Question

Complete the hybridization and bonding scheme for xef4.

Answer 1

The hybridization and bonding scheme for XeF4 involves sp³d² hybridization and results in a square planar molecular structure.

Step-by-step explanation:

The central atom in XeF4 is xenon (Xe). Using the VSEPR model, we find that Xe in XeF4 forms four bonds and has two lone pairs. Therefore, its electron pair geometry is octahedral, and its molecular structure is square planar. To determine the hybridization, we consider the number of electron pairs. In XeF4, there are six electron pairs, indicating that the Xe atom is sp³d² hybridized. The four sp³d² hybrid orbitals form Xe-F bonds, while the remaining two are occupied by lone pairs of electrons.

Answer 2

Xe = VIII = 8 valence electrons
F = VII = 4 (7 ve) = 28 valence electrons

total ve = 8 + 28 = 36 ve

36 - 4(2) = 28 ve
(there are 2 electrons in each bond x 4 bonds)

28 - 4(6) = 4
(We assign the remaining electrons to F atoms)

4 - 2(2) = 0
(Therefore 4 electrons left => we have 2 lone pairs)

The steric number = No. of σ bonds + #lone pairs

= 4 σ bonds + 2 lone pairs = 6 => d²sp³ (6 hybrid orbitals)

4 bonds + 2 lone pairs => square planar