IF5 Lewis Structure
Iodine pentafluoride · A fluorinating agent used in organic synthesis.
The IF5 Lewis structure has 42 valence electrons, drawn as five I–F bonds, with 16 lone pairs in total. The central I is sp3d2 hybridized, which makes it a square pyramidal molecule with a bond angle of under 90 degrees. IF5 is polar: the I-F bonds are polar, and the lone pair on the central atom bends the molecule out of shape, so the dipoles do not cancel.
- 42 valence e−
- square pyramidal
- <90° bond angle
- sp3d2
- polar
IF5 molecular geometry
Square pyramidal, from 6 electron domains on the central I: 5 bonded atoms and 1 lone pair (AX5E1). The electron geometry is octahedral; the shape is what is left when the lone pairs are not counted as atoms. Wedges point toward you, hashed wedges away. Every shape in the VSEPR chart →
Calculated properties
| Total valence electrons | 42 |
|---|---|
| Bonding electrons | 10 (5 shared pairs) |
| Nonbonding electrons | 32 (16 lone pairs) |
| Lone pairs on I | 1 |
| Electron domains (steric number) | 6 |
| VSEPR notation | AX5E1 |
| Electron geometry | octahedral |
| Molecular geometry | square pyramidal |
| Bond angle | <90° |
| Hybridization | sp3d2 |
| Polarity | polar |
| Formal charges | all zero |
| Resonance structures | none |
| Molar mass | 221.89 g/mol |
I Iodine 7 valence e− F Fluorine 5 atoms, 7 valence e− each
How to draw the IF5 Lewis structure
Step through it. The bar shows how much of the 42-electron budget is spent at each stage - it never grows, which is the whole constraint.
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Skeleton. Join every connected pair with one bond. 5 bonds spends 10 electrons.
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Outer octets. Spend electrons as lone pairs on the outer atoms until each has eight. 2 are still unplaced.
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Centre. Everything still unplaced goes on I, which now has 12 electrons around it.
The reasoning at each step
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Count the valence electrons
Add up the outer-shell electrons every atom brings. That total is the budget for the whole structure - every line and every dot has to come out of it, and nothing may be added.
7 (I) + 5 x 7 (F) = 42 valence electrons -
Choose the central atom
I is the least electronegative atom here, so it takes the centre. Hydrogen is never central: it can only form one bond.
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Join everything with single bonds first
Every connection starts as one shared pair. Draw them all before worrying about double bonds - the arithmetic in the next step is what tells you where the double bonds have to go.
5 bonds x 2 = 10 electrons used, 32 left -
Work out how far short you are
Add up what every atom still needs to fill its shell - eight electrons for most atoms, two for hydrogen - minus what its single bonds already give it. Compare that with the electrons you have left. The difference decides everything that follows.
needs 30, has 32, 2 to spare -
Park the spare electrons on the central atom
There are 2 electrons more than the outer atoms can use. They become 1 extra lone pair on I, which is in period 5 and can hold more than eight electrons. A period-2 atom could not do this.
12 electrons around I -
Fill in the lone pairs
Every electron not in a bond sits as a lone pair on an atom. Each atom takes exactly what it needs to finish its shell - there is no choice left at this point.
I: 1, F1: 3, F2: 3, F3: 3, F4: 3, F5: 3 (32 electrons) -
Check the central atom
I ends up with 12 electrons, an expanded octet. Only period 3 and below can do this.
I: 10 bonding + 2 nonbonding = 12 -
Check the formal charges
Formal charge is valence electrons, minus lone-pair electrons, minus the number of bonds. Every atom here comes out at zero, which is the sign of a good structure.
all formal charges = 0 -
Work out the shape
Count the electron domains on I: 5 bonded groups plus 1 lone pair. A double or triple bond still counts as one domain, because it points in one direction. That gives octahedral electron geometry; ignore the lone pairs and the atoms themselves sit in a square pyramidal arrangement. The lone pair occupies one vertex and pushes the remaining bonds down.
steric number 6 -> sp3d2 -> square pyramidal, bond angle under 90 -
Decide whether it is polar
The I-F bonds are polar, and the lone pair on the central atom bends the molecule out of shape, so the dipoles do not cancel.
dipoles do not cancel -> polar
Is IF5 polar or nonpolar?
IF5 is polar. The I-F bonds are polar, and the lone pair on the central atom bends the molecule out of shape, so the dipoles do not cancel.
| Bond | Electronegativity difference | Character |
|---|---|---|
| I–F | 1.32 | polar covalent |
Common questions
How many valence electrons does IF5 have?
IF5 has 42 valence electrons. 10 of them are in bonds and 32 sit in lone pairs.
What is the molecular geometry of IF5?
IF5 is a square pyramidal molecule. The central I has 5 bonded groups and one lone pair, a steric number of 6, which gives octahedral electron geometry and a square pyramidal molecule.
What is the bond angle in IF5?
The bond angle in IF5 is under 90 degrees. The lone pair occupies one vertex and pushes the remaining bonds down.
What is the hybridization of IF5?
The central I in IF5 is sp3d2 hybridized. Steric number 6 means 6 orbitals have to be mixed, which is exactly what sp3d2 gives you.
Is IF5 polar or nonpolar?
IF5 is polar. The I-F bonds are polar, and the lone pair on the central atom bends the molecule out of shape, so the dipoles do not cancel.
Why does IF5 break the octet rule?
The central atom has more than four bonded groups, so it has to hold more than eight electrons. Elements from period 3 downward can do this; period-2 elements like carbon, nitrogen and oxygen cannot.