## Chapter 7

## 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.

## Step-by-Step

## 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