Inversion Center and Improper Rotations

The $g/u$ parity labels that fall directly out of inversion symmetry are the same labels used to classify molecular orbitals in any centrosymmetric diatomic or polyatomic — $\sigma_g$, $\sigma_u^*$, $\pi_g$, $\pi_u$ in $\mathrm{O_2}$, $\mathrm{N_2}$, and $\mathrm{CO_2}$ are textbook applications of exactly the parity analysis given above, and they directly control which electronic transitions are allowed or forbidden in UV-visible spectroscopy of diatomics and small centrosymmetric complexes. In coordination chemistry, the Laporte selection rule — which forbids $d$–$d$ transitions in centrosymmetric complexes like octahedral $[\mathrm{Ti(H_2O)_6}]^{3+}$ unless vibronic coupling relaxes it — is a direct consequence of the same inversion-parity argument used above for IR/Raman mutual exclusion, and it is why octahedral transition-metal complexes are typically pale in color while tetrahedral ones (lacking $i$) are far more intensely colored for a comparable metal–ligand pair. Improper rotation axes underlie the modern definition of molecular chirality used throughout pharmaceutical chemistry: regulatory testing for enantiopurity, and the entire logic of why thalidomide's two enantiomers have such different biological effects, rests on the rigorous $S_n$ criterion rather than the folk-notion "no mirror plane." Materials scientists designing nonlinear optical crystals or ferroelectrics deliberately choose space groups lacking an inversion center, since second-harmonic generation and piezoelectricity are symmetry-forbidden in any centrosymmetric material — a direct, technologically load-bearing consequence of the parity argument worked out in this article. In vibrational spectroscopy instrumentation, the mutual exclusion rule is used routinely to confirm whether a newly synthesized coordination complex retains a centrosymmetric ($D_{4h}$, $D_{3d}$, $O_h$) structure in solution, simply by checking whether the IR and Raman spectra share no common bands.

Ask the tutor: How do I quickly find all the $S_n$ axes in a new molecule without checking every possible axis by brute force?

Begin with what you already accept: a $C_n$ rotation carries the molecule into an indistinguishable arrangement by turning it about an axis, and a $\sigma$ reflection does the same by flipping it through a plane. Both operations are "proper" in the sense that a $C_n$ can be carried out by physically rotating a rigid model, and even $\sigma$, though it inverts handedness, is treated as a single distinct symmetry element. The center of inversion and the improper rotation axis are different: they are defined as sequences of a rotation and a reflection performed one after another, about a common axis, through a common point.

Inversion, $i$. The operation is defined purely by coordinates: every atom at position $(x,y,z)$ relative to the inversion center is moved to $(-x,-y,-z)$. Concretely, draw a straight line from any atom through the central point and continue an equal distance beyond; if there is an identical atom sitting there for every atom in the molecule, the point is a center of inversion. Benzene has one, at the ring centroid. A tetrahedron (methane) does not — draw a line from an H atom through the central carbon, and continue the same distance: you land in empty space, not on another hydrogen.

Improper rotation, $S_n$. The formal definition: $S_n$ = rotate the molecule by $2\pi/n$ about an axis, then reflect through the plane perpendicular to that same axis, passing through the same point where the axis crosses it. Crucially, neither the intermediate rotated structure nor the intermediate reflected structure needs to be a symmetry of the molecule on its own — only the net result of doing both must reproduce an indistinguishable configuration. This is why $S_n$ axes can exist with no matching $C_n$ or $\sigma_h$ individually present; methane's $S_4$ axis (coincident with one of the $C_2$ axes bisecting H–C–H angles) is the classic example, worked out in full below. Why does $i$ turn out to be identical to $S_2$? Follow a point through both operations. Under $S_2$: first rotate by $360°/2 = 180°$ about the $z$-axis, sending $(x,y,z) \to (-x,-y,z)$; then reflect through the $xy$-plane (perpendicular to $z$), sending $z \to -z$. The net result is $(x,y,z) \to (-x,-y,-z)$ — exactly inversion, and notice the answer does not depend on which axis you chose for the intermediate rotation. That axis-independence is a special feature of $n=2$: any $C_2$ combined with $\sigma_h$ perpendicular to it gives the same final map, because $180°$ rotation already reverses two coordinates and the perpendicular reflection reverses the third. Similarly, $S_1$ means "rotate by $360°/1 = 360°$" (i.e., do nothing) "then reflect" — which is just a plain reflection, $\sigma$. So the family $S_n$ for $n=1,2,3,4,\dots$ actually contains reflection and inversion as its two smallest, most degenerate members; $S_3, S_4, S_5,\dots$ are the operations that feel genuinely new because they cannot be reduced to a rotation or a single mirror plane.

The coordinate definitions are the working equations you actually use to test any molecule.

where: the origin is placed at the proposed inversion center; the map must send every atom onto an atom of the same kind for $i$ to be a true symmetry element of the molecule.

where: $\hat{C}_n$ is rotation by $2\pi/n$ about the principal axis (taken as $z$), and $\hat{\sigma}_h$ is reflection through the plane perpendicular to that same axis; the operations are applied in either order (they commute) about the identical axis and point.

where: $(x',y')$ is the in-plane image of $(x,y)$ after rotation by angle $2\pi/n$, given explicitly by

where: this is just the standard 2-D rotation matrix acting on the $(x,y)$ components while $z$ is untouched by $C_n$ and then flipped in sign by $\sigma_h$. Reading these together: $S_n$ is not an independent geometric primitive like $C_n$ or $\sigma$; it is literally the matrix product of a rotation matrix and a reflection matrix, both referenced to the same axis. This is exactly why chemists say $S_n$ is a "composite" or "improper" operation — its matrix has determinant $-1$ (it inverts handedness), while $C_n$'s matrix has determinant $+1$. The determinant is the rigorous fingerprint: any operation with determinant $-1$ turns a right-handed set of coordinate axes into a left-handed one, and molecules or objects that are superimposable on such an image of themselves must be achiral if that image is identical to the original structure. Two more relationships matter for building up the full group of operations generated by an $S_n$ axis:

where: applying $S_n$ twice always gives a pure rotation, since two reflections through the same plane cancel ($\sigma_h^2 = E$), leaving $C_n^2$.

where: $k$ is the number of times $S_n$ is applied in succession; even powers strip away the reflection entirely (proper rotations survive), odd powers retain a net reflection character (improper). This power rule is what tells you, for instance, that an $S_4$ axis automatically implies the presence of a genuine $C_2$ axis coincident with it, because $S_4^2 = C_4^2 = C_2$ — a fact used constantly when tabulating the full operation list of a point group.

Example 1 (foundational) — does trans-1,2-dichloroethylene have a center of inversion? The molecule is planar: $\mathrm{ClHC=CHCl}$ with the two Cl atoms on opposite corners across the C=C double bond (trans arrangement). Place the origin at the midpoint of the C=C bond, with the molecular plane as the $xy$-plane and the C=C bond along $x$. Coordinates (schematic, using a C=C bond length of $1.33$ Å so each carbon sits at $x=\pm0.665$ Å, and placing substituents at representative angles rather than a symmetric guess):

• C(1) at $(-0.665, 0, 0)$, C(2) at $(+0.665, 0, 0)$
• Cl on C(1) at $(-1.42, +1.05, 0)$ (up-left)
• H on C(1) at $(-1.15, -1.20, 0)$ (down-left)
• Cl on C(2) at $(+1.42, -1.05, 0)$ (down-right — trans to the first Cl)
• H on C(2) at $(+1.15, +1.20, 0)$ (up-right)

Test inversion through the origin: send $(x,y,z)\to(-x,-y,-z)$.

Every atom maps onto an atom of the same element at the correct position: the origin is a genuine center of inversion. This is the geometric reason the trans isomer is nonpolar (point group $C_{2h}$) while the cis isomer, lacking this point, is polar (point group $C_{2v}$) — a distinction with real consequences for boiling point and dipole-dependent spectroscopy.

Example 2 (applied) — locating the $S_4$ axis in methane and confirming it is not accompanied by a $C_4$ Methane, $\mathrm{CH_4}$, is a tetrahedron, point group $T_d$. Place carbon at the origin with hydrogens at the four alternating corners of a cube of half-edge $a$:

Choose the $z$-axis as the $S_4$ axis (this coincides with a cube edge direction, e.g. $(0,0,1)$, which bisects the $\mathrm{H_1-C-H_2}$ and $\mathrm{H_3-C-H_4}$ angles). Step A — apply $C_4$ (rotate $90°$ about $z$): the rotation matrix sends $(x,y,z) \to (-y,x,z)$.

This point is not one of the four original hydrogen positions (compare to $\mathrm{H_3}=(-a,a,-a)$: the $z$-coordinate disagrees, $a$ vs. $-a$). So $C_4$ alone is not a symmetry operation of methane — confirmed, since $T_d$ has no $C_4$ in its operation list, only $C_3$ and $C_2$ as proper rotations. Step B — now apply $\sigma_h$ (reflect $z \to -z$) to the Step A result:

This is one of the original hydrogen positions. Checking all four atoms the same way confirms every hydrogen maps onto another hydrogen under the combined operation, and carbon (at the origin, on the axis) maps to itself. So $S_4$ about this axis is a genuine symmetry operation of methane even though neither of its two building steps was individually one. This is the paradigmatic illustration of why $S_n$ must be checked as a single combined operation rather than assumed present merely because a $C_n$ or $\sigma_h$ exists (or absent because they don't).

Example 3 (exam/competition level) — allene ($\mathrm{H_2C=C=CH_2}$): find all improper operations and decide whether it is chiral Allene has the two $\mathrm{CH_2}$ planes perpendicular to each other, joined through the central, linear carbon. Set the central carbon at the origin, the C=C=C axis along $z$, and place substituents with an asymmetric but geometrically correct set of coordinates (bond length $\mathrm{C{=}C} = 1.31$ Å, $\mathrm{C{-}H}=1.09$ Å, $\mathrm{H{-}C{-}H}=118°$, consistent with experimental structure):

• Terminal carbons at $C_a = (0,0,+1.31)$ and $C_b = (0,0,-1.31)$
• $\mathrm{CH_2}$ on $C_a$ lies in the $xz$-plane: $\mathrm{H_1}=(0.93,0,1.87)$, $\mathrm{H_2}=(-0.93,0,1.87)$
• $\mathrm{CH_2}$ on $C_b$ lies in the $yz$-plane (rotated $90°$ from the first): $\mathrm{H_3}=(0,0.93,-1.87)$, $\mathrm{H_4}=(0,-0.93,-1.87)$

Step 1 — test for a center of inversion. Send $(x,y,z)\to(-x,-y,-z)$: $\mathrm{H_1}=(0.93,0,1.87) \to (-0.93,0,-1.87)$. Compare against the actual atom list: no hydrogen sits at $(-0.93,0,-1.87)$ — the H's on $C_b$ are at $y=\pm0.93$, $x=0$, not $x=\mp0.93, y=0$. So allene has no center of inversion. Step 2 — test for a $\sigma_h$ (perpendicular to $z$, through the origin). Reflect $z\to -z$: $\mathrm{H_1}=(0.93,0,1.87)\to(0.93,0,-1.87)$. No hydrogen occupies this position (the $C_b$ hydrogens have $x=0$, not $x=0.93$). So no $\sigma_h$ either. Step 3 — test $S_4$ about the $z$-axis (rotate $90°$, then reflect through $xy$-plane). Rotate $(x,y,z)\to(-y,x,z)$:

Then reflect $z \to -z$:

Match. Repeating for $\mathrm{H_2}=(-0.93,0,1.87)$: rotate to $(0,-0.93,1.87)$, reflect to $(0,-0.93,-1.87) = \mathrm{H_4}$ — match. And the carbons, both lying on the $z$-axis, are unaffected by $C_4$ and swap under $\sigma_h$ ($C_a \to C_b$, but wait — check: $\sigma_h$ sends $(0,0,1.31)\to(0,0,-1.31)=C_b$'s position, and $C_b \to C_a$'s position), consistent since both are carbon. $S_4$ is confirmed as a genuine symmetry operation, even though neither $C_4$ nor $\sigma_h$ individually survived Steps 1–2's logic (indeed Step 2 already ruled out $\sigma_h$ alone, and a direct check shows $C_4$ alone also fails, exactly as in methane). Step 4 — chirality verdict. Allene has $S_4$ but no $i$, no $\sigma$ of any kind, review confirms point group $D_{2d}$. Because $S_4$ is present, the general theorem (any $S_n$, for any $n$, forbids chirality) tells us unsubstituted allene is achiral, despite having no inversion center and no mirror plane individually. This is the subtle case that trips up students who only check for $i$ and $\sigma$: the complete test for chirality is "does any $S_n$ (for any $n$, including the disguised $n=1,2$ cases) exist?" — and here the answer is yes, at $n=4$. Limiting-case sanity check: if the two terminal $\mathrm{CH_2}$ groups instead carried two different substituents each (e.g. $\mathrm{HFC=C=CFH}$ with the two ends not related by the $90°$ stagger symmetry), the $S_4$ mapping in Step 3 would send an F onto a position occupied by H, breaking the match — and such substituted allenes are indeed experimentally chiral, isolable as separable enantiomers, exactly as the loss of $S_4$ predicts.

The Hermann–Mauguin system chemists encounter in crystallography (symbols like $\bar{1}$ for inversion, $\bar{4}$ for $S_4$-type rotoinversion) runs parallel to, but is not identical in convention to, the Schoenflies system ($i$, $S_n$) used almost universally in molecular spectroscopy and the point-group tables of Miessler, Tarr, Housecroft, and Sharpe. Schoenflies notation, developed by the German mathematician Arthur Schoenflies in the 1890s specifically for finite point-symmetry groups (as opposed to the infinite space groups of crystals), is the system this article and virtually all inorganic chemistry courses use for molecules.