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

Answer

To determine which two ions from the given list have the same hybridization of orbitals, we will calculate the hybridization for each ion using the formula for steric number:

\[ \text{Steric Number} = \frac{(V + M + \text{Charge})}{2} \]

Where:
- \(V\) = number of valence electrons of the central atom
- \(M\) = number of monovalent atoms attached to the central atom
- Charge = the charge of the ion (add for anions, subtract for cations)

Let's analyze each ion step by step.

### Step 1: Calculate Hybridization for \(NO_2^-\)

1. **Identify the central atom**: Nitrogen (N).
2. **Valence electrons of N**: 5.
3. **Monovalent atoms**: 0 (Oxygen is divalent).
4. **Charge**: +1 (since it's an anion).
5. **Calculation**:
\[ \text{Steric Number} = \frac{(5 + 0 + 1)}{2} = \frac{6}{2} = 3 \]
- **Hybridization**: \(sp^2\).

### Step 2: Calculate Hybridization for \(NO_3^-\)

1. **Central atom**: Nitrogen (N).
2. **Valence electrons of N**: 5.
3. **Monovalent atoms**: 0.
4. **Charge**: +1.
5. **Calculation**:
\[ \text{Steric Number} = \frac{(5 + 0 + 1)}{2} = \frac{6}{2} = 3 \]
- **Hybridization**: \(sp^2\).

### Step 3: Calculate Hybridization for \(NH_2^-\)

1. **Central atom**: Nitrogen (N).
2. **Valence electrons of N**: 5.
3. **Monovalent atoms**: 2 (2 Hydrogens).
4. **Charge**: +1.
5. **Calculation**:
\[ \text{Steric Number} = \frac{(5 + 2 + 1)}{2} = \frac{8}{2} = 4 \]
- **Hybridization**: \(sp^3\).

### Step 4: Calculate Hybridization for \(NH_4^+\)

1. **Central atom**: Nitrogen (N).
2. **Valence electrons of N**: 5.
3. **Monovalent atoms**: 4 (4 Hydrogens).
4. **Charge**: -1.
5. **Calculation**:
\[ \text{Steric Number} = \frac{(5 + 4 - 1)}{2} = \frac{8}{2} = 4 \]
- **Hybridization**: \(sp^3\).

### Step 5: Calculate Hybridization for \(SCN^-\)

1. **Central atom**: Carbon (C).
2. **Valence electrons of C**: 4.
3. **Monovalent atoms**: 1 (1 Sulfur).
4. **Charge**: +1.
5. **Calculation**:
\[ \text{Steric Number} = \frac{(4 + 1 + 1)}{2} = \frac{6}{2} = 3 \]
- **Hybridization**: \(sp^2\).

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

### Conclusion:
The ions \(NO_2^-\) and \(NO_3^-\) both have \(sp^2\) hybridization, while \(NH_2^-\) and \(NH_4^+\) both have \(sp^3\) hybridization.

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

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