Chemistry · Ch 5 — Chemical Bonding

Formal charge

5.2.6

Formal charge

When a polyatomic species can be drawn as more than one valid Lewis structure — differing only in exactly where its double or triple bond(s) sit, never in which atoms are bonded to which — formal charge is the tool used to decide which of those candidate structures best represents the real molecule. Formal charge on a given atom, within one particular Lewis structure, is defined as $FC = V_E - N_E - \tfrac{1}{2}B_E$ , where $V_E$ is that atom's number of valence electrons as a free (unbonded) atom, $N_E$ is the number of non-bonding (lone-pair) electrons it carries in that structure, and $B_E$ is the number of bonding (shared) electrons around it in that structure (so $B_E/2$ counts its bonds). Formal charge assumes every bonding pair is shared PERFECTLY equally between the two atoms, ignoring their real electronegativity difference — it is a bookkeeping tool, not a claim about the true, physically-realistic distribution of charge. Among several candidate Lewis structures for the same species, the preferred (lowest-energy, most representative) structure is the one whose formal charges are as close to zero as possible on every atom; …

Figure 5.2.6-aFormal charges in the ozone (O3) molecule

What this figure shows. For one Lewis structure of O3 with the three oxygens numbered 1 (central), 2 and 3, and a central O=O(2) double bond plus a central O–O(3) single bond: formal charge on central O(1) $=6-2-\tfrac12(6)=+1$ ; formal charge on the doubly-bonded terminal O(2) $=6-4-\tfrac12(4)=0$ ; formal charge on the singly-bonded terminal O(3) $=6-6-\tfrac12(2)=-1$ . So this canonical form of O3 is written with a +1 formal charge on the central O and a −1 formal charge on one terminal O, net charge zero overall, consistent with O3 being …

Misc Problem 5.2Lewis structure and formal charges for the CO molecule

Worked out. Worked example. Step I: valence electrons = 4 (C, $2s^22p^2$ ) + 6 (O, $2s^22p^4$ ) = 10. Step II: skeletal structure C–O. Step III: a single shared pair completes O's octet but leaves 2 electrons as a lone pair on C with C's octet still short; converting to a triple bond, C≡O, satisfies the octet on both atoms. Formal charges are then compared across three candidate structures: Structure A, C≡O with one lone pair on each atom (16 electrons total: 4 from C, 6 from each O) — formal charge on C $=4-0-\tfrac12(8)=0$ and on O $=6-4-\tfrac12(4)=0$ , i.e. zero on both atoms. Structure B (a double bond C=O with an extra lone pair rearranged onto one O and a positive/negative split) gives formal charges of 0 on C, +1 on one O and −1 on the other. Structure C (the mirror image of B) likewise gives 0 on C and ±1 split oppositely between the two O positions. Since Structure A has zero formal charge on every atom (the lowest overall formal-charge structure), it is judged the correct, lowes …

Figure 5.2.6-bFormal charges on carbon dioxide's three candidate structures

What this figure shows. CO2 can be drawn as three different Lewis structures depending on where the double/triple-bond character sits: Structure A, O=C=O (two ordinary C=O double bonds, each oxygen also carrying two lone pairs) — every atom's formal charge works out to zero, matching the double-bond structure normally taught. Structures B and C instead place a triple bond on one side and a single bond on the other (O≡C–O and O–C≡O), which forces one oxygen to carry a −1 formal charge and the other a +1 formal charge. Since the all-double-bond Structure A has zero formal charge everywhere while B and C both carry a +1/−1 split, Structure A (O=C=O) is the preferred, lowest-energy Lewis structure — matching C …

Misc Problem 5.3Formal charges on S, C, N across three candidate structures

Worked out. Worked example for the linear thiocyanate-type skeleton S–C–N, comparing three ways of placing double/triple bonds: Structure A, S=C=N (two double bonds) — formal charge on S $=6-4-\tfrac12(4)=0$ , on C $=4-0-\tfrac12(8)=0$ , on N $=5-4-\tfrac12(4)=-1$ . Structure B, S–C≡N (single bond to S, triple bond to N) — formal charge on S $=6-6-\tfrac12(2)=-1$ , on C $=0$ , on N $=5-2-\tfrac12(6)=0$ . Structure C, S≡C–N (triple bond to S, single bond to N) — formal charge on S $=6-2-\tfrac12(6)=+1$ , on C $=0$ , on N $=5-6-\tfrac12(2)=-2$ . Comparing the three, Structure A (S=C=N) carries the smallest formal charges overall (0, 0, −1) and is therefore judged the lowest-energy, most representative structure; C (S≡ …

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