The ethoxy radical (CH?CH?O?) is a highly reactive, short-lived reaction intermediate. Derived from ethanol, it contains an unpaired electron on the oxygen atom. The picture below shows its lewis structure.

Step 1: Calculate total valence electrons

C: 4 × 2 = 8; H: 1 × 5 = 5; O: 6

Total valence electrons: 8 + 5 + 6 = 19 e?

The total number of electrons is odd; it is a radical possessing one unpaired electron.

Step 2: Determine atomic connectivity framework

C1 (methyl carbon): bonded to 3 H atoms and C2

C2 (methylene carbon): bonded to C1, 2 H atoms, and O

The terminal O atom bears the unpaired electron; no other bonding exists (all bonds are single; no double bonds).

Step 3: Draw all σ covalent bonds

List of chemical bonds (all single bonds): C1–H (×3), C1–C2, C2–H (×2), C2–O

Total of 7 σ single bonds; bonding electrons: 7 × 2 = 14 e?

Remaining electrons: 19 - 14 = 5 e?

Step 4: Assign electrons to the oxygen atom

The O atom is bonded via one C–O single bond; fill in lone pairs first:

Oxygen needs to satisfy the octet rule as much as possible; fill in 2 lone pairs (4 e?)

Remaining electrons: 5 - 4 = 1 e?

Step 5: Place the unpaired electron

Only 1 electron remains to be placed on the oxygen atom; this is characteristic of the ethoxy radical.

Verify total electron count: Bonding electrons (14) + O lone pairs (4) + O unpaired electron (1) = 19 e?; matches the total valence electron count.