Shapes of s, p and d Orbitals

Introduction

The familiar pictures of orbitals — a sphere for s, a dumbbell for p and a four-leaf clover for most d orbitals — are among the most used images in chemistry. Yet an orbital has no edge: the wavefunction decays exponentially and never quite reaches zero. This page explains what those drawings actually represent, where their shapes come from mathematically and why the colour or sign on each lobe matters as much as the shape itself.

Core explanation

Boundary surfaces. Because ψ extends to infinity, any drawing must choose a cut-off. A boundary surface is a surface on which ψ has a constant value, chosen so that it encloses a fixed fraction of the probability, commonly 90%. Choosing 99% would give a larger surface of the same general shape. The shape is set by the angular function Y, and the size by the radial function R.

Phase. ψ is a signed function. Regions where ψ > 0 and ψ < 0 are drawn in different colours or marked + and −. The sign itself has no direct physical meaning (−ψ describes the same state), but the relative sign of lobes on different atoms decides whether orbitals overlap constructively (bonding) or destructively (antibonding).

s orbitals (l = 0). Y(0,0) = 1/√(4π) is a constant, so s orbitals are spherical . There are no angular nodes. A 2s or 3s boundary surface looks like a sphere, but inside it are one or two spherical radial nodes separating shells of opposite phase.

p orbitals (l = 1). The real angular functions are proportional to x/r, y/r and z/r:

- p z ∝ cos θ - pₓ ∝ sin θ cos φ - p y ∝ sin θ sin φ

Each has two lobes of opposite sign along its axis, separated by one nodal plane through the nucleus. For p z, ψ is positive for z > 0 and negative for z < 0, and the xy-plane is the node.

d orbitals (l = 2). The five real functions are proportional to:

Orbital Angular form Nodes --- --- --- d xy xy/r² xz- and yz-planes d xz xz/r² xy- and yz-planes d yz yz/r² xy- and xz-planes d x²−y² (x² − y²)/r² planes x = ±y d z² (3z² − r²)/r² two cones at 54.7° and 125.3°

The first four are cloverleaves of four lobes with alternating signs. d xy points between the x and y axes, whereas d x²−y² points along them. d z² has two positive lobes along ±z and a negative torus (doughnut) in the xy-plane.

Why the labels work. Multiplying Y by r^l turns the angular function into a polynomial in x, y and z. That polynomial is the Cartesian label: d xy ∝ xy, and so on.

Step-by-step reasoning

To draw any real orbital from its label:

1. Use the label to find where ψ is zero, for example xy = 0 gives the planes x = 0 and y = 0. 2. These planes divide space into regions; each region holds one lobe. 3. Assign signs by substituting a point in each region, such as (1, 1, 0) for d xy, which gives a positive lobe. 4. Add any spherical radial nodes for n > l + 1.

Visual explanation

Imagine p z as two balloons tied end to end at the nucleus, one red (+) above the xy-plane and one blue (−) below. d x²−y² is four balloons lying on the x and y axes, red on ±x and blue on ±y. d z² is a red dumbbell along z pushed through a blue ring around its waist.

Real-world analogy

A boundary surface is like the outline drawn around a city on a map. The houses do not stop at the line, but the line encloses most of the population. Choosing a stricter or looser definition moves the line outward or inward without changing the city's general shape.

Real-world example

In square-planar nickel(II) and platinum(II) complexes, the four ligands lie on the x and y axes. They point straight at the lobes of d x²−y², which is strongly raised in energy and left empty. With all eight d electrons paired in the four lower orbitals, these complexes are usually diamagnetic — a property directly traceable to the orientation of one orbital's lobes.

Why?

Why are there exactly three p and five d orbitals? For a given l there are 2l + 1 values of mₗ, so 2l + 1 independent angular functions. Any real set must contain the same number: three for l = 1 and five for l = 2. That is why there is no sixth d orbital, even though six products of two coordinates can be written.

Common misconception

"The electron in a p orbital travels from one lobe to the other through the nodal plane." The orbital is a stationary standing-wave state; the electron does not follow a path. Probability density is zero on the plane but the state describes both lobes at once, just as a vibrating string has one standing wave spanning its node.

Worked example

Question: Determine the sign of d x²−y² at the points (1, 0, 0), (0, 1, 0) and (1, 1, 0), and interpret the results.

Reasoning: The angular form is proportional to x² − y². At (1, 0, 0): 1 − 0 = +1. At (0, 1, 0): 0 − 1 = −1. At (1, 1, 0): 1 − 1 = 0.

Answer: Positive along x, negative along y and zero on the diagonal, which lies on a nodal plane x = y.

Quick check

1. How many angular nodal planes does a d xz orbital have, and which planes are they? Answer: Two nodal planes, the xy-plane (z = 0) and the yz-plane (x = 0), since the function is proportional to xz.

Exam focus

Be able to sketch s, p and all five d orbitals on labelled axes with correct phases, and to state what the boundary surface represents. A common exam point is that d z² has conical, not planar, nodes, and that d xy lobes lie between the axes.

Advanced insight

The five real d functions are not unique: any orthonormal combination is equally valid. The standard choice, with d z² as the odd-looking one, arises because 2z² − x² − y² is chosen as the fifth independent function. Mathematically d z² is equivalent to a combination of shapes like "d z²−x²" and "d z²−y²", which explains why it behaves like the other four in an octahedral field even though it looks different.

Summary

Orbital drawings are boundary surfaces enclosing about 90% of the probability. Their shape comes from the angular function and their size from the radial function. s orbitals are spherical, p orbitals have two lobes of opposite phase separated by a nodal plane, and d orbitals have four lobes, except d z², which has two lobes and a ring separated by nodal cones. Phase signs are essential for describing bonding.

Practice questions

1. What does a 90% boundary surface mean, and why cannot a 100% surface be drawn? Answer: It encloses 90% of the probability of finding the electron. Because ψ only decays exponentially, a 100% surface would have to be infinitely large. 2. Which d orbital has lobes pointing along the x and y axes, and which points between them? Answer: d x²−y² points along the axes; d xy points between them. 3. Describe the phase pattern of p x. Answer: Positive lobe along +x and negative lobe along −x, separated by the yz nodal plane. 4. Why can an orbital drawing with the phases swapped describe exactly the same state? Answer: ψ and −ψ give the same probability density and the same energy; only relative phases between different orbitals have physical consequences.