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Zorluk: OrtaMolecular Shapes, VSEPR Theory, and Hybridization

Complete the following statement regarding the molecular geometry and atomic orbital hybridization of sulfur hexafluoride (SF6SF_6).

Cevap:In a molecule of sulfur hexafluoride (SF6SF_6), the central sulfur atom has six bonding electron pairs and no lone pairs, resulting in a molecular shape described as 【octahedral】 and a central atom hybridization of 【sp3d2】.

Cevap

The molecular shape of sulfur hexafluoride (SF6SF_6) is octahedral and the hybridization of the central sulfur atom is sp3d2sp^3d^2.
In SF6SF_6, the central sulfur atom shares its 6 valence electrons with six fluorine atoms, creating 6 bonding pairs and 0 lone pairs. VSEPR theory predicts an octahedral geometry to minimize electron repulsions. The combination of six atomic orbitals (one s, three p, and two d) results in sp3d2sp^3d^2 hybridization.

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1
Count the total number of electron pairs around the central sulfur atom in SF6SF_6.
Sulfur contributes 6 valence electrons and each of the six fluorine atoms contributes 1 electron for sharing, forming 6 single covalent bonds (6 bonding pairs) and 0 lone pairs.
According to VSEPR theory, steric number equals the sum of bonding pairs and lone pairs around the central atom.
2
Determine the molecular geometry based on VSEPR theory.
With 6 electron domains and 0 lone pairs, the electron pairs orient themselves toward the corners of a regular octahedron to minimize repulsion, giving an octahedral molecular geometry with bond angles of 9090^\circ.
Symmetrical arrangement of 6 identical bond pairs around a central atom produces an octahedral shape.
3
Determine the hybridization of the central sulfur atom.
Mixing one s orbital, three p orbitals, and two d orbitals yields six equivalent sp3d2sp^3d^2 hybrid orbitals directed toward the vertices of an octahedron.
Six equivalent bonding orbitals require sp3d2sp^3d^2 hybridization.

Anahtar Kavram

VSEPR prediction of molecular shape and valence orbital hybridization for hypervalent molecules.
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