Question

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add lone pairs to compound b.
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/ || ||| c h n o
b.
h₃c c nh₂
h c c n h
h h h

Explanation:

Step1: Identify O's valence electrons

Oxygen has 6 valence electrons.

Step2: Subtract bonded electrons

O forms a double bond (4 electrons) to C: $6 - 4 = 2$ electrons left, so 1 lone pair (already shown, but confirm: $2$ electrons = $1$ lone pair, total lone pairs for O: $2$? Correction: 6 valence e⁻, 4 used in double bond, $6-4=2$ pairs (4 electrons total as lone pairs).

Step3: Identify N's valence electrons

Nitrogen has 5 valence electrons.

Step4: Subtract bonded electrons

N forms 3 single bonds (3 electrons) to C and 2 H: $5 - 3 = 2$ electrons left, so 1 lone pair.

Answer:

The completed structure has:

• Oxygen atom with 2 lone pairs (4 total non-bonding electrons)
• Nitrogen atom with 1 lone pair (2 total non-bonding electrons)

Final structure:

     ..
O = C - N:
     |     |
H3C   H2

(Or in the displayed structure: O has two sets of double dots, N has one set of double dots, matching the skeletal structure provided with all lone pairs added correctly.)