Shib Sagar posted an Question
July 25, 2020 • 03:56 am 30 points
  • IIT JAM
  • Chemistry (CY)

1.why in sp hybridization s and pz orbital is used why not px or py?? 2. in sp3d hybridization dx2-y2 is used why not other d orbitals are used??

1.why in sp hybridization s and Pz orbital is used why not Px or Py?? 2. In sp3d hybridization dx2-y2 is used why no other d orbitals used??

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    Dinesh khalmaniya 1 best-answer

    in sp hybridisation ww can use px,py or pz because all have same energy but we use z internulear axis so pz used

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    The choice of 'd' orbital for a particular type of hybridization depends on the spatial orientation of the orbital and the geometry of the molecule or ion which the hybridized orbitals are expected to form. The dx^2-dy^2 orbital lies in the xy plane with its lobes directed along the x,y axes. The lobes of the px and py orbitals are also directed along the same x,y axes. The dsp2 hybridization favours a planar geometry in which the four hybridized orbitals lie in the equatorial xy plane, as in ICl4- or XeF4. So, the dx^2-dy^2 orbital is used in dsp2 hybridization, along with the s, px and py orbitals to form a square planar geometry. The dz^2 orbital has its main lobes along the z axis, and its cylindrical collar around the same axis. The sp3d hybridization favours either a trigonal bipyramidal geometry using the dz^2 orbital or a square pyramidal geometry using the dx^2-y^2 orbital. In forming a stable trigonal pyramid, the two axial bonds above and below the equatorial xy plane have to be almost equally strong as the three equatorial bonds, as in PCl5. So, the dz^2 orbital is used, along with the 's' and the three 'p' orbitals, to form three equatorial bonds and two equally strong axial bonds for a trigonal bipyramid.

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    got it??

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    please accept the answer if you got it 🙏

  • Suman Kumar best-answer

    Which orbital will participate in hybridization depends on geometry & shape of the molecule. A molecule which has sp hybridization has linear shape. Now why Pz only? Since by convention we have taken internuclear axis is along z-axis. So in order to have linear geometry only Pz can mix.

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    Now participation of d: if geometry & shape of the molecule is square pyramidal then dx2-y2. And if the geometry is trigonal bipyramidal then dz2 will participate.

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    There must be symmetry match with molecule

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    These d have best symmetry match with the geometry so they participate in hybridization accordingly

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    Read this Shib carefully. Please ask if you don't get anything 🙏

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    SHIB did you get this

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    Priyanshu kumar Best Answer

    The choice of 'd' orbital for a particular type of hybridization depends on the spatial orientation of the orbital and the geometry of the molecule or ion which the hybridized orbitals are expected to form. The dx^2-dy^2 orbital lies in the xy plane with its lobes directed along the x,y axes. The lobes of the px and py orbitals are also directed along the same x,y axes. The dsp2 hybridization favours a planar geometry in which the four hybridized orbitals lie in the equatorial xy plane, as in ICl4- or XeF4. So, the dx^2-dy^2 orbital is used in dsp2 hybridization, along with the s, px and py orbitals to form a square planar geometry. The dz^2 orbital has its main lobes along the z axis, and its cylindrical collar around the same axis. The sp3d hybridization favours either a trigonal bipyramidal geometry using the dz^2 orbital or a square pyramidal geometry using the dx^2-y^2 orbital. In forming a stable trigonal pyramid, the two axial bonds above and below the equatorial xy plane have to be almost equally strong as the three equatorial bonds, as in PCl5. So, the dz^2 orbital is used, along with the 's' and the three 'p' orbitals, to form three equatorial bonds and two equally strong axial bonds for a trigonal bipyramid.

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    Mainly by convention we take z axis as internuclear axis so...for sp hybridization s and pz orbital is used for linear overlapping

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    if you have any doubt shib you can ask...and please accept the answer if you got it😊🙏

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    please respond dear shib😊

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    Considering proper symmetry of orbitals these orbitals participates in hybridisation👇

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