Chemistry · Lesson

Hydrogen, hydrides and main-group bonding

Compare hydrogen’s bonding and oxidation-state assignments across different partners. Use paired-electron ledgers and an explicitly invented isotope oscillator to distinguish representation, formal balance and measured behavior.

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By the end, you can

Assign H in selected species using a stated electronegativity ordering.
Compare CH₄, NH₃ and H₂O Lewis pair inventories.
Balance a symbolic hydride/proton-bearing-water net without inferring a preparation.
Explain how reduced mass changes a fixed-force-constant oscillator model.

Hydrogen does not always receive the same assignment

Hydrogen has one neutral-atom valence electron and commonly reaches a duet in ordinary pair-bond Lewis models. Its oxidation-state assignment depends on its partner and the adopted bond-electron convention. For the selected Allen ordering C>H>Si>Na, H is +1 in CH₄ and −1 in SiH₄ and NaH; elemental H₂ gives zero.

Corresponding partner assignments are C −4, Si +4 and Na +1. These formal assignments do not say that every C–H or Si–H bond is a complete ionic separation, nor do they establish a real partial-charge magnitude. Bonds between partners close in electronegativity, including some H–P descriptions, require an explicit convention rather than an unqualified H=+1 shortcut.

Eight total valence electrons can support different shapes

CH₄ has 4+4=8 total valence electrons in four ordinary C–H bonding pairs. NH₃ has 5+3=8, with three bonding pairs and one lone pair at N. H₂O has 6+2=8, with two bonding pairs and two lone pairs at O. Each attached hydrogen has a duet in these selected Lewis representations.

Those central pair inventories give different molecular shapes in a qualitative electron-region model: tetrahedral methane, trigonal-pyramidal ammonia and bent water. Electron-region arrangement and molecular shape are distinct. The inventories do not assert identical measured angles, frequencies or charge distributions. Hydrides can also be ionic, molecular, polymeric or extended, so one discrete molecule’s drawing is not a universal description of every solid.

Conserve charge and distinguish a net ledger from its pathway

For the symbolic net H⁻ + H₂O → H₂ + OH⁻, both sides contain three H atoms, one O and total charge −1. The hydride H changes oxidation-state assignment from −1 to zero, while one water-derived H changes from +1 to zero. The formal electron exchange is one electron for each such hydride/proton pairing.

This does not describe free H⁺ as a bottle of reagent or imply a one-electron radical mechanism. Oxidation-state electron exchange and curved-pair bond bookkeeping are different descriptions of the same net constraints. A full molecular pathway, solvent environment and practical behavior would require additional evidence; the formula ledger is not a handling procedure.

Reduced mass changes the oscillator even with a fixed force constant

A two-body harmonic oscillator has frequency f=(1/(2π))√(k/μ), where k is the force constant and μ=m₁m₂/(m₁+m₂) is reduced mass. Wavenumber is f/c, with unit conversion if displayed in cm⁻¹. This is a model for small harmonic motion rather than a complete real molecular spectrum.

In an invented equal-mass diatomic comparison, choose k=500 N m⁻¹ and each atom mass 1 u or 2 u. Use the supplied atomic mass constant u=1.66053906892× 10⁻²⁷ kg and exact vacuum light speed c=299792458 m s⁻¹. Reduced masses are 0.5 u and 1 u. Model wavenumbers are approximately 4119.788 and 2913.130 cm⁻¹; their ratio is √2.

The atom masses 1 u/2 u and force constant are simplified teaching inputs, not actual measured H and D atom masses or H₂/D₂ bands. The atomic mass constant is an evaluated reference value, while c is exact in SI. Anharmonicity, real isotope masses and the actual potential are absent. Isotope substitution can support an assignment when measured evidence and an appropriate model are compared.

Equal-atom-mass harmonic oscillator with a stipulated force constant

Model wavenumber / cm⁻¹

Stipulated equal atom mass / u

Selected oscillator masses
— Fixed-k harmonic model

Underlying data

Stipulated equal atom mass / u	Model wavenumber / cm⁻¹
1	4119.788108254157 · Selected oscillator masses
2	2913.130108398213 · Selected oscillator masses

Original model with k=500 N m⁻¹, μ=m/2, supplied u=1.66053906892× 10⁻²⁷ kg and vacuum c=299792458 m s⁻¹. Markers represent invented equal-atom masses 1 u and 2 u. The dashed curve varies a mathematical mass parameter continuously between them; it does not represent additional physical isotopes, measured H₂/D₂ spectra or anharmonic motion.

Worked examples

State the ordering before assigning H

Use the selected Allen ordering C>H>Si>Na for H₂, NaH, CH₄ and SiH₄.

H₂ divides the homonuclear bond equally: H zero.
NaH assigns the pair to H: H −1 and Na +1.
CH₄ assigns C–H pairs to C: H +1, C −4.
SiH₄ assigns Si–H pairs to H: H −1, Si +4.

Selected formal assignments, rather than a universal hydrogen rule or actual ionic-charge measurement.

Distribute the eight valence electrons

Compare CH₄, NH₃ and H₂O ordinary Lewis inventories.

CH₄: four bond pairs, no central lone pair.
NH₃: three bond pairs and one central lone pair.
H₂O: two bond pairs and two central lone pairs.

Each selected entity has eight total valence electrons, with different central pair distributions.

Check the symbolic hydride net

Balance H⁻ + H₂O → H₂ + OH⁻.

Hydrogen totals 1+2=3 and 2+1=3.
Oxygen totals one on each side.
Total charge is −1 on each side.

Balanced net ledger; its one-electron oxidation-state exchange does not assign a radical pathway.

Use the equal-mass reduced mass

Compare the invented 1 u and 2 u atoms at the same k.

For identical atoms, μ=m/2.
Reduced mass therefore doubles between the cases.
Frequency and wavenumber scale as 1/√μ, giving ratio √2≈ 1.41421.

Model values approximately 4119.788 and 2913.130 cm⁻¹, rather than measured isotope spectra.

Try it yourself

Is H always +1 in compounds under the stated ionic bond assignment?
Show the answer and reasoning
No.
The selected NaH and SiH₄ cases assign H −1.
In the same fixed-k equal-atom oscillator toy, compare wavenumber for invented atom mass 4 u with that for 1 u.
Show the answer and reasoning
The 4 u value is half the 1 u value, approximately 2059.89 cm⁻¹.
Equal-atom reduced mass is m/2, so quadrupling m halves frequency and wavenumber. The masses are teaching inputs, not measured isotope values.
For ordinary NH₃ with eight total valence electrons, how many bonding and central lone pairs are in the selected Lewis ledger?
Show the answer and reasoning
Three bonding pairs and one central lone pair.
Three N–H pairs use six electrons; the remaining two form the N lone pair.
Are the oscillator’s 1 u/2 u masses and k=500 N m⁻¹ actual H₂/D₂ spectral data?
Show the answer and reasoning
No.
They are explicit teaching inputs for a restricted harmonic model.

What to carry forward

Assign H from its actual partner and a stated convention.
A duet at H complements the central atom’s pair ledger.
Separate a conserved net from evidence for a pathway.
Retain reduced mass in the oscillator equation.
Label model isotope inputs and measured spectra distinctly.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.