Shapes and Energies of Atomic Orbitals

In the previous section, you learned that each set of quantum numbers $(n, l, m_l)$ defines an atomic orbital — a specific wave function $\psi$ that describes an electron. Now we ask: what do these orbitals look like?

The shape of an orbital is determined by the azimuthal quantum number $l$:

• $l = 0$: s orbitals — spherical
• $l = 1$: p orbitals — dumbbell-shaped
• $l = 2$: d orbitals — double dumbbell / clover-leaf

The principal quantum number $n$ controls the size — and introduces radial nodes (spherical shells where $\psi = 0$). Going from $1s \to 2s \to 3s$, the orbital gets larger and gains radial nodes: $1s$ has 0 nodes, $2s$ has 1, $3s$ has 2. The formula is: radial nodes = $n - l - 1$.

s Orbitals ($l = 0$) are spherically symmetric — the probability of finding the electrondepends only on the distance from the nucleus, not the direction. There's no preferredaxis — the electron cloud looks the same from every angle.

As $n$ increases, the s orbital becomes larger, but it also develops radial nodes — spherical shellswhere the probability of finding the electron drops to exactly zero.

p Orbitals ($l = 1$) have a dumbbell shape: two lobes on opposite sides of the nucleus, separated by a nodal plane through the nucleus where $\psi = 0$.

There are three p orbitals in every p subshell, oriented along the three coordinate axes:

• $p_x$: lobes along the x-axis, nodal plane is the yz-plane
• $p_y$: lobes along the y-axis, nodal plane is the xz-plane
• $p_z$: lobes along the z-axis, nodal plane is the xy-plane

The two lobes have opposite signs of $\psi$ (shown as different colours in diagrams). This does not mean one lobe is negative probability — $|\psi|^2$ is positive everywhere. The sign matters when orbitals overlap to form bonds (constructive vs destructive interference).

Like s orbitals, p orbitals grow larger and gain radial nodes as $n$ increases. The $2p$ orbital has 0 radial nodes; $3p$ has 1 radial node ($n - l - 1 = 3 - 1 - 1 = 1$).

d Orbitals ($l = 2$) appear starting from $n = 3$ (the M shell). There are five d orbitals in every d subshell.

Four of them have a four-leaf clover pattern with four lobes:

• $d_{xy}$: lobes between the x and y axes
• $d_{xz}$: lobes between the x and z axes
• $d_{yz}$: lobes between the y and z axes
• $d_{x^2-y^2}$: lobes along the x and y axes (rotated 45° from $d_{xy}$)

The fifth, $d_{z^2}$, has a unique shape — two lobes along the z-axis with a donut-shaped ring (torus) in the equatorial plane. Despite looking different, all five d orbitals are equivalent in energy in an isolated atom.

Each d orbital has two nodal surfaces ($l = 2$) — these are either planes or (in the case of $d_{z^2}$) conical surfaces.

The boundary surface diagram tells you the shape of the orbital — but not where within that shape the electron is most likely to be found. For that, we need the Radial Distribution Function (RDF):

$P(r) = 4\pi r^2 [R(r)]^2$

where $R(r)$ is the radial part of the wave function $\psi$.

$P(r)$ tells you the probability of finding the electron at a distance $r$ from the nucleus, summed over all directions (all angles $\theta$ and $\phi$).

The peak of the $P(r)$ curve is the most probable radius — the distance from the nucleus where you are most likely to find the electron. For the $1s$ orbital, this peak is at exactly $r = a_0 = 0.529$ Å — the Bohr radius! This is a beautiful result: the quantum mechanical calculation gives the same answer as the Bohr model for the ground state.

The dips in the curve (where $P(r) = 0$) correspond to the radial nodes. For example, the $2s$ orbital curve dips to zero at $r = 2a_0$ — the spherical shell where the electron is never found.

An important subtlety: for s orbitals, $|\psi|^2$ is actually maximum at the nucleus ($r = 0$). But $P(r) = 4\pi r^2 |\psi|^2$ is zero at $r = 0$ because of the $r^2$ factor. The $r^2$ accounts for the fact that there is more "volume" available at larger distances. It is like asking: where are most people in a city? Not at the centre (dense but tiny area), but at some middle radius where density × area is maximised.

In hydrogen (one-electron atom), orbital energy depends only on $n$: $1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f$.

All subshells with the same $n$ are degenerate (equal energy) because there is no electron-electron repulsion — the electron feels only the nuclear charge.

In multi-electron atoms, this degeneracy is broken. Electrons in inner shells shield the nuclear charge from outer electrons. The effective nuclear charge $Z_{\text{eff}}$ felt by an electron depends on both $n$ and $l$.

s electrons penetrate closer to the nucleus than p electrons (look at the RDF curves above), so they experience less shielding and feel a stronger $Z_{\text{eff}}$. This means: for the same $n$, $s < p < d < f$ in energy.

The order of orbital energies in multi-electron atoms follows the (n+l) rule:

• Lower $(n + l)$ = lower energy (fills first)
• If two subshells have the same $(n + l)$: lower $n$ fills first

This gives the familiar filling order: $1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f \ldots$