Where Do Lone Pairs Go in a Trigonal Bipyramid?
Start with both lone pairs in ClF₃ on equatorial sites. Move one to an axial site and you'll get fewer 90° contacts between lone pairs and bonds. You'll also bring the two lone pairs 90° apart. That second change matters more in introductory VSEPR. A grid with six boxes lets you see exactly which contacts change.
Draw the six 90° contacts
Five electron domains around a central atom form a trigonal bipyramid in the VSEPR model. The two axial sites, A1 and A2, sit on opposite sides of the central atom. The three equatorial sites, E1 through E3, lie in a plane perpendicular to that axis. Each axial site is 90° from each equatorial site. The two axial sites are 180° apart and any two equatorial sites are 120° apart, so the grid leaves those pairs out. OpenStax's diagrams of the ideal sites and the three ClF₃ layouts
Copy the grid and fill the five blanks in parentheses. Use LP for a lone pair on the central atom and B for a bonding domain. In each box below, combine the labels of the row and column. If A1 is a bond and E2 is a lone pair, write LP–B in the A1/E2 box. Use the same label when the lone pair is axial and the bond is equatorial.
Equatorial sites
E1 (__) E2 (__) E3 (__)
Axial sites
A1 (__) _____ _____ _____
A2 (__) _____ _____ _____
Tally the LP–LP, LP–B, and B–B boxes. The counts always add up to six. In PF₅, for instance, all five sites are B, so every box reads B–B and the atoms form the full trigonal bipyramid.
Compare placements, then name the shape
VSEPR stands for valence-shell electron-pair repulsion. In this model, a lone pair takes up more space around the central atom than a bonding pair does. For the five-domain examples here, where all the attached atoms are fluorine, compare the 90° LP–LP counts first and favor the layout with fewer. If those counts tie, favor fewer 90° LP–B contacts. This is a qualitative comparison of ideal layouts. Actual bond angles can shift away from the values used in the grid. Davidson College's VSEPR placement rules
Try the grid with the one lone pair on sulfur in SF₄. Put it at E1 and the two boxes in that column become LP–B. Move it to A1 and the three boxes in that row become LP–B. The equatorial placement is preferred because it gives one fewer close contact. Read the molecular shape from the sites that still have bonds. In SF₄, two axial and two equatorial bonds give a seesaw.
For each molecule below, the number of axial lone pairs tells you how many are left for the equatorial sites. Swapping labels within the axial sites or within the equatorial sites leaves the totals the same. Bold marks the preferred placements.
| Molecule | Axial lone pairs | LP–LP | LP–B | B–B |
|---|---|---|---|---|
| SF₄ | 0 | 0 | 2 | 4 |
| SF₄ | 1 | 0 | 3 | 3 |
| ClF₃ | 0 | 0 | 4 | 2 |
| ClF₃ | 1 | 1 | 3 | 2 |
| ClF₃ | 2 | 0 | 6 | 0 |
| XeF₂ | 0 | 0 | 6 | 0 |
| XeF₂ | 1 | 2 | 3 | 1 |
| XeF₂ | 2 | 2 | 4 | 0 |
In ClF₃, splitting the two lone pairs between an axial and an equatorial site creates one 90° LP–LP contact. That rules out the split layout even though it has the fewest LP–B contacts. Two axial lone pairs are 180° apart and two equatorial lone pairs are 120° apart, so both layouts avoid a 90° LP–LP contact. The equatorial layout has four LP–B contacts compared with six for the axial layout. Its two axial bonds and one equatorial bond make a T shape.
In XeF₂, any split of the three lone pairs across the axis and the plane creates two 90° LP–LP contacts. Put all three in the plane and those contacts disappear. All six boxes then read LP–B, and the two fluorines occupy opposite axial sites. That makes the molecule linear.
Try four problems with a fresh grid
- A classmate says the lone pair in SF₄ has four neighboring bonds wherever it goes, so both sites should be equally favorable. List the four ideal angles to those bonds for an axial lone pair and then for an equatorial one. Which site is preferred?
- In a trial layout for ClF₃, A1 and E3 hold lone pairs. A student adds the three boxes in row A1 and the two in column E3 and reports five LP–B contacts. Which box did they count twice? Label all six boxes and give the corrected totals.
- A filled grid has two LP–LP boxes and exactly one B–B box. Use the B–B box to work backward. How many bonds are there? How many lone pairs are axial and how many are equatorial?
- CO₂ and XeF₂ both have a linear molecular shape. Count the domains around each central atom. Which molecule fits the grid, and does its pair of bonds appear together in any box?
Check your answers
1. The equatorial site is preferred. The ideal angles from an axial lone pair to the bonds are 90°, 90°, 90°, and 180°. From an equatorial lone pair, they're 90°, 90°, 120°, and 120°. Both sites have four neighboring bonds. The equatorial site has only two at 90°.
2. A1/E3 was counted twice, and it's LP–LP. A1/E1, A1/E2, and A2/E3 are LP–B. A2/E1 and A2/E2 are B–B. The totals are 1 LP–LP, 3 LP–B, and 2 B–B.
3. There are two bonds, one axial lone pair, and two equatorial lone pairs. A B–B box requires one axial bond and one equatorial bond. Any third bond would meet one of those two at 90° and create another B–B box. Fill the remaining axial site and both remaining equatorial sites with lone pairs. The one axial lone pair meets each of the other two at 90°.
4. Carbon has two domains in CO₂. Xenon has five in XeF₂. Each C=O double bond counts as one domain, and carbon has no lone pairs. XeF₂ fits the grid because it has two bonding domains and three lone pairs. In the preferred layout its two bonds are both axial, so they are 180° apart and share no box in the grid. More examples of electron geometry and molecular geometry
Rebuild one layout from memory
In a few days, redraw the two ClF₃ layouts with zero 90° LP–LP contacts. See if you can choose between them before checking the table. Spacing out your attempts to recall an answer can help you remember it longer. Karpicke and Bauernschmidt's study on spaced retrieval
For more practice, the VSEPR molecular geometry deck has 78 English text cards, including why AX₄E favors an equatorial lone pair and how the lone pairs in ClF₃ determine its shape. Fill the grid before flipping a placement card. VSEPR molecular geometry flashcards
AI assisted with research and writing and generated the illustration.