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Electron sp^(3) d^(2) hybridization by suitable example. OR Explain geometry of SF_(6). |
Answer» Solution :Electron configuration of sulphur : S (Z= 16) has `bar(e)` configuration [Ne] `3s^(2) 3p^(4)` . In exited state it has six half filled orbitals are three 3s, TWO 3p and one 3d orbital. `sp^(3) d^(2)` hybridization of orbitals: These six half filled orbitals have less difference in their energies so overlap with each other and form six `sp^(3)d^(2)`, size and SYMMETRY. These `sp^(3) d^(2)`orbitals are arrange in octahedral shape at `90^(@)` angle.  Bond fornmation in `SF_(6)` :  FLUORINE has one half filled 2p orbital which axially overlap with six half filled`sp^(3) d^(2)` orbital & form six S - F sigma bond. Shape : Like six `sp^(3) d^(2)` orbitals all the six bonds are arrange in octahedral shape. All F are at six carbon of octahedron os `SF_(6)` has regular octahedral geometry.  Bond angle and bond length : All F - S - F bond angle is `90^(@)` all S -F bond length is also same. |
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