The C2v Point Group in Water

Introduction

Water is an unusually useful first example of molecular group theory: its equilibrium bent geometry is simple enough to test every operation directly, yet it contains enough symmetry to distinguish several types of orbital behaviour. C₂v has only four operations, and each forms its own conjugacy class. Working through the geometry carefully prepares the more abstract character-table and symmetry-adapted-linear-combination methods that follow.

Core explanation

Place the oxygen at the origin, the two hydrogen atoms at (0,+a,b) and (0,−a,b), and the molecule in the yz plane. The z axis bisects the H–O–H angle. The identity E leaves all coordinates unchanged. C₂(z) rotates 180° about z, mapping (x,y,z) to (−x,−y,z); it exchanges the two hydrogens and leaves oxygen fixed. Reflection in the molecular yz plane, σ(yz), maps (x,y,z) to (−x,y,z); all three nuclei remain at their locations because they have x = 0. Reflection in the perpendicular plane through the bisector, σ(xz), maps (x,y,z) to (x,−y,z); it exchanges the hydrogens.

These four operations are distinct even though some have the same effect on the three nuclear positions. C₂ and σ(xz) both exchange the two H atoms in this coordinate setup, but they act differently on general points and on orbitals with x dependence. Similarly E and σ(yz) each leave the nuclei in place, yet σ(yz) changes the sign of an x-oriented p function while E does not. This is why symmetry analysis cannot be reduced to a permutation of nuclei alone when classifying electronic functions.

There is no inversion operation for bent water. Applying inversion to H at (0,+a,b) would require another H at (0,−a,−b), which is absent. There is also no horizontal mirror plane perpendicular to z: reflection z → −z would not preserve the H positions. No second C₂ axis perpendicular to the bisector maps the structure onto itself, so water belongs to the C₂ rather than D₂ family; its two valid vertical reflections give the suffix v.

The C₂v group is abelian: any pair of its operations commutes. Each nonidentity operation is its own inverse, and combining the two distinct mirrors gives the C₂ rotation. These algebraic facts mean all irreducible representations of C₂v are one-dimensional. A function in a particular symmetry species has a definite plus or minus sign under each operation. The complete sign pattern can later be read from the character table.

Consider the two hydrogen 1s functions h₁ and h₂. Their sum h₊ = h₁ + h₂ remains unchanged when the hydrogens exchange. Their difference h₋ = h₁ − h₂ changes sign under such exchange. In the stated coordinate system, h₊ is unchanged by all four operations and therefore has the totally symmetric A₁ character. h₋ is unchanged by E and σ(yz), but changes sign under C₂ and σ(xz); this matches the transformation of the y coordinate in this convention. The oxygen 2s and 2p z functions can interact with h₊ by symmetry; an oxygen 2p y function can interact with h₋. Energy and overlap determine the actual strength of each allowed interaction.

Water's C₂v symmetry also constrains vibrations. A nonlinear triatomic has 3N − 6 = 3 normal vibrational modes. These include symmetric stretch, bend and antisymmetric stretch, each transformable under C₂v operations. Their labels and spectral activities require careful analysis of atomic displacements and dipole components, which later pages develop. The key here is that an operation acts on vectors as well as on equilibrium atom locations.

Step-by-step reasoning

Fix the coordinate convention before naming planes or p orbitals. Test each mapping on O and both H atoms, then test a generic coordinate to distinguish operations with the same nuclear permutation. Check inversion and horizontal reflection explicitly. Finally form hydrogen sum and difference functions and apply the operations to determine their signs.

Visual explanation

Draw O at the origin with two H atoms on opposite sides of the z bisector in the yz plane. Shade the molecular yz plane and draw an xz plane perpendicular to the page through z. Show C₂ and σ(xz) arrows swapping the H labels, while σ(yz) leaves their positions fixed. Below the structure, write h₁ + h₂ and h₁ − h₂ with plus or minus signs after exchange.

Real-world analogy

Two identical seats can be exchanged without changing a room's layout. A sum of occupants' responses is unchanged by the swap, while their difference reverses sign. Hydrogen 1s sum and difference functions behave similarly. The analogy is incomplete because orbital functions also have spatial phase and can change sign even when nuclei do not move.

Real-world example

In a qualitative molecular-orbital diagram for H₂O, symmetry prevents an oxygen orbital with one transformation pattern from mixing with a hydrogen combination of another. This reduces the apparent three-centre problem into compatible blocks. It helps explain why some oxygen valence-orbital character remains largely nonbonding rather than forcing every O orbital into the same O–H interaction.

Why?

Why specify that water lies in the yz plane? Character tables conventionally refer to x, y and z, so changing axes relabels which p orbital transforms like the hydrogen difference function. The physical molecule and predictions remain the same, but a coordinate-dependent label can shift. Declaring the convention makes an orbital diagram reproducible.

Common misconception

If two operations exchange the same identical nuclei, they are not necessarily the same operation. C₂(z) and σ(xz) differ in their action on a general point and on x-dependent orbitals. It is also incorrect to infer inversion from a central oxygen atom: a central atom alone does not supply opposite matching hydrogen positions.

Worked example

Under C₂(z), h₁ ↔ h₂. Thus C₂(h₁ + h₂) = h₂ + h₁ = h₊, while C₂(h₁ − h₂) = h₂ − h₁ = −h₋. Under σ(yz), each H lies in the plane, so both combinations are unchanged. Under σ(xz), the H atoms exchange again, giving signs + for h₊ and − for h₋. The resulting sign patterns are (+,+,+,+) for h₊ and (+,−,+,−) for h₋ in the operation order E, C₂, σ(yz), σ(xz).

Quick check

1. Which C₂v operation leaves every water nucleus fixed but changes x to −x? Answer: Reflection in the molecular yz plane, σ(yz), under the stated coordinate convention. 2. Does water have an inversion centre at oxygen? Answer: No. Inversion would send each hydrogen to an unoccupied position on the opposite side of oxygen.

Exam focus

Show the four operations explicitly and state the coordinate choice. Distinguish the effect on nuclear positions from the effect on orbitals. For SALCs, demonstrate exchange of h₁ and h₂ rather than memorising signs with an unstated axis convention.

Advanced insight

The C₂v character table has four one-dimensional irreducible representations because it has four classes. Orthogonality requires distinct sign patterns. The totally symmetric representation has every character +1 and is the symmetry required for a nonzero scalar integral. These facts later make selection-rule tests especially transparent for water.

Summary

Ideal bent water has E, C₂(z), σ(yz) and σ(xz), giving C₂v symmetry. Some operations share a nuclear permutation but remain distinct because they transform general points and orbitals differently. Hydrogen sum and difference functions have definite sign patterns, enabling symmetry-based molecular-orbital construction.

Practice questions

1. In the stated coordinate system, what happens to an oxygen 2p x function under reflection in the molecular yz plane? Answer: Its sign changes because the x coordinate changes sign. This differs from the hydrogen 1s sum, which is unchanged under that reflection. 2. Why can the hydrogen sum interact by symmetry with oxygen 2s but the hydrogen difference cannot? Answer: Oxygen 2s is unchanged by every C₂v operation, matching the sum's totally symmetric behaviour. The difference changes sign under operations that exchange the hydrogens, so its interaction with 2s is symmetry-forbidden. 3. How many normal vibrational modes does water have, and why? Answer: A nonlinear molecule with N = 3 atoms has 3N − 6 = 3 vibrational modes after removing three translations and three rotations. Their detailed symmetry labels require transforming displacement vectors.