Atomic Structure and the Periodic Table — NEET Chemistry

Everything in chemistry begins here: the atom's internal structure fixes how it bonds, reacts and where it sits in the periodic table. NEET draws 2–3 questions from this block almost every year, and they are among the most reasoned marks in the paper — once you understand effective nuclear charge, most periodic trends follow without memorising. This chapter builds the atom from its particles up to the quantum model, then uses that model to explain the entire periodic table. Every result is derived, every trend is justified, and the standard exceptions NEET tests are called out explicitly.

Part A — Structure of the Atom

1. Sub-atomic particles and the nuclear atom

An atom is built from three particles:

• Atomic number $Z$ = number of protons = number of electrons in a neutral atom. It fixes the element's identity and its place in the periodic table.
• Mass number $A$ = protons + neutrons (nucleons). So neutrons $=A-Z$.
• Almost all the mass sits in the tiny, dense nucleus (Rutherford's α-scattering experiment); the electrons occupy the vast surrounding volume.

Isotopes, isobars, isotones — a NEET favourite to distinguish:

• Isotopes — same $Z$, different $A$ (same element): ${}_1^1\text{H},{}_1^2\text{H},{}_1^3\text{H}$. Same chemistry, different mass.
• Isobars — same $A$, different $Z$ (different elements): ${}_{18}^{40}\text{Ar},{}_{19}^{40}\text{K},{}_{20}^{40}\text{Ca}$.
• Isotones — same number of neutrons: ${}_6^{14}\text{C}$ and ${}_8^{16}\text{O}$ (both 8 neutrons).

Worked example 1.1. How many protons, neutrons and electrons are in ${}_{17}^{35}{\text{Cl}}^{-}$? Protons $=Z=17$. Neutrons $=A-Z=35-17=18$. Electrons $=17+1=18$ (the −1 charge means one extra electron).

2. The Bohr model of the hydrogen atom

Classical physics failed for the atom: an orbiting electron should spiral into the nucleus radiating energy. Bohr fixed this with quantised orbits — the electron occupies only certain allowed energy levels and radiates only when it jumps between them.

Postulates. (i) Electrons move in fixed circular orbits without radiating. (ii) Angular momentum is quantised, $mvr=n\frac{h}{2\pi }$. (iii) Energy is emitted/absorbed only on transitions, $\Delta E=h\nu$.

From these, two key results for a one-electron species of nuclear charge $Z$:

$r_n=0.529\frac{n^2}{Z}\text{A˚},E_n=-13.6\frac{Z^2}{n^2}\text{eV}$

• Radius grows as $n^2$ — higher orbits are much larger.
• Energy is negative (bound state), scaling as $1/n^2$. The ground state ($n=1$) is the most tightly bound at $-13.6$ eV for hydrogen.
• The energy needed to remove the electron entirely (to $n=\infty$, $E=0$) is the ionisation energy — exactly $+13.6$ eV for hydrogen from its ground state.

Emission spectrum — when an excited electron falls from $n_2$ to $n_1$, it emits a photon whose wavelength follows the Rydberg formula:

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

The hydrogen spectral series are named by the level the electron lands on:

Worked example 2.1. Energy released when a hydrogen electron falls from $n=3$ to $n=2$? $E_2=-13.6/4=-3.4$ eV, $E_3=-13.6/9=-1.51$ eV. Photon energy $=E_3-E_2=-1.51-(-3.4)=1.89$ eV — the red H${}_{\alpha }$ line of the Balmer series.

Worked example 2.2. The radius of the $n=2$ orbit of hydrogen? $r_2=0.529\times 2^2/1=0.529\times 4=2.12$ Å.

Limitations of the Bohr model. It works only for one-electron species (H, He⁺, Li²⁺); it fails for multi-electron atoms, cannot explain fine spectral structure or the splitting of lines in a magnetic field (Zeeman effect), and it violates the uncertainty principle by assigning a definite orbit. It was the essential bridge to the quantum model, not the final word.

3. Toward the quantum model: dual nature and uncertainty

de Broglie (1924) proposed that matter, like light, has a wave nature with wavelength

$\lambda =\frac{h}{mv}=\frac{h}{p}$

Big objects have negligible wavelength; an electron's is significant, and Bohr's quantisation condition drops out naturally if you require a standing wave around the orbit ($2\pi r=n\lambda$).

Heisenberg's uncertainty principle — you cannot know both an electron's position and momentum exactly:

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

This forbids well-defined Bohr orbits. Instead of a path, quantum mechanics gives an orbital — a region of space where the probability of finding the electron is high (its wavefunction $\psi$; the probability density is ${\psi }^2$).

Worked example 3.1. Why don't we notice the wave nature of a cricket ball? For a 0.15 kg ball at 40 m/s, $\lambda =h/mv=6.6\times 10^{-34}/(0.15\times 40)\approx 1.1\times 10^{-34}$ m — unimaginably smaller than the ball, so no wave effects show. For an electron ($m\sim 10^{-30}$ kg) the wavelength is comparable to atomic sizes, so its wave nature dominates.

4. The four quantum numbers

Each electron in an atom is labelled by four quantum numbers — its complete "address":

• Principal quantum number $n$ ($1,2,3,\dots$) — the shell; sets size and energy. Maximum electrons in a shell $=2n^2$.
• Azimuthal (angular momentum) quantum number $l$ ($0$ to $n-1$) — the sub-shell/shape. $l=0,1,2,3\to s,p,d,f$. Orbital angular momentum $=\sqrt{l(l+1)}\frac{h}{2\pi }$.
• Magnetic quantum number $m_l$ ($-l$ to $+l$, including 0) — the orbital orientation; $2l+1$ values per sub-shell (so $p$ has 3, $d$ has 5, $f$ has 7 orbitals).
• Spin quantum number $m_s$ ($+\frac{1}{2}$ or $-\frac{1}{2}$) — the electron's intrinsic spin direction.

Worked example 4.1. How many electrons can have $n=3$? And how many orbitals? Orbitals in $n=3$: $s(1)+p(3)+d(5)=9$ orbitals. Electrons $=2n^2=2\times 9=18$.

Orbital shapes and nodes.

• $s$ orbitals are spherical; $p$ orbitals are dumb-bell shaped along the $x,y,z$ axes; $d$ orbitals are mostly double dumb-bell (four-lobed).
• A node is a surface of zero electron probability. Total nodes $=n-1$; angular nodes $=l$; radial nodes $=n-l-1$.

Worked example 4.2. Number of radial and angular nodes in a $3p$ orbital? $n=3,l=1$. Angular nodes $=l=1$. Radial nodes $=n-l-1=3-1-1=1$. Total $=2=n-1$. ✓

5. Filling the orbitals: the three rules

Electrons fill orbitals to give the lowest-energy (ground state) configuration, governed by three rules.

1. Aufbau principle — fill lowest-energy orbitals first. Energy order follows the $(n+l)$ rule: lower $(n+l)$ fills first; if two orbitals tie, the one with lower $n$ fills first.

$1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s$

(Check: $4s$ has $n+l=4$, $3d$ has $n+l=5$, so $4s$ fills before $3d$.)

2. Pauli exclusion principle — no two electrons in an atom can have all four quantum numbers identical; hence an orbital holds at most two electrons, with opposite spins.

3. Hund's rule of maximum multiplicity — within a sub-shell, electrons occupy orbitals singly first, all with parallel spins, before any pairing begins. This minimises electron–electron repulsion.

Worked example 5.1. Ground-state configuration of oxygen ($Z=8$)? $1s^{2}2s^{2}2p^4$. By Hund's rule the four $2p$ electrons are arranged $\uparrow \downarrow \uparrow \uparrow$ across the three $p$ orbitals — two unpaired electrons (which is why O₂ is paramagnetic).

The famous exceptions — chromium and copper. A half-filled ($d^5$) or completely-filled ($d^{10}$) sub-shell has extra stability (symmetry + exchange energy), so:

• Cr ($Z=24$): expected $[Ar]3d^{4}4s^2$; actual $[Ar]3d^{5}4s^1$.
• Cu ($Z=29$): expected $[Ar]3d^{9}4s^2$; actual $[Ar]3d^{10}4s^1$.

Worked example 5.2. Write the configuration of Fe ($Z=26$) and Fe³⁺. Fe: $[Ar]3d^{6}4s^2$. Forming Fe³⁺ removes the two $4s$ electrons first, then one $3d$: $[Ar]3d^5$ — a stable half-filled $d$ sub-shell, which is why Fe³⁺ is so common.

Key trap. The $4s$ orbital fills before $3d$ (lower energy when empty) but ionises first (once $3d$ is occupied, $3d$ drops below $4s$). Transition-metal ions always lose $s$ electrons before $d$.

Part B — Classification of Elements and Periodicity

6. The modern periodic law and the long form

Mendeleev (1869) arranged elements by increasing atomic mass and grouped by similar properties, boldly leaving gaps for undiscovered elements (correctly predicting eka-silicon = germanium). But mass ordering produced anomalies (Ar before K).

Modern periodic law: the properties of elements are a periodic function of their atomic number (Moseley). Ordering by $Z$ removes Mendeleev's anomalies.

The long form has 7 periods (rows = principal quantum number being filled) and 18 groups (columns of similar valence configuration), organised into four blocks by the sub-shell being filled:

• Period number = highest principal quantum number of the valence shell.
• Group (for main-group elements) relates to the number of valence electrons.

7. Effective nuclear charge — the master idea

Almost every periodic trend follows from one quantity: the effective nuclear charge $Z_{\text{eff}}$ — the net positive pull actually felt by a valence electron after inner electrons shield it:

$Z_{\text{eff}}=Z-S$

where $S$ is the screening (shielding) constant. Two patterns to internalise:

• Across a period (left → right): $Z$ rises by one each step, but the added electron goes into the same shell and shields poorly, so $Z_{\text{eff}}$ increases. Valence electrons are pulled in tighter.
• Down a group: a whole new shell is added each step; the extra inner shells shield strongly, so $Z_{\text{eff}}$ rises only slightly while distance grows a lot. The outer electrons feel a weaker net grip.

Hold these two patterns and you can predict every trend below.

8. Atomic and ionic radius

Atomic radius (covalent/metallic):

• Across a period → decreases. Rising $Z_{\text{eff}}$ pulls the same shell inward.
• Down a group → increases. Each new shell outweighs the modest $Z_{\text{eff}}$ rise.

Ionic radius:

• Cations are smaller than the parent atom (lost a shell and/or increased $Z_{\text{eff}}$/electron ratio): Na⁺ < Na.
• Anions are larger than the parent atom (added electrons increase repulsion): Cl⁻ > Cl.
• Isoelectronic species (same electron count) shrink as $Z$ rises: ${\text{N}}^{3-}>{\text{O}}^{2-}>{\text{F}}^{-}>{\text{Na}}^{+}>{\text{Mg}}^{2+}>{\text{Al}}^{3+}$ (all 10 electrons; more protons pull harder).

Worked example 8.1. Arrange by size: O²⁻, F⁻, Na⁺, Mg²⁺. All are isoelectronic (10 electrons). More protons → smaller: Mg²⁺(12) < Na⁺(11) < F⁻(9) < O²⁻(8). So O²⁻ > F⁻ > Na⁺ > Mg²⁺.

9. Ionisation enthalpy

The ionisation enthalpy (IE) is the energy to remove the most loosely held electron from a gaseous atom, $X(g)\to X^{+}(g)+e^{-}$.

• Across a period → increases (higher $Z_{\text{eff}}$, smaller size hold the electron tighter).
• Down a group → decreases (the outer electron is farther out and better shielded).

Two anomalies NEET always tests — both from sub-shell stability:

• Be > B and Mg > Al: removing an electron from the stable filled $n{s}^2$ of Be/Mg is harder than removing the lone $n{p}^1$ electron of B/Al.
• N > O and P > S: the half-filled $n{p}^3$ of N/P is extra stable, so its IE exceeds that of O/S (where removing an electron actually relieves pairing repulsion).

Successive ionisation energies always rise ($I{E}_1<I{E}_2<I{E}_3$) because removing an electron from an increasingly positive ion is harder; a large jump signals reaching a noble-gas core (e.g. Na's $I{E}_2\gg I{E}_1$ — it "does not want" to lose a second electron).

Worked example 9.1. Which has higher first IE — nitrogen or oxygen? Why? Nitrogen. Its $2p^3$ is a stable half-filled sub-shell, so its electron is held more tightly; oxygen's fourth $2p$ electron is paired and easier to remove. Hence $I{E}_1(\text{N})>I{E}_1(\text{O})$ — a break in the general left-to-right increase.

10. Electron gain enthalpy and electronegativity

Electron gain enthalpy (${\Delta }_{eg}H$) — the energy change when a gaseous atom gains an electron, $X(g)+e^{-}\to X^{-}(g)$. Usually negative (energy released).

• Across a period → becomes more negative (atoms more eager to gain electrons, approaching a noble-gas shell).
• Down a group → becomes less negative (larger atom, added electron less strongly attracted).
• Key exception: chlorine has a more negative ${\Delta }_{eg}H$ than fluorine. Fluorine's tiny 2p sub-shell is so compact that adding an electron brings strong repulsion, so Cl (and often the second-period elements generally) is anomalous. Order: Cl > F > Br > I.

Electronegativity — the tendency of an atom in a bond to attract the shared electron pair. It is a relative property (Pauling scale), not a measured energy.

• Across a period → increases; down a group → decreases.
• Fluorine is the most electronegative element (4.0). The trend drives bond polarity and much of chemical behaviour.

Worked example 10.1. Why is Cl a stronger electron-acceptor (more negative ${\Delta }_{eg}H$) than F, against the group trend? Fluorine is very small, so its 2p electrons are packed tightly; adding an extra electron into that compact shell raises electron–electron repulsion sharply, partly offsetting the nuclear pull. Chlorine's larger 3p shell accommodates the new electron with less repulsion, giving it the more negative electron gain enthalpy.

11. Trends at a glance and other periodic properties

• Metallic character = tendency to lose electrons; strongest at the bottom-left (Cs, Fr), weakest top-right.
• Nature of oxides shifts across a period from basic (metals, Na₂O) through amphoteric (Al₂O₃, ZnO) to acidic (non-metals, SO₃, Cl₂O₇).
• Diagonal relationship: Li–Mg, Be–Al, B–Si resemble each other because rising charge across balances rising size down.

Worked example 11.1. Classify Na₂O, Al₂O₃ and SO₃ as basic, amphoteric or acidic. Na₂O — basic (reacts with acid); Al₂O₃ — amphoteric (reacts with both acid and base); SO₃ — acidic (reacts with base, gives H₂SO₄ with water). This left-to-right basic → amphoteric → acidic shift mirrors metallic → non-metallic character.

12. Common traps NEET sets here

• Isotopes vs isobars vs isotones — same $Z$ / same $A$ / same neutrons respectively. Read carefully.
• $4s$ fills before $3d$ but ionises first. Remove $s$ electrons before $d$ when forming ions.
• Cr and Cu exceptions — half-filled and fully-filled $d$ stability: $3d^{5}4s^1$ and $3d^{10}4s^1$.
• IE anomalies — Be > B, Mg > Al (filled $s$); N > O, P > S (half-filled $p$). Expect one every year.
• ${\Delta }_{eg}H$: Cl > F, not F > Cl — small-size repulsion in fluorine.
• Isoelectronic size — more protons means smaller; N³⁻ is the largest, Al³⁺ the smallest of the 10-electron set.
• Cation smaller, anion larger than the neutral atom — never the reverse.
• Radial nodes $=n-l-1$, angular nodes $=l$; students swap these.

13. Memory aids

• "$(n+l)$ decides, low $n$ breaks a tie" — the Aufbau filling order.
• "Fill $s$ first, lose $s$ first" — $4s$ before $3d$ in, $4s$ before $3d$ out.
• "Half and full are happy" — $d^5$ and $d^{10}$ stability behind Cr, Cu and the N/O, Be/B anomalies.
• "Small and mean fluorine" — most electronegative, but Cl beats it on electron gain enthalpy.
• "Across pulls in, down pushes out" — radius decreases across, increases down (opposite for IE/EN).

14. Exam protocol

1. For particle counts, use $Z$ = protons, $A-Z$ = neutrons, adjust electrons for charge.
2. One-electron species: apply $r_n\propto n^2/Z$, $E_n=-13.6Z^2/n^2$; use Rydberg for spectral lines (Balmer = visible).
3. Assign quantum numbers with $l=0\dots n-1$, $m_l=-l\cdots +l$; nodes: radial $n-l-1$, angular $l$.
4. Write configurations by the $(n+l)$ rule; apply Hund and Pauli; remember Cr, Cu exceptions and ionise $s$ before $d$.
5. Predict trends from $Z_{\text{eff}}$: radius down across/up down; IE and EN up across/down down.
6. Recall the two IE anomalies (Be>B, N>O) and the ${\Delta }_{eg}H$ anomaly (Cl>F).
7. For isoelectronic species, order by proton count — more protons, smaller ion.