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

$4s$ orbital is spherical while a $4p$ is not, even though both have $n=4$. True. ::: Shape is set by $\ell$, not $n$. The $4s$ has $\ell =0$ so its angular part is a constant (same in every direction → a ball); the $4p$ has $\ell =1$, whose angular part $\propto \cos \theta$ varies with direction → a dumbbell.

$s$ orbital has zero nodes of any kind. False. ::: An $s$ orbital has zero angular nodes ($\ell =0$), but it can still have radial nodes: $n-\ell -1=n-1$ of them. A $3s$, for example, has 2 spherical shells where $\psi =0$.

$p$ orbital mean one lobe holds positive charge and the other negative charge. False. ::: Those signs are the sign of the wavefunction $\psi$, not electric charge. Probability is $|\psi {|}^2\ge 0$ everywhere, so both lobes are equally likely to hold the (negative) electron. The sign only matters when two orbitals overlap in bonding.

$2p_z$ orbital and a $3p_z$ orbital have the same shape of nodal surface passing through the nucleus. True. ::: Both have $\ell =1$, so both have exactly one angular node, and for the $z$-oriented $p$ that node is the $xy$-plane (where $\cos \theta =0$). The $3p_z$ merely adds one extra radial node and is larger; the angular geometry is identical.

$d_{z^2}$ is a $d$ orbital, it must have four lobes like $d_{xy}$. False. ::: All $d$ orbitals share $\ell =2$ and therefore two angular nodes, but the geometry differs. For $d_{z^2}$ the two nodes are cones, not planes, giving a single dumbbell along $z$ plus a torus ring — same node count, different node shape.

$n$ while keeping $\ell$ fixed always increases the number of angular nodes. False. ::: Angular nodes depend only on $\ell$. Raising $n$ at fixed $\ell$ adds radial nodes ($n-\ell -1$ grows) and makes the orbital bigger, but the angular pattern (its lobe shape) is unchanged.

$p$ orbital must physically cross the nodal plane to get from the top lobe to the bottom lobe. False. ::: This assumes the electron follows a path, which it does not (see Heisenberg Uncertainty Principle). The orbital is a probability cloud, not a track; talking about "crossing" the node imports a classical trajectory that quantum mechanics forbids.

$1s$ and a $2p$ orbital have the same total number of nodes. True. ::: Total nodes $=n-1$. For $1s$: $1-1=0$. For $2p$: $2-1=1$... so actually they differ. ::: False — re-check: $1s$ has 0 total nodes, $2p$ has 1 (one angular node, the nodal plane). Don't be fooled: equal $|n-\ell |$ is not equal total nodes.

$\ell$ but different $m$ (e.g. $p_x$ and $p_y$) have the same shape. True. ::: Same $\ell$ means the same number and type of angular nodes, so the same shape — they are just rotated into different directions ($x$ vs $y$). $m$ chooses orientation, not shape.

$3d$ orbital has $n-1=2$ nodes, and since it's a $d$ they are all angular. Total = 2, done." The count is right but the classification is lucky-accidental. ::: For $3d$: angular nodes $=\ell =2$, radial nodes $=n-\ell -1=3-2-1=0$. So yes all 2 are angular — but only because the radial count happens to be 0. Never assume "$d$ ⇒ all nodes angular"; a $4d$ has one radial node.

$2s$ orbital is spherical, so it has no nodes at all." Confuses "no angular nodes" with "no nodes." ::: Spherical shape ⇒ zero angular nodes, correct. But $2s$ has $n-\ell -1=2-0-1=1$ radial node — a hollow spherical shell of zero probability inside the outer ball.

$p$ orbitals appear starting at $n=1$ because every shell needs an $s$ and a $p$." Violates the rule $\ell <n$. ::: The allowed $\ell$ for a given $n$ run $0,1,\dots ,n-1$. For $n=1$ only $\ell =0$ ($1s$) exists — no $1p$. The first $p$ subshell is $2p$ (see Quantum Numbers n, l, m, s).

$d$ orbitals because $d$ has five lobes." Confuses orbital count with lobe count. ::: There are five $d$ orbitals because $m$ takes $2\ell +1=2(2)+1=5$ values. Each individual $d$ orbital usually has four lobes (and $d_{z^2}$ has a dumbbell + torus). Lobe count ≠ orbital count.

$d_{z^2}$ torus is a radial node, so $d_{z^2}$ has one radial and one angular node." Mislabels the torus. ::: The torus is a feature of the angular pattern, and $d_{z^2}$'s two angular nodes are two cones. A $3d_{z^2}$ still has 0 radial nodes ($3-2-1=0$). The ring is where the lobes pinch, not a spherical radial shell.

$|\psi {|}^2$ is always positive, orbitals never have signs, so the whole $+/-$ lobe talk is meaningless." Overcorrects a real subtlety. ::: $|\psi {|}^2$ is indeed always $\ge 0$, but $\psi$ itself can be positive or negative, and that sign of $\psi$ is physically real: it decides in-phase (bonding) vs out-of-phase (antibonding) overlap. Ignoring it breaks all of molecular orbital theory.

$s$ orbital the only shape guaranteed spherical, no matter the element or $n$? Because its angular part $Y_0^0=\frac{1}{\sqrt{4\pi }}$ is a constant with no $\theta$ or $\varphi$ dependence. ::: A function that is identical in every direction can only trace out a sphere. Every other $\ell$ has an angular function that varies with direction, so it must have directional lobes.

$\ell$ produce a more complicated shape (more lobes)? Because $Y_{\ell }^m$ is a directional polynomial of degree $\ell$, so it can cross zero on up to $\ell$ surfaces. ::: Each angular node ("cut") slices existing lobes in two, so more cuts → more lobes: 0 cuts (ball), 1 cut (dumbbell), 2 cuts (cloverleaf), 3 cuts (flower). The origin of these functions is in Spherical Harmonics.

Because we draw a surface enclosing ~90% of the probability $|\psi {|}^2$. ::: The shape is not the electron's path but the outline of the probability cloud — the region where the electron is overwhelmingly likely to be found. It is a statistical portrait, consistent with Heisenberg Uncertainty Principle.

$p_z$ node fall exactly on the $xy$-plane rather than somewhere tilted? Because $p_z\propto \cos \theta$, and $\cos \theta =0$ precisely at $\theta =90^{\circ }$. ::: $\theta =90^{\circ }$ is the flat $xy$-plane (straight sideways from the $z$-axis), so the wavefunction vanishes on exactly that plane — the one place equidistant from the two lobes.

$p$, five $d$, seven $f$ orbitals — always odd numbers? Because $m$ runs over $-\ell ,\dots ,0,\dots ,+\ell$, giving $2\ell +1$ values. ::: $2\ell +1$ is always odd (an even number plus one). Each $m$ labels one spatial orientation, so the orientation count is forced to be odd (linked to Aufbau, Hund and Pauli filling).

Because radial nodes live in $R_{n,\ell }(r)$, which depends only on distance $r$, not direction. ::: A shell at fixed radius is symmetric in all directions, so carving it out cannot break the angular pattern set by $Y_{\ell }^m$. It nests an inner shell inside the same overall shape (see Radial Distribution Function).

$\ell =0$ — is it even a function of angle? It is a constant function of angle. ::: With $\ell =0$ the only allowed $m$ is $0$, and $Y_0^0$ has no $\theta ,\varphi$ terms. This is the degenerate limit where "angular dependence" collapses to a single number — the reason the sphere is the base case.

$n\to \infty$ for fixed $\ell$, what happens to the orbital's shape and its radial nodes? The shape (angular pattern) stays fixed; the number of radial nodes $n-\ell -1\to \infty$ and the orbital swells outward. ::: The electron is spread over ever-larger, ever-more-shelled but identically-shaped clouds — approaching the ionisation (free-electron) limit.

$r=0$) a node for a $2p$ orbital? Yes — at $r=0$ every $p$ orbital has $\psi =0$. ::: The angular factor of a $p$ vanishes along its nodal plane, and the radial factor $R_{2,1}$ vanishes at $r=0$, so the origin sits on the nodal plane. This is why $p$ orbitals have zero density right at the nucleus, unlike $s$.

Yes. ::: A $2p$ ($n=2,\ell =1$) has 1 angular node and $n-\ell -1=0$ radial nodes. Similarly $3d$ has 2 angular, 0 radial. Whenever $n=\ell +1$ the orbital is "purely angular" in its nodes.

$d_{xy}$ and $d_{x^2-y^2}$ have the same node count despite pointing differently? Yes — both have exactly 2 angular nodes (two planes). ::: $d_{xy}$'s nodal planes are the $xz$- and $yz$-planes, so its lobes sit between the axes; $d_{x^2-y^2}$'s nodal planes bisect those, so its lobes sit on the $x$ and $y$ axes. Same $\ell$, same count, rotated orientation.

$g$" subshell ($\ell =4$) have, extrapolating the rules? $2\ell +1=9$ orbitals, 4 angular nodes, and radial nodes $=n-\ell -1$ (needs $n\ge 5$). ::: The rules never break: each higher $\ell$ just adds one more angular cut. A $5g$ would have $5-4-1=0$ radial nodes and a very multi-lobed angular pattern.