Atomic orbitals: shapes, nodes and radial probability

From orbit to orbital. Bohr's orbit specified an exact position and velocity for the electron at every moment — deterministic, like a planet on a track. That picture cannot survive once the electron is treated as a wave obeying the Heisenberg uncertainty principle: position and momentum cannot both be known exactly. The Schrödinger wave equation replaces the fixed path with a wavefunction $\psi$, and $|\psi|^2$ gives the probability of finding the electron at a point. An orbital is conventionally the boundary surface that encloses about 90% of that total probability — a fuzzy, three-dimensional region, not a line.

s orbitals ($l=0$). Perfectly spherical — probability is identical in every direction at a given distance from the nucleus. $1s$ is a single solid sphere. $2s$ is also spherical but larger, and it contains an internal spherical shell of zero probability (a node) before the density rises again further out. Every increase in $n$ for an $s$ orbital adds one more such internal spherical node.

p orbitals ($l=1$). Dumbbell-shaped: two lobes on opposite sides of the nucleus, separated by a flat nodal plane that passes through the nucleus itself — probability is zero exactly on that plane. Three degenerate $p$ orbitals exist per shell ($p_x$, $p_y$, $p_z$), one along each perpendicular axis, corresponding to $m_l = -1, 0, +1$.

d orbitals ($l=2$). Generally four-lobed clover shapes (the exception, $d_{z^2}$, has two lobes along the $z$-axis plus a doughnut-shaped ring around the middle). Five degenerate $d$ orbitals exist per shell: $d_{xy}$, $d_{yz}$, $d_{xz}$, $d_{x^2-y^2}$, $d_{z^2}$.

Counting nodes. A node is a surface where the electron probability is exactly zero. Two kinds exist: Radial (spherical) nodes are concentric spherical shells inside an orbital where probability drops to zero, coming from the radial part of the wavefunction:

where: $n$ is the principal quantum number and $l$ is the azimuthal quantum number of the orbital. Angular (planar) nodes are flat planes (or cones) through the nucleus, coming from the orbital's angular shape:

Adding the two always gives a clean result independent of $l$:

Radial probability distribution. Instead of asking for the probability density at one exact point, chemists ask a more useful question: what is the probability of finding the electron somewhere within a thin spherical shell at distance $r$ from the nucleus, as $r$ varies? Plotted against $r$, this curve rises and falls, with a dip to zero at each radial node, and — counter-intuitively — it is zero exactly at the nucleus ($r=0$) for every orbital, rising to a maximum at a small non-zero radius (close to the Bohr radius for $1s$) before tailing off at large $r$.

Question. Calculate the number of radial nodes and angular nodes in a $3p$ orbital, state the total, and then compare it against $2s$ and $3s$ to show how the same total node count can come from different combinations.

Step 1. For $3p$: $n=3$, and $p$ means $l=1$.

Step 2. Radial nodes $= n-l-1 = 3-1-1 = 1$.

Step 3. Angular nodes $= l = 1$.

Step 4. Total nodes $= n-1 = 3-1 = 2$, matching Step 2 + Step 3 $= 1+1=2$. So $3p$ has one radial node and one angular node, two nodes total.

Step 5. For $2s$: $n=2, l=0$, giving radial nodes $=2-0-1=1$, angular nodes $=0$, total $=1$. For $3s$: $n=3,l=0$, giving radial nodes $=3-0-1=2$, angular nodes $=0$, total $=2$.

Step 6. Comparison: $3s$ and $3p$ both total 2 nodes (as expected since both have $n=3$, so $n-1=2$), but $3s$'s two nodes are both radial shells, keeping it spherically symmetric, while $3p$ splits its two nodes as one radial shell plus one nodal plane — and it is that nodal plane that breaks the sphere into two separate lobes. $3p$: 1 radial + 1 angular = 2 nodes total.

Competitive stretch

A $4d$ orbital has $n=4, l=2$: radial nodes $=4-2-1=1$, angular nodes $=2$, total $=4-1=3$. So among $2p, 3d, 4d$, and $3p$, the orbital with the most angular (planar) nodes for the least total is the one with $l$ closest to $n-1$ — here $3d$ ($n=3,l=2$: radial $=0$, angular $=2$, total $2$) has zero radial nodes, the purest example of an orbital shape built entirely from angular structure.

• Confusing "orbit" (Bohr's outdated fixed circular path with a definite position and velocity) with "orbital" (a modern 3D probability region) — they sound alike but describe entirely different models of the atom.
• Assuming the three $2p$ orbitals (or five $3d$ orbitals) differ in energy in an isolated atom — they are degenerate, differing only in spatial orientation, not energy.
• Forgetting the checksum total nodes $=n-1$ always holds regardless of $l$, which is the fastest way to catch an arithmetic slip in radial or angular node counts.
• Believing electron probability is highest exactly at the nucleus for an $s$ orbital — it is actually zero at $r=0$ and only reaches its maximum at a small non-zero radius, as the radial probability distribution curve shows.

Ask the tutor: want a step-by-step walkthrough of how a nodal plane splits a $p$ orbital into two separate lobes, or how radial nodes look as internal shells inside an $s$ orbital — just ask the AI tutor to sketch it out.