Which statement about the 2pₓ orbital is correct?

It has a nodal plane passing through the nucleus

The electron travels along a fixed circular path within it

It has a nodal plane passing through the nucleus

It can hold up to six electrons

Its boundary surface encloses 100 % of the electron probability

Atomic Structure — Quantum Numbers, Orbitals and Configurations

Atomic structure is the chapter that decides how well you will cope with the whole of inorganic chemistry, because periodicity, bonding, colour, magnetism and oxidation states are all consequences of where the electrons sit. It is also the chapter where the paper most often tests conceptual rather than numerical ability: a typical question gives four statements about orbitals and asks which is correct. The central shift to make is from the Bohr picture of an electron on a track to the quantum-mechanical picture of an electron as a standing wave whose square gives a probability. Once that shift is made, the shapes of orbitals, the existence of nodes and the impossibility of a definite path all stop being facts to memorise and become things you can reconstruct.

What Bohr Got Right, and Where It Breaks

Bohr postulated that angular momentum is quantised, mvr = nh/2π, which forces the energy of the hydrogen atom to take only the values Eₙ = −13.6/n² eV and the radius to be rₙ = 0.529 n² Å. For a one-electron species with nuclear charge Z these become Eₙ = −13.6 Z²/n² eV and rₙ = 0.529 n²/Z Å. These formulae are still examined heavily because they are exact for H, He⁺, Li²⁺ and Be³⁺ — any species with exactly one electron.

How Bohr quantities scale with n and Z
Quantity	Dependence	Consequence worth remembering
Radius rₙ	∝ n²/Z	The n = 2 shell of H is four times the n = 1 shell
Energy Eₙ	∝ −Z²/n²	Levels crowd together as n rises and converge at 0
Velocity vₙ	∝ Z/n	Electrons move slower in outer shells
Angular momentum	∝ n only	Independent of Z — the same for H and Li²⁺ at equal n
Time period	∝ n³/Z²	Outer orbits are far slower per revolution

⚠️ Bohr fails the moment there are two electrons

The model cannot explain the spectrum of helium, the splitting of lines in a magnetic field, or the fine structure of hydrogen itself — and, most fundamentally, it violates the uncertainty principle by assigning an electron both a definite radius and a definite momentum. So a question that asks for "the radius of the 3rd orbit of lithium atom" is testing whether you noticed that neutral Li has three electrons and the Bohr formula does not apply; the formula works for Li²⁺, not Li. Read the charge before you reach for 0.529 n²/Z.

Bohr does, however, give the spectral series correctly through the Rydberg relation 1/λ = R_H Z²(1/n₁² − 1/n₂²). The Lyman series (n₁ = 1) lies in the ultraviolet, Balmer (n₁ = 2) in the visible — the only series you can see — Paschen (n₁ = 3) and Brackett (n₁ = 4) in the infrared. The number of distinct spectral lines emitted when an electron falls from level n to the ground state is n(n − 1)/2, a small combinatorial result that appears almost every year.

The Four Quantum Numbers

What each quantum number fixes
Symbol and name	Allowed values	What it determines
n — principal	1, 2, 3, …	Shell, size and (for hydrogen) the energy entirely
l — azimuthal	0 to n − 1	Subshell and shape: 0 = s, 1 = p, 2 = d, 3 = f
m — magnetic	−l to +l, so 2l + 1 values	Orientation in space — why p has three orbitals and d has five
s — spin	+½ or −½	Spin orientation; caps every orbital at two electrons

Counting follows directly. A subshell holds 2(2l + 1) electrons, a shell holds 2n². The number of radial nodes is n − l − 1, the number of angular nodes is l, and the total is always n − 1. So a 3d orbital (n = 3, l = 2) has zero radial nodes and two angular ones, while 3s has two radial and none angular — both totalling 2. Questions that ask for "the number of nodes in 4f" are asking you to apply n − 1 = 3 and then split it as l = 3 angular and 4 − 3 − 1 = 0 radial.

🎯 Why the 2p orbital has no radial node but 3p has one

A node is a surface where the wavefunction changes sign, and a standing wave needs one extra sign-change for each extra half-wavelength it fits into the atom. Going from 2p to 3p keeps the shape (still dumb-bell, so still one angular node through the nucleus) but adds one more radial oscillation, hence one radial node. The formula n − l − 1 is just this bookkeeping: n − 1 total sign-changes, of which l are used up by the shape, leaving the rest for the radial direction. Understanding it this way means you never need to memorise node counts for individual orbitals.

⚠️ An orbital is not an orbit, and an electron does not have a path

Heisenberg's principle Δx·Δp ≥ h/4π forbids simultaneous exact knowledge of position and momentum, so the very idea of a trajectory is meaningless for an electron. What ψ² gives is a probability density, and an "orbital" is the region enclosing about 90 % of that probability. Three consequences are examined: the 2pₓ orbital has a nodal plane through the nucleus where the probability is exactly zero (yet an electron gets from one lobe to the other, because it was never travelling along a path); the boundary surface is a convention, not a wall; and the electron is not "spinning" in any literal sense — spin is an intrinsic property with no classical analogue.

De Broglie ties the two pictures together: λ = h/mv, or λ = h/√(2mKE), and for a charged particle accelerated through V volts, λ = √(150/V) Å for an electron. The reason wave behaviour is invisible for a cricket ball is simply that h is 6.626 × 10⁻³⁴ — a ball of 100 g at 10 m/s has a wavelength of about 10⁻³⁴ m, unimaginably smaller than any aperture it could diffract through. Bohr's quantisation condition follows from de Broglie: an orbit is allowed only if its circumference fits a whole number of wavelengths, 2πr = nλ.

Filling the Orbitals — and the Famous Exceptions

Three rules govern filling. Aufbau: electrons occupy the lowest available energy orbital, and for multi-electron atoms the order follows increasing (n + l), with the lower n breaking a tie. Pauli: no two electrons in an atom share all four quantum numbers, which caps an orbital at two. Hund: within a degenerate subshell, electrons occupy separate orbitals with parallel spins before any orbital is doubly filled.

🧠 The (n + l) rule replaces memorising the filling order

Compare 4s (n + l = 4 + 0 = 4) with 3d (3 + 2 = 5): 4s has the lower sum and fills first, which is exactly why potassium is [Ar]4s¹ and not [Ar]3d¹. Compare 3d (5) with 4p (4 + 1 = 5): the sums tie, so the lower n wins and 3d fills before 4p. Applying this generates the whole sequence 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s without any diagonal diagram. Note that once the 3d electrons are present they are lower in energy than 4s, which is why transition metals lose 4s electrons first when they ionise — Fe²⁺ is [Ar]3d⁶, not [Ar]3d⁴4s².

The configuration exceptions the paper expects by name
Element	Expected	Actual
Cr (Z = 24)	[Ar]3d⁴4s²	[Ar]3d⁵4s¹ — half-filled d
Cu (Z = 29)	[Ar]3d⁹4s²	[Ar]3d¹⁰4s¹ — fully filled d
Mo (Z = 42)	[Kr]4d⁴5s²	[Kr]4d⁵5s¹
Ag (Z = 47)	[Kr]4d⁹5s²	[Kr]4d¹⁰5s¹
Pd (Z = 46)	[Kr]4d⁸5s²	[Kr]4d¹⁰5s⁰ — both s electrons move in

🎯 Why half-filled and fully filled subshells are extra stable

Two effects add up. Symmetry: a d⁵ or d¹⁰ set has spherically symmetric charge distribution, which lowers repulsion. Exchange energy: electrons with parallel spins in different orbitals can swap places, and every such swap lowers the energy. The number of exchange pairs is n(n − 1)/2 for n parallel electrons, so d⁵ with five parallel spins gives 10 exchange pairs against only 6 for d⁴ — a gain large enough to outweigh the cost of promoting one 4s electron. This is also the reason Mn²⁺ (d⁵) is unusually stable and why Fe³⁺ (d⁵) is a stronger oxidising target than Fe²⁺ is a reducing one.

Two derived quantities are asked directly. Number of unpaired electrons follows from Hund's rule applied to the actual configuration; a species with none is diamagnetic and one with any is paramagnetic. Spin-only magnetic moment is μ = √(n(n + 2)) Bohr magnetons for n unpaired electrons, giving 1.73 for one, 2.83 for two, 3.87 for three, 4.90 for four and 5.92 for five — a short table worth knowing outright, because coordination-chemistry questions run the calculation backwards from a measured μ to the number of unpaired electrons and hence to the geometry.

Key takeaways

Bohr is exact only for one-electron species: Eₙ = −13.6Z²/n² eV, rₙ = 0.529n²/Z Å. Check the charge first.
Angular momentum depends on n alone, not on Z — the same for H and Li²⁺ in the same shell.
An electron falling from level n emits n(n − 1)/2 distinct spectral lines; only the Balmer series is visible.
l fixes shape, m fixes orientation (2l + 1 orbitals), s caps each orbital at two electrons.
Radial nodes = n − l − 1, angular nodes = l, total always n − 1.
Orbitals fill by increasing (n + l), with the lower n breaking a tie — 4s before 3d, 3d before 4p.
Cr is 3d⁵4s¹ and Cu is 3d¹⁰4s¹ because symmetry plus exchange energy favours half-filled and filled d.
Spin-only moment μ = √(n(n + 2)) BM: 1.73, 2.83, 3.87, 4.90, 5.92 for one to five unpaired electrons.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The total number of nodes in a 3p orbital is:
1. 1
2. 2
3. 3
4. 0
Show answer
Answer: B — 2
Total nodes = n − 1 = 3 − 1 = 2, made up of l = 1 angular node (the nodal plane through the nucleus) and n − l − 1 = 1 radial node. The answer 1 is what you get by quoting only the radial count, which is the commonest slip here — the question asked for the total, and angular nodes are nodes too.
The ground-state electronic configuration of chromium (Z = 24) is:
1. [Ar]3d⁴4s²
2. [Ar]3d⁵4s¹
3. [Ar]3d⁶4s⁰
4. [Ar]3d³4s²4p¹
Show answer
Answer: B — [Ar]3d⁵4s¹
Chromium promotes one 4s electron to give the half-filled [Ar]3d⁵4s¹, because a spherically symmetric d⁵ set plus the extra exchange energy from five parallel spins (10 exchange pairs against 6 for d⁴) more than repays the promotion cost. The naive Aufbau answer [Ar]3d⁴4s² is offered first precisely because it is what the filling rule predicts before stability corrections.
For which of the following would the Bohr expression rₙ = 0.529 n²/Z Å be valid?
1. He atom
2. Li atom
3. Li²⁺ ion
4. H₂ molecule
Show answer
Answer: C — Li²⁺ ion
The Bohr model applies only to species with exactly one electron, since it ignores electron–electron repulsion entirely. Li²⁺ has lost two of its three electrons and so qualifies, as do H, He⁺ and Be³⁺. Neutral He and Li have two and three electrons respectively, and H₂ is a molecule with two nuclei — none of these can be handled by a one-electron formula.
The spin-only magnetic moment of Fe²⁺ ([Ar]3d⁶) in an octahedral weak field is closest to:
1. 1.73 BM
2. 2.83 BM
3. 4.90 BM
4. 5.92 BM
Show answer
Answer: C — 4.90 BM
In a weak field the d⁶ set is high-spin, so Hund's rule fills all five orbitals singly and the sixth electron pairs up, leaving four unpaired electrons. Then μ = √(4 × 6) = √24 = 4.90 BM. The 5.92 option corresponds to five unpaired electrons, which would be d⁵ (Fe³⁺) not d⁶ — a switch of oxidation state that questions use to test whether you counted the electrons for the ion actually named.
An electron in a hydrogen atom drops from n = 4 to the ground state. The maximum number of distinct spectral lines observed is:
1. 3
2. 4
3. 6
4. 10
Show answer
Answer: C — 6
Every pair of levels from 1 to 4 can give a transition, so the count is n(n − 1)/2 = 4 × 3/2 = 6: the jumps 4→3, 4→2, 4→1, 3→2, 3→1 and 2→1. The answer 3 counts only the direct drops to the ground state and misses the cascade, which is the usual error — the electron need not fall in one step.
Which orbital fills before 3d according to the (n + l) rule?
1. 4p
2. 4s
3. 4d
4. 4f
Show answer
Answer: B — 4s
For 4s, n + l = 4 + 0 = 4; for 3d, n + l = 3 + 2 = 5. The smaller sum fills first, so 4s precedes 3d — which is exactly why potassium is [Ar]4s¹. For 4p the sum is 5, tying with 3d, and the tie is broken by the smaller n, so 3d fills before 4p rather than after it.
The maximum number of electrons that can occupy the subshell with n = 4 and l = 3 is:
1. 10
2. 14
3. 18
4. 32
Show answer
Answer: B — 14
l = 3 is the f subshell, with 2l + 1 = 7 orbitals and therefore 2 × 7 = 14 electrons. The answer 32 is 2n² for n = 4, which is the capacity of the entire fourth shell (4s + 4p + 4d + 4f = 2 + 6 + 10 + 14), not of the single subshell the question specified.
Which statement about the 2pₓ orbital is correct?
1. The electron travels along a fixed circular path within it
2. It has a nodal plane passing through the nucleus
3. It can hold up to six electrons
4. Its boundary surface encloses 100 % of the electron probability
Show answer
Answer: B — It has a nodal plane passing through the nucleus
A p orbital has l = 1 and therefore one angular node — for 2pₓ this is the yz plane through the nucleus, where the probability density is exactly zero. The path option contradicts the uncertainty principle, six electrons is the capacity of the whole 2p subshell rather than one orbital, and the boundary surface is drawn to enclose about 90 % of the probability by convention, since the wavefunction never falls to exactly zero at any finite distance.