Past exam of the chemistry A course of the University of Cambridge / 2024 / Paper 1 / A4

The molecular symmetry operations must be closed under composition. Write the proposed nonidentity operations as $R=C_2$ and $S=\sigma$. Their product $RS$ cannot equal any proposed member:Each conclusion contradicts the distinct operations stipulated. Therefore at least a fourth symmetry operation must exist. For a mirror plane containing the twofold axis, the product generates another mirror plane; for a plane perpendicular to it, the product is inversion. A tilted plane may generate still more operations, but cannot evade group closure.

One choice is the C2v molecular point group, $\{E,C_2(z),{\sigma }_{xz},{\sigma }_{yz}\}$. The additional operation is a second reflection, sinceA bent water molecule has this symmetry. Choose the $z$ axis along its bond-angle bisector and put the molecule in the $xz$ plane: reflection in $yz$ exchanges its two hydrogen atoms, while reflection in $xz$ fixes their positions.

With $R=C_2(z)$, $S={\sigma }_{xz}$ and $T={\sigma }_{yz}$, composition givesEvery entry belongs to the set, proving closure. $E$ is the identity element; every element is its own inverse. Associativity follows because these are actual spatial transformations: composing functions is associative. Thus all group axioms hold, and the group is isomorphic to the Klein four-group.

For a concrete matrix check, use $E=\mathrm{diag}(1,1,1)$, $R=\mathrm{diag}(-1,-1,1)$, $S=\mathrm{diag}(1,-1,1)$ and $T=\mathrm{diag}(-1,1,1)$. Their matrix products reproduce the table.

A symmetry-closed orbital basis, often called a complete basis in this context, is a linearly independent set whose span is carried into itself by every group operation. No transformed basis function requires a function outside that span. This is completeness for the chosen group representation, not a claim that a small atomic-orbital set spans the full electronic Hilbert space.

A representation matrix $D(g)$ specifies the linear transformation produced by a molecular symmetry operation $g$. With the column convention,Its character of a representation is the matrix trace,The character is unchanged by a simultaneous change of basis and is constant on a conjugacy class, which is why character tables have one column per class.

Use the ordered basis $(p_x,p_y,p_z)$ at the origin. Choose $z$ as the principal improper-rotation axis, $x,y$ along the two $C_2^{\prime }$ axes, and the diagonal mirror planes as $x=\pm y$. Use active operations with $\hat{U}(g)f(\mathbf{r})=f(g^{-1}\mathbf{r})$, so the matrix columns give the transformed orbitals. For these real p orbitals the matrices are the ordinary vector transformation matrices.

One representative from each D2d molecular point group class isFor example $S_4^{+}$ sends $p_x\mapsto p_y$, $p_y\mapsto -p_x$ and $p_z\mapsto -p_z$. Choosing the opposite sense of $S_4$ transposes its xy block but leaves every character and the final decomposition unchanged.

Taking the matrix trace of those matrices givesThe one-dimensional $p_z$ subspace has characters $(1,-1,1,-1,1)$, identifying $B_2$. The pair $(p_x,p_y)$ has characters $(2,0,-2,0,0)$, identifying $E$. ThereforeThe dimensions add to three, as required. The $p_x,p_y$ pair must be retained together: a ninety-degree improper rotation mixes them, so either one alone is not a complete symmetry basis.^[ref]

Ideal gaseous phosphorus pentachloride is a trigonal bipyramid with D3h molecular point group symmetry. There are five chlorine sites and three 3p orbitals per site, hence a fifteen-dimensional basis. A moved atom contributes zero to the character; each fixed chlorine contributes the trace of the three-dimensional polar-vector matrix acting on its local p orbitals.For example, a $C_2^{\prime }$ axis through one equatorial chlorine fixes that site but rotates its p-vector basis with trace $1-1-1=-1$. Counting that atom as $+1$ would be incorrect.

Apply molecular character reduction with class sizes $(1,2,3,1,2,3)$ and group order twelve:For $A_1^{\prime }$, this gives $(15-3+3+9)/12=2$; for $E^{\prime }$, $(30+6)/12=3$. The full result isThe dimension check is $2+1+6+2+4=15$. As an additional transparent separation, the two axial chlorines supply $A_1^{\prime }+E^{\prime }+A_2^{\prime \prime }+E^{\prime \prime }$, while the three equatorial ones supply $A_1^{\prime }+A_2^{\prime }+2E^{\prime }+A_2^{\prime \prime }+E^{\prime \prime }$. These are all chlorine 3p orbitals, including both radial and tangential functions; a five-orbital ligand-sigma basis would answer a different question.

The highest expected orbital degeneracy is five. Ideal buckminsterfullerene has Ih molecular point group symmetry. Its irreducible orbital representations include five-dimensional $H_g$ and $H_u$ representations; no single-valued orbital irrep has higher dimension.

This counts distinct spatial molecular orbitals. It does not multiply by electron spin, and it excludes accidental coincidences between unrelated irreps. In a real environment, symmetry lowering can split these degeneracies.^[ref]