--%>

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 Elevation in boiling point? The

    The boiling of a liquid may be defused by the temperature at which its vapour pressure which is equal to atmospheric pressure. The effect of addition in a non-volatile solute on the boiling point shown and its solution containing non-volatile solute with tempe

  • Q : Solubility of a gas The solubility of a

    The solubility of a gas in water depends on: (a) Nature of the gas (b) Temperature (c) Pressure of the gas (d) All of the above. Can someone help me in finding out the right answer.

  • Q : Problem on relative volatility In

    In vapor-liquid equilibrium the relative volatility αij is defined to be the ratio of the separation or K factor for species i to that for species j, that is,  αij = Ki/Kj

  • Q : F-centres If a electron is present in

    If a electron is present in place of anion in a crystal lattice, then it is termed as: (a) Frenkel defect  (b) Schottky defect  (c) Interstitial defects (d) F-centre Answer: (d) When electrons are trapped in anion vacancies, thes

  • Q : Mole fraction in vapours Choose the

    Choose the right answer from following. If two substances A and B have P0A P0B= 1:2 and have mole fraction in solution 1 : 2 then mole fraction of A in vapours: (a) 0.33 (b) 0.25 (c) 0.52 (d) 0.2

  • Q : Electrons present in a benzene How

    How many electrons are present in a benzene?

  • Q : Molecular weight of solute Select right

    Select right answer of the question. A dry air is passed through the solution, containing the 10 gm of solute and 90 gm of water and then it pass through pure water. There is the depression in weight of solution wt by 2.5 gm and in weight of pure solvent by 0.05 gm. C

  • Q : Dependcy of colligative properties

    Colligative properties of a solution depends upon: (a) Nature of both solvent and solute (b) The relative number of solute and solvent particles (c) Nature of solute only (d) Nature of solvent only

  • Q : Polyhalogen compounds introduction for

    introduction for polyhalogen compound

  • Q : Molarity what is the molarity of the

    what is the molarity of the solution prepared by dissolving 75.5 g of pure KOH in 540 ml of solution