e The final Lewis structure for the cyanate ion (OCN⁻) can be represented as: ``` .. O: C ≡ N: .. ``` Where: - Oxygen has 3 lone pairs and a single bond with carbon. - Carbon has a triple bond with nitrogen. - Nitrogen has 1 lone pair.2d:T7fa,To draw the Lewis structure for the ion OCN⁻ (cyanate ion), we can follow these steps:
### Step 1: Count the Valence Electrons
- Oxygen (O) has 6 valence electrons.
- Carbon (C) has 4 valence electrons.
- Nitrogen (N) has 5 valence electrons.
- Since the ion has a negative charge (-1), we add one additional electron.
**Total valence electrons = 6 (O) + 4 (C) + 5 (N) + 1 (negative charge) = 16 electrons.**
### Step 2: Arrange the Atoms
- The typical arrangement for OCN⁻ is to place carbon (C) in the center, with oxygen (O) and nitrogen (N) on either side. Thus, we have the arrangement: O - C - N.
### Step 3: Draw Single Bonds
- Connect the atoms with single bonds:
- O - C
- C - N
This uses 4 of the 16 total electrons (2 for each bond).
### Step 4: Distribute Remaining Electrons
- After using 4 electrons for the bonds, we have 12 electrons left (16 - 4 = 12).
- Start by placing lone pairs on the outer atoms (O and N) to satisfy their octets:
- Oxygen can have 6 electrons (3 lone pairs).
- Nitrogen can have 6 electrons (3 lone pairs).
Now we have used 12 electrons (6 for O and 6 for N).
### Step 5: Check Octets and Adjust
- At this stage, carbon only has 2 electrons (from the single bonds), so we need to form multiple bonds to satisfy the octet rule.
- Form a triple bond between carbon (C) and nitrogen (N) by moving one lone pair from nitrogen to form a bond with carbon:
- This results in a structure of O = C ≡ N.
### Step 6: Add the Negative Charge
- The negative charge is typically placed on the oxygen atom, which now has 7 electrons (6 from lone pairs and 1 from the bond with carbon).
### Final Lewis Structure
The final Lewis structure for th

Lewis -structure `OCN^(-)`

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