Atomic Structure

4s fills before 3d. So when iron ionises, which electrons leave first?

Most say the 3d ones — last in, first out.

The 4s electrons leave first. Fe${}^{2+}$ is $[\mathrm{Ar}]3d^6$, not $[\mathrm{Ar}]3d^{4}4s^2$. Filling order and removal order are different questions, and this one catches almost everyone once.

Students usually meet this chapter as a sequence of models to memorise. That framing hides the single thread running through all of it:

• Atomic spectra are discrete, and every model here is an attempt to explain why. Bohr does it by quantising angular momentum. Quantum mechanics does it by abandoning the idea of a path altogether. Each step buys explanatory power by surrendering a classical certainty.
• The payoff comes at the end. Once the quantum mechanical model is in place, an electron is completely specified by four quantum numbers, and every filling rule in chemistry is a constraint on which combinations are allowed.

1. The Fact That Needed Explaining

Heat hydrogen in a discharge tube and pass the light through a prism. You do not get a rainbow. You get a handful of sharp lines against blackness, always at exactly the same wavelengths, for every sample of hydrogen anywhere.

Classical physics cannot produce this. An orbiting electron radiating continuously would sweep through all frequencies as it spiralled inward, giving a continuous band and then a dead atom within about $10^{-8}$ s.

$c=\nu \lambda ,E=h\nu =\frac{hc}{\lambda },h=6.626\times 10^{-34}\text{J s}$

Planck's proposal — forced on him by the black body spectrum — was that energy is exchanged in packets. The photoelectric effect then confirmed the packet is real rather than an accounting device: light below a threshold frequency ejects nothing, however intense.

$K{E}_{max}=h\nu -h{\nu }_0=h\nu -\varphi$

The work function $\varphi$ is the minimum energy to free an electron from that metal. Raising the intensity sends more photons, so more electrons come out — each still with the same maximum kinetic energy.

Memorise $hc=1240$ eV nm. It turns any wavelength in nanometres into a photon energy in electronvolts in one division, and removes most of the arithmetic from this chapter.

Illustration 1

A sodium street lamp emits 20 W at 589 nm. How many photons leave it each second?

$E=\frac{1240}{589}=2.105\text{eV}=3.37\times 10^{-19}\text{J}$

$n=\frac{P}{E}=\frac{20}{3.37\times 10^{-19}}\Rightarrow n=5.9\times 10^{19}{\text{photons s}}^{-1}$

Sixty million million million per second. That is exactly why light seems continuous to the eye — the graininess is real, but averaged over numbers this large it is entirely invisible.

2. The Hydrogen Spectrum

$\frac{1}{\lambda }=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right),n_2>n_1,R_H=1.097\times 10^7{\text{m}}^{-1}$

Found empirically, before anyone could explain it.

Only Balmer falls in the visible, which is why it was found first. Lyman always has the shortest wavelengths, because it involves the largest energy drop.

Excitation energy lifts an electron to a specified higher level; ionisation energy removes it entirely, the jump to $n=\infty$. For hydrogen: 10.2 eV to reach $n=2$, 13.6 eV to ionise from the ground state — but only 3.4 eV to ionise from $n=2$, since the electron is already most of the way out.

Each series has a limiting line at $n_2=\infty$. Lyman's limit is 91.2 nm, whose photon energy is exactly 13.6 eV. Beyond the limit the spectrum turns continuous, because a freed electron may carry away any surplus kinetic energy. Sharp lines exist only while the electron is bound.

Illustration 2

The Lyman series limit of He⁺ is observed at 22.8 nm. Find its ionisation energy and check the result against the Bohr prediction.

The series limit is the $n=\infty \to 1$ transition, so its photon energy is the whole ionisation energy from the ground state:

$E=\frac{1240}{22.8}=54.4\text{eV}$

Bohr predicts $|E_1|=13.6Z^2$, and He⁺ has $Z=2$:

$13.6\times 4=54.4\text{eV}$

Exact agreement, because He⁺ has only one electron and that is the sole case Bohr describes.

Try the same on neutral helium and it fails badly: the measured first ionisation energy is 24.6 eV, not 54.4, because the second electron screens the nucleus. The gap between those two numbers is a direct measure of the electron-electron repulsion the Bohr model has no way to represent.

$N=\frac{n(n-1)}{2}\text{distinct lines from level }n$

which is simply the number of ways to choose two levels out of $n$.

Illustration 3

Hydrogen atoms are excited to $n=4$. How many distinct spectral lines can the sample emit, and how many of them are visible?

$N=\frac{4(3)}{2}=6\text{lines}$

The visible ones belong to the Balmer series, which ends on $n=2$: the jumps $4\to 2$ and $3\to 2$. So only two of the six can be seen. Three end on $n=1$ and are ultraviolet, and the last, $4\to 3$, is infrared.

Group the jumps by the level each lands on and the total confirms itself: $3+2+1=6$.

Note that this counts the lines a whole sample produces, not what one atom does. A single atom makes one cascade and emits at most three photons on the way down. The six lines appear because different atoms in the sample take different routes.

Illustration 4

Find the wavelength of the $n=3\to 2$ transition in hydrogen.

$\frac{1}{\lambda }=1.097\times 10^7\left(\frac{1}{4}-\frac{1}{9}\right)=1.097\times 10^7\times \frac{5}{36}=1.524\times 10^6{\text{m}}^{-1}$

$\lambda =656\text{nm}$

The red H-alpha line, first member of the Balmer series, and the reason hydrogen discharge tubes glow pink.

3. The Bohr Model

Bohr's move was to accept the classical orbit but forbid most of them.

The third breaks classical physics outright, and Bohr offered no justification beyond the fact that it works.

$r_n=0.529\frac{n^2}{Z}\text{A˚},E_n=-13.6\frac{Z^2}{n^2}\text{eV},v_n=2.18\times 10^6\frac{Z}{n}{\text{m s}}^{-1}$

Read the dependences rather than the numbers. Radius grows as $n^2$ and shrinks with $Z$; energy deepens as $Z^2$ and shallows as $1/n^2$.

Why the energy is negative. A free electron at rest infinitely far away is assigned zero. Any bound electron has less than that. So $|E_n|$ is the energy needed to remove it — which is why hydrogen's ionisation energy is exactly 13.6 eV.

$K=-E_n,U=2E_n$

Pulling the electron inward makes the potential energy fall twice as fast as the kinetic energy rises.

Trap. These formulas are exact for one-electron species only — H, He${}^{+}$, Li${}^{2+}$, Be${}^{3+}$ — and wrong for everything else, because they ignore electron-electron repulsion. Bohr also fails on Zeeman and Stark splitting, on line intensities, and on bonding. Most fundamentally it assumes a definite orbit of definite radius, which the uncertainty principle forbids.

Illustration 5

Find the radius, energy and speed of the $n=2$ level of Be${}^{3+}$.

Beryllium has $Z=4$, and Be${}^{3+}$ has a single electron, so Bohr applies exactly:

$r_2=0.529\times \frac{4}{4}=0.529\text{A˚},E_2=-13.6\times \frac{16}{4}=-54.4\text{eV}$

$v_2=2.18\times 10^6\times \frac{4}{2}=4.36\times 10^6{\text{m s}}^{-1}$

The radius has come out exactly the Bohr radius — the $n^2$ growth and the $1/Z$ contraction have cancelled. Whenever $n^2=Z$ this happens, and spotting it saves the arithmetic entirely.

4. Matter Waves and Uncertainty

$\lambda =\frac{h}{p}=\frac{h}{mv},\lambda =\frac{12.27}{\sqrt{V}}\text{A˚}\text{(electron through }V\text{ volts)}$

Wavelength is inversely proportional to mass, which is why the effect is invisible for everyday objects: a cricket ball's de Broglie wavelength is around $10^{-34}$ m, far below any structure it could diffract from.

De Broglie rescues Bohr's second postulate

Bohr's quantisation looked arbitrary. De Broglie made it inevitable. If the electron is a wave circling the nucleus, the orbit must hold a whole number of wavelengths, or the wave interferes destructively with itself and vanishes:

$2\pi r=n\lambda =\frac{nh}{mv}⟹mvr=\frac{nh}{2\pi }$

Bohr's postulate, recovered from a standing-wave condition. It is the most satisfying derivation in the chapter.

Illustration 6

Verify numerically that the $n=1$ orbit of hydrogen holds exactly one de Broglie wavelength.

$2\pi r_1=2\pi (0.529\times 10^{-10})=3.32\times 10^{-10}\text{m}$

$\lambda =\frac{h}{mv}=\frac{6.626\times 10^{-34}}{(9.11\times 10^{-31})(2.18\times 10^6)}=3.34\times 10^{-10}\text{m}$

They agree to better than one per cent — the residual is rounding in the constants. The ground state is literally the smallest orbit in which the electron wave closes on itself.

The uncertainty principle

$\Delta x\cdot \Delta p\ge \frac{h}{4\pi }$

Trap. This is not a statement about clumsy measurement disturbing the electron. It is a property of the particle itself, which simply does not possess a simultaneously definite position and momentum.

The consequence for chemistry is decisive: a Bohr orbit specifies radius and speed exactly, making the uncertainty product zero, which is forbidden. The orbit has to go.

For a proton or a cricket ball the bound is numerically irrelevant, because the mass in the denominator of $\Delta v$ is so large. It bites only for something as light as an electron confined to something as small as an atom.

Illustration 7

An electron is known to lie within an atom, so take $\Delta x=1.0\times 10^{-10}$ m. Find the minimum uncertainty in its velocity.

$\Delta p=\frac{h}{4\pi \Delta x}=\frac{6.626\times 10^{-34}}{4\pi (1.0\times 10^{-10})}=5.27\times 10^{-25}{\text{kg m s}}^{-1}$

$\Delta v=\frac{5.27\times 10^{-25}}{9.11\times 10^{-31}}=5.8\times 10^5{\text{m s}}^{-1}$

That uncertainty is the same order as the Bohr orbital speed itself. Speaking of "the electron's velocity" in an atom as a definite number is therefore meaningless — which is precisely why the orbit had to be abandoned rather than merely refined.

Illustration 8

Light of 300 nm falls on a metal of work function 2.5 eV. Find the maximum kinetic energy of the ejected electrons and their de Broglie wavelength.

$E=\frac{1240}{300}=4.13\text{eV}\Rightarrow K{E}_{max}=4.13-2.5=1.63\text{eV}$

Since $K{E}_{max}$ in electronvolts is numerically the accelerating voltage that would produce it:

$\lambda =\frac{12.27}{\sqrt{1.63}}=9.6\text{A˚}=0.96\text{nm}$

The ejected electron's wavelength is about 300 times shorter than that of the light which ejected it.

Now change the intensity rather than the colour. More electrons come out, each still carrying 1.63 eV, so their wavelength does not shift at all — only their number does. Frequency sets the energy, intensity sets the count, and the de Broglie wavelength follows the energy.

5. The Quantum Mechanical Model

What replaces the orbit is a wave function $\psi$, from solving the Schrodinger equation.

$\psi$ itself has no physical meaning and can be negative. The meaningful quantity is ${\psi }^2$, the probability density.

An orbital is a one-electron wave function. It is not a path, and not a region with a hard edge, since ${\psi }^2$ only approaches zero asymptotically. What is drawn as an orbital boundary is conventionally the surface enclosing 90 per cent of the probability.

Trap. The ${\psi }^2$ plot and the radial distribution curve are different things, and this is where the chapter is most often misread. The radial distribution gives the probability of finding the electron in a thin shell at radius $r$, and because that shell's volume grows as $r^2$, the curve is zero at the nucleus, peaks, and then decays — even though ${\psi }^2$ is largest at the nucleus for 1s.

For 1s hydrogen the radial distribution peaks at 0.529 Å — exactly the Bohr radius. Bohr's orbit survives as the most probable distance rather than the only one.

Illustration 9

Compare the nodes of 3s, 3p and 3d.

The split between radial and angular shifts across the row, but the total is always $n-1$. Given any two of the three numbers you can write the third down without thinking, which is what these questions are really testing.

6. The Four Quantum Numbers

$l=0,1,2,3$ are labelled s, p, d, f. A given $l$ has $2l+1$ orbitals — one s, three p, five d, seven f. A shell holds $n^2$ orbitals and therefore at most $2n^2$ electrons.

Four of the five d orbitals are cloverleaves; $d_{z^2}$ has two lobes and a doughnut. Spin is the odd one out — it has no classical picture and the electron is not literally rotating. It is an intrinsic property with exactly two values.

Illustration 10

How many electrons in an atom can have (a) $n=4$, (b) $n=3$ and $l=1$, (c) $n=3$, $l=1$, $m_l=0$?

$\text{(a)}2n^2=32\text{(b)}2(2l+1)=6\text{(c)}2$

Each quantum number specified cuts the count. Fixing $n$ leaves a shell, fixing $l$ leaves a subshell, fixing $m_l$ leaves one orbital — and an orbital holds two electrons, which is the last thing left to vary.

Illustration 11

Which of these sets of quantum numbers are impossible, and why?

• (a) $n=3,l=3,m_l=0,m_s=+\frac{1}{2}$
• (b) $n=2,l=1,m_l=-2,m_s=-\frac{1}{2}$
• (c) $n=4,l=0,m_l=0,m_s=+\frac{1}{2}$
• (d) $n=0,l=0,m_l=0,m_s=-\frac{1}{2}$

Test them against the chain of constraints in order, since each one depends on the value above it.

(a) Impossible. $l$ runs from 0 to $n-1$, so $n=3$ allows $l=0,1,2$ only. There is no 3f subshell.

(b) Impossible. $m_l$ runs from $-l$ to $+l$, so $l=1$ allows only $-1,0,+1$.

(c) Legal. This is a 4s electron.

(d) Impossible. $n$ is a positive integer and starts at 1. A zeroth shell would have zero radius and zero energy spacing.

Work down the chain rather than checking the four numbers independently. Each constraint is set by the one before it, so a set can only fail at the first place it breaks — and spotting that place is the whole question.

7. Filling the Orbitals

Aufbau. Electrons occupy the lowest available energy orbital first, and orbital energies follow the $(n+l)$ rule: lower $(n+l)$ fills first; among equal $(n+l)$, the lower $n$ fills first.

In hydrogen — one electron, no repulsion — all subshells of the same $n$ are degenerate and the rule is unnecessary.

Pauli. No two electrons may share all four quantum numbers. Since an orbital fixes three, it follows that an orbital holds at most two electrons, with opposed spins.

Hund. Within degenerate orbitals, electrons occupy separate orbitals with parallel spins before any pairing. Pairing costs repulsion energy, so it is postponed while an empty orbital remains.

The half-filled and fully filled exceptions

$\mathrm{Cr}=[\mathrm{Ar}]3d^{5}4s^1(\text{not }3d^{4}4s^2),\mathrm{Cu}=[\mathrm{Ar}]3d^{10}4s^1(\text{not }3d^{9}4s^2)$

Two effects drive this: exactly half-filled and completely filled subshells have symmetrical distributions, and they maximise the number of parallel-spin pairs, lowering energy through the exchange interaction. The 4s and 3d levels are close enough that this small gain is decisive. Chromium and copper are the two exceptions JEE actually asks about.

Trap. Ions lose electrons from the highest $n$ first, not in the reverse of the filling order. Iron loses 4s before 3d, so Fe${}^{2+}$ is $[\mathrm{Ar}]3d^6$. Once 3d is occupied it drops below 4s in energy, so the two questions genuinely have different answers.

Illustration 12

Write the configurations of Mn${}^{2+}$ and Cu${}^{2+}$ and count the unpaired electrons in each.

Manganese has 25 electrons, $[\mathrm{Ar}]3d^{5}4s^2$. Remove the two 4s electrons:

$\mathrm{M}{\mathrm{n}}^{2+}=[\mathrm{Ar}]3d^5\Rightarrow \text{5 unpaired}$

Copper has 29, $[\mathrm{Ar}]3d^{10}4s^1$. Remove 4s first, then one 3d:

$\mathrm{C}{\mathrm{u}}^{2+}=[\mathrm{Ar}]3d^9\Rightarrow \text{1 unpaired}$

Mn${}^{2+}$ with its half-filled d shell is the reason manganese(II) is so stubbornly stable, and five unpaired electrons make it strongly paramagnetic. Note that Cu${}^{2+}$ required taking one electron out of the filled 3d shell — the 4s runs out first.

Beyond the JEE Main Syllabus

Thomson's plum pudding model and Rutherford's nuclear model, with their limitations, were removed from JEE Main in the 2023 revision and remain out for 2026. So was the discovery of the fundamental particles as a separate topic.

Both remain in JEE Advanced and in most textbooks. Rutherford's alpha-scattering experiment is worth reading once, because it established that the atom has a tiny dense nucleus at all, and the instability objection to his model is the cleanest statement of the problem Bohr was solving. If you are sitting only Main, read for context and do not drill.

Summary

• Discrete spectra are the fact the chapter exists to explain; classical physics cannot produce them.
• $E=hc/\lambda$, and $hc=1240$ eV nm removes most of the arithmetic.
• Rydberg: $1/\lambda =R_H(1/n_1^2-1/n_2^2)$; only Balmer is visible; $n(n-1)/2$ lines from level $n$.
• Each series limit is the ionisation energy from that level; beyond it the spectrum turns continuous.
• Bohr: $mvr=nh/2\pi$ gives $r_n=0.529n^2/Z$ Å and $E_n=-13.6Z^2/n^2$ eV, with $K=-E$ and $U=2E$.
• Exact for one-electron species only — H, He${}^{+}$, Li${}^{2+}$, Be${}^{3+}$.
• $\lambda =h/p$ turns Bohr's postulate into a standing-wave condition: $2\pi r=n\lambda$.
• $\Delta x\Delta p\ge h/4\pi$ is a property of the particle, not of the measurement, and it kills the definite orbit.
• $\psi$ has no meaning; ${\psi }^2$ is probability density. An orbital has no hard edge — the drawn boundary encloses 90 %.
• ${\psi }^2$ and the radial distribution are different plots. The latter is zero at the nucleus because shell volume vanishes there.
• Radial nodes $n-l-1$, angular $l$, total $n-1$.
• Four quantum numbers specify an electron completely; a shell holds $2n^2$ electrons.
• Aufbau fills by the $(n+l)$ rule, Pauli caps an orbital at two opposed spins, Hund keeps degenerate orbitals singly occupied.
• Cr is $3d^{5}4s^1$ and Cu is $3d^{10}4s^1$ — half-filled and filled stability plus exchange energy.
• Ions lose the highest $n$ first, so Fe${}^{2+}$ is $3d^6$, not $3d^{4}4s^2$.