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A bond-dissociation energy specifies the energy needed for a stated dissociation into separated fragments. An energy from the lowest vibrational state, an energy from the bottom of a potential energy well, and a finite-temperature molar dissociation enthalpy require different conventions; do not silently mix them in an electronegativity formula.

Which fragments, in which quantum states, and at which temperature? For a neutral gaseous molecule with a specified two-electron covalent bond, the usual thermochemical convention is homolysis,Each fragment receives one of the two bonding Electrons; the dots denote unpaired Electrons. The final fragments are isolated from each other, with their own relaxed molecular geometry and specified electronic quantum states. The energy cost includes this relaxation, rather than merely stretching the initial chemical bond while freezing everything else. For a neutral parent, heterolysis instead produces ${\mathrm{A}}^{+}(g)+{\mathrm{B}}^{-}(g)$, or the opposite assignment of electric charges; those are different final states and different energies. A value for gaseous neutral fragments cannot be transferred to ions, a chemical solution, or a solid without accounting for those changed conditions.^[ref]

The distinction is useful beyond terminology. Pauling electronegativity compares consistently defined bond-dissociation energies; photodissociation can reveal a specified ground-state threshold; a thermal chemical reaction balance requires compatible enthalpies. A threshold is also different from an activation energy: it compares initial and final quantum states, whereas a barrier can obstruct the path between them.

A bond-dissociation enthalpy is the molar enthalpy change for the specified homolysis at temperature $T$, with reactant and products in specified standard states. Its usual units are $\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The superscript $\circ$ specifies standard states; it does not fix $T$ to room temperature.^[ref]

Write $\delta h_i(T)=H_{m,i}^{\circ }(T)-H_{m,i}^{\circ }(0)$ for each species' molar thermal enthalpy increment. With $D_0$ an energy per parent molecule,Here $N_{\mathrm{A}}$ is the Avogadro constant. The thermal increments include translational kinetic energy, molecular rotation, thermally excited molecular vibration and electronic states when populated, and the $pV$ contribution to enthalpy. They are additional to the zero-point energies already included in $D_0$. If values of standard enthalpy of formation at the same $T$ are available, Hess's law gives the same quantity as ${\Delta }_f{H}^{\circ }(A)+{\Delta }_f{H}^{\circ }(B)-{\Delta }_f{H}^{\circ }(AB)$.

Units alone do not identify the convention. Multiplying either $D_e$ or $D_0$ by the Avogadro constant gives molar energy, but does not add thermal enthalpy. One electronvolt per dissociation corresponds to $96.485332\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. A spectroscopic wavenumber $\tilde{\nu }$ represents energy $hc\tilde{\nu }$, using the Planck constant $h$ and speed of light $c$; $1\mathrm{c}{\mathrm{m}}^{-1}$ corresponds to $0.011962657\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$.^[ref]

Heterolysis assigns both bonding Electrons to one fragment. For an initially neutral molecule, the resulting cation and anion define a different dissociation channel from homolysis; their isolated gas-phase energy differs from that of solvated ions.

Homolysis splits a covalent bond with one bonding Electron assigned to each fragment. A neutral closed-shell molecule thereby produces fragments with unpaired Electrons, each a radical. The usual bond-dissociation enthalpy uses this product convention.

Light initiates methane chlorination by photodissociation of $\mathrm{C}{\mathrm{l}}_2$ through homolysis. The elementary bookkeeping isHere $h\nu$ denotes the photon energy. The first propagation step replaces a chlorine radical with a methyl radical; the second regenerates a chlorine radical. Adding the two propagation equations cancels both free radicals and gives $\mathrm{C}{\mathrm{H}}_4+\mathrm{C}{\mathrm{l}}_2⟶\mathrm{C}{\mathrm{H}}_3\mathrm{Cl}+\mathrm{HCl}$: chloromethane and hydrogen chloride are made while the chain carrier is regenerated.^[ref]

For example, recombination $2{\mathrm{C}{\mathrm{H}}_3}^{\bullet }⟶{\mathrm{C}}_2{\mathrm{H}}_6$ of methyl radicals makes ethane and a covalent bond from two unpaired Electrons, removing two carriers. Recombination of two ${\mathrm{Cl}}^{\bullet }$ also terminates carriers; in the gas phase, stabilization of the product requires disposal of released energy, for example to a third collision partner. The figure omits that partner. Termination competes with propagation, and further chlorination can consume chloromethane, so this cycle alone does not predict a pure product or a measured reaction rate.^[ref]

A radical has one or more unpaired Electrons. In a simple atomic orbital or molecular orbital occupancy picture, an unpaired Electron occupies a spatial orbital without an opposite-spin partner in that orbital. A dot such as ${\mathrm{H}}^{\bullet }$ marks this electronic property; it does not mean a missing Electron or positive electric charge. Free radicals can be neutral or charged. Terminology also has a boundary: metal ions exhibiting paramagnetism are not normally called free radicals just because they have unpaired Electrons.^[ref]

Compare three species containing the same single proton. The hydrogen atom ${\mathrm{H}}^{\bullet }$ has one Electron in its $1s$ atomic orbital, and is a neutral radical. The hydrogen cation ${\mathrm{H}}^{+}$ has no Electrons, so no unpaired Electron. The hydrogen anion ${\mathrm{H}}^{-}$ has two Electrons paired in its $1s$ atomic orbital. It is charged but is not a radical in its ground state. With $N_p$ protons, $N_e$ Electrons, and positive elementary charge $e$, the independent electric charge count isThe electron pair and electric charge questions require different information. An odd $N_e$ ensures that not all Electrons can be paired; an even $N_e$ alone does not ensure pairing, as the molecular orbital description of molecular oxygen illustrates.

The distinction matters whenever a chemical reaction mechanism uses dots and charge signs together. Homolysis gives one bonding Electron to each fragment, while heterolysis assigns both to one fragment. A radical chain reaction can pass an unpaired Electron from one species to another without generating net electric charge. Detection then exploits the unpaired Electron's spin and magnetic moment, rather than merely measuring whether the sample carries an electric charge.