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Structure 1. Models of the particulate nature of matter - Electron configurations

Grade 12IBChemistry

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

🔑Concepts

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The electron configuration of an atom describes the distribution of electrons in its atomic orbitals. Main energy levels are denoted by the principal quantum number nn (n=1,2,3,…n = 1, 2, 3, \dots).

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Each main energy level contains sublevels: ss (1 orbital), pp (3 orbitals), dd (5 orbitals), and ff (7 orbitals).

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The Aufbau Principle states that electrons fill the lowest energy orbitals first (e.g., 4s4s is filled before 3d3d because it is lower in energy).

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The Pauli Exclusion Principle states that an atomic orbital can hold a maximum of two electrons, and they must have opposite spins.

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

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Full-shell and half-shell stability leads to exceptions in the 3d3d series: Chromium ([Ar]4s13d5[Ar] 4s^1 3d^5) and Copper ([Ar]4s13d10[Ar] 4s^1 3d^{10}).

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When transition metals form ions, they lose electrons from the 4s4s subshell before the 3d3d subshell.

📐Formulae

2n2 (Maximum number of electrons in a main energy level n)2n^2 \text{ (Maximum number of electrons in a main energy level } n\text{)}

E=hν=hcλ (Energy of a photon associated with electron transitions)E = h\nu = \frac{hc}{\lambda} \text{ (Energy of a photon associated with electron transitions)}

c=λν (Speed of light relationship)c = \lambda \nu \text{ (Speed of light relationship)}

💡Examples

Problem 1:

Write the full electron configuration for a Neutral Phosphorus atom (Z=15Z = 15) and its ion P3−P^{3-}.

Solution:

P:1s22s22p63s23p3P: 1s^2 2s^2 2p^6 3s^2 3p^3 P3−:1s22s22p63s23p6P^{3-}: 1s^2 2s^2 2p^6 3s^2 3p^6

Explanation:

Phosphorus has 15 electrons. Following the Aufbau sequence: 1s→2s→2p→3s→3p1s \rightarrow 2s \rightarrow 2p \rightarrow 3s \rightarrow 3p. The P3−P^{3-} ion gains 3 electrons to achieve a stable noble gas configuration (isoelectronic with Argon).

Problem 2:

Explain the electron configuration of Copper (Z=29Z = 29).

Solution:

Cu:1s22s22p63s23p64s13d10 or [Ar]4s13d10Cu: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^{10} \text{ or } [Ar] 4s^1 3d^{10}

Explanation:

According to the general rule, it should be [Ar]4s23d9[Ar] 4s^2 3d^9. However, a full 3d3d subshell (3d103d^{10}) is more stable than a partially filled one. One electron from the 4s4s orbital is promoted to the 3d3d orbital to achieve this stability.

Problem 3:

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

Solution:

Fe:[Ar]4s23d6Fe: [Ar] 4s^2 3d^6 Fe2+:[Ar]3d6Fe^{2+}: [Ar] 3d^6

Explanation:

When forming positive ions (cations) of transition metals, electrons are removed from the 4s4s subshell before the 3d3d subshell. Thus, the two electrons are lost from 4s24s^2, leaving 3d63d^6.