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Structure of Atom - Quantum Mechanical Model

Grade 11ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Dual Nature of Matter: Proposed by de Broglie, suggesting that moving particles like electrons exhibit both wave and particle properties. The de Broglie wavelength is given by λ=hmv\lambda = \frac{h}{mv}.

Heisenberg's Uncertainty Principle: It states that it is impossible to determine simultaneously the exact position and exact momentum (or velocity) of an electron with absolute accuracy: ΔxΔph4π\Delta x \cdot \Delta p \geq \frac{h}{4\pi}.

Schrödinger Wave Equation: A fundamental equation in quantum mechanics that describes the behavior of an electron in an atom. The solution gives the wave function ψ\psi, where ψ2|\psi|^2 represents the probability density of finding an electron.

Quantum Numbers: Four numbers used to describe the state of an electron: Principal (nn), Azimuthal (ll), Magnetic (mlm_l), and Spin (msm_s).

Principal Quantum Number (nn): Defines the main energy shell and size of the orbital. n=1,2,3,n = 1, 2, 3, \dots.

Azimuthal Quantum Number (ll): Defines the shape of the orbital. Values range from 00 to (n1)(n-1). l=0(s)l=0 (s), l=1(p)l=1 (p), l=2(d)l=2 (d), l=3(f)l=3 (f).

Magnetic Quantum Number (mlm_l): Describes the orientation of orbitals in space. Values range from l-l to +l+l including zero.

Spin Quantum Number (msm_s): Describes the direction of electron spin, either +12+\frac{1}{2} or 12-\frac{1}{2}.

Pauli Exclusion Principle: No two electrons in an atom can have the same set of all four quantum numbers. An orbital can hold a maximum of 2 electrons with opposite spins.

Aufbau Principle: Electrons are filled in orbitals in the increasing order of their energies, following the (n+l)(n+l) rule.

Hund's Rule of Maximum Multiplicity: Pairing of electrons in degenerate orbitals (p,d,fp, d, f) does not occur until each orbital is singly occupied with parallel spins.

Nodes: Regions where the probability of finding an electron is zero. Radial nodes =nl1= n - l - 1, Angular nodes =l= l, Total nodes =n1= n - 1.

📐Formulae

λ=hmv=hp\lambda = \frac{h}{mv} = \frac{h}{p}

ΔxΔph4π\Delta x \cdot \Delta p \geq \frac{h}{4\pi}

ΔxmΔvh4π\Delta x \cdot m\Delta v \geq \frac{h}{4\pi}

Orbital Angular Momentum=l(l+1)h2π\text{Orbital Angular Momentum} = \sqrt{l(l+1)} \frac{h}{2\pi}

Maximum electrons in a shell=2n2\text{Maximum electrons in a shell} = 2n^2

Number of orbitals in a subshell=2l+1\text{Number of orbitals in a subshell} = 2l + 1

Total Nodes=n1\text{Total Nodes} = n - 1

💡Examples

Problem 1:

Calculate the de Broglie wavelength of an electron moving with a velocity of 2.05×107 m/s2.05 \times 10^7 \text{ m/s}. (Mass of electron me=9.1×1031 kgm_e = 9.1 \times 10^{-31} \text{ kg}, h=6.626×1034 J sh = 6.626 \times 10^{-34} \text{ J s})

Solution:

Using λ=hmv\lambda = \frac{h}{mv}: λ=6.626×10349.1×1031×2.05×1073.55×1011 m\lambda = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 2.05 \times 10^7} \approx 3.55 \times 10^{-11} \text{ m}.

Explanation:

The de Broglie relation relates the momentum of a particle to its wavelength, demonstrating the wave nature of matter.

Problem 2:

Determine the set of four quantum numbers for the last electron of a Sodium (NaNa, Z=11Z=11) atom.

Solution:

Configuration: 1s22s22p63s11s^2 2s^2 2p^6 3s^1. The last electron enters 3s13s^1. Thus: n=3n = 3, l=0l = 0 (for ss orbital), ml=0m_l = 0, ms=+12m_s = +\frac{1}{2}.

Explanation:

Sodium has 11 electrons. Following the Aufbau principle, the 11th electron occupies the 3s3s subshell.

Problem 3:

How many radial and angular nodes are present in a 4d4d orbital?

Solution:

For 4d4d: n=4,l=2n=4, l=2. Radial nodes =nl1=421=1= n - l - 1 = 4 - 2 - 1 = 1. Angular nodes =l=2= l = 2. Total nodes =1+2=3= 1 + 2 = 3.

Explanation:

Radial nodes depend on both nn and ll, whereas angular nodes are determined solely by the azimuthal quantum number ll.

Quantum Mechanical Model - Revision Notes & Key Formulas | ICSE Class 11 Chemistry