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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.

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.

First construct the energy cost for heterolysis from the molecular potential minimum. Dissociation to neutral atoms costs $D_e$, ionizing caesium costs $I$, and attachment of the electron to iodine releases EA. ThereforeFor the ground-state reference convention used above, the initial molecule already has vibrational zero-point energy. Thus its ground-state dissociation energy is $D_0=D_e-\frac{1}{2}{k}_B{\theta }_{\mathrm{vib}}$ andAt $3000\mathrm{K}$, $k_{B}T=0.258520\mathrm{eV}$, givingAlternatively, retaining the molecular minimum as reference gives $\Delta {\epsilon }_{\min }=4.36\mathrm{eV}$ and a Boltzmann factor $4.7372\times 10^{-8}$. That convention must use the full vibrational partition function, including $e^{-{\theta }_{\mathrm{vib}}/(2T)}$, in the next part. The ground-state convention instead uses $q_{\mathrm{vib}}^0$. Either consistent convention gives the same final equilibrium constant; using the ground-state reaction energy with the full zero-point factor would count the correction twice. The distinction between $D_e$ and $D_0$ is a distinction between initial states, not a change in the dissociation products.^[ref]

The primary chloride undergoes an SN2 reaction; the triphenylmethyl chloride undergoes an SN1 reaction. The primary site is accessible to back-side attack, whereas three phenyl groups make back-side attack at the triphenylmethyl site extremely hindered and stabilize its carbocation through resonance.

For the SN2 reaction, a suitable transition state isThe entering C and leaving Cl lie on opposite sides of the reacting carbon, with partial C–C and C–Cl bonds. The total charge is $-1$. This transition state is not a discrete reaction intermediate.

For the SN1 reaction, the approximately planar $\mathrm{P}{\mathrm{h}}_3{\mathrm{C}}^{+}$ carbocation is a genuine reaction intermediate. Assuming its formation by C–Cl heterolysis is rate determining and trapping is rapid,Thus the limiting rate law is independent of ethanol concentration. The successive attack and proton transfer occur after the slow ionization; a cation-stabilizing environment and efficient trapping underpin this idealized SN1 description.

The products are hexanenitrile, $\mathrm{C}{\mathrm{H}}_3(\mathrm{C}{\mathrm{H}}_2{)}_4\mathrm{C}\equiv \mathrm{N}$, and ethyl triphenylmethyl ether, $\mathrm{P}{\mathrm{h}}_3\mathrm{C}-\mathrm{O}-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$. The nitrile carbon comes from cyanide, so replacing chloride in the five-carbon chain gives a six-carbon product.

For the primary chloride, the carbon-end lone pair of $\mathrm{C}{\mathrm{N}}^{-}$ attacks the carbon carrying Cl from the back side while the C–Cl bond pair moves onto Cl. The single concerted nucleophilic substitution step forms the C–C bond and releases $\mathrm{C}{\mathrm{l}}^{-}$. Attack through the carbon end gives the nitrile, not an isocyanide.

For the triphenylmethyl chloride, the C–Cl bond first undergoes heterolysis to $\mathrm{P}{\mathrm{h}}_3{\mathrm{C}}^{+}$ and $\mathrm{C}{\mathrm{l}}^{-}$. An ethanol oxygen lone pair then attacks the carbocation, giving $\mathrm{P}{\mathrm{h}}_3\mathrm{C}-{\mathrm{O}}^{+}(\mathrm{H})\mathrm{Et}$. Finally pyridine accepts that O–H proton, with the O–H bond pair returning to oxygen, giving the neutral ether and pyridinium chloride.

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.