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Atomic Structure - Electron configuration

Grade 11IBChemistry

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

🔑Concepts

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The principal quantum number nn defines the main energy levels; the maximum number of electrons per level is given by 2n22n^2.

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Sublevels are denoted as s,p,d, and fs, p, d, \text{ and } f, containing 1,3,5, and 71, 3, 5, \text{ and } 7 orbitals respectively. Each orbital can hold a maximum of 22 electrons.

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The Aufbau Principle states that electrons occupy the orbitals of lowest energy first (1s→2s→2p→3s→3p→4s→3d1s \rightarrow 2s \rightarrow 2p \rightarrow 3s \rightarrow 3p \rightarrow 4s \rightarrow 3d).

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The Pauli Exclusion Principle states that no two electrons in an atom can have the same four quantum numbers; effectively, an orbital holds a maximum of two electrons with opposite spins.

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Hund's Rule states that for degenerate orbitals (orbitals of the same energy), electrons fill them singly first with parallel spins to minimize inter-electron repulsion.

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Notable exceptions to the Aufbau Principle occur in Chromium ([Ar]3d54s1[Ar] 3d^5 4s^1) and Copper ([Ar]3d104s1[Ar] 3d^{10} 4s^1) because half-filled and fully-filled dd-subshells offer extra stability.

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When transition metals form positive ions, electrons are removed from the 4s4s orbital before the 3d3d orbital (e.g., Ti2+Ti^{2+} is [Ar]3d2[Ar] 3d^2 not [Ar]4s2[Ar] 4s^2).

📐Formulae

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

c=νλc = \nu \lambda

E=hνE = h \nu

E=hcλE = \frac{hc}{\lambda}

💡Examples

Problem 1:

Write the full electron configuration for a Neutral Phosphorus atom (Z=15Z = 15) and identify the number of unpaired electrons.

Solution:

1s22s22p63s23p31s^2 2s^2 2p^6 3s^2 3p^3

Explanation:

Following the Aufbau principle, we fill 1s1s, 2s2s, 2p2p, and 3s3s completely (12 electrons). The remaining 3 electrons enter the 3p3p subshell. According to Hund's Rule, these 3 electrons will occupy the three 3p3p orbitals individually, resulting in 3 unpaired electrons.

Problem 2:

Determine the electron configuration of the Fe2+Fe^{2+} ion (Z=26Z = 26).

Solution:

1s22s22p63s23p63d61s^2 2s^2 2p^6 3s^2 3p^6 3d^6 or [Ar]3d6[Ar] 3d^6

Explanation:

The neutral FeFe atom is [Ar]4s23d6[Ar] 4s^2 3d^6. When ionizing to Fe2+Fe^{2+}, the two electrons are removed from the 4s4s orbital first, even though the 3d3d subshell was filled 'after' the 4s4s in the Aufbau sequence.

Problem 3:

Calculate the energy of a photon with a frequency of 6.0×1014 s−16.0 \times 10^{14} \, s^{-1}. (Use h=6.63×10−34 J sh = 6.63 \times 10^{-34} \, J \, s)

Solution:

E=3.98×10−19 JE = 3.98 \times 10^{-19} \, J

Explanation:

Using the formula E=hνE = h\nu: E=(6.63×10−34 J s)×(6.0×1014 s−1)=3.978×10−19 JE = (6.63 \times 10^{-34} \, J \, s) \times (6.0 \times 10^{14} \, s^{-1}) = 3.978 \times 10^{-19} \, J.

Electron configuration - Revision Notes & Key Formulas | IB Grade 11 Chemistry