The Lewis structure for the molecule XeF4 (Xenon Tetrafluoride) can be drawn and explained as follows:
1. Determine the total number of valence electrons: Xenon (Xe) is in Group 18 of the periodic table and has 8 valence electrons. Fluorine (F) is in Group 17 and has 7 valence electrons. Since there are 4 fluorine atoms, the total number of valence electrons is calculated as:
Total valence electrons = 8 (Xe) + 4 × 7 (F) = 8 + 28 = 36
2. Place the least electronegative atom in the center: Xenon is less electronegative than fluorine, so it will be placed in the center of the Lewis structure.
3. Connect the outer atoms to the central atom with single bonds: Each fluorine atom will be connected to the xenon atom with a single bond. This uses 8 electrons (4 bonds × 2 electrons).
4. Distribute the remaining electrons: After forming the single bonds, there are 28 electrons left (36 – 8 = 28). These electrons are placed as lone pairs on the fluorine atoms. Each fluorine atom will have 3 lone pairs, using 24 electrons (4 F atoms × 6 electrons).
5. Place the remaining electrons on the central atom: The remaining 4 electrons are placed as lone pairs on the xenon atom.
6. Check the octet rule: Xenon has 12 electrons around it (4 bonding pairs and 2 lone pairs), which is an exception to the octet rule. This is because xenon can have an expanded octet due to its d-orbitals.
The final Lewis structure for XeF4 is as follows:
F
|
F — Xe — F
|
F
In this structure, xenon is surrounded by four fluorine atoms, each connected by a single bond. Each fluorine atom has three lone pairs of electrons, and xenon has two lone pairs of electrons.