--%>

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 6σd 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 : What is solvent dielectric effect?

    Ionic dissociation depends on the dielectric constant of the solvent.The Arrhenius that ions are in aqueous solutions in equilibrium with parent molecular species allows many of the properties of ionic solutions to be understood. But difficulties began to

  • Q : Theory of three dimensional motion

    Partition function; that the translational energy of 1 mol of molecules is 3/2 RT will come as no surprise. But the calculation of this result further illustrates the use of quantized states and the partition function to obtain macroscopic properties. The partition fu

  • Q : Units of Measurement Unit of

      Unit of measurement- These are also some systems for units:      (1) 

  • Q : Cations Explain how dissolving the

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

  • Q : Molar mass lculwhat is the equation for

    lculwhat is the equation for caating molar mass of non volatile solute

  • 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 : Problem on making solution Select the

    Select the right answer of the question. The weight of H2C2O42H2O required to prepare 500ml of 0.2N solution is : (a) 126g (b) 12.6g (c) 63g (d) 6.3g

  • Q : Problem on solutions The 2N aqueous

    The 2N aqueous solution of H2S04 contains: (a) 49 gm of H2S04 per litre of solution (b) 4.9 gm of H2S04 per litre of solution (c) 98 gm of H2S04

  • Q : Meaning of molality of a solution The

    The molality of a solution will be: (i) Number of moles of solute per 1000 ml of solvent (ii) Number of moles of solute per 1000 gm of solvent (iii) Number of moles of solute per 1000 ml of solution (iv) Number of gram equivalents of solute per 1000 m

  • Q : Mole fraction of water and ethanol Give

    Give me answer of this question. A solution contains 1 mole of water and 4 mole of ethanol. The mole fraction of water and ethanol will be: (a) 0.2 water + 0.8 ethanol (b) 0.4 water + 0.6 ethanol (c) 0.6 water + 0.8 ethanol (d) 0.8 water + 0.2 ethanol