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Atomic Structure - Electron configuration

Grade 12IBChemistry

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

🔑Concepts

Main energy levels are defined by the principal quantum number nn. Each level can hold a maximum of 2n22n^2 electrons.

Energy levels are split into sub-levels (s,p,d,fs, p, d, f), which consist of orbitals. An orbital is a region of space where there is a high probability of finding an electron.

The Aufbau Principle states that electrons occupy the orbitals of lowest energy first (e.g., 1s<2s<2p<3s<3p<4s<3d1s < 2s < 2p < 3s < 3p < 4s < 3d).

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

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

Exceptions to the Aufbau Principle: Chromium (CrCr) has the configuration [Ar]3d54s1[Ar] 3d^5 4s^1 and Copper (CuCu) has [Ar]3d104s1[Ar] 3d^{10} 4s^1. This is due to the increased stability of half-filled and fully-filled dd-subshells.

When transition metals form positive ions (cations), electrons are removed from the 4s4s subshell before the 3d3d subshell because the 4s4s electrons are in a higher principal energy level (n=4n=4 vs n=3n=3).

Continuous spectra are produced when all wavelengths of light are present, whereas line emission spectra are produced by excited electrons falling to lower energy levels, emitting specific frequencies of photons defined by ΔE=hν\Delta E = h\nu.

📐Formulae

2n22n^2

E=hνE = h\nu

c=νλc = \nu\lambda

E=hcλE = \frac{hc}{\lambda}

ΔE=EhigherElower\Delta E = E_{higher} - E_{lower}

💡Examples

Problem 1:

Write the full electron configuration for the Ni2+Ni^{2+} ion (Z=28Z=28 for Nickel).

Solution:

1s22s22p63s23p63d81s^2 2s^2 2p^6 3s^2 3p^6 3d^8

Explanation:

The neutral Nickel atom has the configuration [Ar]4s23d8[Ar] 4s^2 3d^8. When it forms a 2+2+ ion, it loses two electrons. For transition metals, electrons are always removed from the 4s4s orbital before the 3d3d orbital. Thus, the 4s24s^2 electrons are removed, leaving 3d83d^8.

Problem 2:

Explain why the ground state electron configuration of Chromium (CrCr) is [Ar]3d54s1[Ar] 3d^5 4s^1 instead of [Ar]3d44s2[Ar] 3d^4 4s^2.

Solution:

The 3d54s13d^5 4s^1 configuration provides extra stability.

Explanation:

By having a half-filled dd-subshell (3d53d^5) and a half-filled ss-subshell (4s14s^1), electron-electron repulsion is minimized, and the exchange energy is maximized, creating a more stable, lower-energy state than the predicted [Ar]3d44s2[Ar] 3d^4 4s^2.

Problem 3:

Calculate the energy of a photon (in Joules) emitted when an electron transition produces light with a frequency of 6.1imes1014 Hz6.1 imes 10^{14} \text{ Hz}. (Use h=6.63imes1034 J sh = 6.63 imes 10^{-34} \text{ J s})

Solution:

E=4.04imes1019 JE = 4.04 imes 10^{-19} \text{ J}

Explanation:

Using the formula E=hνE = h\nu, we multiply Planck's constant by the frequency: E=(6.63imes1034 J s)×(6.1imes1014 s1)=4.0443imes1019 JE = (6.63 imes 10^{-34} \text{ J s}) \times (6.1 imes 10^{14} \text{ s}^{-1}) = 4.0443 imes 10^{-19} \text{ J}.

Electron configuration - Revision Notes & Key Formulas | IB Grade 12 Chemistry