Valence Bond Theory And Hybridisation

Sigma and Pi Bonding: 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 ($\sigma$-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 $\sigma$-bond. A $\sigma$-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 p_x may undergo "head –on" overlap to form a $\sigma$-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 p_y orbitals, and the other by the two p_z orbitals. These overlaps are different from the type of overlap in a $\sigma$-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 $\sigma$-bond; instead, it possess a plane of symmetry. For the overlap of the p_z atomic orbital, the xy plane provides this plane of symmetry; for the overlap of the p_y atomic orbitals, the zx plane serves the purpose. Bonds arising out of such orientation of the bonding electron cloud are designated as $\pi -$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 $\pi -$bond. A $\pi -$bond possess a plane of symmetry, often referred to as the nodal plane.

$\sigma$-Bond

$\pi -$ Bond: This type of bond is formed by the sidewise or lateral overlapping of two half filled atomic orbitals.

HYBRIDIZATION

The tetravalency shown by carbon is actually due to excited state of carbon which is responsible for carbon bonding capacity.

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.

(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

(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.

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 $\pi$ bond as in $H_{2}C=C{H}_2$

(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.

The remaining two unhybridised p orbitals overlap with another unhybridised p orbital leading to the formation of triple bond as in HC $\equiv$ CH.

Shape Hybridisation

Linear sp

Trigonal planar sp²

Tetrahedral sp³

Trigonal bipyramidal sp³d

Octahedral sp³d²

Pentagonal bipyramidal sp³d³

Example 1. 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.

Rule for determination of total number of hybrid orbitals

$\bullet$ Detect the central atom along with the peripheral atoms.

$\bullet$ Count the valence electrons of the central atom and the peripheral atoms.

$\bullet$Divide the above value by 8. Then the quotient gives the number of $\sigma$ bonds and the remainder gives the non-bonded electrons. So number of lone pair

=$\frac{non\text{ bonded electrons }}{2}$.

$\bullet$ The number of $\sigma$ 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

$\therefore$ total number of valence electrons = 6+(4 x7) = 34

Now 34/8= 4$\frac{2}{8}$$\therefore$ Number of hybrid orbitals = 4$\sigma$ bonds + 1 lone pair)

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 $\sigma$ bonds + 1 lone pair

Hybridization = sp³

Shape = pyramidal

2. BBr₃ Total valence electron = 24

Requirement = 3$\sigma$ bonds

Hybridization = sp²

Shape = planar trigonal

3. SiCl₄ Total valence electrons = 32

Requirement = 4$\sigma$ bonds

Hybridization = sp³

Shape = Tetrahedral

4. CI₄ Total valence electron = 32

Requirements = 4 $\sigma$ bonds

Hybridization = sp³

Shape = Tetrahedral

5. SF₆ Total valence electrons = 48

Requirement = 6 $\sigma$ bonds

hybridization = sp³d²

shape = octahedral / square bipyramidal

6. BeF₂ Total valence electrons : 16

Requirement : 2 $\sigma$ bonds

Hybridization : sp

Shape : Linear

F – Be – F

7. ClF₃ Total valence electrons : 28

Requirement : 3 $\sigma$ 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 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 $\sigma$ bonds

Hybridization : sp³d

Shape : Trigonal bipyramidal (TBP)

9. XeF₄ Total valence electrons : 36

Requirement:4$\sigma$ 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.

10. XeF₂ Total valence electrons : 22

Requirements : 2$\sigma$ bonds + 3 lone pairs

Hybridisation: sp³d

Shape : Linear

[$\therefore$ 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 $\sigma$ 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.

12. $P{O}_4^{3-}$ Total valence electrons : 32

Requirement : 4 $\sigma$ bonds

Hybridisation: sp³

Shape: tetrahedral

13. NO₂^– Total valence electron: 18

Requirement : 2$\sigma$ bonds + 1 lone pair

Hybridisation: sp²

Shape: angular

14. CO₃^2– Total valence electrons: 24

Requirement = 3 $\sigma$ bonds

Hybrdisation = sp²

Shape: planar trigonal

But C has 4 valence electrons of these 3 form $\sigma$ bonds $\therefore$ the rest will form a $\pi$ bond.

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$\sigma$ bonds

Hybridisation: sp

Shape: linear O = C = O

16. $B{F}_4^{-}$ Total valence electrons = 32

Requirement= 4 $\sigma$ bonds

Hybridisation: sp³

Shape: Tetrahedral

17. $Cl{O}_3^{-}$ Total valence electron = 26

Requirement = 3 $\sigma$ bond + 1 lone pair

Hybridization: sp³

Shape: pyramidal

18. XeO₂F₂ Total valence electrons : 34

Requirement: 4 $\sigma$ bonds +1 lone pairs

Hybridization : sp³d

Shape: Distorted TBP (sea-saw geometry)

19. XeO₃ Total valence electrons : 26

Requirement: 3 $\sigma$ bonds + 1 lone pair

Hybridization: sp³

Shape: Pyramidal

20. XeOF₄ Total valence electrons : 42

Requirement: 5 $\sigma$ bonds + 1 lone pair

Hybridization: sp³d²

Shape: square pyramidal.

Example 2. Predict the hybridization for the central atom in$POC{l}_3$, $OS{F}_4$, $\text{OI}{\text{F}}_{\text{5}}$

Solution: $POC{l}_3$Total No. of V.E. = $\frac{5+6+21}{8}=\frac{32}{8}=4$

So, hybridization = $s{p}^3$

$OS{F}_4$= $\frac{6+6+28}{8}=\frac{40}{8}=5$

So, hybridization of s = $ds{p}^3$

\text{OI}{{\text{F}}_{\text{5}}}\]=\dfrac{6+7+35}{8}=\dfrac{48}{8}=6$

So, hybridization of I = $d^{2}s{p}^3$