--%>

Hybridization

Atomic orbitals can be combined, in a process called hybridization, to describe the bonding in polyatomic molecules.

Descriptions of the bonding in CH4 can be used to illustrate the valence bond procedure. We must arrive at four bonds projecting from the carbon atom in procedure. We must arrive at four bonds projecting from the carbon atom in tetrahedral directions.

Lithus Pauling pointed out that the 2s and 3p orbitals of the carbon atom could be used to form new orbitals better suited to the description of the bonds. This procedure of combining orbitals to form new ones is called hybridization, and the new sets are called hybrid orbitals. The most suitable set can be found, according to Pauling, by forming wave functions which project out farthest from the central atom. When the four orbitals that they are concentrated along tetrahedral directions. Thus the sp3 hybrid orbitals are tetrahedrally oriented and are suitable for describing the bonding in CH4.

Other combinations of s, p and d orbitals can be constructed to provide orbitals suitable for molecules of other shapes, hybrid orbitals that project in linear, trigonal, tetrahedral and octahedral directions are produced by the combinations. The trigonal and linear hybrids, which leave one p and two p orbitals of the atom unchanged, are the basis for descriptions of double and triple bonds. The p orbitals form bonds and supplement the σ bonds, to notice that σ and bonds are similar to those constructed for homonuclear diatomic molecules.

Hybrid orbitals from symmetry: the hybrid orbitals constructed by Pauling led to the geometry, or symmetry, of the molecule for which they were constructed. If the geometry of the molecule is taken as known, the approximate hybrid orbitals can be deduced from symmetry consideration alone. Consider the four tetrahedrally arranged carbon atom bond orbitals needed in this approach to describe the bonding in methane. For these orbitals the characters for the various symmetry operations of the Td group can be seen by calculating the number of unchanged bond orbitals, or bond lines, for each operation. We obtain:

Td E 8C3 3C2 d 6S4
σorb 4 1 0 2 0

                    
Thus we need atomic orbitals that transform as A1 and T2 to provide the basis for the tetrahedrally directed hybrid orbitals. The totally symmetric s atomic orbital transforms according to A1. In a similar way, the hybrid combinations of table can be deduced from the symmetry of the bonding situation for which they are to be used.

Some Hybridization used in describing σ bonding:

Number of orbitals Shape Atomic-Orbital Combinations
Example
2 Linear sp CH≡CH
3 Trigonal sp2 CH2 1851_Hybridization.png CH2, BF3
4 Tetrahedral sp3 or sd3 CH4, MnO4-
  Square planner dsp2 PtCl24-, Ni(CN)24-
5 Trigonal bipyramid dsp3 PCl5, Fe (CO)5
6 Octahedron d2sp3 PtF6, CoF26-

   Related Questions in Chemistry

  • Q : Equimolar solutions Select the right

    Select the right answer of the question. Equimolar solutions in the same solvent have : (a)Same boiling point but different freezing point (b) Same freezing point but different boiling poin (c)Same boiling and same freezing points (d) Different boiling and differe

  • Q : Describe various systems for

    Common system According to this system, the individual members are named according to alkyl groups att

  • Q : Molal elevation constant of water The

    The boiling point of 0.1 molal aqueous solution of urea is 100.18oC  at 1 atm. The molal elevation constant of water is: (a) 1.8    (b) 0.18   (c) 18    (d) 18.6Answer: (a) Kb

  • Q : Problem on melting of ice A) It has

    A) It has been suggested that the surface melting of ice plays a role in enabling speed skaters to achieve peak performance. Carry out the following calculation to test this hypothesis. Suppose that the width of the skate in contact with the ice has been reduced by sh

  • Q : Question based on vapour pressure and

    Give me answer of this question. The vapour pressure of water at 20degreeC is 17.54 mm. When 20g of a non-ionic, substance is dissolved in 100g of water, the vapour pressure is lowered by 0.30 mm. What is the molecular weight of the substances: (a) 210.2 (b) 206.88

  • Q : Dipole moment Elaborate a dipole moment

    Elaborate a dipole moment?

  • Q : Which solution will have highest

    Which solution will have highest boiling point:(a) 1% solution of glucose in water  (b) 1% solution of sodium chloride in water  (c) 1% solution of zinc sulphate in water  (d) 1% solution of urea in waterAnswer: (b) Na

  • Q : Problem on endothermic or exothermic At

    At low temperatures, mixtures of water and methane can form a hydrate (i.e. a solid containing trapped methane). Hydrates are potentially a very large source of underground trapped methane in the pole regions but are a nuisance when they form in pipelines and block th

  • Q : Problem on Neutralization What weight

    What weight of hydrated oxalic acid should be added for complete neutralisation of 100 ml of 0.2N - NaOH solution? (a) 0.45 g  (b)0.90 g  (c) 1.08 g  (d) 1.26 g      Answer

  • Q : Chem Explain how dissolving the Group

    Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid.