Is Atomic Structure part of the EAPCET Chemistry syllabus?

Yes. Atomic Structure is taught in Intermediate first year Chemistry, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.

How many Chemistry questions are there in EAPCET?

40 of the 160 questions in the Engineering paper are Chemistry, spread across 24 chapters.

What mistakes do students make in Atomic Structure?

Writing Cr as 3d⁴4s² and Cu as 3d⁹4s² — use the stable half-filled and filled forms 3d⁵4s¹ and 3d¹⁰4s¹. Allowing l = n (for example n = 2, l = 2) in a quantum-number set — l can only run from 0 to n − 1. Counting total nodes as radial nodes — radial nodes are n − l − 1 and angular nodes are l, so a 3p orbital has one of each. Swapping n₁ and n₂ in the Rydberg formula and getting a negative 1/λ — n₁ is always the lower level.

EAPCET Chemistry · Intermediate first year

Atomic Structure

Atomic Structure covers the discovery of subatomic particles, the Bohr model, the quantum-mechanical model, quantum numbers and electronic configuration.

Subject: Chemistry
Taught in: Intermediate first year
Chemistry in the paper: 40 of 160 questions

Key formulas and results

Energy of the nth Bohr orbit: $E_n = -2.18 \times 10^{-18}\left(\frac{Z^2}{n^2}\right)\ \text{J}$
de Broglie wavelength $\lambda = \frac{h}{mv}$
Heisenberg: $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$
An orbital with quantum number $l$ holds $2(2l + 1)$ electrons in its subshell
Filling follows the Aufbau $(n + l)$ rule, Pauli’s exclusion principle and Hund’s rule
Radial nodes $= n - l - 1$ ; angular nodes $= l$
Radius of the nth Bohr orbit: $r_n = 0.529\left(\frac{n^2}{Z}\right)\ \text{\AA} = 52.9\left(\frac{n^2}{Z}\right)\ \text{pm}$
Velocity of the electron in the nth orbit: $v_n = 2.18 \times 10^{6}\left(\frac{Z}{n}\right)\ \text{m/s}$
Bohr energy in other units: $E_n = -13.6\left(\frac{Z^2}{n^2}\right)\ \text{eV}$ per atom $= -1312\left(\frac{Z^2}{n^2}\right)\ \text{kJ/mol}$
Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ , with $R = 1.097 \times 10^{7}\ \text{m}^{-1} = 109677\ \text{cm}^{-1}$
Photon energy shortcut: $E\,(\text{eV}) \approx \frac{12400}{\lambda\,(\text{\AA})}$
de Broglie wavelength of an electron accelerated through V volts: $\lambda = \frac{h}{\sqrt{2meV}} = \frac{12.27}{\sqrt{V}}\ \text{\AA}$
Photoelectric equation: $h\nu = h\nu_0 + \tfrac{1}{2}mv_{\max}^2$ , and stopping potential $V_0 = \frac{\text{KE}_{\max}}{e}$
Number of spectral lines when an electron falls from $n_2$ to $n_1$ : $\frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$ ; down to the ground state from $n$ : $\frac{n(n - 1)}{2}$

What this chapter covers

Subatomic particles and atomic models
Bohr’s model of the hydrogen atom
Dual nature and the uncertainty principle
Quantum numbers
Shapes of orbitals
Electronic configuration

How it is usually asked

Valid sets of quantum numbers
Electronic configurations and exceptions
Number of nodes
Energy and wavelength calculations

Understanding Atomic Structure

An atom is a small, dense nucleus of protons and neutrons surrounded by electrons. Bohr’s model explained the hydrogen spectrum by allowing the electron only certain orbits with fixed energies. When an electron drops from a higher orbit to a lower one, the energy difference is released as a photon, and each such jump gives one line of the spectrum: Lyman (to n = 1), Balmer (to n = 2, visible) and Paschen (to n = 3).

Bohr’s model fails for atoms with more than one electron, so the quantum-mechanical model replaces fixed orbits with orbitals — regions where the electron is likely to be found. Each electron is described by four quantum numbers: n (shell and energy), l (subshell and shape, 0 to n − 1), mₗ (orientation, −l to +l) and mₛ (spin, +½ or −½). De Broglie’s wave nature and Heisenberg’s uncertainty principle are the reasons fixed orbits cannot exist.

Electrons fill orbitals in order of increasing (n + l), and for equal (n + l) the lower n fills first. Pauli allows two electrons per orbital with opposite spins, and Hund’s rule spreads electrons singly across degenerate orbitals before pairing. Half-filled and fully filled d subshells are extra stable, which is why Cr is 3d⁵4s¹ and Cu is 3d¹⁰4s¹. EAPCET questions typically test valid quantum-number sets, nodes, and spectral-line calculations.

Worked example

EAPCET-style question

What is the wavelength of the photon emitted when the electron in a hydrogen atom falls from $n = 3$ to $n = 2$ ? ( $R = 1.097 \times 10^{7}\ \text{m}^{-1}$ )

(1)656 nm
(2)486 nm
(3)122 nm
(4)1875 nm

Solution

Use the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ , with $Z = 1$ , $n_1 = 2$ and $n_2 = 3$ .
$\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{1}{4} - \frac{1}{9} = \frac{5}{36}$ .
$\frac{1}{\lambda} = 1.097 \times 10^{7} \times \frac{5}{36} = 1.524 \times 10^{6}\ \text{m}^{-1}$ .
$\lambda = \frac{1}{1.524 \times 10^{6}} = 6.56 \times 10^{-7}\ \text{m} = 656\ \text{nm}$ , the red Hα line of the Balmer series.
The distractors are other transitions: 486 nm is $4 \to 2$ , 122 nm is $2 \to 1$ and 1875 nm is $4 \to 3$ .

Answer: 656 nm

Shortcuts and results to remember

In any Bohr orbit $\text{KE} = -E$ and $\text{PE} = 2E$ , so total energy $E = -\text{KE} = \frac{\text{PE}}{2}$ .
Bohr proportionalities: $r \propto \frac{n^2}{Z}$ , $v \propto \frac{Z}{n}$ , $E \propto -\frac{Z^2}{n^2}$ , time period $\propto \frac{n^3}{Z^2}$ and orbital frequency $\propto \frac{Z^2}{n^3}$ .
The first (longest-wavelength) line of a series has $n_2 = n_1 + 1$ and the series limit has $n_2 = \infty$ : Lyman $\lambda_{\max} = \frac{4}{3R}$ , $\lambda_{\min} = \frac{1}{R}$ ; Balmer $\lambda_{\max} = \frac{36}{5R}$ , $\lambda_{\min} = \frac{4}{R}$ .
For proton, deuteron and α-particle: with equal kinetic energy $\lambda \propto \frac{1}{\sqrt{m}}$ gives $2 : \sqrt{2} : 1$ ; accelerated through the same voltage $\lambda \propto \frac{1}{\sqrt{mq}}$ gives $2\sqrt{2} : 2 : 1$ .
Isotopes share atomic number, isobars share mass number, isotones share neutron number, isoelectronic species share electron count, and isosters share both atom count and electron count.

Common mistakes to avoid

Writing Cr as 3d⁴4s² and Cu as 3d⁹4s² — use the stable half-filled and filled forms 3d⁵4s¹ and 3d¹⁰4s¹.
Allowing l = n (for example n = 2, l = 2) in a quantum-number set — l can only run from 0 to n − 1.
Counting total nodes as radial nodes — radial nodes are n − l − 1 and angular nodes are l, so a 3p orbital has one of each.
Swapping n₁ and n₂ in the Rydberg formula and getting a negative 1/λ — n₁ is always the lower level.

How to prepare it

Learn the configuration exceptions (chromium and copper) — they appear as options every year.

Once the formulas feel familiar, test them the way the exam does: practise Atomic Structure questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Chemistry chapters

See every chapter in the EAPCET Chemistry syllabus.

Practise Atomic Structure chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Atomic Structure part of the EAPCET Chemistry syllabus?: Yes. Atomic Structure is taught in Intermediate first year Chemistry, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Atomic Structure for EAPCET?: Learn the configuration exceptions (chromium and copper) — they appear as options every year.
How many Chemistry questions are there in EAPCET?: 40 of the 160 questions in the Engineering paper are Chemistry, spread across 24 chapters.
What mistakes do students make in Atomic Structure?: Writing Cr as 3d⁴4s² and Cu as 3d⁹4s² — use the stable half-filled and filled forms 3d⁵4s¹ and 3d¹⁰4s¹. Allowing l = n (for example n = 2, l = 2) in a quantum-number set — l can only run from 0 to n − 1. Counting total nodes as radial nodes — radial nodes are n − l − 1 and angular nodes are l, so a 3p orbital has one of each. Swapping n₁ and n₂ in the Rydberg formula and getting a negative 1/λ — n₁ is always the lower level.