CBSE Class 11 Chemistry Structure Of Atom Notes Set 10

This is the second part of the Class 11 CBSE Chemistry notes on Chapter 2 Structure of Atom, Set 10. It explains the shift from Bohr's orbits to the quantum mechanical model: de Broglie's matter waves, Heisenberg's uncertainty principle, orbitals and their shapes, the four quantum numbers, nodes, the rules for filling electrons and the electronic configurations of atoms and ions, including the extra stability of half-filled and completely filled subshells. Seven solved examples show the calculations.

Important Notes and Formulas

Dual Behaviour of Matter: de Broglie Relation

Louis de Broglie (1924) proposed that, as light behaves as both wave and particle, every moving material particle also has a wave associated with it. The wavelength is

\[ \lambda = \frac{h}{mv} = \frac{h}{p} \]

1. Here m is the mass, v the speed and p = mv the momentum of the particle. The wavelength is inversely proportional to momentum.
2. For a particle of kinetic energy K.E. the wavelength is λ = h/√(2m·K.E.); for a charge q accelerated through a potential V, λ = h/√(2mqV).
3. The wavelength of a large object (a cricket ball, a car) is so small (about 10⁻³⁴ m) that wave properties cannot be detected. For electrons and other sub-atomic particles the wavelength is comparable to atomic dimensions and can be observed.
4. The wave nature of electrons was confirmed by electron diffraction (Davisson and Germer). The particle nature is seen in scintillation experiments.

Combining Bohr's postulate mvr = nh/2π with de Broglie's relation gives 2πr = nλ: the circumference of an allowed orbit holds a whole number of electron wavelengths.

Heisenberg's Uncertainty Principle

It is impossible to determine simultaneously the exact position and the exact momentum (or velocity) of an electron or any other microscopic particle. The product of the uncertainties is never less than h/4π:

\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \quad \text{or} \quad \Delta x \cdot \Delta v \geq \frac{h}{4\pi m} \]

1. Measuring the position of an electron needs light, and the photon disturbs its momentum. The more exactly one quantity is fixed, the less exactly the other is known.
2. The principle rules out definite paths or trajectories for electrons, so Bohr's fixed circular orbits cannot be correct. It has a noticeable effect only for sub-atomic particles; for macroscopic objects the uncertainty is far too small to matter.
3. If Δx = 0, then Δp is infinite, and vice versa.

Both the dual nature of matter and the uncertainty principle are failures of Bohr's model that the quantum mechanical model puts right.

Quantum Mechanical Model of the Atom

Quantum mechanics studies the motion of microscopic particles that show both wave and particle behaviour. It is based on the Schrödinger equation, which is solved for the electron in an atom to give a set of allowed energy values and a wave function ψ for each.

1. The wave function ψ describes the electron wave. It has no direct physical meaning, but ψ² at a point is the probability density: the probability of finding the electron there.
2. Only certain solutions are allowed, each corresponding to a definite energy, so energy is quantised without extra assumptions.
3. An orbital is a one-electron wave function. Its boundary surface encloses the region around the nucleus in which the electron is found about 90% of the time. An orbital is not a path.
4. An orbit (Bohr) is a flat circular path; an orbital is a three-dimensional probability region.

Quantum Numbers

Three quantum numbers (n, l, m_l) specify an orbital; all four specify an electron. A shell n has n subshells, and a subshell l has 2l + 1 orbitals.

Orbitals of the same subshell have equal energy and are called degenerate orbitals. In hydrogen the energy depends only on n (1s < 2s = 2p < 3s = 3p = 3d). In many-electron atoms it depends on both n and l, because inner electrons shield outer ones from the nucleus (the s electron penetrates closest, so the order is s < p < d < f in a shell).

Shapes of Atomic Orbitals

1. s orbitals are spherical and non-directional. 1s and 2s have the same shape, but 2s is larger.
2. p orbitals (l = 1) have a dumb-bell shape with two lobes either side of a nodal plane through the nucleus. The three orbitals p_x, p_y and p_z lie along the x, y and z axes at right angles. There is no 1p orbital.
3. d orbitals (l = 2) are five in number: d_xy, d_yz, d_zx and d_x²−y² have four lobes (cloverleaf), the first three lying between the axes and the last along them; d_z² has two lobes along the z axis with a ring in the xy plane. There is no d orbital for n = 1 or 2.

Nodes. A node is a region where the probability density is zero. Radial (spherical) nodes = n − l − 1; angular nodes (nodal planes or cones) = l; total nodes = n − 1. For example 2s has one radial node, 3p has one radial and one angular node, and 3d has two angular nodes and no radial node.

Filling of Orbitals: Aufbau Principle, Pauli Principle and Hund's Rule

1. Aufbau principle: in the ground state, electrons occupy the available orbitals of lowest energy first.
2. (n + l) rule: the lower the value of (n + l), the lower the energy of the subshell. If two subshells have the same (n + l), the one with the lower n has the lower energy.
3. Order of filling: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s ...
4. Pauli exclusion principle: no two electrons in an atom can have the same set of four quantum numbers. An orbital holds a maximum of two electrons, with opposite spins.
5. Hund's rule of maximum multiplicity: electrons are not paired in degenerate orbitals (p, d, f) until each orbital of the subshell has one electron. The singly occupied orbitals have parallel spins.

Maximum electrons: s 2, p 6, d 10, f 14.

Electronic Configuration of Atoms

Configurations are written in the s^ap^bd^c notation, e.g. oxygen (Z = 8) is 1s² 2s² 2p⁴, or with a noble gas core, e.g. K is [Ar] 4s¹. In the orbital diagram, each orbital is a box and each electron an arrow (↑ or ↓).

Stability of Half-filled and Completely Filled Subshells

Subshells that are exactly half-filled (p³, d⁵, f⁷) or completely filled (p⁶, d¹⁰, f¹⁴) are more stable. Two reasons are given.

1. Symmetrical distribution of electrons: symmetry gives stability, and electrons in such subshells shield each other less, so they are held more tightly by the nucleus.
2. Exchange energy: electrons of the same spin in degenerate orbitals can exchange positions, releasing energy. The number of exchanges is largest for half-filled and filled subshells.

This is why chromium is [Ar] 3d⁵ 4s¹ (not 3d⁴ 4s²) and copper is [Ar] 3d¹⁰ 4s¹ (not 3d⁹ 4s²).

Configuration of Ions and Magnetic Behaviour

1. To form a cation, remove electrons from the outermost shell (highest n) first: Fe is [Ar] 3d⁶ 4s², so Fe²⁺ is [Ar] 3d⁶ and Fe³⁺ is [Ar] 3d⁵.
2. To form an anion, add electrons to the lowest empty orbitals: Cl⁻ is [Ne] 3s² 3p⁶.
3. Species with unpaired electrons are paramagnetic; those with all electrons paired are diamagnetic.
4. Spin-only magnetic moment: μ = √[n(n + 2)] B.M., where n is the number of unpaired electrons (n = 1: 1.73; 2: 2.83; 3: 3.87; 4: 4.90; 5: 5.92 B.M.).

Solved Examples

Example 1. Estimate the de Broglie wavelength of (a) an electron moving at 7.28 × 10⁷ m s⁻¹ and (b) a 100 kg motorbike moving at 6.63 m s⁻¹.
Answer:
(a) λ = h/mv = 6.626 × 10⁻³⁴ ÷ (9.109 × 10⁻³¹ × 7.28 × 10⁷) = 1.0 × 10⁻¹¹ m.
(b) λ = 6.626 × 10⁻³⁴ ÷ (100 × 6.63) = 1.0 × 10⁻³⁶ m.
The electron wave is detectable; the motorbike wave is far too small to observe.

Example 2. What accelerating potential must be applied to a proton beam to give it a de Broglie wavelength of 0.005 nm? (m_p = 1.673 × 10⁻²⁷ kg)
Answer:
K.E. = qV = p²/2m and p = h/λ, so V = h²/(2 m q λ²).
V = (6.626 × 10⁻³⁴)² ÷ (2 × 1.673 × 10⁻²⁷ × 1.602 × 10⁻¹⁹ × (5 × 10⁻¹²)²) = 32.8 V (about 33 V).

Example 3. Calculate the uncertainty in the position of a dust particle of mass 1 mg if the uncertainty in its velocity is 5.5 × 10⁻²⁰ m s⁻¹.
Answer:
Mass = 1 mg = 10⁻⁶ kg. Δx = h/(4π m Δv).
Δx = 6.626 × 10⁻³⁴ ÷ (4 × 3.14 × 10⁻⁶ × 5.5 × 10⁻²⁰) = 9.6 × 10⁻¹⁰ m.

Example 4. How many electrons in a sulphur atom (Z = 16) have n + l = 3?
Answer:
S: 1s² 2s² 2p⁶ 3s² 3p⁴.
(n + l): 1s = 1, 2s = 2, 2p = 3, 3s = 3, 3p = 4.
Subshells with n + l = 3 are 2p and 3s, holding 6 + 2 = 8 electrons.

Example 5. Write the electronic configurations of Cu⁺, Cu²⁺, Cr³⁺ and Co³⁺ (Z: Cu 29, Cr 24, Co 27) and give the number of unpaired electrons.
Answer:
Cu: [Ar] 3d¹⁰ 4s¹. Cu⁺: [Ar] 3d¹⁰, 0 unpaired electrons (diamagnetic).
Cu²⁺: [Ar] 3d⁹, 1 unpaired electron.
Cr: [Ar] 3d⁵ 4s¹. Cr³⁺: [Ar] 3d³, 3 unpaired electrons.
Co: [Ar] 3d⁷ 4s². Co³⁺: [Ar] 3d⁶, 4 unpaired electrons.

Example 6. What are the four quantum numbers of the valence electron of potassium (Z = 19) in the ground state?
Answer:
K: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹. The valence electron is the 4s electron.
n = 4, l = 0, m_l = 0, m_s = +1/2 (or −1/2).

Example 7. Calculate the orbital angular momentum of an electron in (a) a 2s orbital, (b) a 2p orbital and (c) a 4f orbital.
Answer:
Angular momentum = √[l(l + 1)] × h/2π.
(a) 2s: l = 0, so the angular momentum is 0.
(b) 2p: l = 1, so it is √2 h/2π.
(c) 4f: l = 3, so it is √12 h/2π = 3.46 h/2π (= 2√3 h/2π).