Structure of Atom
Structure of Atom for JEE Main & Advanced
Atomic Models and Quantum Theory
Discovery of Subatomic Particles and Atomic ModelsTopic 1
Discovery of Subatomic Particles:
| Particle | Discovered by | Year | Charge | Mass |
|---|---|---|---|---|
| Electron | J.J. Thomson (cathode rays) | 1897 | $-1.6 \times 10^{-19}$ C | $9.11 \times 10^{-31}$ kg |
| Proton | Goldstein (anode/canal rays); Rutherford named | 1886/1919 | $+1.6 \times 10^{-19}$ C | $1.67 \times 10^{-27}$ kg |
| Neutron | James Chadwick | 1932 | $0$ | $\approx$ proton mass |
Thomson's Plum Pudding Model (1898): Atom = uniform positive sphere with electrons embedded like raisins. Couldn't explain Rutherford's scattering.
Rutherford's Nuclear Model (1911):
- Fired α particles at thin gold foil
- Observations: most α passed through; some deflected; few ($1/8000$) bounced back
- Conclusions: atom mostly empty space; small dense positive nucleus at centre; electrons revolve around it (planetary)
Limitations of Rutherford's Model:
- According to classical EM theory, an accelerating electron should radiate energy continuously and spiral into the nucleus — atom would be unstable
- Cannot explain discrete (line) atomic spectra
Atomic Number and Mass Number:
- Atomic number ($Z$) = number of protons = number of electrons (neutral atom)
- Mass number ($A$) = protons + neutrons
- Notation: $^A_Z X$
Isotopes: Same $Z$, different $A$ (e.g., $^1$H, $^2$H, $^3$H). Same chemistry. Isobars: Same $A$, different $Z$ (e.g., $^{40}$Ar, $^{40}$Ca). Isotones: Same number of neutrons ($A-Z$). e.g., $^{14}$C and $^{16}$O. Isoelectronic species: Same number of electrons (e.g., Na⁺, Mg²⁺, F⁻, O²⁻, Ne — all $10\,e^-$).
Calculate number of protons, neutrons, electrons in $^{40}_{20}$Ca²⁺.
Show solution
- Protons = $Z = 20$
- Neutrons = $A - Z = 40 - 20 = 20$
- Electrons = $20 - 2 = 18$ (lost 2 from $+2$ charge)
Final Answer: $20 p, 20 n, 18 e$.
Identify isoelectronic among: N³⁻, O²⁻, F⁻, Ne, Na⁺, Mg²⁺.
Show solution
All have $10$ electrons → all isoelectronic. (N: $Z=7$, gains $3$ → $10$. O: $8 + 2 = 10$. F: $9+1 = 10$. Ne: $10$. Na: $11 - 1 = 10$. Mg: $12 - 2 = 10$.)
Final Answer: All six are isoelectronic.
The electron was discovered by:
Neutrons were discovered in:
Mass number = ?
Isobars have:
Which is NOT a limitation of Rutherford's model?
Electromagnetic Radiation, Planck's Theory and Bohr ModelTopic 2
Electromagnetic Radiation: Travels at $c = 3 \times 10^8$ m/s. Related: $c = \nu\lambda$. Wavenumber $\bar\nu = 1/\lambda$.
Planck's Quantum Theory (1900): Energy is emitted/absorbed in discrete packets (quanta): $$E = h\nu = \frac{hc}{\lambda}$$ $h = 6.626 \times 10^{-34}$ J·s.
Photoelectric Effect (Einstein 1905): $$h\nu = \phi + KE_{\max}$$ $\phi = h\nu_0$ = work function. $KE_{\max} = eV_0$ (stopping potential).
Atomic Spectra (Hydrogen): Discrete lines, given by: $$\bar\nu = \frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right), \quad n_2 > n_1$$ $R_H = 1.097 \times 10^7$ m⁻¹.
Hydrogen Spectral Series:
| Series | $n_1$ | $n_2$ | Region |
|---|---|---|---|
| Lyman | $1$ | $2, 3, \ldots$ | UV |
| Balmer | $2$ | $3, 4, \ldots$ | Visible |
| Paschen | $3$ | $4, 5, \ldots$ | IR |
| Brackett | $4$ | $5, 6, \ldots$ | IR |
| Pfund | $5$ | $6, 7, \ldots$ | Far IR |
Bohr's Postulates:
- Electron revolves only in certain stationary orbits without radiating
- Angular momentum quantized: $mvr = nh/(2\pi)$, $n = 1,2,3,\ldots$
- Energy emitted/absorbed only on transition: $h\nu = E_i - E_f$
Bohr Atom Formulae (for H-like atom, $Z$ = nuclear charge):
| Quantity | Formula | For H ($Z = 1$) at $n = 1$ |
|---|---|---|
| Radius | $r_n = 0.529 \cdot n^2/Z$ Å | $0.529$ Å |
| Velocity | $v_n = (2.18 \times 10^6) Z/n$ m/s | $2.18 \times 10^6$ m/s |
| Energy | $E_n = -13.6\,Z^2/n^2$ eV | $-13.6$ eV |
Number of spectral lines from $n_2 \to n_1$ transition: $N = (n_2 - n_1)(n_2 - n_1 + 1)/2$. From $n$ to ground: $N = n(n-1)/2$.
Limitations of Bohr's model:
- Works only for one-electron atoms
- Cannot explain fine structure (spin-orbit)
- Cannot explain Zeeman/Stark effects
- Violates Heisenberg's uncertainty principle (fixed orbit + momentum)
Find wavelength of $H_\alpha$ line ($n_2 = 3 \to n_1 = 2$) in Balmer series.
Show solution
$$\bar\nu = R_H(1/4 - 1/9) = R_H(5/36)$$ $$\lambda = 36/(5R_H) = 36/(5 \times 1.097 \times 10^7) = 6.563 \times 10^{-7} \text{ m} = 656.3 \text{ nm}$$
Final Answer: $656.3$ nm.
Find energy of electron in second orbit of $Li^{2+}$ ($Z = 3$).
Show solution
$$E_2 = -13.6 \times 9/4 = -30.6 \text{ eV}$$
Final Answer: $-30.6$ eV.
The energy of $n$th orbit in H atom:
Balmer series of hydrogen lies in:
Bohr's angular momentum:
Number of spectral lines from $n = 4$ to $n = 1$:
The Bohr radius of $H$ atom in ground state:
Quantum Mechanical Model
Dual Nature, Heisenberg, Quantum NumbersTopic 1
de Broglie Hypothesis (1924): Matter has wave properties. $$\lambda = \frac{h}{p} = \frac{h}{mv}$$
For accelerated charged particle through $V$ volts: $\lambda = h/\sqrt{2mqV}$. For electron: $\lambda = 12.27/\sqrt{V}$ Å.
Heisenberg Uncertainty Principle (1927): Position and momentum cannot be measured simultaneously with arbitrary accuracy: $$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$$
This made Bohr's model untenable (electrons cannot have fixed orbits and momenta).
Quantum Mechanical Model (Schrödinger, 1926): Electrons described by wave function $\psi$. $|\psi|^2$ gives probability density of finding electron. Solutions for H-atom yield four quantum numbers.
Quantum Numbers:
| Quantum number | Symbol | Allowed values | Significance |
|---|---|---|---|
| Principal | $n$ | $1, 2, 3, \ldots$ | Size and energy (shell): K, L, M, N |
| Azimuthal (subsidiary) | $l$ | $0, 1, \ldots, (n-1)$ | Subshell and shape: s, p, d, f |
| Magnetic | $m_l$ (or $m$) | $-l, \ldots, 0, \ldots, +l$ ($2l+1$ values) | Orientation of orbital |
| Spin | $m_s$ | $+1/2, -1/2$ | Spin of electron |
Subshell ↔ $l$: s → 0, p → 1, d → 2, f → 3.
Number of orbitals:
- in subshell of value $l$: $(2l+1)$
- in shell $n$: $n^2$
Number of electrons:
- in subshell: $2(2l+1)$
- in shell $n$: $2n^2$
Subshell capacity: s = 2, p = 6, d = 10, f = 14.
Find de Broglie wavelength of an electron moving with $10^6$ m/s. ($m_e = 9.1 \times 10^{-31}$ kg)
Show solution
$$\lambda = h/(mv) = (6.626 \times 10^{-34})/(9.1 \times 10^{-31} \times 10^6)$$ $$= 7.28 \times 10^{-10} \text{ m} = 7.28 \text{ Å}$$
Final Answer: $7.28$ Å.
Give all four quantum numbers for the last electron of nitrogen ($Z = 7$).
Show solution
Electronic configuration: $1s^2\,2s^2\,2p^3$. Last electron in $2p$. Using Hund's rule, $2p^3$ has each $p$-orbital singly occupied with parallel spins. $n = 2$, $l = 1$, $m_l = +1$ (last orbital), $m_s = +1/2$.
Final Answer: $(2, 1, +1, +1/2)$.
The number of orbitals in $n = 3$ shell:
Allowed values of $l$ for $n = 3$:
Magnetic quantum number for $p$-subshell:
Maximum number of electrons in $n = 4$ shell:
Uncertainty principle implies:
Orbitals and Electronic ConfigurationTopic 2
Atomic Orbital: Region in space where probability of finding an electron is high (typically $> 90\%$). Each orbital can hold maximum $2$ electrons (with opposite spin: Pauli).
Shapes of Orbitals:
| Orbital | Shape | Nodal planes (angular) |
|---|---|---|
| s ($l = 0$) | Spherical | $0$ |
| p ($l = 1$) | Dumbbell along axis | $1$ |
| d ($l = 2$) | Cloverleaf (4 types: $d_{xy}, d_{yz}, d_{zx}, d_{x^2-y^2}, d_{z^2}$) | $2$ |
| f ($l = 3$) | Complex | $3$ |
Nodes:
- Total nodes for orbital of $(n, l)$: $n - 1$
- Angular (nodal planes) = $l$
- Radial nodes = $n - l - 1$
Energy Order of Orbitals (Aufbau): Determined by $(n + l)$ rule:
- Lower $(n+l)$ filled first
- For same $(n+l)$, lower $n$ filled first
Filling order: $1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p\ldots$
Three Rules for Filling Electrons:
| Principle | Statement |
|---|---|
| Aufbau | Electrons fill orbitals of lower energy first |
| Pauli Exclusion | No two electrons in an atom can have all four quantum numbers same |
| Hund's rule | In degenerate orbitals, electrons singly occupy with parallel spins before pairing |
Stability of Half-filled and Fully filled subshells: Extra stability for configurations like $3d^5, 3d^{10}, 4s^1$ etc. due to:
- Symmetrical distribution
- Higher exchange energy
Anomalous configurations:
- Cr ($Z = 24$): $[Ar]\,3d^5\,4s^1$ (not $3d^4\,4s^2$)
- Cu ($Z = 29$): $[Ar]\,3d^{10}\,4s^1$ (not $3d^9\,4s^2$)
- Mo, Ag, Au follow similar patterns
Magnetic Moment: $\mu = \sqrt{n(n+2)}$ BM, where $n$ = number of unpaired electrons.
- Paramagnetic: has unpaired electrons (attracted to magnetic field)
- Diamagnetic: all paired
Write electronic configuration of $Fe$ ($Z = 26$) and $Fe^{3+}$.
Show solution
Fe: $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^6\,4s^2$ or $[Ar]\,3d^6\,4s^2$. Fe³⁺: Remove 3 e⁻ — first from $4s$ (higher n), then from $3d$. $[Ar]\,3d^5$.
Final Answer: Fe: $[Ar]\,3d^6\,4s^2$; Fe³⁺: $[Ar]\,3d^5$.
Calculate magnetic moment of Mn²⁺ ($Z(Mn) = 25$).
Show solution
Mn: $[Ar]\,3d^5\,4s^2$. Mn²⁺: $[Ar]\,3d^5$. Unpaired electrons = $5$. $$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ BM}$$
Final Answer: $\mu \approx 5.92$ BM.
Number of radial nodes in $3p$ orbital:
Anomalous configuration of Cu ($Z = 29$):
Hund's rule applies to:
Magnetic moment of an atom with $4$ unpaired electrons:
The shape of $d_{z^2}$ orbital:
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