Write the Lewis structure of the nitrite ion, NO2^-
The Correct Answer and Explanation is:
To write the Lewis structure of the nitrite ion (NO₂⁻), we need to follow a few steps:
Step 1: Count the Total Valence Electrons
Nitrogen (N) is in group 15 of the periodic table, so it has 5 valence electrons. Oxygen (O) is in group 16, so each oxygen atom has 6 valence electrons. The nitrite ion has a negative charge, meaning it has gained one extra electron. Therefore, the total number of valence electrons is:5 (from N)+2×6 (from two O)+1 (from the negative charge)=18 electrons.5 \, (\text{from N}) + 2 \times 6 \, (\text{from two O}) + 1 \, (\text{from the negative charge}) = 18 \, \text{electrons}.5(from N)+2×6(from two O)+1(from the negative charge)=18electrons.
Step 2: Determine the Central Atom
In this case, nitrogen (N) will be the central atom because it is less electronegative than oxygen and can form more bonds.
Step 3: Connect the Atoms with Single Bonds
We place single bonds between nitrogen and each oxygen atom. A single bond consists of two electrons, so for two bonds, we have already used 4 electrons.N−OandN−ON – O \quad \text{and} \quad N – ON−OandN−O
Step 4: Distribute the Remaining Electrons
After forming the single bonds, we have 18 total electrons, and we’ve used 4 for the bonds. This leaves us with 14 electrons to distribute. Place lone pairs on the oxygen atoms to satisfy their octet rule. Each oxygen atom needs 6 more electrons to complete its octet, so we place 3 lone pairs (6 electrons) on each oxygen atom.
Step 5: Check Nitrogen’s Octet
At this point, nitrogen only has 4 electrons in its valence shell (one from each bond). Since nitrogen can hold more than 8 electrons, we will form a double bond with one of the oxygen atoms. This will move two electrons from the lone pairs of that oxygen into the bonding region, creating a double bond.
Step 6: Final Structure
Now we have:
- A double bond between nitrogen and one oxygen.
- A single bond between nitrogen and the other oxygen.
- 2 lone pairs on the oxygen with the single bond.
- 3 lone pairs on the oxygen with the double bond.
- The negative charge is placed on the oxygen with the single bond.
The final structure of NO₂⁻ is:O=N−O−O = N – O^-O=N−O−
Step 7: Verify
- Each oxygen atom has a full octet of electrons.
- Nitrogen has an expanded octet (it can have up to 10 electrons in its valence shell).
- The total number of electrons used (18) matches the count from step 1.
- The negative charge is correctly placed on one of the oxygen atoms.
Thus, the Lewis structure for the nitrite ion is correct.
