Boron, carbon, silicon and related element families

A pair acceptor need not carry positive formal charge

In the selected simple BF₃ Lewis representation, three B–F single bonds give six bonding electrons around B, total molecular valence electrons 3+3× 7=24 and B formal charge zero. Fluoride pair acceptance gives BF₄⁻ with four bonds around B and 32 total valence electrons. Its B formal charge is −1 in the all-single-bond representation, while B oxidation state remains +3 with F assigned −1.

Lewis acidity describes pair acceptance, rather than a formal-charge sign. The simple BF₃ ledger does not claim the actual bonds have no additional electronic character, and formal charge does not equal measured density. The assigned oxidation state and pair-bond inventory answer different questions.

Pair donation completes a new ordinary bond ledger

For the symbolic NH₃–BF₃ adduct, nitrogen contributes its lone pair to the new B–N connection. Total valence electrons are 8+24=32. In the completed all-single-bond Lewis representation, N has four bonds and no lone pair, formal charge +1; B has four bonds and formal charge −1. The overall adduct is neutral.

A donor arrow can show the origin of the bonding pair in a proposed formation ledger. Once the bond is formed, that origin is not a separate permanent category of electrons distinguishable from every other bonding pair. The formal-charge split neither makes this adduct two disconnected ions nor gives actual population charges.

Elemental carbon can have different networks

Diamond has a four-coordinate extended carbon network, while graphite has three-coordinate layers with delocalized π character. Molecular C₆₀ instead forms a closed carbon cage. Each is elemental carbon with oxidation state zero, but their connections and electronic descriptions differ.

An elemental formula or oxidation-state assignment therefore cannot establish hardness, conductivity or molecular versus extended structure. Phase, defects and environment can also matter. The selected connectivity descriptions are useful before any specific measured property comparison is attempted.

Shared oxygen vertices give composition without making molecules

In a selected common low-pressure silica network, each Si has four oxygen neighbors and each bridging O links two Si centers. Ten Si centers give 10× 4=40 Si–O incidences. Sharing each O between two Si gives 20 O and empirical ratio Si:O=1:2. This is a periodic or boundary-corrected network count, not an isolated finite fragment with unspecified surface groups.

Ordinary solid CO₂ instead contains discrete molecules, so equal dioxide formula style does not imply equal connectivity. Silica has crystalline polymorphs and amorphous forms, and its coordination can change at other pressures. A low-pressure tetrahedral network is a declared structural class rather than a description of every possible SiO₂ phase.

A topology calculation has restrictive premises

For the selected C₆₀ cage, take 60 vertices, degree three at every vertex, closed spherical topology and only pentagonal or hexagonal faces. Counting edge incidences gives 2E=3× 60, so E=90. Euler’s relation V−E+F=2 gives F=32. If P and H are pentagon and hexagon counts, P+H=32 and 5P+6H=180. Thus P=12 and H=20.

Those equations constrain this chosen graph class. They do not give measured bond lengths, cage geometry, electron localization, formation conditions or a general result for every carbon allotrope. A drawing’s graph and a substance’s actual structure remain distinct evidence levels.

Worked examples

Try it yourself

1. Must a Lewis pair acceptor have positive formal charge before accepting the pair?

   Show the answer and reasoning

   No.

   The selected BF₃ acceptor has formal charge zero at B in the simple ledger.

2. Count total valence electrons in BF₄⁻ and compare B formal charge with oxidation state using F=−1.

   Show the answer and reasoning

   32 electrons; B formal charge −1 in the selected Lewis drawing, oxidation state +3.

   3+4× 7+1=32. Four ordinary B–F bonds give B formal charge 3−4=−1, while x+4(−1)=−1 gives oxidation state +3.

3. In the stipulated periodic tetrahedral sharing model, 20 Si centres each contact 4 O atoms and each O is shared by 2 Si. How many distinct O sites are counted?

   Show the answer and reasoning

   40 O sites, giving Si:O=1:2.

   20× 4=80 Si–O incidences; dividing by two shares gives 40 O. This is periodic counting rather than an unpassivated finite cluster.

4. Does the C₆₀ topology calculation establish actual C–C bond lengths?

   Show the answer and reasoning

   No.

   Vertex, edge and face constraints do not supply measured geometry.

What to carry forward

• Describe pair acceptance separately from charge signs.
• Use a completed adduct’s connected bond ledger.
• Elemental composition does not fix connectivity or properties.
• Count shared network atoms with explicit boundary assumptions.
• State the premises of a cage-topology derivation.