Lewis structure of IF₇
56 valence electrons. Seven I-F single bonds, with no lone pair on I, so I has 14 electrons around it (an expanded octet).
Lewis structure
56 valence electronsFormal charges: none
Why this shape
- I has 7 valence electrons. It uses seven in seven single bonds to F, so none are left over.
- Steric number = seven sigma bonds + no lone pairs = 7, so I is sp³d³ hybridised.
- Seven electron pairs point to the corners of a pentagonal bipyramid (pentagonal bipyramidal electron geometry). With no lone pairs the atoms take the same shape: pentagonal bipyramidal.
- All the bond pairs repel each other equally, so the angle is the ideal 72° and 90°.
- The bond dipoles cancel in the symmetric pentagonal bipyramidal shape, so IF₇ is non-polar.
How to draw it
- Count the valence electrons: I 7 + 7 × F 7 = 56 valence electrons, 28 pairs.
- Put I in the middle and join it to the seven outer atoms with single bonds.
- Give each outer atom lone pairs until it has an octet (H needs only two electrons), then put any electrons left over on the central atom.
- Result: seven I-F single bonds, with no lone pair on I, so I has 14 electrons around it (an expanded octet).
- Formal charge = valence electrons − lone pair electrons − number of bonds. Here: none.
Class 11, Chapter 4 Chemical Bonding and Molecular Structure.
Lewis structure of IF7: quick answers
How many valence electrons are in IF7?
I 7 + 7 × F 7 = 56 valence electrons, 28 pairs.
What are the formal charges in the Lewis structure of IF7?
Every atom has a formal charge of zero.
How many lone pairs are in the Lewis structure of IF7?
21 in all, none on the central I.