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Describe hybridization in the case of PCl_(5), and SF_(6) . The axial bonds are longer as compared to equatorial bonds in PCl_(5) whereas in SF_(6) both axial bonds and equatorial bonds have the same bond length Explain. |
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Answer» Solution :Formation of `PCl_(5)` Electronic configuration of `""_(15)P` (ground state) Electronic configuration of `""_(15)P`( excited state )  In `PCl`, phosphorus is `sp^(3)`d hybridised to produce a set of FIVE `sp^(3)` d hybrid orbitals which are directed towards the five comers of trigonal bipyramidal . These five `sp^(3)` d hybrid orbitals. overlap with singly occupied p-orbitals of CL -atoms to form five P - Cl sigma bonds. Three P - Cl bonds are in one plane and make an angle of `120^(@)` with each other. These bonds are called equatorial bonds.  The remaining TWO P - Cl bonds one lying above and other lying below the plane make an angle of `90^(@)` with the equatorial plane. These bonds are called axial bonds. Axial bonds are slightly longer than equatorial bonds because axial bond pairs SUFFER more repulsive interaction from the equatorial bond pairs. Formation of `SF_(6)` : Electronic configuration of `""_(16)S`(ground state)s(excitedstate)  In `SF_(6)`. sulphur is `sp^(3) d^(2)` hybridised to produce a set of six `sp^(3) d^(2)` hybrid orbitals which are directed towards the six comers of a regular octahedron. These six `sp^(3) d^(2)` hybrid orbitals overlap with singly occupied orbitals of fluorine atoms to form six S - F sigma bonds. Thus, `SF_(6)` molecule has a regular octahedral geometry and all S - F bonds have same bond LENGTH. |
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