To determine the hybridization of the nitrogen atom in the ions \( NO_3^- \), \( NO_2^+ \), and \( NH_4^+ \), we will follow these steps:
### Step 1: Determine the Valence Electrons for \( NO_3^- \)
- Nitrogen (N) has 5 valence electrons.
- Each oxygen (O) contributes 6 valence electrons, but since we are considering the overall charge, we will not add them directly.
- The negative charge on \( NO_3^- \) adds 1 additional electron.
- Total count:
\[
5 \, (\text{from N}) + 0 \, (\text{from O}) + 1 \, (\text{from charge}) = 6 \, \text{electrons}
\]
### Step 2: Calculate the Hybridization for \( NO_3^- \)
- Divide the total number of electrons by 2 to find the number of orbitals:
\[
\frac{6}{2} = 3 \, \text{orbitals}
\]
- Since we have 3 orbitals, the hybridization is \( sp^2 \).
### Step 3: Determine the Valence Electrons for \( NO_2^+ \)
- Again, nitrogen has 5 valence electrons.
- Each oxygen contributes 6, but we will not add them directly.
- The positive charge on \( NO_2^+ \) removes 1 electron.
- Total count:
\[
5 \, (\text{from N}) + 0 \, (\text{from O}) - 1 \, (\text{from charge}) = 4 \, \text{electrons}
\]
### Step 4: Calculate the Hybridization for \( NO_2^+ \)
- Divide the total number of electrons by 2:
\[
\frac{4}{2} = 2 \, \text{orbitals}
\]
- Since we have 2 orbitals, the hybridization is \( sp \).
### Step 5: Determine the Valence Electrons for \( NH_4^+ \)
- Nitrogen has 5 valence electrons.
- Each hydrogen (H) contributes 1 valence electron, and there are 4 hydrogens.
- The positive charge on \( NH_4^+ \) removes 1 electron.
- Total count:
\[
5 \, (\text{from N}) + 4 \, (\text{from H}) - 1 \, (\text{from charge}) = 8 \, \text{electrons}
\]
### Step 6: Calculate the Hybridization for \( NH_4^+ \)
- Divide the total number of electrons by 2:
\[
\frac{8}{2} = 4 \, \text{orbitals}
\]
- Since we have 4 orbitals, the hybridization is \( sp^3 \).
### Final Answer
The hybridization of the nitrogen atom in \( NO_3^- \), \( NO_2^+ \), and \( NH_4^+ \) are respectively:
- \( NO_3^- \): \( sp^2 \)
- \( NO_2^+ \): \( sp \)
- \( NH_4^+ \): \( sp^3 \)
