1.3 Electron configurations

Read every spectral transition in two directions: absorption raises an electron by ΔE, while a downward transition emits a photon with ΔE = hf = hc/λ. A shorter-wavelength line therefore represents a larger energy gap, not a higher line intensity.

Orient the spectrum before comparing lines: radio → microwave → infrared → visible → ultraviolet → X-ray → gamma is increasing frequency and photon energy, and decreasing wavelength. Within visible light, red has longer wavelength and lower photon energy than violet.

Compare lines by their energy gaps. Transitions ending at n = 2 form the visible series, and lines crowd together as the starting level rises because adjacent high-n levels are closer in energy. The convergence limit represents removal of the electron, not one more bound-state transition.

For n = 3, the theoretical capacity is 2(3²) = 18 electrons. This is a capacity, not a claim that every third shell is full: the actual occupancy depends on the atom and the relative energies of available sublevels.

Use the hierarchy as a classification test: a p sublevel contains three orbitals, and each orbital can hold two electrons. An orbital describes a probability region with a characteristic shape; it is not a circular route travelled by an electron.

Build an orbital diagram by checking electron total, energy order, single occupation of equal-energy orbitals, then opposite-spin pairing. For transition-metal ions remove 4s electrons before 3d, and verify Cr and Cu against the stated exceptions rather than forcing the simple filling pattern.

Worked ion check: Fe has 26 electrons and condensed configuration [Ar] 4s² 3d⁶. To form Fe³⁺, remove the two electrons from the highest principal level, 4s, before removing one 3d electron, giving [Ar] 3d⁵. The final superscripts total 23 electrons, matching 26 − 3.

Worked example — hydrogen convergence limit

The local course book gives $\lambda=9.12\times10^{-8}\,\mathrm{m}$. First, $f=c/\lambda=(3.00\times10^8\,\mathrm{m\,s^{-1}})/(9.12\times10^{-8}\,\mathrm{m})=3.29\times10^{15}\,\mathrm{s^{-1}}$. Then $E_{\text{photon}}=hf=(6.63\times10^{-34}\,\mathrm{J\,s})(3.29\times10^{15}\,\mathrm{s^{-1}})=2.18\times10^{-18}\,\mathrm{J}$. Convert one-photon energy to one mole and joules to kilojoules: $IE=(2.18\times10^{-18})(6.02\times10^{23})/1000=1.31\times10^3\,\mathrm{kJ\,mol^{-1}}$. This is the molar energy for the first ionization process $\mathrm{H(g)\rightarrow H^+(g)+e^-}$.

At the convergence limit, convert wavelength or frequency to energy per photon with E = hf, then multiply by the Avogadro constant and convert J mol⁻¹ to kJ mol⁻¹. Across-period trends are general patterns; subshell energy and electron pairing explain the named dips.

Locate the first order-of-magnitude jump before naming the group. A large jump after the second electron is removed shows two outer-shell electrons; the third electron would come from a lower shell, supporting a group 2 assignment.