Question

question 11 of 49
draw the lewis structure of h₂so₄ with minimized formal charges. include lone pairs.
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s o h

Explanation:

1. First, calculate total valence electrons: S has 6, each O has 6 (4 O = 24), each H has 1 (2 H = 2), total = 6+24+2=32 valence electrons.
2. Place S as the central atom. Attach two -OH groups (single bonds from S to 2 O atoms, each of those O atoms bonded to one H) and two double-bonded O atoms directly to S.
3. Assign lone pairs: Each double-bonded O has 2 lone pairs (4 electrons total), each O in the -OH groups has 2 lone pairs (4 electrons total).
4. Verify formal charges:

• Formal charge formula: $FC = V - N - \frac{B}{2}$, where V=valence electrons, N=non-bonding electrons, B=bonding electrons
• For central S: $FC = 6 - 0 - \frac{12}{2} = 0$
• For double-bonded O: $FC = 6 - 4 - \frac{4}{2} = 0$
• For O in -OH: $FC = 6 - 4 - \frac{4}{2} = 0$
• For H: $FC = 1 - 0 - \frac{2}{2} = 0$

All formal charges are minimized to 0, and total electrons used match the 32 valence electrons.

Answer:

The Lewis structure has a central sulfur atom bonded to two oxygen atoms via double bonds (each double-bonded O has 2 lone electron pairs) and two oxygen atoms via single bonds. Each of the single-bonded oxygen atoms is also bonded to one hydrogen atom, and each of these single-bonded O atoms has 2 lone electron pairs. All atoms have a formal charge of 0.

Structural summary (text representation):

      O
      ||
H-O-S-O-H
      ||
      O

(With 2 lone pairs on each oxygen atom)