To determine the Lewis structure of XeF4, we need to follow a few steps:
Find the total number of valence electrons: Xe (Xenon) is in Group 18 (Group 8A or noble gases), so it has 8 valence electrons. Fluorine (F) is in Group 17 (Group 7A), so each fluorine atom contributes 7 valence electrons. There are four fluorine atoms in XeF4, so the total number of valence electrons is 8 (Xe) + 4 (4 × F) = 40.
Determine the central atom: In XeF4, xenon (Xe) is the central atom since it is less electronegative than fluorine and can accommodate more than four bonds.
Connect the atoms with single bonds: Connect the central xenon atom to each of the four fluorine atoms using single bonds. Each bond will be represented by a pair of shared electrons, contributing a total of 8 electrons (4 × 2) to the structure.
Distribute the remaining electrons around the atoms: We subtract the electrons used for bonds from the total number of valence electrons (40 - 8 = 32). Place the remaining 32 electrons around the atoms to satisfy the octet rule. Start by placing lone pairs on the fluorine atoms, as fluorine is more electronegative than xenon. Each fluorine atom requires 3 lone pairs to complete its octet, which uses up a total of 24 electrons (4 × 3 × 2).
Place any remaining electrons on the central atom: With 8 electrons already accounted for in the xenon-fluorine bonds and 24 electrons in the lone pairs of the fluorine atoms, we have 32 - 8 - 24 = 0 electrons remaining. Since there are no more electrons left, the Lewis structure is complete.
The Lewis structure of XeF4, using the atomic number Z, can be represented as:
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