Exemplars of Hybridization in Chemistry

Hybridization is a very frequently used word of chemistry. Though in simple words it means intermixing, but the exact definition of Hybridization is as follows:

Hybridization is the mixing of two orbital molecules to form new orbital molecules. This is basically aimed at offering maximum stability to the molecule. There can be a variety of hybridization’s like:

• sp
• sp2
• sp3
• sp3d
• sp3d2
• sp3d3

  

We list here some examples of sp hybridization in detail:

1)    Beryllium Chloride (BeCl₂)

‘Be’ has electronic configuration as 1s2 2s2 in the ground state. Due to the absence of any unassociated electron, excitation takes place fostering its one 2s electron into the empty 2p orbital. Thus, in this phase, the electronic configuration is  1s2 2s1 2p1.

BeCl2 is linear in shape. In case the beryllium atom formed bonds using pure orbitals, the shape of the molecule might have been angular. Hence for this reason, the below given sp hybridization was proposed.

During the sate of excitation, the atom undergoes sp hybridization involving the mixing of 2s and 2p orbitals. Thus two half filled ‘sp’ hybrid orbitals are formed, which are arranged linearly. These half filled Sp-orbitals form two σ bonds with two ‘Cl’ atoms and hence this leads to the linear shape of Thus BeCl2 with the bond angle of 180^o.

2)    Acetylene (C₂H₂)

The ground state configuration of ‘C’ being 1s2 2s2 2px12py1 has only two unassociated electrons. But carbon makes four bonds since its valency is four. For this 4 unpaired electrons are required. Hence it promotes its 2s electron into the empty 2p_z orbital in the excited state.

Thus in the excited state, the electronic configuration of carbon is 1s² 2s¹       2p_x¹2p_y¹2p_z¹. Each carbon atom undergoes ‘sp’ hybridization by using a 2s and one 2p orbitals in the excited state to give two half filled ‘sp’ orbitals, which are arranged linearly.

By the use of Sp-orbitals, a σsp-sp bond is formed between the two carbon atoms. At the same time there are two unhybridized p orbitals i.e., 2p_y and 2p_z on each carbon atom which being perpendicular to the Sp-orbitals form two π_p-p bonds between the two carbon atoms. Hence this results in the formation of a triple bond between the carbon atoms. Each carbon also forms a σ_sp-s bond with the hydrogen atom.

Result: Acetylene molecule has bond angle as 180 o and is linear in shape.

 

sp² HYBRIDIZATION

We now discuss some examples of sp2 hybridization

1)    Boron trichloride (BCl₃)

Ground state configuration being 1s² 2s² 2p¹, it has only one unpaired electron. Since three unassociated electrons are required so there is promotion of one of 2s electron into the 2p sublevel by absorbing energy.

Thus the configuration becomes 1s² 2s² 2p_x¹2p_y¹.

However to account for the trigonal planar shape of this BCl₃ molecule, sp² hybridization before bond formation was advocated.

Boron experiences sp² hybridization by using a 2s and two 2p orbitals in the excitation state to give three half-filled sp² hybrid orbitals which are oriented in trigonal planar symmetry. It then forms three σ_sp-p bonds with three chlorine atoms by using its half-filled sp² hybrid orbitals. For the σ-bond formation, the half-filled p-orbital are used.

Result: with bond angle 120o , the shape is trigonal.

2)    Ethylene (C₂H₄)

In the state of excitation, each carbon atom experiences sp² hybridization by mixing 2s and two 2p orbitals to give three half-filled sp² hybrid orbitals oriented in trigonal planar symmetry. Another unhybridized 2p_z orbital is also present on each carbon perpendicular to the plane of sp² hybrid orbitals in half-filled state.

The carbon atoms form a σ_sp2_-sp2 bond with each other by using sp² hybrid orbitals. Lateral overlapping between the unhybridized 2pz orbitals also leads to the formation of a Ï€_p-p bond. Thus there is a double bond (σ_sp2_-sp2 &  Ï€_p-p) between two carbon atoms. Each carbon atom also forms two σ_sp2_-s bonds with two hydrogen atoms.

Result: Hence it is planar and has the ∠HCH & ∠HCC bond angles equal to 120^o

 SP³ HYBRIDIZATION

1) Methane (CH₄)

In the state of excitation, the carbon atom undergoes sp³ hybridization which involves the mixing of one ‘2s’ and three 2p orbitals to furnish four half-filled sp³ hybrid orbitals, which are oriented in tetrahedral symmetry in space around the carbon atom. Each one of these sp³ hybrid orbitals forms a σ_sp3_-s bond with one hydrogen atom. Thus carbon forms four σ_sp3_-s bonds with four hydrogen atoms. Result: The shape is tetrahedral and it has bond angle of 109^o28’.

 2) Ethane (C₂H₆)

The case is parallel to that of methane. It experiences sp³ hybridization in the excited state which yields four sp³ hybrid orbitals in tetrahedral geometry.

The overlapping of sp3 hybrid orbitals along the inter-nuclear axis leads to the formation of a σ_sp3_-sp3bond. Each carbon atom also forms three σ_sp3_-s bonds with hydrogen atoms.

Result: Hence ethane has tetrahedral symmetry around each carbon with ∠HCH & ∠HCC bond angles equal to 109^o28′.

3) Ammonia (NH₃)

The configuration in ground state being 1s2 2s2 2px12py12pz1 shows that there are three unassociated electrons in the 2p sublevel and hence the nitrogen atom can form three bonds with three hydrogen atoms. This results in the bond angle of 90^o differing from the generally reported angle of 107^o48′.

Hence it was advocated for the Nitrogen to undergo sp³ hybridization of a 2s and three 2p orbitals to give four sp³ orbitals, which henceforth are organized in a tetrahedral symmetry. Since it reduces the number of repulsions so the molecule is bound to become more stable. It includes one full filled and three half filled.

Nitrogen atom forms 3 σ_sp3_-s bonds with three hydrogen atoms by using three half-filled sp³ hybrid orbitals. A lone pair also exists on the nitrogen atom which requires more space than the bond pairs and belongs to full filled sp³ hybrid orbital.

Nevertheless, ∠HNH bond angle does not equal the normal tetrahedral angle: 109^o28′ which is also different from the reported angle of 107^o48′. The reduction in the bond angle is caused by the repulsion of lone pair over the bond pairs.

Result: The ammonia is trigonal pyramidal and has a lone pair on nitrogen atom.

 

4) Water molecule (H₂O)

Oxygen atom has two unpaired electrons which are free to combine with the hydrogen atoms. But the only restriction is that the resulting bond angle should be 90^o. The electronic configuration of oxygen is 1s2 2s2 2px22py12pz1. The reported bond angles were 104^o28′. To account this, sp³ hybridization before the bond formation was proposed.

Formation of water molecule involves the sp3 hybridization of the oxygen atom by mixing of a 2s and three 2p orbitals to furnish four sp3 hybrid orbitals oriented in tetrahedral geometry. This comprises two half-filled and two completely filled. The half-filled hybrid orbitals are then used by the oxygen atoms to form two σ_sp3_-s bonds. The reported bond angle is 104^o28′ instead of regular tetrahedral angle: 109^o28′. It is again due to repulsions caused by two lone pairs on the bond pairs.

Result: Water molecule possesses angular shape (V shape).

Sp³d HYBRIDIZATION

1) Phosphorus pentachloride(PCl₅)

The ground state electronic configuration of phosphorus atom is: 1s² 2s²2p⁶ 3s²3p_x¹3p_y¹3p_z¹. Five unassociated electrons will suffice for the formation of PCl₅ molecule. Hence the phosphorus atom undergoes excitation to promote one electron from 3s orbital to one of empty 3d orbital. As a result, in the state of excitation, the electronic configuration of ‘P’ becomes 1s² 2s²2p⁶ 3s²3p_x¹3p_y¹3p_z¹ 3d¹.

In the state of excitation, the formation of five half-filled sp3d hybrid orbitals involves the intermixing of one 3s, three 3p and one 3d orbitals. They are then organized in a trigonal bipyramidal symmetry. While three orbitals are organized in trigonal planar symmetry, the rest two are kept perpendicularly above and below this plane. These half-filled sp³d orbitals aid phosphorous in forming five σ_sp3_d-p bonds with chlorine atoms. Each chlorine atom makes use of half-filled 3p_z orbital for the bond formation.

Result: PCl₅ molecule has a trigonal bi pyramidal shape with 120^o and 90^o of ∠Cl – P – Cl bond angles.

SP3d2 HYBRIDIZATION

1) Sulfur hexaflouride (SF₆)

The ground state configuration is 1s² 2s²2p⁶ 3s²3p_x²3p_y¹3p_z¹. It must have 6 unpaired electrons as the sulphur atom forms six bonds. But, the ground state of sulphur has just two unpaired electrons and hence those two only are promoted into two of the 3d orbitals (one from 3s and one from 3p_x). Thus the electronic configuration of ‘S’ in its 2^nd excited state is 1s² 2s²2p⁶ 3s¹3p_x¹3p_y¹3p_z¹3d².

In the next state of excitation, sulfur under goes sp³d² hybridization by mixing a 3s, three 3p and two 3d orbitals. The half-filled sp³d² hybrid orbitals which are six in number are arranged in octahedral symmetry. Sulphur atom forms six σ_sp3_d2_-p bonds with 6 fluorine atoms by using these sp³d² orbitals. The half-filled 2p_z orbitals are utilized by each fluorine atom for the bond formation.

Result: SF₆ is octahedral in shape with bond angles equal to 90^o.

sp3d3 HYBRIDIZATION

1) Iodine heptafluoride (IF₇):

The ground state electronic configuration of the Iodine atom is [Kr]4d¹⁰5s²5p⁵. Since 7 unpaired electrons are vital for the formation of IF₇ , the iodine atom promotes three of its electrons (one from 5s orbital and two from 5p sublevel) into empty 5d orbitals.

This is called as the third excited state. The electronic configuration of Iodine in the third excited state can be written as: [Kr]4d¹⁰5s¹5p³5d³. In this state, the sp³d³ hybridization of iodine atom gives 7 half-filled sp³d³ hybrid orbitals in pentagonal bi-pyramidal symmetry. These will form 7 σ_sp3_d3_-p bonds with fluorine atoms.

Result: IF₇ is pentagonal bipyramidal. The ∠F-I-F bond angles in the pentagonal plane are equal to 72^o, while the two fluorine lie perpendicularly to the pentagonal plane above and below.

This post was written by Aditya Singhal, managing director of askIITians.