Exercises — Orbital shapes — s (spherical), p (dumbbell), d (cloverleaf), f

The letter is just a nickname for a value of $\ell$:

• $s\Rightarrow \ell =0\Rightarrow$ sphere (round ball).
• $p\Rightarrow \ell =1\Rightarrow$ dumbbell (two lobes, like a peanut).
• $d\Rightarrow \ell =2\Rightarrow$ cloverleaf (four lobes) — except $d_{z^2}$.
• $f\Rightarrow \ell =3\Rightarrow$ complex flower (eight or more lobes).

Why: the shape lives entirely in the angular part $Y_{\ell }^m$, and $\ell$ is the only quantum number that changes how many "cuts" (angular nodes) slice the cloud.

Rule: number of orbitals $=2\ell +1$. For $d$, $\ell =2$, so $2(2)+1=5$. Why the $+1$: the orientation number $m$ runs over the whole numbers from $-\ell$ to $+\ell$ including zero. Counting $-2,-1,0,+1,+2$ gives five values, one orbital each.

Read off $n=3$ (the "3") and $\ell =1$ (the "$p$").

• Angular nodes $=\ell =1$. What it looks like: one flat plane through the nucleus (a $p$ dumbbell always has exactly one such plane).
• Radial nodes $=n-\ell -1=3-1-1=1$. What it looks like: one spherical shell of zero probability nested inside.
• Total $=n-1=2$. Check: $1+1=2$. ✔

$4d\Rightarrow n=4,\ell =2$. $\text{radial nodes}=n-\ell -1=4-2-1=1.$ Meaning: one spherical shell where the electron is never found, sitting between the inner and outer parts of the cloverleaf.

$5f\Rightarrow n=5,\ell =3$.

• Total $=n-1=4$.
• Angular $=\ell =3$.
• Radial $=n-\ell -1=5-3-1=1$. Check: $3+1=4$. ✔

First, what is $\theta$? Stand at the nucleus and point straight up along $+z$; that direction is $\theta =0^{\circ }$. Point straight down along $-z$; that is $\theta =180^{\circ }$. The flat "equator" plane (the $xy$-plane) is $\theta =90^{\circ }$. See the figure.

The cloud density follows $(\cos \theta {)}^2$, because probability is ${\psi }^2$ and $\psi \propto \cos \theta$.

• Densest: where $\cos \theta =\pm 1$, i.e. $\theta =0^{\circ }$ and $\theta =180^{\circ }$ → straight up ($+z$) and straight down ($-z$). This is why the dumbbell points along $z$.
• Node: where $\cos \theta =0$, i.e. $\theta =90^{\circ }$ → the entire $xy$-plane. The electron is never found in that flat plane.

Why $\cos$ and not $\sin$? We wanted the function biggest along the axis and zero at the equator. $\cos \theta$ does exactly that: it starts at $1$ on the axis and falls to $0$ at $90^{\circ }$. That is the whole reason $\cos \theta$ (not $\sin \theta$) is the fingerprint of $p_z$.

Split the question into shape vs size.

• Shape comes from $\ell$. For any $s$, $\ell =0$, so angular nodes $=0$ → no cuts → one solid ball. This is true for $1s,2s,3s$ alike. Roundness is untouched by $n$.
• Size and internal structure come from $n$. Radial nodes $=n-\ell -1$. For $3s$: $3-0-1=2$ spherical shells of zero probability inside; for $1s$: $1-0-1=0$. More radial nodes push the outer cloud further out → $3s$ is larger.

So $n$ changes how far out and how many hollow shells, never the roundness. That is the difference between "size" and "shape."

Both facts are true; the node count and the type of node are separate things.

• Node count: $\ell =2$ forces exactly two angular nodes. No exception.
• Node shape: the other four $d$ orbitals put their two nodes as flat planes, which slices four lobes. $d_{z^2}$ instead puts its two nodes as cones (the angular part $\propto 3{\cos }^2\theta -1$ vanishes on two cones symmetric about the $z$-axis).

The two cones open at the angle where $3{\cos }^2\theta -1=0$, i.e. $\cos \theta =\pm \frac{1}{\sqrt{3}}$, giving $\theta \approx 54.7^{\circ }$ and $\theta \approx 125.3^{\circ }$ (the "magic angle"). Resulting shape: one big lobe up the $+z$ axis, one big lobe down the $-z$ axis, and a torus (doughnut) ring hugging the $xy$-plane between the cones. Same two-node budget, spent on cones instead of planes → different-looking but fully consistent geometry.

Aufbau fills lowest energy first: $1s,2s,2p,3s,3p,4s,3d,\dots$ (see Aufbau, Hund and Pauli). Each orbital holds 2 electrons; a subshell holds $2(2\ell +1)$. Running total of electrons after each subshell:

• $1s$: 2 (total 2)
• $2s$: 2 (total 4)
• $2p$: 6 (total 10)
• $3s$: 2 (total 12)
• $3p$: 6 (total 18)
• $4s$: 2 (total 20)
• $3d$: next up — electron 21 is the first electron of $3d$.

So the 21st electron enters the $3d$ subshell. Its shape: $\ell =2\Rightarrow$ cloverleaf (four lobes) for four of the five orbitals, or the dumbbell-plus-torus for $d_{z^2}$. (Element 21 is scandium — its first $d$ electron.)

The sign on a lobe is the sign of the wavefunction $\psi$ in that region — a mathematical bookkeeping value, like the $\pm$ of a wave crest vs trough. It is not electric charge.

• Probability is always positive: the chance of finding the electron is $|\psi {|}^2={\psi }^2$, and squaring a negative number gives a positive one. Both lobes are equally "full" of electron probability. There is no negative-charge lobe.
• The electron is a single particle of charge $-e$ smeared over both lobes; you cannot split its charge by sign of $\psi$.

Where the sign truly matters: in bonding overlap (see Bonding — Sigma and Pi Overlap). When two orbitals meet, a $(+)$ lobe overlapping a $(+)$ lobe adds in phase → a bonding (constructive) combination; a $(+)$ meeting a $(-)$ cancels out of phase → an antibonding (destructive) combination. So the sign controls whether atoms stick together, never how charge is distributed within one atom.

Start from the definitions: $\text{radial nodes}=n-\ell -1.$ A count of nodes cannot be negative (you can't have $-1$ nodal shells). So: $n-\ell -1\ge 0⟹\ell \le n-1.$

• Interpretation of the bound: the biggest $\ell$ allowed is ${\ell }_{\max }=n-1$, and there the orbital spends all of its $n-1$ nodes on angular ones, leaving zero radial nodes.
• Since angular nodes $=\ell$ and total nodes $=n-1$, angular can at most equal total (when $\ell =n-1$), never exceed it. ∎

Consequence: for $n=1$, only $\ell =0$ (just $1s$). For $n=2$: $\ell =0,1$ ($2s,2p$). For $n=3$: up to $\ell =2$ ($3s,3p,3d$) — but no $3f$, because $f$ needs $\ell =3>n-1=2$. This is why $3f$ and $2p$-and-below patterns exist exactly as the periodic table shows.

• 1.2: $d$ subshell has 5 orbitals.
• 2.1: $3p$ → angular $1$, radial $1$, total $2$.
• 2.2: $4d$ → radial nodes $=1$.
• 2.3: $5f$ → total $4$, angular $3$, radial $1$.
• 3.1: $2p_z$ node at $\theta =90^{\circ }$; maxima at $0^{\circ },180^{\circ }$.
• 4.1: $d_{z^2}$ cones at $\theta \approx 54.7^{\circ },125.3^{\circ }$.
• 4.2: 21st electron enters $3d$.
• 5.2: ${\ell }_{\max }=n-1$.