--%>

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 : Explain methods for industrial

    The important methods for the preparation of alcohol on large-scale are given below:    

  • Q : Molality of a glucose solution What

    What will be the molality of a solution containing 18g of glucose (having mol. wt. = 180) dissolved in 500g of water: (i) 1m  (ii) 0.5m  (iii) 0.2m  (iv) 2m

  • Q : Ions in solution The accuracy of your

    The accuracy of your written English will be taken into account in marking. 1.    (a)   Identify the spectator ions in the following equation                    &nb

  • Q : Problem on moles of solution The number

    The number of moles of a solute in its solution is 20 and total no. of moles are 80. The mole fraction of solute wil be: (a) 2.5 (b) 0.25 (c) 1 (d) 0.75

  • Q : Basicity order order of decreasing

    order of decreasing basicity of urea and its substituents

  • Q : Law of multiple proportions and Law of

    Describe the difference between law of multiple proportions and law of definite proportions?

  • Q : Molarity Give me answer of this

    Give me answer of this question. If 20ml of 0.4N, NaoH solution completely neutralises 40ml of a dibasic acid. The molarity of the acid solution is:(a) 0.1M (b) 0.2M (c)0.3M (d)0.4M

  • Q : Hydroxide is highly insoluble in

     : 1) Chromium(III) hydroxide is highly insoluble in distilled water but dissolves readily in either acidic or basic solution. Briefly explain why the compound can dissolve in acidic or in basic but not in neutral solution. Write appropriate equations to

  • Q : Calculating value of molar solution

    Choose the right answer from following. An X molal solution of a compound in benzene has mole fraction of solute equal to 0.2. The value of X is: (a)14 (b) 3.2 (c) 4 (d) 2

  • Q : Show your calculations Superphosphate

    Superphosphate has the formulae: CaH4 (PO4)2H2).  Calculate the percentage of phosphorus in this chemical.  Show your calculations  (around ten lines);  also Work out how to make up a nutrient mixtur