Question

What would be the correct molecular geometry for XeI4? Please

explain your answer.
(hint:the answer is square planar)

Answer 1

The correct molecular geometry for XeI4 is square planar.

Step-by-step explanation:

Using the VSEPR model, we find that Xe in XeF4 forms four bonds and has two lone pairs, so its structure is square planar and it has six electron pairs. The six electron pairs form an octahedral arrangement, so the Xe must be sp³d² hybridized. With 12 electrons around Xe, four of the six sp³d² hybrid orbitals form Xe-F bonds, and two are occupied by lone pairs of electrons.

Answer 2

XeI₄ has a square planar molecular geometry, resulting from an octahedral electron-pair arrangement with four iodine atoms bonded to the central xenon atom and two lone pairs directly across from one another.

Step-by-step explanation:

The correct molecular geometry for XeI₄ is square planar. This can be determined using the Valence Shell Electron Pair Repulsion (VSEPR) model. The central xenon (Xe) atom has eight valence electrons, and it forms four bonds with iodine (I) atoms. This leaves the central atom with two lone pairs. In terms of electron-pair geometry, XeI4 would have an octahedral arrangement if all positions were occupied by bonds, but two of the positions are indeed occupied by the lone pairs.

The geometry that minimizes the repulsion between these lone pairs and the bonds is square planar, with the lone pairs directly across from one another. Similar to the molecular structure of XeF₄, XeI₄ adopts this arrangement to maintain a stable configuration.