Question

The d-orbitals used in hybridization of $SF_{6}$ are:-
1> $d_{xy} , d_{yz} , d_{zx}$
2>$d_{z^{2}} , d_{x^{2}-y^{2}}$
3>$d_{z^{2}} , d_{xz}$
4>$d_{xy}} , d_{x^{2}-y^{2}}$

answer with correct explanation will be marked as brainliest

Answer 1

★ HYBRIDIZATION AND ORBITAL
STRUCTURES ★

Structure of SF₆

SF₆ : It's formation is due to sp³d² Hybridization , aslike , it's hexavalency is responsible for its electronic charge distribution in orbitals , it has octahedral structure because it's spd hybridized orbitals overlaps with 2 of its p orbitals axially , means , with 2p orbital of alternative flourine which results in S - F bond formation ,

Orbitals involved : s + pₓ + py + pz + dz² + dx²-y²

dz² and dx²-y² are axially forming electronic clouds , which helps to overlap with px , py or pz axially , because non axial electronic clouds can't overlap easily to form any bond between S and F , to form the bond , axial symmetrical geometry should be available , which gives only dz² and dx²-y²

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Answer 2

Sulfur:  1s2, 2s2 sp6 3s2 3p4 3d0

    d-orbitals used in the hybridization of SF6:

  The answer is 2)  $d_{x^2-y^2} \: and \: d_{z^2}$  orbitals.

   The answer lies in the geometric orientation of the orbitals of Sulphur atom and the direction in which Fluorine atoms are bonded. The molecule has the shape Octahedron. One Fluorine atom each is along +z axis and -z axis. One F atoms is located along +x, +y , -x and -y axes. The bond angles are 90°. 

   We know that d_z² orbitals have electron lobes along the z axis (+ve & -ve).  The d_x²-y² orbitals have 4 lobes along +x, +y, -x & -y axes.  So they bond with six F atoms.

   d_xy, d_yz and d_xz orbitals of S are located in xy, yz and xz planes. They are inclined at 45° to the axes. So they do not bond with F atoms.

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  The splitting of d orbitals into 5 degenerate equal energy orbitals happens first. Later in presence of Octahedral electric fields (weak) they get reorganized into two levels.  dxy, dxz, dyz are at lower level than reorganization. dx²-y² and dz² orbitals are at higher level than before reorganization.

Hybridization in SF6 :  Molecular Orbital theory

    H = (V+M-C+A) /2  
          V = 6 (Sulfur),  M = 6 (six fluorine atoms)
          C = A = 0 (charge=0).

   H = 6 So it is a sp3d2 Hybridization.

   So two lower energy level 3d Orbitals in valence shell of Sulfur participate in the bonds with Fluorine atoms.