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## Q. 7.4

Draw Lewis structures of $XeF_{4}$.

STRATEGY

If AE < NE, follow the process described in Figure 7.3.
If AE = NE, your skeleton is correct; add electrons as unshared pairs to form octets around the atoms.
If AE > NE, follow the process described in Figure 7.9.

## Verified Solution

Skeleton            $\begin{matrix} &\underset{\mid }{F} \\F- & \underset{\mid }{Xe} – & F \\& F & \end{matrix}$
VE        4(7 (for each F)) + 8 (for Xe) = 36
AE         AE = VE – 2(bonds) = 36 – 2(4 bonds) = 28
NE          4(6 (for each F to have an octet)) + 0 (Xe has an octet) = 24

AE = NE ?         No; AE > NE. There are 4 extra electrons.
Satisfy the octet rule     $\begin{matrix} & :\overset{..}{\underset{\mid }{F} } : \\ :\overset{..}{\underset{..}{F}- } & \underset{\mid }{Xe} – & \overset{..}{\underset{..}{F}: } \\ &:{\underset{..}{F}: } & \end{matrix}$

Lewis structure