Question

apply the vsepr model to xef₄. which of the following molecular shapes is consistent with the model? (hint: start with a lewis structure for the molecule) disphenoidal square planar tetrahedral square pyramidal

Explanation:

Step1: Determine the Lewis structure of XeF₄

Xenon (Xe) has 8 valence electrons, and each fluorine (F) has 7 valence electrons. In XeF₄, there are 4 F atoms, so the total number of valence electrons is \(8 + 4\times7 = 8 + 28 = 36\) valence electrons. The Lewis structure of XeF₄ has Xe as the central atom bonded to 4 F atoms, and there are 2 lone pairs of electrons on Xe.

Step2: Apply VSEPR theory

The VSEPR (Valence Shell Electron Pair Repulsion) theory considers the number of bonding pairs and lone pairs around the central atom. For XeF₄, the central Xe atom has 4 bonding pairs (with F atoms) and 2 lone pairs. The electron - pair geometry (considering both bonding and lone pairs) is octahedral (since the total number of electron pairs \(= 4 + 2=6\), and the arrangement for 6 electron pairs is octahedral).

When we consider the molecular geometry (the arrangement of the atoms), the two lone pairs in an octahedral electron - pair geometry will occupy positions that are opposite to each other (to minimize repulsion). The four F atoms will then be in a square planar arrangement around the Xe atom.

Answer:

B. Square planar