11 Facts On IF5 Lewis Structure: Detailed Explanations

IF5 or iodine pentafluoride is an interhalogen compound having a molar mass of 221.89 g/mol. Let us focus on some molecular properties of IF5 in detail.

IF5 is prepared by the reaction between iodine and fluorine in a ratio of 1:5. Both iodine and fluorine are halogen elements belonging to group 17 and for this reason, this type of molecule refers to an interhalogen compound. It can rapidly react with water to form hydrofluoric and iodic acid.

The compound has a monoclinic crystal structure in its lattice form. Now we will discuss the hybridization, lewis structure, bond angle, and shape of the IF5 with proper explanation in the following part of the article.

1. How to draw the IF5 lewis structure

With the help of the octet rule, valency, molecular orientation, and central atom we can draw the lewis structure in many steps. Let us draw the lewis structure of IF5.

Counting the total valence electrons

The IF5 are 42, where both Iodine and F have seven valence electrons each, and just adding them together to get total valence electrons.

Choosing the central atom

For the lewis structure construction, we need a central atom because all the atoms are connected by a suitable number of bonds with that particular atom. Based on the size and less electronegativity we have to select the central atom.

Satisfying the octet

Every atom whether it belongs to the s or p block needs to be filled with its valence orbital by accepting a suitable number of electrons for the successive bond formation. To complete the octet I and F both need 1 more electron as they belong to the 17th element of p blocks respectively.

Satisfying the valency

During the octet formation, each atom should be aware that it can form that number of stable bonds which is their stable valency. As per octet electrons required 8*6 = 48 for the IF5 formation, but the valence electrons are 42, so the remaining electrons should be filled by suitable bonds of each atom.

Assign the lone pairs

After bond formation, if electrons are left in the valence shell of each atom, then those electrons exist as lone pairs over that particular atom in a molecule. In IF5, both I and F have lone pairs and we just add them together to get total lone pairs over the molecule. Iodine has 1 and F has 3 pairs of lone pairs.

2. IF5 valence electrons

Electrons are present in valence shell of each atom and responsible for its chemical property and are called valence electrons. Let us count the valence electrons of IF5.

The total number of valence electrons for the IF5 molecule is 42. There are 7 valence electrons from the iodine site and 7 electrons from each F site, so we just count the individual atoms’ valence electrons and just added them together to get the total valence electrons for the IF5 molecule.  

  • The valence electrons for the Iodine are 7 (5s25p5)
  • The valence electrons for the F are 7 (2s22p5)
  • So, the total number of valence electrons for IF5 is 7+(7*5) = 32 electrons.

3. IF5 lewis structure lone pairs

The electrons that exist in the [paired form in the valence shell after the bond formation in excess are called lone pairs. Let us predict the lone pairs over IF5.

There are 16 pairs of lone pairs present in IF5 which means 32 electrons are present in the valence shell which has no contribution to the bond formation. Those electrons form the iodine site as well as F because both have excess electrons in their valence shell after the bond formation and exist as lone pairs.

  • We can predict the lone pairs over each atom by using the formula, lone pairs = electrons present in the valence orbital – electrons involved in the bond formation
  • So, the lone pairs are present over the Iodine atom, 7-5 = 2
  • The lone pairs present over the F atom, 7-1 = 6
  • So, the total lone pairs present over the IF5 molecule is, 1+(5*3) =16 pairs or 32 electrons.

4. IF5 lewis structure octet rule

To complete the valence orbital of each atom every atom accepts a suitable number of electrons is called the octet rule. Let us see the octet of the IF5 molecule.

IF5 follows the octet rule because both Iodine and F are not completed their valence orbital yet. So, they try to complete their valence electrons through the bond formation. F needs one more electron to complete the octet, as because it belongs to the p block element so it needs 8 electrons in its valence orbital.

Iodine is the group 17th element and it forms five bonds with five F atoms and one lone pair so it also needs one more electron to complete the octet. But during the IF5 molecule formation, Iodine share 10 electrons in five bonds and one lone pair, so it violated the octet and exceeds the octet also.

5. IF5 lewis structure shape

The molecular shape is the proper arrangement of the elements by substituent atoms to gain a perfect geometric structure. Let us predict the shape of IF5.

IF5 is a square pyramidal structure without its lone pairs and if we involved the lone pairs over the iodine then it exists as octahedral geometry as per the following table,

No. of
bond pairs
No. of
lone pairs
Shape   Geometry    
AX 1 0 Linear   Linear
AX2         2 0 Linear   Linear  
AXE        1 1 Linear   Linear  
AX3 3 0 Trigonal
AX2E      2 1 Bent Trigonal
AXE2      1 2 Linear   Trigonal
AX4 4 0 Tetrahedral Tetrahedral
AX3E      3 1 Trigonal
AX2E2                  2 Bent Tetrahedral
AXE3                      1 3 Linear   Tetrahedral
AX5 5 0 trigonal
AX4E      4 1 seesaw trigonal
AX3E2     3 2 t-shaped          trigonal
AX2E3     2 3 linear    trigonal
AX6 6 0 octahedral octahedral
AX5E      5 1              square
AX4E2                     4 2 square
Screenshot 2022 11 19 201715
IF5 Molecular Shape

The geometry or shape of a molecule is predicted by the VSEPR (Valence Shell Electrons Pair Repulsion) theory, and the theory stated that if a molecule has type AX5E and there is one lone pair present then it doesn’t adopt perfect octahedral and turns to square pyramidal.

6. IF5 lewis structure angle

A bond angle is made by the central and substituent atoms for proper orientation in a particular geometry. Let us calculate the bond angle of IF5.

The bond angle between F-I-F is near about 720 because it adopts the square pyramidal and for the Penta-coordinated molecule the better bond angle is 720. The size of iodine is too large that it can easily hold five F atoms without any steric repulsion or lone pairs -bond pairs repulsion.

Screenshot 2022 11 19 201603
IF5 Bond Angle
  • The bond angle value can be calculated by the hybridization value.
  • The bond angle formula according to Bent’s rule is COSθ = s/(s-1).
  • Here the central atom iodine is sp3d hybridized, so the p character is 1/5th
  • So, the bond angle is, COSθ = {(1/5)} / {(1/5)-1} =-(1/4)
  • Θ = COS-1(-1/4) = 720
  • So, from the hybridization value, the bond angle for the calculated and theoretical value is the same.

7. IF5 lewis structure formal charge

The formal charge is a hypothetical concept, where the electronegativity of all atoms is equal and predict the charge of atom. let us calculate the formal charge of IF5.

The net formal charge of the IF5 is 0 because the net charge over central iodine is 0 due to the utilization of all electrons in the bond formation along with lone pairs.

  • The formal charge of the IF5 can be calculated by the formula, F.C. = Nv – Nl.p. -1/2 Nb.p
  • The formal charge possesses by the iodine is, 7-2-(10/2) = 0
  • The formal charge possesses by thefluorine is, 7-6-(2/2) = 0
  • So, both iodine and fluorine individually show zero formal charges and for this reason, the overall formal charge for the molecule is 0.

8. IF5 hybridization

Due to the different energy of the orbitals central atom undergoes hybridization to form a hybrid orbital of equivalent energy. Let us predict the hybridization of IF5.

The central iodine is sp3d hybridized to form a covalent bond in the IF5 molecule which can be discussed below.

Structure    Hybridization
State of
of central
1.Linear          2          sp /sd / pd 1800
3 sp2                    1200
3.Tetrahedral  4 sd3/ sp3 109.50
5 sp3d/dsp3 900 (axial),
5.Octahedral           sp3d2/
7 sp3d3/
Hybridization Table
  • We can calculate the hybridization by the convention formula, H = 0.5(V+M-C+A),
  • So, the hybridization of central iodine is, ½(5+5+0+0) = 5 (sp3d)
  • One s orbital, three p orbitals, and one d orbital of Iodine are involved in the hybridization.
  • The lone pairs over central iodine are also involved in the hybridization.

9. IF5 lewis structure resonance

Resonance is the process of delocalization of electronic clouds between different skeleton forms of the molecule. Let us see the resonating structure of IF5.

The molecule IF5 shows resonance due to the presence of more electron density over F atoms.The electron density over each F atom can delocalize to the iodine site and form different skeleton forms of the IF5 structure. IF5 has two resonating structures which are drawn below –

Screenshot 2022 11 19 201628
IF5 Resonating Structures

Structure I and structure II both have the same contribution because the first structure has no charge present over it, but in structure II there are a higher number of covalent bonds are present along with a positive charge also present over electronegative F atoms, for this reason, both have the same contribution.

10. Is IF5 ionic or covalent?

A molecule is covalent or ionic it depends on the nature of bond formation between the cation and anion of that molecule. Let us see whether IF5 is ionic or covalent.

IF5 is a covalent molecule because,

  • In IF5 the central atom makes a bond by sharing electrons with surrounding atoms.
  • In IF5 bond between Iodine and F is non-polar
  • In IF5The central atom undergoes hybridization to minimize the energy level of required orbitals.
  • In IF5 polarizability of F is very low and the polarizing power of the ionic potential of iodine is also bad so it was unable to form an ionic bond.

As per Fajan’s rule, no molecule is 100% ionic or covalent it depends on the theory of polarizability and in the case of IF5, it is more covalent and has less ionic character.

11. Is IF5 stable?

IF5 is an unstable molecule because it is an interhalogen compound, and every interhalogen compound has an electronegativity difference. For those electronegativity differences, the sigma electron density will be dragged away to the most electronegative halogen atom, the bond becomes weaker and easily cleaved.

12. IF5 uses

  • IF5 is used as a fluorinating agent – by using this reagent we can incorporate fluorine into another molecule.
  • IF5 is also used as a solvent of a different non-polar molecule.


IF5 is the most common interhalogen compound and it can be easily prepared in the laboratory. By the reaction with fluorine, we can get iodine heptafluoride. Interhalogen compounds are more reactive than normal halogen atoms and for this reason, they can use in many rreactionsand participate in many organic reactions where nucleophile and electrophile both are required.

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