Polar & Spherical Coordinates — Maths for Quantum Chemistry

Round problems want round coordinates

Where orbitals come from

Instead of $x$ and $y$ , polar coordinates locate a point by a distance r from the origin and an angle θ. The conversion between the two systems is a direct application of SOHCAHTOA:

x = r cos θ, y = r sin θ, r = x^{2} + y^{2}, θ = arctan (\frac{y}{x})

In 3D, polar coordinates extend to spherical coordinates $(r, θ, ϕ)$ where $r$ is the radial distance, $θ$ is the polar angle from the z-axis (0 to π), and $ϕ$ is the azimuthal angle in the xy-plane (0 to 2π). For anything centred on a point — like an electron orbiting a nucleus — this is the natural description, because the Coulomb potential depends on $r$ alone.

Solving the Schrödinger equation for the hydrogen atom in spherical coordinates splits the wavefunction into a radial part $R_{n l} (r)$ and an angular part $Y_{l}^{m} (θ, ϕ)$ (a spherical harmonic). The angular shapes — $cos θ$ for a p-orbital, $cos 2 θ$ for a d-orbital — are exactly the polar curves you can plot in the explorer below. The volume element in spherical coordinates is $d V = r^{2} sin θ d r d θ d ϕ$ — that extra $r^{2} sin θ$ factor determines where electrons are most likely to be found.

Key relations at a glance

Polar ↔ Cartesian (2D)

x = r cos θ, y = r sin θ

Spherical coordinates (3D)

x = r sin θ cos ϕ, y = r sin θ sin ϕ, z = r cos θ

Volume element

d V = r^{2} sin θ d r d θ d ϕ

p-orbital angular part

Y_{1}^{0} \propto cos θ

— two lobes along z

d-orbital angular part

Y_{2}^{0} \propto (3 cos^{2} θ - 1)

— four lobes

s-orbital (spherical)

Y_{0}^{0} = const

— uniform in all directions

Nodes and angular structure of atomic orbitals

Angular nodes: number = l

s (l = 0): 0 angular nodes. p (l = 1): 1 angular node (a plane). d (l = 2): 2 angular nodes (two planes or one cone). Total nodes = n − 1; radial nodes = n − l − 1.

p-orbital angular nodes

2 p_{z}

Y \propto cos θ

, node at

θ = 9 0^{\circ}

(xy-plane).

2 p_{x}

Y \propto sin θ cos ϕ

, node at

x = 0

(yz-plane).

d-orbital angular nodes

3 d_{z^{2}}

Y \propto (3 cos^{2} θ - 1)

, nodes where

cos θ = \pm 1/ 3

(two conical surfaces).

3 d_{x y}

Y \propto sin^{2} θ sin 2 ϕ

, nodes at

x = 0

and

y = 0

Why 4πr² in P(r)

Integrating

∣ ψ ∣^{2}

over a thin shell at radius r:

\int_{0}^{π} \int_{0}^{2 π} r^{2} sin θ d θ d ϕ = 4 π r^{2}

. That is why the radial distribution function is

P (r) = 4 π r^{2} ∣ R (r) ∣^{2}

(for s orbitals).

Common pitfalls

Swapping θ and φ conventions. In physics, $θ$ is the polar (zenith) angle from the z-axis and $ϕ$ is the azimuthal angle in the xy-plane. In some engineering and geography texts these are reversed. Always check: z = r cos θ confirms the physics convention. Using the wrong angle in $z = r cos θ$ mislabels which orbital lobe points along z.
Forgetting r²sin θ in 3D integrals. The volume element is $d V = r^{2} sin θ d r d θ d ϕ$ , not $d r d θ d ϕ$ . Omitting it when normalising a wavefunction or computing ⟨r⟩ gives a result that peaks at $r = 0$ — physically wrong for any orbital except the nucleus itself.
Degrees vs radians in angular integration limits. The polar angle θ runs from 0 to π radians (not 360°) and φ from 0 to 2π. Using 0 → 180 (degrees) in a numerical integral without converting introduces an error of factor π/180 in every angular element.
Confusing angular nodes with radial nodes. The number of angular nodes equals l (azimuthal quantum number) and the number of radial nodes equals n − l − 1. For the 3d orbital (n=3, l=2): 2 angular nodes and 0 radial nodes — not the other way round.

Orbitals

Solving the Schrödinger equation in spherical coordinates gives radial functions

R (r)

and angular spherical harmonics

Y (θ, ϕ)

. The angular parts —

cos θ

cos 2 θ

and friends — are exactly the lobed shapes of p- and d-orbitals. Try the polar plotter to see them.

Going deeper

The connection between the spherical volume element and orbital probability is one of the most instructive results in quantum chemistry. Because $d V = r^{2} sin θ d r d θ d ϕ$ , the angular part integrates to $4 π$ for a spherically symmetric function (the $\int_{0}^{π} sin θ d θ = 2$ and $\int_{0}^{2 π} d ϕ = 2 π$ together). This is why the radial distribution function for any orbital with quantum numbers (n, l, m) is $P (r) = r^{2} ∣ R_{n l} (r) ∣^{2}$ when the angular part is already normalised — the $r^{2}$ comes directly from the geometry of spherical shells, not from the physics of the electron itself. Forgetting it is the single most common error in calculating ⟨r⟩, ⟨r²⟩, or the probability of finding the electron inside a given radius.

Polar curves & orbitals

Select a curve to see how r(θ) sculpts it. The p and d orbital shapes are exactly the angular parts of the hydrogen wavefunctions.

Curve:

r = ∣ cos θ ∣

x = r cos θ, y = r sin θ

Two lobes along an axis — the angular shape of a p-orbital.

Worked example 1Converting the 2p_z orbital's node to Cartesian form

The angular part of the $2 p_{z}$ orbital is proportional to $cos θ$ . Where is the nodal plane (where $cos θ = 0$ ), and what is its Cartesian equation?

Worked example 2Polar-to-Cartesian conversion of a point on a bond

An electron is found at $r = 120 pm$ , $θ = 6 0^{\circ}$ in the plane of an H₂O molecule. Find its Cartesian coordinates $(x, y)$ .

Worked examples

Polar and spherical coordinates applied to orbitals and probability — try each before revealing the solution.

Worked example 3Volume element in spherical coordinates — the r² sin θ factor

Show that the volume element in spherical coordinates is $d V = r^{2} sin θ d r d θ d ϕ$ , and confirm that integrating over all space gives an infinite volume.

Worked example 4Most probable radius of the hydrogen 1s orbital

The 1s radial wavefunction is $R_{10} (r) = (2/ a_{0}^{3/2}) e^{- r / a_{0}}$ where $a_{0} = 52.9 pm$ . The radial distribution function is $P (r) = r^{2} ∣ R_{10} ∣^{2}$ . Find the most probable radius by setting $d P / d r = 0$ .

Worked example 5Cartesian-to-polar: locating an electron in a 2D cross-section

A molecular orbital calculation places an electron at Cartesian coordinates $(x, y) = (- 80, 60) pm$ in the plane of the molecule. Convert this to polar coordinates and identify which quadrant the point lies in.

Worked example 6Angular nodes of p and d orbitals

For each orbital, state how many angular nodes it has, find the angle(s) where the angular wavefunction is zero, and give the nodal surface in Cartesian form: (a) $2 p_{z}$ with $Y \propto cos θ$ ; (b) $3 d_{z^{2}}$ with $Y \propto (3 cos^{2} θ - 1)$ .

Worked example 7Why 4πr² appears in the radial distribution function

Starting from the volume element $d V = r^{2} sin θ d r d θ d ϕ$ , show that integrating $∣ ψ_{1 s} ∣^{2}$ over all angles gives a factor of $4 π r^{2}$ , so the probability of finding the electron between r and r + dr is $P (r) d r = 4 π r^{2} ∣ R_{10} (r) ∣^{2} d r$ .

ChallengeNode count for 3d and 4s orbitals

For the $3 d$ (n = 3, l = 2) and $4 s$ (n = 4, l = 0) orbitals, state (a) the number of angular nodes, (b) the number of radial nodes, and (c) the total number of nodes. Then explain why the $4 s$ orbital penetrates closer to the nucleus than the $3 d$ , making it lower in energy in multi-electron atoms despite the higher n.

Check yourself

Four quick questions on polar coordinates, orbitals and spherical symmetry.

Question 1 of 4 · Score 0

Polar coordinates describe a point by:

Choose an answer.