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IF4− Valence Electrons and Lewis Structure Electron Count

What is the total number of valence electrons in IF4−, and how are those electrons assigned in the Lewis structure and VSEPR geometry?

Subject: General Chemistry Chapter: Chemical Bonds Topic: Lewis Structure of Molecules with Expanded Octet Answer included
if4- valence electrons IF4- valence electron count iodine tetrafluoride anion IF4− Lewis structure IF4− formal charge VSEPR AX4E2 square planar geometry expanded octet iodine
Accepted answer Answer included

The total for if4- valence electrons is set by counting valence electrons from iodine and fluorine and then adjusting for the overall negative charge on IF4. That total controls how many bonding pairs and lone pairs appear in the Lewis structure and predicts the VSEPR molecular shape.

Valence-electron sources in IF4

Iodine is a Group 17 element with 7 valence electrons. Each fluorine atom is also Group 17 with 7 valence electrons. The overall −1 charge contributes one additional electron to the total count.

\[ \text{Total valence electrons} = \underbrace{7}_{\text{I}} + \underbrace{4 \times 7}_{\text{4 F}} + \underbrace{1}_{\text{charge}} = 7 + 28 + 1 = 36 \]

IF4 contains 36 total valence electrons.

Electron placement consistent with octets and an expanded octet

Four I–F single bonds require \(4 \times 2 = 8\) electrons. After bonding, fluorine atoms are completed to octets with three lone pairs each (\(6\) electrons per F, \(24\) electrons total for four F). The remaining electrons reside on iodine as lone pairs.

\[ 36 \;-\; 8 \;=\; 28 \quad (\text{after 4 bonds}) \qquad 28 \;-\; 24 \;=\; 4 \quad (\text{after filling F octets}) \qquad 4 \;=\; 2 \text{ lone pairs on I} \]

Iodine therefore carries two lone pairs in addition to four bonding pairs, giving six electron domains around iodine. This exceeds an octet on the central atom and is categorized as an expanded-octet Lewis structure, which is permitted for iodine (period 5).

Formal charges in the dominant Lewis structure

Formal charge bookkeeping confirms the placement: each fluorine is neutral, and the negative charge resides on iodine in the simplest single-bond structure.

Atom Valence electrons Nonbonding electrons Bonding electrons Formal charge \(=\;V - (N + \tfrac{B}{2})\)
I 7 4 (two lone pairs) 8 (four single bonds) \(7 - (4 + \tfrac{8}{2}) = 7 - (4 + 4) = -1\)
F (each) 7 6 (three lone pairs) 2 (one single bond) \(7 - (6 + \tfrac{2}{2}) = 7 - (6 + 1) = 0\)

VSEPR geometry and molecular shape

Six electron domains around iodine correspond to an octahedral electron-domain geometry. With four bonding domains and two lone-pair domains, the VSEPR classification is AX4E2. The two lone pairs occupy opposite axial positions (trans), leaving the four I–F bonds in a single plane.

Electron-domain geometry: octahedral (6 domains). Molecular shape: square planar (AX4E2).

IF₄⁻ valence electrons and square-planar molecular shape Left panel shows the 36 valence-electron count from iodine, four fluorines, and the negative charge. Right panel shows iodine in the center with four fluorines in a square plane and two lone pairs placed on iodine above and below the plane (axial positions). Valence-electron count (total = 36) Group numbers give valence electrons; charge adds 1 electron. Iodine (I): 7 e⁻ Group 17 → 7 valence electrons Fluorine (4 × F): 4 × 7 e⁻ = 28 e⁻ Each F is Group 17 → 7 valence electrons Total from fluorine atoms: 28 valence electrons Charge (−1): +1 e⁻ Anions contain one extra electron 7 + 28 + 1 = 36 valence electrons Lewis/VSEPR picture (AX₄E₂) Square planar bonds; two lone pairs occupy opposite axial sites. I F F F F 4 I–F single bonds 8 bonding electrons 2 lone pairs on I (axial) 4 nonbonding electrons IF₄⁻ is square planar
The lone pairs are drawn on iodine in the two axial positions (above and below the square plane). The four fluorine atoms occupy the planar positions, giving a square-planar molecular shape for IF4.

Common counting errors

A missing electron from the −1 charge shifts the total from \(36\) to \(35\) and forces an incorrect odd-electron Lewis structure. Another frequent mistake is placing three lone pairs on iodine after completing fluorine octets; the remaining-electron count after the four fluorine octets is \(4\) electrons, which corresponds to exactly two lone pairs on iodine.

Direct conclusion

IF4 contains \(36\) valence electrons in total; the Lewis structure places four single I–F bonds, three lone pairs on each fluorine, and two lone pairs on iodine, giving an octahedral electron-domain geometry and a square-planar molecular shape (AX4E2).

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