---
title: "A model of the bonding and orbital overlap in propadiene (allene, \\(\\text{C}_3\\text{H}_4\\)) is shown above, where unhybridized \\(p\\)-orbitals are depicted forming \\(\\pi\\) bonds between adjacent carbon atoms. The \\(\\text{C-C-C}\\) bond angle is \\(180^\\circ\\).  Based on the orbital representation, which of the following best describes the geometry and spatial arrangement of the hydrogen atoms in the molecule?"
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url: "https://nerd-notes.com/ubq/123493/"
date_modified: "2026-09-28T11:59:56+00:00"
---

# A model of the bonding and orbital overlap in propadiene (allene, \(\text{C}_3\text{H}_4\)) is shown above, where unhybridized \(p\)-orbitals are depicted forming \(\pi\) bonds between adjacent carbon atoms. The \(\text{C-C-C}\) bond angle is \(180^\circ\).

Based on the orbital representation, which of the following best describes the geometry and spatial arrangement of the hydrogen atoms in the molecule?

A model of the bonding and orbital overlap in propadiene (allene, \(\text{C}_3\text{H}_4\)) is shown above, where unhybridized \(p\)-orbitals are depicted forming \(\pi\) bonds between adjacent carbon atoms. The \(\text{C-C-C}\) bond angle is \(180^\circ\).

Based on the orbital representation, which of the following best describes the geometry and spatial arrangement of the hydrogen atoms in the molecule?

![A grayscale schematic diagram illustrating the orbital overlap in propadiene along a horizontal axis. Three carbon atoms are arranged linearly and labeled from left to right as \(\text{C}_1\), \(\text{C}_2\), and \(\text{C}_3\). The central carbon \(\text{C}_2\) displays two mutually perpendicular unhybridized \(p\)-orbitals: a vertical pair of lobes (oriented along the \(y\)-axis) and a perpendicular pair of lobes projecting forward and backward (oriented along the \(z\)-axis). The left terminal carbon \(\text{C}_1\) has a single vertical unhybridized \(p\)-orbital whose lobes overlap side-by-side (indicated by parallel dashed lines) with the vertical lobes of \(\text{C}_2\). The right terminal carbon \(\text{C}_3\) has a single unhybridized \(p\)-orbital oriented along the \(z\)-axis whose lobes overlap side-by-side with the \(z\)-axis lobes of \(\text{C}_2\). Two \(\text{C-H}\) bonds on \(\text{C}_1\) lie in the horizontal plane (one wedge, one dash), while two \(\text{C-H}\) bonds on \(\text{C}_3\) lie in the vertical plane (solid lines pointing up-right and down-right). No other particles, labels, text, or annotations appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1790596795-0wRUOL.jpg)

- **A.** The four hydrogen atoms all lie in the same plane because the \(\pi\) bonds are formed from parallel \(p\)-orbitals aligned along the same axis across all three carbon atoms.
- **B.** The four hydrogen atoms form a tetrahedral arrangement around the central carbon atom because the central carbon atom is \(sp^3\) hybridized.
- **C.** The two \(\text{CH}_2\) groups lie in mutually perpendicular planes because the central carbon atom uses two mutually perpendicular \(p\)-orbitals to form the two separate \(\pi\) bonds.
- **D.** The two \(\text{CH}_2\) groups are free to rotate relative to each other at room temperature because the \(\sigma\) bonds along the \(\text{C-C-C}\) axis permit unrestricted rotation.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123493/*