VSEPR Theory - Geometry of Organic Molecules - Chemistry Steps - Free Printable

Let's solve each part of this problem step by step. We are asked to determine the following for the central atom marked with an asterisk (*) in each molecule:

- # Lone Pairs
- Steric Number
- Electron Geometry
- Molecular Geometry
- Bond Angle
- Hybridization

We'll go through each molecule (a–l) and analyze the central atom.

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🔹 General Rules:

1. Steric Number (SN) = number of atoms bonded to the central atom + number of lone pairs on the central atom.
2. Electron Geometry: Based on steric number (e.g., SN=4 → tetrahedral, SN=3 → trigonal planar, etc.).
3. Molecular Geometry: Based on electron geometry and number of lone pairs.
4. Hybridization:
- SN=2 → sp
- SN=3 → sp²
- SN=4 → sp³
- SN=5 → sp³d
- SN=6 → sp³d²
5. Bond Angles:
- Ideal angles depend on geometry (e.g., 109.5° for tetrahedral, 120° for trigonal planar), but lone pairs reduce bond angles slightly.

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Let’s now analyze each structure.

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a)

Structure: CH₃–CH₂–O–CH₃
Central atom: Oxygen (*)

- O is bonded to two atoms: C (from CH₂) and C (from CH₃)
- Oxygen has 2 lone pairs (since it has 6 valence electrons; 2 bonds use 2 electrons → 4 left as 2 lone pairs)
- # Lone Pairs: 2
- Steric Number: 2 (bonds) + 2 (lone pairs) = 4
- Electron Geometry: Tetrahedral
- Molecular Geometry: Bent (due to 2 lone pairs)
- Bond Angle: ~104.5° (less than 109.5° due to lone pair repulsion)
- Hybridization: sp³

✔ Answer:
| # Lone Pairs | Steric Number | Electron Geometry | Molecular Geometry | Bond Angle | Hybridization |
|--------------|---------------|-------------------|--------------------|------------|---------------|
| 2 | 4 | Tetrahedral | Bent | ~104.5° | sp³ |

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b)

Structure: CH₃–CH₂–CH=CH–CH₃
Central atom: The carbon in the double bond (C=C), specifically the *CH* group (marked *)

- This is a carbon in a double bond: C=C–CH₃
- It is bonded to:
- One H
- One C (from CH₂)
- One C (from =C–CH₃) via double bond
- Double bond counts as one region of electron density
- So total bonding regions: 3 (H, CH₂, and C=C)
- Carbon has no lone pairs (valence = 4; all used in bonds)
- # Lone Pairs: 0
- Steric Number: 3
- Electron Geometry: Trigonal Planar
- Molecular Geometry: Trigonal Planar
- Bond Angle: ~120°
- Hybridization: sp²

✔ Answer:
| # Lone Pairs | Steric Number | Electron Geometry | Molecular Geometry | Bond Angle | Hybridization |
|--------------|---------------|-----------------------|------------------------|------------|---------------|
| 0 | 3 | Trigonal Planar | Trigonal Planar | ~120° | sp² |

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c)

Structure: A tertiary amine-like group, NH₂ attached to a carbon chain
Central atom: Nitrogen (*)

- N is bonded to:
- Two H atoms
- One C atom (from the chain)
- Nitrogen has 1 lone pair (N has 5 valence e⁻; 3 bonds use 3 e⁻ → 2 e⁻ left = 1 lone pair)
- # Lone Pairs: 1
- Steric Number: 3 (bonds) + 1 (lone pair) = 4
- Electron Geometry: Tetrahedral
- Molecular Geometry: Trigonal Pyramidal
- Bond Angle: ~107° (slightly less than 109.5° due to lone pair)
- Hybridization: sp³

✔ Answer:
| # Lone Pairs | Steric Number | Electron Geometry | Molecular Geometry | Bond Angle | Hybridization |
|--------------|---------------|-------------------|--------------------------|------------|---------------|
| 1 | 4 | Tetrahedral | Trigonal Pyramidal | ~107° | sp³ |

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d)

Structure: Cyclohexane-like ring? But appears to be a cyclopentane derivative
But actually: It's a five-membered ring, with a methyl group and a starred carbon
Wait — let’s interpret:
It looks like a cyclopropane or cyclobutane ring? Actually, it's likely methylcyclobutane, but the star is on a carbon in the ring.

But looking at the drawing:
```
H₂
C
H₂C-C-CH₃
C
H₂
```

So it's a four-membered ring (cyclobutane), with a methyl group on one carbon, and the star is on that carbon.

So the starred carbon is:
- Part of a ring
- Bonded to:
- Two ring carbons (each single bond)
- One CH₃ group
- One H (implied, since carbon must have 4 bonds)
- So four single bonds → sp³ hybridized
- No lone pairs
- # Lone Pairs: 0
- Steric Number: 4
- Electron Geometry: Tetrahedral
- Molecular Geometry: Tetrahedral
- Bond Angle: Slightly compressed (~88° in cyclobutane due to ring strain)
- Hybridization: sp³

✔ Answer:
| # Lone Pairs | Steric Number | Electron Geometry | Molecular Geometry | Bond Angle | Hybridization |
|--------------|---------------|-------------------|--------------------|------------|---------------|
| 0 | 4 | Tetrahedral | Tetrahedral | ~88° | sp³ |

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e)

Structure: CH₃–CH₂–C≡C–CH₃
Starred carbon is the triple-bonded carbon

- Carbon in ≡C–CH₃
- Bonded to:
- One C (via triple bond)
- One C (via single bond, CH₃)
- Triple bond counts as one region of electron density
- So two bonding regions
- No lone pairs (carbon uses all 4 valence electrons in bonds)
- # Lone Pairs: 0
- Steric Number: 2
- Electron Geometry: Linear
- Molecular Geometry: Linear
- Bond Angle: 180°
- Hybridization: sp

✔ Answer:
| # Lone Pairs | Steric Number | Electron Geometry | Molecular Geometry | Bond Angle | Hybridization |
|--------------|---------------|-------------------|--------------------|------------|---------------|
| 0 | 2 | Linear | Linear | 180° | sp |

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f)

Structure: CH₃–NH–CH₂–CH₃
Starred atom: Nitrogen

- N bonded to:
- CH₃
- H
- CH₂CH₃
- Three bonds → uses 3 electrons
- Nitrogen has 1 lone pair (5 valence – 3 used = 2 electrons = 1 lone pair)
- # Lone Pairs: 1
- Steric Number: 3 + 1 = 4
- Electron Geometry: Tetrahedral
- Molecular Geometry: Trigonal Pyramidal
- Bond Angle: ~107°
- Hybridization: sp³

✔ Answer:
| # Lone Pairs | Steric Number | Electron Geometry | Molecular Geometry | Bond Angle | Hybridization |
|--------------|---------------|-------------------|--------------------------|------------|---------------|
| 1 | 4 | Tetrahedral | Trigonal Pyramidal | ~107° | sp³ |

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g)

Structure: A five-membered ring with nitrogen and double bonds — looks like pyrrole?

Structure:
```
H
C
HC=C
N
CH
```

But labeled as:
```
H
C
HC=C
N*
CH
```

And double bonds between C–C and C=N

So the nitrogen is:
- In a ring
- Double bonded to C
- Single bonded to C
- And has one H (from diagram: "H" above N)
- So bonds:
- One double bond (counts as 1 region)
- One single bond to C
- One single bond to H
- That’s three bonding regions
- Nitrogen has one lone pair — but in aromatic systems like pyrrole, the lone pair is delocalized into the ring.

However, in pyrrole, the nitrogen has:
- Two sigma bonds (to C and H)
- One pi bond (double bond)
- One lone pair in p orbital, delocalized into the ring

But for VSEPR purposes, we count:
- Three electron domains:
- Two sigma bonds
- One pi bond (but pi bond is not counted separately — it's part of the double bond)
- But the lone pair is in a p orbital and is delocalized, so it doesn't occupy a domain in the same way.

Wait — better approach: in VSEPR, we consider electron domains:
- Each single bond = 1 domain
- Double bond = 1 domain (pi bond is included)
- Lone pair = 1 domain

So for nitrogen in pyrrole:
- Bonds: 2 sigma bonds (to C and H), and one double bond (to C) → that’s 3 bonding domains
- But nitrogen has only 3 electrons used in bonding? Wait:

Actually:
- Nitrogen has 5 valence electrons
- In pyrrole, it forms:
- 1 single bond to C
- 1 single bond to H
- 1 double bond to C → 4 electrons used in bonding
- 1 lone pair (2 electrons) → but this lone pair is in p orbital and delocalized

In VSEPR, the lone pair is still considered if it's localized.

But in pyrrole, the lone pair is delocalized and not localized, so it's not treated as a lone pair domain in VSEPR.

Thus, nitrogen has:
- 3 electron domains: two single bonds and one double bond
- No lone pairs (because the lone pair is in the ring system)

But wait — this is tricky.

Standard treatment: in pyrrole, the nitrogen is sp² hybridized, with:
- 3 sigma bonds (to two C and one H)
- One lone pair in p orbital, delocalized into the ring
- So for VSEPR, we count:
- 3 sigma bonds → 3 electron domains
- The lone pair in p orbital is not in the plane, so it's not counted in VSEPR geometry

So steric number = 3
- Electron geometry: Trigonal Planar
- Molecular geometry: Trigonal Planar
- Bond angle: ~120°
- Hybridization: sp²
- Lone pairs: 0 (for VSEPR, because the lone pair is delocalized and not local)

But technically, there is a lone pair, but it's not localized, so it does not affect geometry.

So for VSEPR:
- # Lone Pairs: 0 (because delocalized)
- Steric Number: 3
- Electron Geometry: Trigonal Planar
- Molecular Geometry: Trigonal Planar
- Bond Angle: ~120°
- Hybridization: sp²

✔ Answer:
| # Lone Pairs | Steric Number | Electron Geometry | Molecular Geometry | Bond Angle | Hybridization |
|--------------|---------------|-----------------------|------------------------|------------|---------------|
| 0 | 3 | Trigonal Planar | Trigonal Planar | ~120° | sp² |

> Note: This is a common exception where lone pair is delocalized.

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h)

Structure: CH₃–CH₂–C(=O)–CH₃
Starred atom: Carbonyl carbon

- Carbon double bonded to O, single bonded to CH₂ and CH₃
- So:
- One double bond to O
- One single bond to CH₂
- One single bond to CH₃
- Total bonding regions: 3
- No lone pairs (carbon has 4 bonds)
- # Lone Pairs: 0
- Steric Number: 3
- Electron Geometry: Trigonal Planar
- Molecular Geometry: Trigonal Planar
- Bond Angle: ~120°
- Hybridization: sp²

✔ Answer:
| # Lone Pairs | Steric Number | Electron Geometry | Molecular Geometry | Bond Angle | Hybridization |
|--------------|---------------|-----------------------|------------------------|------------|---------------|
| 0 | 3 | Trigonal Planar | Trigonal Planar | ~120° | sp² |

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i)

Structure: Tertiary ammonium ion: N⁺(CH₃)₃ with CH₂ groups around it
Starred atom: Nitrogen

- Nitrogen is positively charged
- Bonded to three methyl groups (CH₃)
- Also bonded to one CH₂ group (from ring?)
- Wait: Structure shows:
```
H₂C–CH₂
H₂C–N⁺–CH₃
H₂C–CH₃
```
But it's probably a quaternary ammonium ion, with N bonded to four carbon atoms:
- Three CH₃
- One CH₂ (part of ring?)

Yes: N⁺ bonded to:
- CH₃
- CH₃
- CH₃
- CH₂ (from ring)

So four single bonds, no lone pairs
- Charge: +1 → means it lost one electron → so only 4 bonds, no lone pair
- # Lone Pairs: 0
- Steric Number: 4
- Electron Geometry: Tetrahedral
- Molecular Geometry: Tetrahedral
- Bond Angle: ~109.5°
- Hybridization: sp³

✔ Answer:
| # Lone Pairs | Steric Number | Electron Geometry | Molecular Geometry | Bond Angle | Hybridization |
|--------------|---------------|-------------------|--------------------|------------|---------------|
| 0 | 4 | Tetrahedral | Tetrahedral | ~109.5° | sp³ |

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j)

Structure: CH₃–C⁺–CH₃
Starred atom: Carbon with positive charge

- Central carbon:
- Bonded to:
- CH₃ (left)
- CH₃ (right)
- H (explicitly shown)
- Only three bonds → positive charge
- So three bonding regions, no lone pairs
- # Lone Pairs: 0
- Steric Number: 3
- Electron Geometry: Trigonal Planar
- Molecular Geometry: Trigonal Planar
- Bond Angle: ~120°
- Hybridization: sp²

✔ Answer:
| # Lone Pairs | Steric Number | Electron Geometry | Molecular Geometry | Bond Angle | Hybridization |
|--------------|---------------|-----------------------|------------------------|------------|---------------|
| 0 | 3 | Trigonal Planar | Trigonal Planar | ~120° | sp² |

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k)

Structure: CH₃–CH₂–CH₂⁻
Starred atom: Carbon with negative charge

- Terminal carbon: CH₂⁻
- Bonded to:
- CH₂ (left)
- H
- H
- And carries negative charge
- So carbon has:
- Three single bonds
- One lone pair (since carbon has 4 valence electrons; 3 bonds use 3 electrons, extra electron makes a lone pair)
- # Lone Pairs: 1
- Steric Number: 3 (bonds) + 1 (lone pair) = 4
- Electron Geometry: Tetrahedral
- Molecular Geometry: Trigonal Pyramidal
- Bond Angle: ~107°
- Hybridization: sp³

✔ Answer:
| # Lone Pairs | Steric Number | Electron Geometry | Molecular Geometry | Bond Angle | Hybridization |
|--------------|---------------|-------------------|--------------------------|------------|---------------|
| 1 | 4 | Tetrahedral | Trigonal Pyramidal | ~107° | sp³ |

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l)

Structure: CH₃–CH₂–O⁻–CH₃
Starred atom: Oxygen with negative charge

- Oxygen bonded to:
- CH₃ (left)
- CH₂ (right)
- Negative charge → oxygen has extra electron
- Oxygen has 6 valence electrons
- 2 bonds → 2 electrons used
- 2 lone pairs (4 electrons) + 1 extra electron → 3 lone pairs?
- Wait:
- Each bond uses 1 electron from O → 2 electrons used
- Remaining: 6 - 2 = 4 electrons → 2 lone pairs
- Plus negative charge → adds 1 more electron → total 5 electrons? No

Better: Oxygen has 6 valence electrons. With a negative charge, it has 7 electrons.
- 2 bonds → 2 electrons used in bonding → 5 electrons remain → 2 lone pairs (4 e⁻) and 1 unpaired electron? No — lone pairs are pairs.

Actually:
- In alkoxide ion: R–O⁻–R
- Oxygen has:
- Two single bonds → 2 bonding pairs
- Three lone pairs → 6 electrons
- Total: 2×2 + 3×2 = 10 electrons → too many

Correct:
- Oxygen has 6 valence electrons
- Negative charge → adds 1 electron → total 7 electrons
- But in Lewis structure: O has two single bonds → 2 bonding electrons
- Then 5 nonbonding electrons → but must be in pairs

Standard: Alkoxide ion has:
- O bonded to two carbons
- Two lone pairs (4 electrons)
- One lone pair and one unpaired electron? No — better: three lone pairs and one bond?

No: correct Lewis structure:
- O has two single bonds to two carbons
- And three lone pairs (6 electrons) → total electrons: 2 (bonds) + 6 = 8 → but oxygen should have 8 electrons

Wait: each bond is shared → so O contributes 2 electrons (one per bond)
- So O has 2 electrons in bonds
- Plus 6 electrons in lone pairs → total 8 electrons
- But with negative charge: O has 7 valence electrons? No — negative charge means extra electron

So:
- O has 6 valence electrons + 1 from charge = 7 electrons
- Used in 2 bonds → 2 electrons used
- Remaining 5 electrons → cannot form full pairs

But in reality, alkoxide ions have:
- O with two bonds and three lone pairs → 6 electrons in lone pairs, 2 in bonds → total 8 electrons
- So the negative charge is satisfied by having three lone pairs (6 electrons) and two bonds → total 8 electrons

So yes:
- # Lone Pairs: 3
- Steric Number: 2 (bonds) + 3 (lone pairs) = 5
- Electron Geometry: Trigonal Bipyramidal
- Molecular Geometry: Linear? No — bent?

Wait: SN=5 → trigonal bipyramidal electron geometry
- But with 3 lone pairs and 2 bonding pairs → molecular geometry is linear (lone pairs occupy equatorial positions)

But actually: for X₂E₃ → linear molecular geometry

Example: I₃⁻ is linear with 3 lone pairs

So:
- # Lone Pairs: 3
- Steric Number: 5
- Electron Geometry: Trigonal Bipyramidal
- Molecular Geometry: Linear
- Bond Angle: 180°
- Hybridization: sp³d

✔ Answer:
| # Lone Pairs | Steric Number | Electron Geometry | Molecular Geometry | Bond Angle | Hybridization |
|--------------|---------------|---------------------------|--------------------|------------|---------------|
| 3 | 5 | Trigonal Bipyramidal | Linear | 180° | sp³d |

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✔ Final Table Summary:

| Letter | # Lone Pairs | Steric Number | Electron Geometry | Molecular Geometry | Bond Angle | Hybridization |
|--------|--------------|---------------|-------------------------|--------------------------|------------|---------------|
| a) | 2 | 4 | Tetrahedral | Bent | ~104.5° | sp³ |
| b) | 0 | 3 | Trigonal Planar | Trigonal Planar | ~120° | sp² |
| c) | 1 | 4 | Tetrahedral | Trigonal Pyramidal | ~107° | sp³ |
| d) | 0 | 4 | Tetrahedral | Tetrahedral | ~88° | sp³ |
| e) | 0 | 2 | Linear | Linear | 180° | sp |
| f) | 1 | 4 | Tetrahedral | Trigonal Pyramidal | ~107° | sp³ |
| g) | 0 | 3 | Trigonal Planar | Trigonal Planar | ~120° | sp² |
| h) | 0 | 3 | Trigonal Planar | Trigonal Planar | ~120° | sp² |
| i) | 0 | 4 | Tetrahedral | Tetrahedral | ~109.5° | sp³ |
| j) | 0 | 3 | Trigonal Planar | Trigonal Planar | ~120° | sp² |
| k) | 1 | 4 | Tetrahedral | Trigonal Pyramidal | ~107° | sp³ |
| l) | 3 | 5 | Trigonal Bipyramidal | Linear | 180° | sp³d |

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