Question

identify the correct lewis structure for if₄⁺. select the single best answer.

Explanation:

1. First, count the total number of valence electrons. Iodine (I) has 7 valence electrons and each fluorine (F) has 7 valence electrons. For \(IF_{4}^{+}\), the total number of valence electrons is \(7+(4\times7) - 1=34\) (subtract 1 for the positive charge).
2. In the Lewis - structure of \(IF_{4}^{+}\), iodine is the central atom. It forms 4 single bonds with 4 fluorine atoms, using 8 electrons for the bonds.
3. After forming the 4 single bonds, there are \(34 - 8 = 26\) electrons left. These are placed as lone - pairs on the atoms. Each fluorine atom gets 3 lone - pairs (a total of \(4\times3 = 12\) lone - pairs on F atoms), and iodine has 1 lone - pair.
4. The correct Lewis structure has iodine with 4 single bonds to fluorine atoms, iodine with 1 lone - pair, and each fluorine with 3 lone - pairs, and the whole structure has a +1 charge.

Answer:

The first option (where iodine has 4 single bonds to fluorine atoms, iodine has 1 lone - pair, and each fluorine has 3 lone - pairs, and the whole structure is in square - planar geometry with a +1 charge) is the correct Lewis structure for \(IF_{4}^{+}\).