Final answer:
To draw the Lewis structure for XeF4, calculate the number of valence electrons, arrange the atoms, add lone pairs on the fluorine atoms, and finally place the remaining lone pairs on the central xenon atom.
Step-by-step explanation:
The student is asking to draw the Lewis structure for XeF4 (xenon tetrafluoride) and include any resonance structures if necessary. When drawing the Lewis structure, we must remember that xenon is a noble gas, but it can form stable compounds with fluorine due to its ability to accommodate more than eight electrons (an exception to the octet rule) using empty valence shell d orbitals.
To draw the Lewis structure of XeF4:
Calculate the total number of valence electrons: For XeF4, 8 (for Xe) + (4 x 7) for four F atoms = 36.
Place Xe in the center and draw single bonds to four F atoms.
Each F atom will have three lone pairs of electrons to complete its octet, which accounts for 32 electrons (8 being bond pairs and 24 being lone pairs on F atoms).
The remaining four electrons (two pairs) are lone pairs placed on the xenon atom.
The molecular geometry of XeF4 determined using VSEPR theory is square planar, with xenon having sp³d² hybridization. This structure has two lone pairs and four bonding pairs of electrons around the xenon atom, resulting in six regions of high electron density that form an octahedral arrangement of electron pairs.
Resonance structures are not possible for XeF4 because there are no multiple bonds or adjacent atoms that can share the terminal electrons differently.