The oxidation state of iodine in tetrabutylammonium iodide ([(CH₃CH₂CH₂CH₂)₄N]I) is -1.

Here's why:

* Tetrabutylammonium ([(CH₃CH₂CH₂CH₂)₄N]⁺) is a cation with a +1 charge.

* Iodide (I⁻) is an anion with a -1 charge.

* To form a neutral compound, the charges must balance. Therefore, the iodine must have a -1 oxidation state.