Question 8.11: Draw the Lewis structure for ICl4^-.
Draw the Lewis structure for ICl^{-}_{4}.
Iodine (Group 17) has seven valence electrons. Each chlorine atom (Group 17) also has seven. An extra electron is added to account for the 1- charge of the ion. Therefore, the total number of valence electrons is 7 + (4 × 7) + 1 = 36. The I atom is the central atom in the ion. Putting eight electrons around each Cl atom (including a pair of electrons between I and each Cl to represent the single bond between these atoms) requires 8 × 4 = 32 electrons. We are thus left with 36 – 32 = 4 electrons to be placed on the larger iodine:
\begin{bmatrix}:\overset{..}{\underset{..}Cl}&&\overset{..}{\underset{..}Cl}:\\&\overset{\diagdown_{\large..}\diagup }{\underset{\diagup^{\large..}\diagdown }I}&\\:\overset{..}{\underset{..}Cl}&&\overset{..}{\underset{..}Cl}:\end{bmatrix}^-Iodine has 12 valence electrons around it, four more than needed for an octet.
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