Which of the following statements are true and which are false? A sigma molecular orbital may

Question

1) have delocalized electrons in conjugated compounds.

2) not result from two atomic p orbitals.

3) not have a single lobe of electron density.

4) be bonding or antibonding.

5) result from an atomic p orbital and a sp² orbital.

Answer 1

1) False  2) True  3) True  4) True  5) True

Before the individual items, fix the definition, because it settles most of them at once. A sigma MO is one that is cylindrically symmetric about the internuclear axis — equivalently, it has no nodal plane containing the bond axis. That is the entire criterion. It says nothing about which atomic orbitals were combined, nothing about how many lobes appear, and nothing about bonding versus antibonding. Students who classify by "which orbitals went in" instead of "what symmetry came out" get items 2, 3, and 5 wrong.

2) "may not result from two atomic p orbitals" — TRUE

Read this one carefully; the "not" is doing real work. It asks whether a sigma MO can arise from something other than two p orbitals. It certainly can: s + s (H₂), s + p (HF), sp³ + sp³ (C–C in ethane), and so on. Two p orbitals overlapping head-on also give a sigma MO — but they are not required. Both statements are compatible because the stem says "may."

3) "may not have a single lobe of electron density" — TRUE

The single-lobe picture only fits a bonding sigma from s + s, where electron density pools in one region between the nuclei. But:

• A σ* antibonding MO has a nodal plane perpendicular to the bond axis, splitting density into two lobes, one on each nucleus. It is still sigma — that node does not contain the axis.
• A sigma from head-on p + p overlap has the large bonding lobe between the nuclei plus small back lobes pointing outward.

So a sigma MO is not obliged to be one lobe.

4) "may be bonding or antibonding" — TRUE

The straightforward one. In-phase combination gives σ, out-of-phase gives σ*. Both are sigma by symmetry. MO theory always produces the same number of MOs as the atomic orbitals you fed in, so the antibonding partner is not optional — it exists whether or not electrons occupy it.

5) "may result from an atomic p orbital and an sp² orbital" — TRUE

Nothing requires the two contributing orbitals to match. An sp² hybrid overlapping head-on with an unhybridised p orbital is cylindrically symmetric about the axis, so the result is sigma. A real example: the C–Cl bond in vinyl chloride, where the sp² carbon overlaps a chlorine 3p orbital.

1) "may have delocalized electrons in conjugated compounds" — FALSE (as your course means it)

Conjugation is a π phenomenon. In butadiene or benzene, the delocalised electrons occupy π MOs built from parallel, side-on p orbitals above and below the molecular plane. The sigma framework is treated as a set of localised two-centre bonds holding the skeleton together, and it is not what conjugation refers to.

One honest caveat, since it trips up students who read ahead: in a full canonical MO calculation, sigma orbitals are formally delocalised over the whole molecule, and hyperconjugation is a real interaction between sigma bonds and adjacent π systems. But in the vocabulary this question is using, "delocalized electrons in conjugated compounds" means the π system. Answer False, and keep the nuance in your pocket for later.

Takeaway: classify orbitals by symmetry about the bond axis, never by lobe count or by the identity of the parent orbitals. Sigma = no nodal plane containing the axis. Pi = exactly one nodal plane containing the axis.