ChemGem

How to Predict Molecular Shape with VSEPR (Step-by-Step, with Practice)

Molecules are three-dimensional. Count the electron regions around a central atom, and you can predict how they spread out: that gives the shape.

Spot it, solve it, practice it: 4 steps, 1 worked example, 5 practice problems. Free, no account.

ChemGemUpdated Section 8.3

The route · 4 steps

  1. Draw the Lewis structure; pick the central atom.
  2. Count all its regions: bonds plus lone pairs.
  3. Separate lone pairs from bonded atoms.
  4. Name the geometry and shape.
Try one now See a worked example
Spot it

How to recognize molecular shape questions

Look for bonds and lone pairs around a central atom, "electron regions," or a question asking for a molecule's shape.

Example wording, with the clue words marked:

A central atom has one double bond, one single bond, and one lone pair. How many VSEPR electron regions does it have?

Not sure it’s this type? Compare the neighbors:

Step-by-step

Worked example: molecular shape

The idea

What counts as one electron region
  • Each bond is one region: single, double or triple.
  • Each lone pair on the central atom is one region.
  • VSEPR (valence-shell electron-pair repulsion): regions spread as far apart as possible.
Two names for every molecule
Electron geometry
All regions, lone pairs included. Water: tetrahedral.
Molecular shape
Bonded atoms only; lone pairs push but are not named. Water: bent.
The six shapes you will meet most
XXALinear2 regions · 0 lone pairs180°XXXATrigonal planar3 regions · 0 lone pairs120°XXABent3 regions · 1 lone paira little under 120°XXXXATetrahedral4 regions · 0 lone pairs109.5°XXXATrigonal pyramidal4 regions · 1 lone paira little under 109.5°XXABent4 regions · 2 lone pairsabout 104.5° (water)
A = central atom, X = bonded atom, shaded lobe = lone pair. Wedge points toward you; dashes point away.
VSEPR: regions → electron geometry → shape → angle
RegionsLone pairsElectron geometryMolecular shapeIdeal angleExample
20LinearLinear180°CO2
30Trigonal planarTrigonal planar120°BF3
31Trigonal planarBent< 120°SO2
40TetrahedralTetrahedral109.5°CH4
41TetrahedralTrigonal pyramidal< 109.5°NH3
42TetrahedralBent≈ 104.5°H2O
50Trigonal bipyramidalTrigonal bipyramidal90°, 120°PCl5
51Trigonal bipyramidalSeesaw< 90°, < 120°SF4
52Trigonal bipyramidalT-shaped< 90°ClF3
53Trigonal bipyramidalLinear180°XeF2
60OctahedralOctahedral90°SF6
61OctahedralSquare pyramidal< 90°BrF5
62OctahedralSquare planar90°XeF4
5 regions: lone pairs take the middle-ring (equatorial) positions first.

Lone pairs push bonds a little closer, so angles shrink slightly below the ideal value.

On the periodic table: On the periodic table, the group gives main-group valence electrons: C (14) 4, N (15) 5, O (16) 6, H 1. Helium has 2.

On the periodic table: The table builds the Lewis structure; it does not show the shape.

Worked example · 4 steps

Use electron regions to find the shapes of CO2 and H2O.

Main-group columns help count valence electrons. Gold badges give main-group valence electrons; gray numbers beneath are group numbers. Helium has 2 valence electrons, not 8. Carbon (group 14) contributes 4; nitrogen (15) contributes 5; oxygen (16) contributes 6.
Valence electrons for the Lewis structures: C 4, O 6, H 1.
  1. 1
    Count regions around C in CO2

    O=C=O. Each double bond is one region.

    Around CRegions
    C=O double bond1
    C=O double bond1
    Lone pairs0
    Total2
  2. 2
    Name the CO2 shape

    The O atoms sit on opposite sides of C.

    2 regions, 0 lone pairs → linear, 180°
  3. 3
    Count regions around O in H2O

    Lone pairs affect the arrangement, so count them.

    Around ORegions
    O–H bond1
    O–H bond1
    Lone pairs2
    Total4
  4. 4
    Name the H2O shape

    Geometry counts lone pairs; shape names only atoms.

    CountsName
    Electron geometry4 regionsTetrahedral
    Molecular shape2 bonded atomsBent
  5. ✓
    Answer

    CO2 is linear; H2O is bent.

Your turn

Molecular shape practice problems

Answer first, then check. Hints open one at a time; the answer and full solution open after you try.

Problem 1 of 5 · warm-up

A central atom has one double bond, one single bond, and one lone pair. How many VSEPR electron regions does it have?

Hint 1 · the clue

Count each bond once, even a double bond, then add one region for each lone pair.

Hint 2 · the next move

Bonds and a lone pair on a central atom are listed and VSEPR regions are asked: count each bond and lone pair once.

Your path:

  1. Each bond counts once, even double
  2. Add one region per lone pair
Answer

Answer: 3

Full worked solution

What you know: The central atom has 1 double bond, 1 single bond and 1 lone pair. What you want: The number of VSEPR electron regions.

  1. Count each bonded neighbor once, whatever the bond type: double bond = 1, single bond = 1. That is 2 regions.
  2. Add the lone pair as 1 more region: 2 + 1 = 3.

Check: 3 regions with 1 lone pair gives a bent shape, like the central O in O3.

Watch out: Counting the double bond as 2 gives 4, which is wrong. Regions count directions, not electron pairs in bonds.

Your route: Count each bonded group once → Add lone pairs.

Answer: 3

Problem 2 of 5

A central atom has 6 bonding regions and no lone pairs. What is its ideal molecular shape?

Hint 1 · the clue

Bond-region and lone-pair counts are supplied.

Hint 2 · the next move

Bonding regions and lone pairs on a central atom are given and shape is asked: match region count to its geometry.

Your path:

  1. Count regions: bonds + lone pairs
  2. Six regions → octahedral arrangement
  3. No lone pairs, so shape matches arrangement
Answer

Answer: Octahedral

Full worked solution

What you know: 6 bonding regions, 0 lone pairs. What you want: The molecular shape.

  1. Count regions: 6 bonds + 0 lone pairs = 6 regions.
  2. Six regions point to the corners of an octahedron (up, down, and four around the middle), neighboring corners 90° apart.
  3. With no lone pairs, every position holds an atom, so the molecular shape equals the electron arrangement: Octahedral.

Check: For 6 regions, the shapes are octahedral (0 lone pairs), square pyramidal (1) and square planar (2). Zero lone pairs picks octahedral.

Your route: Count regions → Verify no lone pairs → Name region arrangement.

Answer: Octahedral

Problem 3 of 5

A central atom has 2 bonds and 3 lone pairs, all counted as separate regions. What molecular shape results?

Hint 1 · the clue

Both bond and lone-pair counts are given; each one is a separate region.

Hint 2 · the next move

Bonds and lone pairs on a central atom are given and shape is asked: count regions, place lone pairs, then name atom positions.

Your path:

  1. Count regions: 5 → trigonal bipyramid
  2. Lone pairs take middle (equatorial) spots first
  3. Name shape from atom positions only
Answer

Answer: Linear

Full worked solution

What you know: 2 bonds and 3 lone pairs on the central atom. What you want: The molecular shape.

  1. Count regions: 2 + 3 = 5, so the regions form a trigonal bipyramid.
  2. Lone pairs take the middle ring (equatorial) positions first. All 3 middle positions fill with lone pairs.
  3. The 2 bonded atoms sit at the top and bottom, 180° apart. Naming only the atoms gives Linear.

Check: For 5 regions, 0, 1, 2 or 3 lone pairs give trigonal bipyramidal, seesaw, T-shaped or linear. 3 lone pairs → linear.

Watch out: 2 bonded atoms does not mean you can skip the region count. Here the lone pairs decide where the atoms go.

Your route: Count bonds plus lone pairs → Arrange regions → Name bonded-atom shape.

Answer: Linear

Problem 4 of 5

A central atom has two double bonds and no lone pairs. How many electron regions surround it?

Hint 1 · the clue

Each double bond counts as one region; then add any lone pairs.

Hint 2 · the next move

Double bonds and lone pairs on a central atom are given and electron regions are asked: each double bond counts once.

Your path:

  1. Each double bond counts as one region
  2. Add lone pairs (here none)
Answer

Answer: 2

Full worked solution

What you know: The central atom has 2 double bonds and 0 lone pairs. What you want: The number of electron regions.

  1. Count each bonded group once: each double bond is 1 region, so 2 double bonds give 2 regions.
  2. Add lone pairs: 2 + 0 = 2.

Check: This is the carbon in CO2 (O=C=O): 2 regions, linear, 180°.

Watch out: Counting each double bond as 2 gives 4 regions and a wrong tetrahedral shape.

Your route: Count each bonded group once → Add lone pairs.

Answer: 2

Problem 5 of 5

Three bonds and one lone pair around the central atom: shape?

Hint 1 · the clue

Add the bonds and lone pairs to get the electron regions around the central atom. Then name only the positions of the bonded atoms.

Hint 2 · the next move

Bonds and a lone pair on a central atom are given and shape is asked: count regions, then name only atom positions.

Your path:

  1. Count regions: bonds + lone pairs
  2. Four regions → tetrahedral arrangement
  3. Name shape from atoms only
Answer

Trigonal pyramidal; four electron regions.

Full worked solution

What you know: The central atom has 3 bonds and 1 lone pair. What you want: The molecular shape.

  1. Count regions: 3 bonds + 1 lone pair = 4 regions, arranged tetrahedrally.
  2. Atoms sit at only 3 of the 4 corners. The lone pair sits on top and pushes the 3 bonds down.
  3. Naming only the atom positions gives a pyramid with a triangle base: Trigonal pyramidal; four electron regions.

Check: NH3 fits this pattern: three N–H bonds and one lone pair on N make it trigonal pyramidal.

Watch out: Trigonal planar has 3 regions and no lone pair. One lone pair adds a 4th region and lifts the shape out of the plane.

Answer: Trigonal pyramidal; four electron regions.

Watch out

Common molecular shape mistakes