Draw the Lewis structure for xenon tetrafluoride, please include the following: What is the formula? How many valence electrons are available? Place the Xe in the center and make 4 bonds to F Complete the octets on the F atoms How many electrons remain? Place lone pairs of electrons on Xe. How many lone pairs are on Xe?
The Correct Answer and Explanation is:
Formula: The chemical formula is XeF₄ (xenon tetrafluoride).
Step-by-Step Lewis Structure:
- Count Valence Electrons:
- Xenon (Xe) is in group 18, so it has 8 valence electrons.
- Each fluorine (F) atom is in group 17 and has 7 valence electrons.
- There are 4 fluorine atoms, so total from fluorine = 4 × 7 = 28 electrons.
- Total valence electrons = 8 (Xe) + 28 (F) = 36 electrons.
- Structure Setup:
- Place Xe in the center since it is the least electronegative.
- Attach 4 fluorine atoms to Xe with single bonds (each bond uses 2 electrons).
- 4 bonds × 2 electrons = 8 electrons used.
- Complete the Octets on Fluorine:
- Each fluorine needs 6 more electrons to complete its octet.
- 4 fluorines × 6 electrons = 24 electrons.
- Electrons Remaining:
- Total available: 36 electrons
- Used so far: 32 electrons
- Remaining: 36 − 32 = 4 electrons
- Place Lone Pairs on Xe:
- Place the remaining 4 electrons (or 2 lone pairs) on Xe.
Final Structure:
- Xe is bonded to four F atoms.
- Each F has 3 lone pairs (6 electrons) to complete its octet.
- Xe has 2 lone pairs on it.
- Total regions of electron density around Xe = 4 bonding + 2 lone pairs = 6 regions → square planar geometry.
Summary:
- Formula: XeF₄
- Total valence electrons: 36
- Electrons used in bonds: 8
- Electrons used on F atoms: 24
- Remaining electrons on Xe: 4
- Lone pairs on Xe: 2
Xenon expands its octet and holds more than 8 electrons, which is allowed for elements in period 3 and beyond.
