Write the Lewis structure for XeF[latex]_{2}[/latex].

Begin by writing the skeletal structure. Since fluorine is so electronegative, put the fluorine atoms in terminal positions. F Xe F Calculate the total number of electrons for the Lewis structure by summing the number of valence electrons for each atom. Total number of electrons for Lewis structure = (number of valence e[latex]^{-}[/latex] in Xe) + 2(number of valence e[latex]^{-}[/latex] in F) = 8 + 2(7) = 22 Place two bonding electrons between each pair of atoms. F:Xe:F (4 of 22 electrons used) Distribute the remaining electrons to give octets to as many atoms as possible, beginning with terminal atoms and finishing with the central atom. Arrange additional electrons (beyond an octet) around the central atom, giv-ing it an expanded octet of up to 12 electrons. [latex]:\overset{..}{\underset{..}{F}}:Xe:\overset{..}{\underset{..}{F}} :[/latex] (16 of 22 electrons used) (22 of 22 electrons used)

Question 9.10: Write the Lewis structure for XeF2.

Principles of Chemistry: A Molecular Approach

Write the Lewis structure for XeF $_{2}$ .

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Begin by writing the skeletal structure. Since fluorine is so electronegative, put the fluorine atoms in terminal positions.	F Xe F
Calculate the total number of electrons for the Lewis structure by summing the number of valence electrons for each atom.	Total number of electrons for Lewis structure = (number of valence e $^{-}$ in Xe) + 2(number of valence e $^{-}$ in F) = 8 + 2(7) = 22
Place two bonding electrons between each pair of atoms.	F:Xe:F (4 of 22 electrons used)
Distribute the remaining electrons to give octets to as many atoms as possible, beginning with terminal atoms and finishing with the central atom. Arrange additional electrons (beyond an octet) around the central atom, giv-ing it an expanded octet of up to 12 electrons.	$:\overset{..}{\underset{..}{F}}:Xe:\overset{..}{\underset{..}{F}} :$ (16 of 22 electrons used) (22 of 22 electrons used)