Hybridization

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What is a VBT or Valence Bond Theory?

Pauli and Slater introduced VBT.

The theory considers two bonds, i.e., σ and π-bonds. 

So, what are σ and π- bonds?

σ-bonds form by axial/head-on overlapping whereas π- bonds by parallel/side-by-side overlapping.

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Let’s take an example of CH4 (Methane) to understand VBT:

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The electronic configuration of C = 1s2 2s2 2p2, and H = 1s  

The carbon has: 

1⇂

1

1


|--2s-| |---------- 2p ------------|            

One electron from 2s migrates to 2p to get a stable configuration. This is the first excited state.

1

1

1

|--2s-| |---------- 2p ------------|    

Each H has one electron, i.e., 

1

  1s

Since opposite spins make bonds, so the first bond forms between 1s and 2s, second, third and fourth bonds between one of the spins of 2p and 1s of H-atoms.

Four bonds formed of CH4 are s-s, s-p, s-p, and s-p. 

The VBT says the two bonds, s-s and s-p are different in their bond length and energies while hybridization considers all four bonds equivalent in terms of bond length and energy. Here, VBT failed to explain how these two are equal. This led to the birth of hybridization. 


What is Hybridization?

Hybridization says that all s-s, s-p, s-p, s-p bonds are similar. 

It states that atomic orbitals of the same energy or close to similar energy combine or intermix to form a new orbital of the same energy.

The present orbitals are of CH4:

H                                

1

1s

C

1

1

1

|--2s-| |---------- 2p ------------|   

Electronic configuration of CH4 =  1s and 3p so, sp3.

After intermixing, four new (hybrid) orbitals formed with each having the same energy: 

1

1

1

1

    sp3        sp3        sp3        sp3

Hydrogen combines with each sp3 orbital. The geometry so formed is shown below                                                 

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The carbon formed all four orbitals as sp3. Therefore, the bond formed after intermixing of C and H is sp3-s, sp3-s, sp3-s, and sp3-s. 

Hybridization is responsible for the geometry of the compound.

If we look at Fig.1, it forms the tetrahedral geometry for the minimum repulsion between these orbitals in the space.

Here, sp3 is a hybridization. 

Let’s discuss the other types.

The shape of hybrid orbitals

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Types of Hybridization

Hybridization can occur in other types of orbitals, that is:

  • Half-filled orbitals,

  • Empty orbitals, and

  • Filled orbitals.


1. Sp Hybridization

2s  + 2p =  sp  sp (two new sp orbitals)  

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Let’s take an example of  BeF2

Be = 1s2 2s2, and F = 1s2 2s22p5

Atomic number of Be = 4, the electronic configuration:   

1⇂

1⇂

    1s2                2s2

An atomic number of F = 9, the electronic configuration:

1⇂

1⇂

1⇂

1⇂

1

    1s2      2s2    |-----------2p5-------------|

F is ready with 2p orbital, i.e.,

1⇂

1⇂

1

|-----------2p5-------------|         

Be has a paired electron in the s-shell; how does it bond with F?

As the 2p-orbital of Be is empty, one electron from 2s migrates to 2p, for a p-shell to form a stable configuration.

1

1⇂

    2s          2p

So, one bond forms between the s of Be and p of, and we are not sure how the second bond between Be and F will form.

So, the 2s and 2p of Be combine with itself to form two new orbitals, i.e., 

1

1

    sp         sp

So, two bonds form between sp of Be and p of F, as shown below:

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Geometry: Linear

The same happens in C2H2, Be2Cl2, and BeBr2.


2. Sp2 hybridization

1s

p

p

1s + p + p = Three new orbitals

sp2

sp2

sp2

Let’s take an example of BF3.

An atomic number of B = 5, the electronic configuration:  

1⇂

1



      2s|-----------2p1-------------|              

For F,

1⇂

1⇂

1

|-----------2p5-------------|

B needs three unpaired electrons, while F already has one unpaired electron.

In the first excited state, one electron from 2s to 2p

1

1

1


    2s        |-----------2p-------------| 

  1. One bond of 2s of Be with 2p of F, and

  2. Two bonds of 2p of Be with 2p of F.

Three bonds formed are s-p, p-p, and p-p, where s-p and p-p are different.

So, intermixing of orbitals B, gives three hybrid orbitals, i.e.,

1

1

1

    sp2         sp2                sp2

Geometry: Triangular planar

Another example of C2H4

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3. Sp3 Hybridization

1s + p + p + p = Four orbitals of sp3.

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An example of sp3 has already been explained above.


4. Sp3d Hybridization

s + p + p + p + d = 5 orbitals of sp3d.

P has a configuration as 3s2 3p3

1⇂

1

1

1






      3s2|-----------  3p3 ----------||-----------          3d0              -------------| 

Configuration of Cl = 3s2 3p5

1⇂

1⇂

1⇂

1






      3s  |-----------3p-------------|       |-----------          3d            -----------|

In the first excited state, Cl forms:

1

1⇂

1

1





|-----------3p-------------||-----------          3d          ---------|

In the second excited state, it becomes:

1

1

1

1

1




|-----------3p-------------||-----------          3d          ---------|

So, now Cl has five unpaired electrons.

Now, s, p, and d-orbitals of P combine with itself to form unpaired electrons to form a new hybrid orbital, i.e.,

1

1

1

1

1

    sp3d    sp3d     sp3d    sp3d      sp3d

As you can see in Fig.a

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Geometry: Trigonal bipyramidal (because of two pyramids).

FAQ (Frequently Asked Questions)

Q1: Write an Example for sp3d2 Hybridization.

Ans: We can express the hybridization for sp³d² as:

1s + 3p + 3d = 6 orbitals of sp³d².

For example, SF₆

The configuration of F is 1s²2s²2p⁵:

1⇂

1⇂

1⇂

1

    2s      |------    2p            --------| 

Configuration of S = 3s² 3p⁴,i.e.,

1⇂

1⇂

1

1






3s²   |-----------  3p³ ----------||-----------          3d⁰              -------------|    

 After intermixing, the configuration will be:

1

1

1

1

1

1

 sp³d²  sp³d²     sp³d²      sp³d²    sp³d²       sp³d²

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Geometry: Octahedral

Q2: Is Triple Bond an Sp?

Ans: Yes, a triple bond is sp. An example of it is C₂H₂.

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Q3: Is Sp² Stronger than sp³?

Ans: Yes. It’s because the bond between sp³ and sp² is stronger than the bond between sp³ and sp³. Therefore, sp² hybridized orbitals contain 33.33% s-character, while sp³ contains 25% s-character.

Q4: Write an Example of sp³d³  Hybridization.

Ans: Let’s take an example of IF₇.

1s + 3p + 3d =  7 orbitals of sp³d³ 

The electronic configuration of I = 4d¹⁰ 5s² 5p⁵.

1⇂

1⇂

1⇂

1






5s²   |-----------  5p⁵ ----------||-----------          5d⁰              -------------|   

First excited state 

1⇂

1

1⇂

1

1





5s   |-----------  5p ----------||-----------          5d⁰              -------------| 

Second excited state

1⇂

1

1

1

1

1




  5s   |-----------  5p ----------||-----------          5d⁰              -------------| 

 Third excited state

1

1

1

1

1

1

1

  sp³d³   sp³d³      sp³d³     sp³d³     sp³d³         sp³d³        sp³d³