorine atoms approach phosphorous atom, at a certain distance between the nuclei, it shares one valence electron each with the three chlorine atoms and forms three shared pairs of electrons with three chlorine atoms. The shared pair belongs to each atom equally. As only one pair is shared between the atoms, they are joined to each other by a single covalent bond.

Formation of Methane Molecule:

Electronic configuration of carbon C (Z = 6) is (2, 4). Carbon has 4 electrons in its valence shell. It can acquire stable configuration (2, 8) of nearest inert gas (Ne) by sharing one electron each with four hydrogen atoms.

When four hydrogen atoms approach phosphorous atom, at a certain distance between the nuclei, it shares one valence electron each with the four hydrogen atoms and forms four shared pairs of electrons with four hydrogen atoms. The shared pair belongs to each atom equally. As only one pair is shared between the atoms, they are joined to each other by a single covalent bond.

Formation of Ethane Molecule:

Formation of Oxygen Molecule:

Consider the formation of the oxygen molecule. Electronic configuration of oxygen O (Z = 8) is (2, 6). Oxygen has 6 electrons in its valence shell. It can acquire a stable configuration (2, 8) of nearest inert gas (Ne) by sharing two electrons with another oxygen atom. The two atoms share two pairs of electrons between them completing octet of each. two shared pairs constitute a double bond.

O        +      O        →   O=O

(2,6)      (2,6)       (2,8) (2,8)

Formation of Nitrogen Molecule:

Consider the formation of the nitrogen molecule. Electronic configuration of nitrogen (Z = 7) is (2, 5). Nitrogen has 5 electrons in its valence shell. It can acquire a stable configuration (2, 8) of nearest inert gas (Ne) by sharing three electrons with another nitrogen atom. The two atoms share three pairs of electrons between them completing octet of each. Three shared pairs constitute a triple bond.

 N        +      N        →   N≡N

(2,5)      (2,5)       (2,8) (2,8)

Formation of Ethene (Ethylene) Molecule:

Formation of Ethyne (Acetylene) Molecule:

Science > Chemistry > Physical Chemistry > Nature of Chemical Bond > Formation of Covalent Bonds

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