--%>

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 : Question 6 A student was analyzing an

    A student was analyzing an unknown containing only Group IV cations. When the unknown was treated with 3M (NH4)2CO3 solution, a white precipitate formed. Because the acetic acid bottle was empty, the student used 6M HCl to dissolve the precipitate. Following the procedure of this experiment, the stu

  • Q : Problem on Redlich-Kwong equation i)

    i) Welcome to Beaver Gas Co.! Your first task is to calculate the annual gross sales of our superpure-grade nitrogen and oxygen gases. a) The total gross sales of N2 is 30,000 units. Take the volume of the cylinder to be

  • Q : Polyhalogen compounds we need 10

    we need 10 examples for the polyhalogen compounds....please help me....need it urgently...

  • Q : Isotonic Solutions Which one of the

    Which one of the following pairs of solutions can we expect to be isotonic at the same temperature:(i) 0.1M Urea and 0.1M Nacl  (ii) 0.1M Urea and 0.2M Mgcl2  (iii) 0.1M Nacl and 0.1M Na2SO4  (iv) 0.1M Ca(NO3<

  • Q : Concentration of urea Help me to go

    Help me to go through this problem. 6.02x 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is: (a) 0.02 M (b) 0.01 M (c) 0.001 M (d) 0.1 M (Avogadro constant, N4= 6.02x 1023mol -1)<

  • Q : Depression in the freezing point When

    When 0.01 mole of sugar is dissolved in 100g of a solvent, the depression in freezing point is 0.40o. When 0.03 mole of glucose is dissolved in 50g of the same solvent, depression in the freezing point will be:(a) 0.60o  (b) 0.80o

  • Q : Calculating total number of moles

    Choose the right answer from following. While 90 gm of water is mixed with 300 gm of acetic acid. The total number of moles will be: (a)5 (b)10 (c)15 (d)20

  • Q : Problem on vapour pressure and mole

    Provide solution of this question. The vapour pressure of a solvent decreased by 10 mm of mercury, when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent, if decrea

  • Q : Ionization Potential Second ionization

    Second ionization potential of Li, Be and B is in the order (a)Li>Be>B (b)Li>B>Be (c)Be>Li>B (d)B>Be>Li

  • Q : Determining maximum Osmotic pressure

    Which of the following would have the maximum osmotic pressure (assume that all salts are 90% dissociated): (a) Decimolar aluminium sulphate (b) Decimolar barium chloride (c) Decimolar sodium sulphate (d) A solution obtained by mix