Exercises — Formal charge calculation — best resonance structure

WHAT we have: $V=5$ (N is group 15), $L=2$ (one lone pair $=2$ electrons), $B=6$ (three single bonds $=3\times 2$). WHY the numbers: each single bond contributes 2 bonding electrons; the atom owns half of those. ${\text{FC}}_N=5-2-\frac{1}{2}(6)=5-2-3=0$ Answer: $0$. Neutral — exactly what we expect for N in ammonia.

$V=6$, $L=6$ (three lone pairs), $B=2$ (one single bond). ${\text{FC}}_O=6-6-\frac{1}{2}(2)=6-6-1=-1$ Answer: $-1$. This lone $-1$ is the whole charge of ${\text{OH}}^{-}$ — checksum holds since H is $0$.

$V=4$, $L=0$, $B=8$ (four single bonds). ${\text{FC}}_C=4-0-\frac{1}{2}(8)=4-0-4=0$ Answer: $0$.

$V=5$, $L=0$, $B=8$. ${\text{FC}}_N=5-0-\frac{1}{2}(8)=5-0-4=+1$ Answer: $+1$. Checksum: $+1$ (N) $+4\times 0$ (H) $=+1=$ ion charge. ✔

$V=6$, $L=2$, $B=6$. ${\text{FC}}_O=6-2-\frac{1}{2}(6)=6-2-3=+1$ Answer: $+1$. Checksum: $+1$ matches the ion charge. Oxygen went positive by giving up an "owned" electron to make an extra bond — see the picture below.

Central O: $V=6$, $L=2$, bonds $=$ single(2)+double(4) $\Rightarrow B=6$. $6-2-\frac{1}{2}(6)=6-2-3=+1$ Double-terminal O: $V=6$, $L=4$, $B=4$. $6-4-\frac{1}{2}(4)=6-4-2=0$ Single-terminal O: $V=6$, $L=6$, $B=2$. $6-6-\frac{1}{2}(2)=6-6-1=-1$ Answers: $+1,0,-1$. Checksum: $+1+0-1=0=$ neutral ${\text{O}}_3$. ✔ This is why ozone is drawn with Resonance Structures — the $+1/-1$ swap between the two ends.

Structure A — from the parent note: C $=0$, each O $=0$. Sum $=0$. ✔ Structure B:

• Triple-bonded O: $V=6$, $L=2$, $B=6$ → $6-2-3=+1$.
• Single-bonded O: $V=6$, $L=6$, $B=2$ → $6-6-1=-1$.
• C: $V=4$, $L=0$, $B=8$ (triple 6 + single 2) → $4-0-4=0$. Sum $=+1-1+0=0$. ✔ (checksum passes for both). Ranking: Rule 1 (most zeros) — A has three zeros, B has only one zero plus a $+1$ and $-1$. Structure A is better.

Form P:

• O: $6-4-\frac{1}{2}(4)=6-4-2=0$
• C: $4-0-\frac{1}{2}(8)=0$
• N: $5-4-\frac{1}{2}(4)=5-4-2=-1$ Sum $=-1$ ✔; charge on N. Form Q:
• O: $6-6-\frac{1}{2}(2)=6-6-1=-1$
• C: $4-0-\frac{1}{2}(8)=0$
• N: $5-2-\frac{1}{2}(6)=5-2-3=0$ Sum $=-1$ ✔; charge on O. Both have a single $-1$ and pass checksum → tie on Rules 1 & 2. Rule 3 decides: negative goes on the more electronegative atom. O ($3.4$) > N ($3.0$), so Form Q (charge on O) dominates. (Contrast with SCN⁻, where N beat S — the electronegative atom always wins the $-1$.)

N: $V=5$, $L=0$, $B=$ double(4)+single(2)+single(2) $=8$. $5-0-\frac{1}{2}(8)=5-0-4=+1$ Double-bonded O: $V=6$, $L=4$, $B=4$ → $6-4-2=0$. Each single-bonded O: $V=6$, $L=6$, $B=2$ → $6-6-1=-1$ (there are two of these). Checksum: $(+1)+0+(-1)+(-1)=-1=$ ion charge. ✔ Why equivalent forms: the double bond can be placed on any of the three O's; by symmetry each gives the same FC pattern ($+1$ on N, one O at $0$, two O at $-1$). Since all three are identical in energy, the real ion is an equal blend — the $-1$ is spread over all three O's ($-\frac{1}{3}$ each in reality, but FC only tracks the bookkeeping).

N: $V=5$, $L=0$, $B=8$ (two doubles). $5-0-\frac{1}{2}(8)=5-0-4=+1$ Each O: $V=6$, $L=4$, $B=4$ → $6-4-2=0$. Checksum: $(+1)+0+0=+1=$ ion charge. ✔ The $+1$ sits on N, the least electronegative atom present — consistent with Rule 4 (positive on less electronegative). A good, low-magnitude structure.

All-single structure

• S: $V=6$, $L=0$, $B=8$ (four singles) → $6-0-4=+2$.
• Each O: $V=6$, $L=6$, $B=2$ → $6-6-1=-1$ (four of them). Checksum: $(+2)+4(-1)=+2-4=-2$. ✔ Two-double structure
• S: $V=6$, $L=0$, $B=$ two doubles(8) + two singles(4) $=12$ → $6-0-6=0$.
• Each double-O: $V=6$, $L=4$, $B=4$ → $6-4-2=0$ (two of them).
• Each single-O: $V=6$, $L=6$, $B=2$ → $6-6-1=-1$ (two of them). Checksum: $0+2(0)+2(-1)=-2$. ✔ Which is better? Rule 1 (most zeros): the two-double structure has three zeros (S + two O) versus one in the all-single. Rule 2 (small magnitudes): two-double caps everything at $\pm 1$ while all-single puts a $+2$ on S. The two-double (expanded-octet) structure is the better contributor. This is why S is drawn with more than 8 electrons — the payoff is lower formal charges.

Terminal C: $V=4$, $L=0$, $B=$ two singles(4) + one double(4) $=8$ → $4-0-4=0$. Central N: $V=5$, $L=0$, $B=8$ (two doubles) → $5-0-4=+1$. Terminal N: $V=5$, $L=4$, $B=4$ (one double) → $5-4-2=-1$. Checksum (H's are $0$): $0+(+1)+(-1)=0$ ✔ (neutral fragment). Reading it: the $+1$ on inner N and $-1$ on terminal N with C neutral is the classic diazo pattern — small magnitudes, adjacent opposite charges. A textbook example of formal charge revealing real molecular character. Compare with Oxidation Number, which would assign charges by full electronegativity transfer instead of equal sharing.

• The three inputs to $\text{FC}=V-L-\frac{1}{2}B$ are ::: valence electrons $V$, lone-pair electrons $L$, bonding electrons $B$.
• $\text{FC}$ of N in ${\text{NH}}_4^{+}$ ::: $+1$.
• $\text{FC}$ of O in ${\text{H}}_3{\text{O}}^{+}$ ::: $+1$.
• Best sulfate structure caps charges at ::: $\pm 1$ (via expanded octet on S, giving three zeros).
• When two forms tie on Rules 1–2, the decider is ::: electronegativity — negative on the more electronegative atom.