Draw and Explain the Lewis Structure for XeF2: Does This Molecule Have a Central Atom That Violates the Octet Rule?

The Lewis structure for XeF₂ (xenon difluoride) can be drawn by following these steps:

1. Count the total number of valence electrons: Xenon (Xe) has 8 valence electrons, and each fluorine (F) atom has 7 valence electrons. Since there are two fluorine atoms, the total number of valence electrons is 8 + (7 × 2) = 22.
2. Place the least electronegative atom in the center: Xenon is less electronegative than fluorine, so it will be the central atom.
3. Connect the central atom to the surrounding atoms with single bonds: Draw a single bond between xenon and each fluorine atom. This uses 4 electrons (2 bonds × 2 electrons each).
4. Distribute the remaining electrons: After forming the bonds, 18 electrons remain (22 – 4). These electrons are placed as lone pairs on the fluorine atoms. Each fluorine atom will have 3 lone pairs, using up all 18 electrons.
5. Check the octet rule: Each fluorine atom has 8 electrons (1 bond + 3 lone pairs), satisfying the octet rule. However, xenon has 10 electrons (2 bonds + 3 lone pairs), which means it violates the octet rule.

Explanation: Xenon difluoride (XeF₂) is an example of a molecule where the central atom, xenon, violates the octet rule. Xenon is a noble gas and can expand its valence shell to accommodate more than 8 electrons. In XeF₂, xenon has 10 electrons around it, which is possible due to the availability of d-orbitals in its valence shell. This allows xenon to form more than the typical number of bonds and lone pairs, leading to an expanded octet.

In summary, the Lewis structure of XeF₂ shows xenon as the central atom with two single bonds to fluorine atoms and three lone pairs of electrons. This structure results in xenon having 10 electrons, which violates the octet rule but is allowed due to the expanded valence shell capability of xenon.