When a chemistry problem asks you to identify the orbitals forming “the indicated bond,” it’s asking you to name the two specific orbitals (one from each atom) that overlap to create that bond. The answer always takes the form of “[orbital on atom A] – [orbital on atom B],” such as sp3–sp3 or sp2–1s. Getting there requires just a few steps: figure out each atom’s hybridization, then determine whether the bond is a sigma or pi bond.
How to Determine Each Atom’s Hybridization
Hybridization tells you what type of orbitals an atom uses for its sigma bonds and lone pairs. The quickest way to find it is to count the atom’s “steric number,” which is the number of atoms bonded to it plus the number of lone pairs on it. A steric number of 2 means sp hybridization, 3 means sp2, and 4 means sp3. This works for carbon, nitrogen, oxygen, and most atoms you’ll encounter in organic chemistry.
Carbon is the most straightforward. A carbon with four single bonds (like in ethane) is sp3. A carbon involved in one double bond (like in ethylene) is sp2. A carbon involved in a triple bond (like in acetylene) is sp. You can also verify your answer using bond angles: sp3 centers have angles near 109.5°, sp2 centers sit at about 120°, and sp centers are linear at 180°.
Nitrogen and oxygen follow the same counting rules, but you need to include their lone pairs. In ammonia, nitrogen bonds to three hydrogens and carries one lone pair, giving a steric number of 4 and sp3 hybridization. In water, oxygen bonds to two hydrogens and has two lone pairs, also giving a steric number of 4 and sp3 hybridization. The lone pairs occupy hybrid orbitals just like the bonding pairs do.
One important exception: when a lone pair sits next to a pi bond (as on the nitrogen in an amide or aniline), it typically occupies an unhybridized p orbital instead. This allows the lone pair to participate in resonance with the neighboring pi system. In that situation, the atom’s hybridization drops by one level. For example, the nitrogen in an amide is sp2, not sp3, because its lone pair needs to be in a p orbital for resonance overlap.
Sigma Bonds vs. Pi Bonds
Every single bond is a sigma bond, formed by end-to-end overlap of orbitals along the axis between two nuclei. A double bond consists of one sigma bond plus one pi bond. A triple bond consists of one sigma bond plus two pi bonds. So when an arrow points to a bond, you first need to decide which component of that bond is being indicated.
Sigma bonds form from hybrid orbitals (or from hydrogen’s 1s orbital). Pi bonds form from unhybridized p orbitals overlapping side by side, above and below the bond axis. This distinction is critical because the orbital labels you write down are completely different for each type.
Naming the Orbitals in a Sigma Bond
For a sigma bond, identify the hybridization of each atom, then write both orbital types separated by a dash. Here are the most common examples:
- C–H bond in methane: Carbon is sp3, hydrogen always uses its 1s orbital. The bond is sp3–1s.
- C–C bond in ethane: Both carbons are sp3. The bond is sp3–sp3.
- C=C in ethylene (the sigma component): Both carbons are sp2. The sigma bond is sp2–sp2.
- C≡C in acetylene (the sigma component): Both carbons are sp. The sigma bond is sp–sp.
- C–H bond in ethylene: Carbon is sp2, hydrogen is 1s. The bond is sp2–1s.
- C–N bond in methylamine: Both atoms are sp3. The bond is sp3–sp3.
- O–H bond in water: Oxygen is sp3, hydrogen is 1s. The bond is sp3–1s.
Hydrogen is simple because it only has one orbital available: the 1s. Every bond to hydrogen is always [hybrid orbital]–1s.
Naming the Orbitals in a Pi Bond
If the arrow points to a pi bond (often drawn as the second line in a double bond or indicated explicitly), both orbitals are unhybridized p orbitals. In ethylene, each sp2 carbon has one leftover p orbital perpendicular to the molecular plane. These two p orbitals overlap sideways to form the pi bond. The answer is simply p–p.
In acetylene, each sp carbon has two unhybridized p orbitals. One pair overlaps to form one pi bond, and the other pair forms the second pi bond. Both pi bonds are p–p, oriented in perpendicular planes.
Putting It All Together: A Step-by-Step Method
When you see “what orbitals are used to form the indicated bond,” work through these steps:
- Step 1: Identify the two atoms the bond connects.
- Step 2: Decide if the indicated bond is a sigma bond or a pi bond. If it’s a single bond, it’s sigma. If an arrow points to a double bond without specifying, assume the sigma component unless the question says otherwise. If the arrow clearly points to the “extra” bond in a double or triple bond, it’s pi.
- Step 3: For each atom, count the steric number (bonded atoms plus lone pairs) to determine hybridization. Remember: steric number 4 is sp3, 3 is sp2, 2 is sp.
- Step 4: If the bond is sigma, write the hybrid orbital for each atom (or 1s for hydrogen). If the bond is pi, write p for each atom.
Common Molecules You’ll See in These Problems
Ethylene (C₂H₄)
Both carbons are sp2. The C–C sigma bond uses sp2–sp2 overlap. The C–C pi bond uses p–p overlap. Each C–H bond is sp2–1s.
Acetylene (C₂H₂)
Both carbons are sp. The C–C sigma bond is sp–sp. The two pi bonds are both p–p. Each C–H bond is sp–1s.
Formaldehyde (CH₂O)
Carbon is sp2 (bonded to three atoms). Oxygen in the double bond is also sp2 (one sigma bond to carbon, two lone pairs would suggest sp3, but one lone pair sits in a p orbital due to the pi bond, leaving a steric number of 3). The C=O sigma component is sp2–sp2. The pi component is p–p. Each C–H bond is sp2–1s.
Methanol (CH₃OH)
Carbon is sp3. Oxygen is sp3 (two bonds plus two lone pairs). The C–O bond is sp3–sp3. The O–H bond is sp3–1s. Each C–H bond is sp3–1s.
Hydrogen Cyanide (HCN)
Carbon is sp (two bonded atoms). Nitrogen is also sp (one bond to carbon plus one lone pair in the other sp orbital, with two unhybridized p orbitals forming pi bonds). The C≡N sigma bond is sp–sp. The two pi bonds are p–p. The C–H bond is sp–1s.
Mistakes to Avoid
The most common error is forgetting to count lone pairs when determining hybridization. Nitrogen in ammonia has three bonds but is not sp2. That lone pair counts, making it sp3. Similarly, oxygen in water has only two bonds but is sp3 because of its two lone pairs.
Another frequent mistake is labeling a pi bond with hybrid orbitals. Hybrid orbitals always point directly at another atom and form sigma bonds. Pi bonds exclusively come from unhybridized p orbitals. If you ever write something like “sp2–sp2” for a pi bond, that’s the signal to reconsider.
Finally, watch for atoms next to pi systems where resonance changes the hybridization. The nitrogen in pyridine, for instance, is sp2 with its lone pair in an sp2 orbital (perpendicular to the ring’s pi system). The nitrogen in pyrrole is also sp2, but its lone pair is in the p orbital, participating in the aromatic ring. The bonding context determines which orbital holds the lone pair, and that affects the hybridization and the bonds around it.