# Hybridisation

Hybridisation

Pauling (1931) introduced the revolutionary concept of hybridization. The redistribution of energy of orbitals of individual atoms to give new orbitals of equivalent energy is called hybridisation. The new orbitals formed are known as hybrid orbitals.

Different types of hybridisation along with hybrid orbitals and structures are given below:

Before discussing the examples, we must mention here the hybridisation rules, which are as follows:

(i) Orbitals of a central atom only would undergo hybrodisatoin.

(ii) The orbitals of almost same energy level can be mixed to form hybrid orbitals.

(iii) The numbers of atomic orbitals mixed together are always equal to the number of hybrid orbitals.

(iv) During hybridisation, the mixing of number of orbitals is as per requirement.

(v) The ybrid orbitals are distributed in space and tend to the farthest apart.

(vi) Hybrid bonds are stronger than the non-hybridised bonds.

(vii) If once an orbital has been used to build a hybrid orbital it is no longer available to hold electrons in its ‘pure’ form. s- and p- orbitals can be hybridized in three ways, which are discussed below:

• Sp- Hybridisation: In such hybridisation once s- and one p-orbital are mixed to form two sp-hybrid orbitals, having

linear structure with bond angle $180^0$. For example in the formation of $BeCl_2$, first be atom comes in excited state $(2s^1 2p^1)$, then hybridized to form two sp-hybrid orbitals. These hybrid orbitals overlap with the two p-orbitals of two chlorine atoms to form $BeCl_2$ molecule. This is represented in figure given above:

be (excited state) $\dfrac{\uparrow}{2s} \dfrac{\uparrow}{2p_x} \dfrac{}{2p_y} \dfrac{}{2p_z} \dfrac{\text{sp hy bri-}}{\text{disation}} \dfrac{\uparrow}{sp} \dfrac{\uparrow}{sp} \dfrac{}{2p_y2p_z}$

its other examples are: $CO, CO_2, C_2H_2, HCN, CN^-, N^3_3$ etc.

• sp2-Hybridisation: In such hybridisation one s- and tow p-orbitals are mixed form three $sp^2$- hybrid orbitals, having planar triangular structure with bond angle $120^0$. The formation of $BCl_3$ molecule is shown on next page.

$\underset{\text{excited state}}{B} \dfrac{\uparrow}{2s}\dfrac{\uparrow}{2p_x} \dfrac{\uparrow}{2p_y} \dfrac{}{2p_z} \dfrac{sp^2 \text{hybrid}- \uparrow}{\text{disation} sp^2} \dfrac{\uparrow}{sp^2}\dfrac{\uparrow}{sp^2} \dfrac{}{2p_z} \\[3mm] \text{its other examples are} CO^{2-}_3, SO_2, SO_3, C_2H_4$ etc

• sp3-Hybridisation: In such hybridisation one s- and three p-orbitals are mixed to form four $sp^3$-hybrid orbitals having tetrahedral structure with bond angle $109^0 28$‘ I.e., $109.5^0$.the formation of $CH_4$ molecule is shown below:
$\underset{\text{excited state}}{C} \dfrac{\uparrow}{2s}\dfrac{\uparrow}{2p_x} \dfrac{\uparrow}{2p_y} \dfrac{}{2p_z} \dfrac{sp^3 \text{hybrid}- \uparrow}{\text{disation} sp^3} \dfrac{\uparrow}{sp^3}\dfrac{\uparrow}{sp^3} \dfrac{\uparrow}{sp^3}$

Its other examples are $C_2H_6, H_2O, NH_3, NH^+_4, SO^{2-}_4, ClO^-_4$ etc.

Now we discuss some other interesting examples:

Formation of NH3 and  H2O molecules

In $NH_2$ molecule nitrogen atom is $sp^3$-hybridised and one hybrid orbital contains two electrons. Now three 1s- orbitals of three hydrogen atoms overlap with three $sp^3$ hybrid orbitals to form $Nh_3$ molecule. Although the angle HNH should be $109.5^0$, but due to the presence of one occupied $sp^3$- hybrid orbital the angle decreases to $107.8^0$. Hence the bond angle in $NH_3$ molecule is $107.8^0$.

$\dfrac{\downarrow \uparrow}{2s} \dfrac{\uparrow}{2p_x} \dfrac{\uparrow}{2p_y} \dfrac{\uparrow}{2p_z} \dfrac{sp^3 \text{hybrid}-\downarrow \uparrow}{sp^3} \dfrac{\uparrow}{sp^3} \dfrac{\uparrow}{sp^3} \dfrac{\uparrow}{sp^3}$

Formation of NH3  and H2O  molecules by sp2 hybridization

Similarly in $H_2O$ molecule, oxygen atom is $sp^3$- hybridized and has tow occupied orbitals. Due to this the bond angle in water molecule is $105.5^0$.

$O \dfrac{\downarrow \uparrow}{2s} \dfrac{\downarrow \uparrow}{2p_x} \dfrac{\uparrow}{2p_y} \dfrac{\uparrow}{2p_z} \dfrac{sp^3 \text{hybrid-} \downarrow \uparrow}{sp^3} \dfrac{\uparrow}{sp^3} \dfrac{\uparrow}{sp^3} \dfrac{\uparrow}{sp^3}$

Formation of C2H4  and C2H2 molecules

In $C_2H_4$ molecule carbon atoms are $sp^2$-hybridised and one 2p-orbital remains out to hybridisation. This forms p-bond while $sp^2$ –hybrid orbitals form sigma- bonds as shown below:

$\underset{\text{excited state}}{C} \dfrac{\uparrow}{2s}\dfrac{\uparrow}{2p_x} \dfrac{\uparrow}{2p_y} \dfrac{}{2p_z} \dfrac{sp^2 \text{hybrid}- \uparrow}{\text{disation} sp^2} \dfrac{\uparrow}{sp^2} \dfrac{\uparrow}{sp^2} \dfrac{\uparrow}{sp^2} \dfrac{\uparrow}{2p_z}$

Formation of C2H4 molecule by sp2 hybridisation

Similarly, in $C_2H_2$ molecule there is sp-hybridisation and two 2p-orbitals remain out or hybridisation. Therefore two $\pi$ –bonds are formed in $C_2H_2$ as shown ahead:

Formation of C2H2 molecule by sp hybridisation

$dsp^2$ – Here intermixing of $d_{x^2- y^2}, s, p_x \text{and} p_y$ orbital takes place to give four new $dsp^2$ hybrid orbitals.

• Shape is square planar.

Ex.

$[Ni(CN)_4]^-2$
• Mainly for complexes with a co-ordination no. 4.

$sp^3d$- Here intermixing of $s, p_x, p_y, p_z \text{and} dz^2$. Takes place to form 5 new hybrids which are $sp^3$ hybridized. Shape is Trigonal bipyramidal.

 Steps $CH_4$ $SO_4$ $CO_2$ $NH_3$ $H_2 O$ $SO_4$ $NO_3$ 1. No. of Valence Electrons 8 18 16 8 8 32 24 2. No. of required orbitals 4 2 2 3 2 4 3 3. Required electrons for duplex/ octet 8 16 16 6 4 32 24 4. No. of lone pairs of electrons [(i) –(iii)]/2 0 1 0 1 2 0 0 5. No. of orbitals 4 3 2 4 4 4 3

 1 Hybridisation $SP^3$ $SP^2$ SP $SP^3$ $SP^3$ $SP^3$ $SP^2$ 2 Structure Thetrahedral Triangular Linear Tetrahedral Tetrahedral Tetrahedral Triangular 3 Geometry Tetrahedral Angular Linear Triangular Pyramidal Angular Tetrahedral Triangular

E.g., $PCl_5, XeF_2, I^-_3$

$sp^3d^2$- Here intermixing of $s, p_x, p_y, p_z, d_{z2}, d_{x^2- y^2}$ takes place to give 6 new $sp^3d^2$ hybridized orbitals or hybrids.

$\text{Shape} \to \text{Octahedral}$

E.g., $SF_6, XeF_4$ etc.

Rules for the Calculation of Hybridisation

The following rules are observed to know the type of hybridisation in a compound or an ion.

(i) Calculate total number of valence electrons.

(ii) Calculate the number of duplex or octet.

$= \dfrac{\text{Total valence electrons}}{2}$

Or $\text{Total valence electrons}{8}$

(iii) Number of lone pairs of electrons

$\text{Total number of electrons} = \dfrac{-8 \times \text{Number of duplex}}{2}$ $\text{Total number of electrons} = \dfrac{-8 \times \text{Number of octet}}{2}$

(iv) NO. of used orbital = No. of duplex or octet + No. of lone pairs of electrons

(v) If there is no lone pair of electrons then geometry of orbitals and molecule is different.

For example some molecules and ions are considered

Alternative method to find hybridisation

• Count total valence $e^- \pm$ chare if present

Ex.

$SO^{-2}_4 \to 6 + 4 \times 6 + 2 = 32 \\ 4HN^+_4 \to 5 + 4 = 8$

Now, if this toal is:

$\begin{bmatrix}2 \text{to} 8 \div 2 \\ 10 \text{to} 56 \div 8 \\ 58 \text{to more} \div 18 \end{bmatrix}$ Quotient (X) [Gives bp]

• Only for covalent molecules or ions

If remainder is left divide again similarly to get quotient Y (Y is equal $e^-$)

$\text{Now, if X or} X + Y \to 2 (sp) \\ \hspace{20 mm} \to 2(sp^2) \\ \hspace{20 mm} \to 4 (sp^3) \\ \hspace{20 mm} \to 5(sp^3d) \\ \hspace{20 mm} \to 6(sp^3 d^2) \\ \hspace{20 mm} \to 7 (sp^3d^3) \\ e.g. I^-_3 \to 3 \times 7 + 1 = 22 \\ \hspace{20 mm} 2 + 3 = 5 \\ \hspace{30 mm} = sp^3d$

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