Question

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

Explanation:

Step1: Determine 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).

Step2: Arrange atoms and bond them

Iodine is the central atom. Each F is bonded to I with a single - bond. Each single - bond uses 2 electrons, so 8 electrons are used for the 4 I - F bonds.

Step3: Distribute remaining electrons

We have \(34 - 8=26\) electrons left. These are distributed as lone - pairs on the F atoms and I atom. Each F atom gets 3 lone - pairs (6 electrons per F), using \(4\times6 = 24\) electrons. The remaining \(26-24 = 2\) electrons are placed as a lone - pair on the I atom.

Step4: Check formal charges

The formal charge on I is \(7-(2 + 4)=+1\) (7 valence electrons initially, 2 non - bonding electrons and 4 electrons in bonds), and the formal charge on each F is \(7-(6 + 1)=0\).

The correct Lewis structure has I as the central atom, 4 single bonds to F atoms, 3 lone - pairs on each F atom, and 1 lone - pair on the I atom.

Answer:

The second structure from the top (where I has a lone - pair and each F has 3 lone - pairs and is singly bonded to I)