Question

the skeleton of the bicarbonate ion, hco₃⁻, is shown here. draw the complete structure. add nonbonding electrons and formal charges where appropriate.

Explanation:

Step1: Calculate valence electrons

Total valence electrons: $4(\text{C}) + 3\times6(\text{O}) + 1(\text{H}) + 1(\text{charge}) = 4+18+1+1=24$

Step2: Assign bonding electrons

Skeleton has 4 bonds (C=O, C-O, C-OH, C-H), using $4\times2=8$ electrons. Remaining electrons: $24-8=16$

Step3: Add lone pairs to O atoms

• Double-bonded O: 2 lone pairs ($4$ electrons)
• Single-bonded O (not bonded to H): 3 lone pairs ($6$ electrons), formal charge: $6 - (6 + 1) = -1$
• O in OH: 2 lone pairs ($4$ electrons), H bonded to this O
• Remaining 2 electrons: none, total used: $8+4+6+4+2=24$

Step4: Verify formal charges

C: $4 - (4 + 0) = 0$; H: $1 - (1 + 0) = 0$; Double-bonded O: $6 - (4 + 2) = 0$; OH O: $6 - (4 + 2) = 0$; Single-bonded O: $6 - (6 + 1) = -1$; Total charge: $0+0+0+0-1=-1$ (matches ion charge)

Answer:

The correct Lewis structure is:

      ..
O = C - O⁻
     |  ..
    O-H
    ..

(Or with spatial arrangement: Carbon is central, double-bonded to one O (with 2 lone pairs), single-bonded to one O (with 3 lone pairs and a -1 formal charge), single-bonded to an -OH group (the O has 2 lone pairs, bonded to H). All electrons and charges balance to match the $\text{HCO}_3^-$ ion.)