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 Incoming links: Symmetry operation

Past exam of the chemistry A course of the University of Cambridge / 2021 / Paper 1 / A4 / a / iii / Solution 2026-10-20

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Use the classes

(E, 2 C_{3}, 3 σ_{v})

of the C3v point group. For the three hydrogen

1 s

atomic orbitals, the character of a representation counts orbitals fixed by each symmetry operation. Identity fixes three, a nontrivial threefold rotation fixes none, and a vertical mirror fixes one. Therefore

χ_{H} = (3, 0, 1) .

(1)

The character inner product gives multiplicities

n_{A_{1}} = \frac{3 + 3}{6} = 1, n_{A_{2}} = \frac{3 - 3}{6} = 0, n_{E} = \frac{2 \times 3}{6} = 1.

(2)

Thus $Γ_{H} = A_{1} \oplus E$ , with the

E

contribution accounting for two independent symmetry orbitals.

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Past exam of the chemistry A course of the University of Cambridge / 2021 / Paper 1 / A4 / a / ii / Solution 2026-10-20

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In the C3v point group, the nitrogen valence atomic orbitals transform as

2 s : A_{1}, 2 p_{z} : A_{1}, (2 p_{x}, 2 p_{y}) : E .

(1)

The

s

orbital is invariant under every symmetry operation;

p_{z}

is also invariant because each operation preserves the

z

coordinate. The other two components mix as a two-dimensional irreducible representation.

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Past exam of the chemistry A course of the University of Cambridge / 2021 / Paper 1 / A5 / b / ii / Solution 2026-10-20

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For the tetrahedral symmetry classes

(E, 8 C_{3}, 3 C_{2}, 6 S_{4}, 6 σ_{d})

, each set of four

s

atomic orbitals has

χ_{Li} = χ_{Na} = (4, 1, 0, 0, 2) .

(1)

The counts are the fixed sites of each symmetry operation: a threefold rotation fixes one vertex of each tetrahedron, a twofold or improper fourfold operation fixes none, and a diagonal mirror fixes two. Character orthogonality reduces each set to

Γ_{Li} = Γ_{Na} = A_{1} \oplus T_{2} .

(2)

The dimensions

1 + 3 = 4

provide a check. Thus the complete eight-orbital model contains two copies of each of

A_{1}

and

T_{2}

, which can mix across the two species.

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Past exam of the chemistry A course of the University of Cambridge / 2021 / Paper 1 / A5 / b / i / Solution 2026-10-20

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The $z$ axis is a $C_{2}$ axis through the centers of opposite cube faces: a half turn permutes atoms within each species. A body diagonal through sites 1 and 8 is a $C_{3}$ axis: a third turn fixes those two opposite vertices and cycles the other three vertices of each tetrahedral sublattice. These are symmetry operations of the tetrahedral symmetry group

T_{d}

; a quarter turn about

z

would interchange the two chemical species and is not a symmetry of the specified cluster.

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Past exam of the chemistry B course of the University of Cambridge / 2021 / Paper 1 / B3 / a / Solution 2026-10-20

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There are two, two and four stereoisomers, respectively; all eight individual stereoisomers are chiral. The drawings below give the mirror pairs. Solid and hashed wedges distinguish opposite faces; ordinary changes between conformers are not counted again.

The benzyl-substituted alkylidenecyclohexane has an axial stereogenic element. At one end the exocyclic alkene presents Ph and H; at the other end the substituted ring carbon presents benzyl and H. The ring constrains the intervening paths, so exchanging one end pair gives the nonsuperposable mirror image. There is one enantiomeric pair, rather than an independent ordinary $E / Z$ pair multiplied by a carbon-center count.
The diphenylspiro compound also has one axial chirality. The two four-membered rings occupy different, approximately perpendicular planes. Their remote Ph/H substitutions distinguish the two handed arrangements. It gives one enantiomeric pair and no meso form. axial chirality of this spiro framework is experimentally established, for example by resolution of its 2,6-dicarboxylic-acid derivatives.^[ref]
The phospholane has a stereogenic carbon at the COOH-bearing position and a pyramidal phosphorus stereocenter. The phosphorus has three different carbon ligands and a lone pair: its two ring paths differ because only one encounters the substituted carbon immediately. The four configurations are $(R_{P}, R_{C})$ , $(R_{P}, S_{C})$ , $(S_{P}, R_{C})$ and $(S_{P}, S_{C})$ . The first and last form a mirror pair, as do the middle two. carbon and phosphorus cannot be interchanged by a symmetry operation, so there is no meso reduction in the count.

Unlike the usual rapidly inverting trialkylamine, a pyramidal phosphine can retain a phosphorus configuration on an ordinary observational timescale; its two phosphorus configurations must therefore be included.

Figure 1.
Eight stereoisomers grouped into four mirror pairs, with axial and phosphorus stereochemistry shown
. **Details**.

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