Hybridization , V S E P R Theory

We can explain the formation of four covalent bonds by an atom of carbon by considering promotion of a 2s electron to a 2p orbital. Let us now consider the formation of a molecule of methane, CH4, by such an excited carbon atom.

There will be three C-H bonds formed by overlap of the three 2p-orbitals of carbon with the 1s orbitals of three hydrogen atoms. The sp bonds will be mutually perpendicular to one another. The fourth C-H bond would be formed by overlap of the 2s orbital of carbon with an 1s orbital of hydrogen atom.

Since the 2s orbital is spherically symmetrical, the direction of the hydrogen atom held by this bond cannot be directly ascertained. At the same time, one should expect this s-s bond to be of lower strength than the other three s-p bonds. But it is contrary to our experience.

We know that all four C-H bonds in methane are all alike and they are arranged symmetrically around the central carbon atom directed along the four corners of a tetrahedron.

This necessitates a concept of mixing the s-orbital of carbon with its three p-orbitals before overlap. The resulting equivalent orbitals, each having one-fourth s-character and three – fourth p-character, may now undergo overlap with four hydrogen atoms to form four equivalent C-H bonds.

We may reasonably extend this concept to interpret the equivalence of the two bonds in BeF2 or the three bonds in BCl3. This procedure of prior mixing of the orbitals has been given rigorous mathematical formulation.

Hybridization is a concept of mixing different atomic orbitals of comparable energy resulting in an equal number of orbitals with mixed character.

The resulting hybrid orbitals undergo better overlap and form stronger bonds than the pure orbitals in conformity with the most stable geometry for a molecule.

The formation of four equivalent C-H bonds by carbon in forming methane may then be conceived of in terms of the following successive steps:

At first, a 2s electron of the carbon atom gets unpaired and promoted to a 2p orbital. The 2s and the three 2p orbitals are now hybridized to give four equivalent orbitals, each possessing one part s character to three parts p character in their wave function, directed to the corners of a regular tetrahedron.

These sp3 hybrid orbitals now form four equivalent C-H bonds in the methane molecule; the bonds are distributed tetrahedrally around the carbon atom.

Shape……………….. Hybridisation

Linear………………… sp

Trigonal planar……… sp2

Tetrahedral………….. sp3

Trigonal bipyramidal… sp3d

Octahedral………….. sp3d2

Pentagonal bipyramidal….. sp3d3



This theory starts from the general principle that valence shell electrons occupy essentially localised orbitals. Mutual interaction among the electrons orient the orbitals in space to an equilibrium position where repulsion becomes minimum. The extent of repulsive interaction then follows the order.

Lone pair – lone pair > lone pair – bond pair> bond pair – bond pair:

A lone pair is concentrated around the central atom while a bond pair is pulled out between two bonded atoms. As such repulsion becomes greater when a lone pair is involved.

Let’s take an example to illustrate this theory. CH4 contains no lone pairs. The bond pair – bond pair interactions brings about the most stable equilibrium bond angle of 109°28′ the angle predicted from sp3 hybridisation

Illustration : Why the bond angle of H – C – H in methane (CH4) is 109° 28′ while H – N – H bond angle in NH3 is 107° though both carbon and nitrogen are sp3 hybridized.

Solution: In CH4 there are 4 bond pair of electrons while in NH3 are 3 bond pair of electrons and 1 lone pair of electrons. Since bond pair bond pair repulsion is less than lone pair bond pair repulsion, in NH3 bond angle is reduced from 109°28′ to 107°

Illustration :  Why bond angle in NH3 is 107° while in H2O it is 104.5° . Though the central atom in both the molecules is hybridized.

Solution: In NH3 , central nitrogen atom bears only one lone pair of electrons whereas in H2O central oxygen atom bears two lone pair of electrons.

Since the repulsion between lone pair – lone pair and lone pair – bond pair is more than that between bond pair – bond pair, the repulsion in H2O is much greater than that in NH3 which results in contraction of bond angle from 109°28″ to 104.5° in water while in NH3 contraction is less i.e. from 109°28″ to 107°.

” If the electronegativity of the peripheral atoms is more, then the bond angle will be less “. For example – if we consider NH3 and NF3 , F – N – F bond angle will be lower than H – N – H bond angle. This is because in NF3 the bond pair is displaced more towards F and in NH3 it is displaced more towards N. So accordingly the b.p. – b.p. interaction is less in NF3 and more in NH3.

Illustration : The bond angle of H2O is 104° while that that of F2O is 102° .

Solution: Both H2O and F2O have a lone pair of electrons. But fluorine being highly electronegative, the bond pair electrons are drawn more towards F in F2O, whereas in H2O it is drawn towards O.

So in F2O the bond pairs being displaced away from the central atom has very little tendency to open up the angle. But in H2O this opening up is more as the bond pair electrons are closer to each other. So bond ∠ of F2O is less than H2O.

Exercise : Among PCl3 & PF3 which is having greater bond angle and why ?

” If the electronegativity of central atom is more then bond angle will be more ” .

For example in NH3 and PH3 , the H – N – H bond angle is more then H – P – H bond angle. The reason for it can be explained in the same way as above.

Illustration : Out of H2O and H2S which is having greater bond angle and why ?

Solution : H2O is having greater bond angle than H2S since oxygen is more electronegative than S and draws the shared pair of electrons toward itself more than S. Therefore the bond pair interaction is more in case of H2O.

Rules for determination of total no. of hybrid orbitals

∎ Detect the central atom along with the peripheral atoms.

∎ Count the valence electron of the central atom and the peripheral atoms.

∎ Divide the above value by 8. Then the quotient gives the number of σ bonds and the remainder gives the non-bonded electrons. So

$\large Number of lone pair = \frac{Non \;bonded \ electrons}{2}$

∎ The number of σ 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 x 7) = 34

Number of hybrid orbital = 4σ bonds + 1 lone pair

So 5 hybrid orbitals are necessary and hybridisation mode is sp3d and it is trigonal bipyramidal (TBP)

Both the structures are TBP. But the lone pair is placed in different position. In B it is placed in equatorial position and in A it is in axial.

Now when a lone pair is in equatorial position the repulsion are minimized. So structure (B) is correct.

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 σ bonds + 1 lone pair

Hybridsation = sp3

Shape = pyramidal

2. BBr3    Total valence electron = 24

Requirement = 3σ bonds

Hybridisation = sp2

Shape = planar trigonal

3. SiCl4    Total valence electrons = 32

Requirement = 4σ bonds

Hybridisation = sp3

Shape = Tetrahedral

4. CI4 Total valence electron = 32

Requirements = 4 σ bonds

Hybridisation = sp3

Shape = Tetrahedral

5. SF6 Total valence electrons = 48

Requirement = 6 σ bonds

hybridisation = sp3d2
shape = octahedral / square bipyramidal

6. BeF2 Total valence electrons : 16

Requirement : 2 σ bonds

Hybridisation : sp

Shape : Linear

7. ClF3 Total valence electrons : 28

Requirement : 3 σ bonds + 2 lone pairs

Hybridisation : sp3d

Shape : T – shaped

Also Read :

Maximum Covalency & Resonance
Deviation from ideal behaviour & FAJAN’S RULE
Role of φ ( ionic Potential )
Hydrogen Bonding
Intermolecular Forces

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