Question

The point group of \(\mathrm{CH_2=C=CH_2}\) is

• A. \(D_{2h}\)
• B. \(C_{2h}\)
• C. \(C_{2v}\)
• D. \(D_{2d}\)

Hint:

Draw allene in 3D: the two terminal $CH_2$ planes are perpendicular because the central carbon uses two orthogonal p orbitals for its two $\pi$ bonds; look for an $S_4$ axis along $C=C=C$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept.
\(\mathrm{CH_2=C=CH_2}\) is allene (propa-1,2-diene). The central carbon is \(sp\) hybridized and forms two mutually perpendicular \(\pi\) bonds, one to each terminal \(CH_2\) carbon. Because the two \(\pi\) bonds use orthogonal p-orbitals on the central carbon, the two \(=CH_2\) planes at the two ends of the molecule are twisted 90 degrees relative to each other.

Step 2: List the symmetry elements.
Take the \(C=C=C\) axis as the principal axis. Because the two \(CH_2\) planes are perpendicular, rotating the molecule by 90 degrees about this axis and then reflecting through a plane perpendicular to the axis maps the molecule onto itself, so there is an \(S_4\) axis along \(C=C=C\). There are also two \(C_2\) axes perpendicular to the \(S_4\) axis, each one bisecting the \(H-C-H\) angle of one terminal \(CH_2\) group, and two dihedral mirror planes \(\sigma_d\), each containing the \(S_4\) axis and one \(CH_2\) group.
There is no horizontal mirror plane \(\sigma_h\) perpendicular to the \(S_4\) axis (that would require both \(CH_2\) planes to be coplanar, which they are not), and there is no center of inversion.

Step 3: Identify the point group.
A molecule with a principal \(S_{2n}\) axis, \(n\) perpendicular \(C_2\) axes, \(n\) dihedral mirror planes \(\sigma_d\), and no \(\sigma_h\) belongs to the \(D_{nd}\) family. Here \(n=2\) (the \(S_4\) axis plus two \(C_2\) axes plus two \(\sigma_d\) planes), so the point group is \(D_{2d}\).

Step 4: Rule out the other options.
\(D_{2h}\) and \(C_{2h}\) both require a center of inversion and/or a \(\sigma_h\), which allene does not have because its two \(CH_2\) ends are perpendicular, not coplanar. \(C_{2v}\) would apply if the whole molecule were planar with a single mirror plane containing all the atoms, which is also not the case here since the two ends are twisted out of a common plane.

Final Answer:
Allene, \(\mathrm{CH_2=C=CH_2}\), belongs to the \(D_{2d}\) point group. \[ \boxed{\text{(D) } D_{2d}} \]