Which of the two lons from the list given have the geometry that is explained by the same hybridization of orbitals `NO_(2)^(-),NO_(3)^(-) ,NH_(2)^(-) NH_(4)^(+) SCN^(-)`?

Options

A. `NO_(2)^(-) and NH_(2)^(-)`
B. `NO_(2)^(-) and NO_(3)^(-)`
C. `NH_(4)^(+) and NO_(3)^(-)`
D. `SCN^(-)` and `NH_(2)`

Explanation

To determine which of the two ions from the given list have the same hybridization of orbitals, we will analyze each ion one by one using the formula for calculating the hybridization based on the electron pair count (Z):

\[ Z = \frac{1}{2} \left( \text{Valence Electrons of Central Atom} + \text{Monovalent Atoms} - \text{Cationic Charge} + \text{Anionic Charge} \right) \]

1. **For \( NO_2^- \)**:
- Valence electrons of nitrogen = 5
- Monovalent atoms = 0 (no hydrogen or similar atoms)
- Cationic charge = 0
- Anionic charge = 1 (since it is \( NO_2^- \))
- Calculation:
\[ Z = \frac{1}{2} (5 + 0 - 0 + 1) = \frac{1}{2} \times 6 = 3 \]
- Hybridization: \( sp^2 \)

2. **For \( NO_3^- \)**:
- Valence electrons of nitrogen = 5
- Monovalent atoms = 0
- Cationic charge = 0
- Anionic charge = 1 (since it is \( NO_3^- \))
- Calculation:
\[ Z = \frac{1}{2} (5 + 0 - 0 + 1) = \frac{1}{2} \times 6 = 3 \]
- Hybridization: \( sp^2 \)

3. **For \( NH_2^- \)**:
- Valence electrons of nitrogen = 5
- Monovalent atoms = 2 (2 hydrogen atoms)
- Cationic charge = 0
- Anionic charge = 1 (since it is \( NH_2^- \))
- Calculation:
\[ Z = \frac{1}{2} (5 + 2 - 0 + 1) = \frac{1}{2} \times 8 = 4 \]
- Hybridization: \( sp^3 \)

4. **For \( NH_4^+ \)**:
- Valence electrons of nitrogen = 5
- Monovalent atoms = 4 (4 hydrogen atoms)
- Cationic charge = 1 (since it is \( NH_4^+ \))
- Anionic charge = 0
- Calculation:
\[ Z = \frac{1}{2} (5 + 4 - 1 + 0) = \frac{1}{2} \times 8 = 4 \]
- Hybridization: \( sp^3 \)

5. **For \( SCN^- \)**:
- Valence electrons of sulfur = 6
- Monovalent atoms = 1 (1 carbon atom)
- Cationic charge = 0
- Anionic charge = 1 (since it is \( SCN^- \))
- Calculation:
\[ Z = \frac{1}{2} (6 + 1 - 0 + 1) = \frac{1}{2} \times 8 = 4 \]
- Hybridization: \( sp^3 \)

### Summary of Hybridizations:
- \( NO_2^- \): \( sp^2 \)
- \( NO_3^- \): \( sp^2 \)
- \( NH_2^- \): \( sp^3 \)
- \( NH_4^+ \): \( sp^3 \)
- \( SCN^- \): \( sp^3 \)

### Conclusion:
The ions that have the same hybridization are:
1. \( NO_2^- \) and \( NO_3^- \) (both \( sp^2 \))
2. \( NH_2^- \), \( NH_4^+ \), and \( SCN^- \) (all \( sp^3 \))

### Final Answer:
The two ions with the same hybridization are \( NO_2^- \) and \( NO_3^- \).