Match the species in Column 1 with the type of hybrid orbitals in Column II.
`{:("Column I", "Column II"),((a)SF_(4),(i)sp^(3)d^(2)),((b)lF_(5),(ii)d^(2)sp^(3)),((c)NO_(2)^(+),(iii)sp^(3)d),((d)NH_(4)^(+),(iv)sp^(3)),((v),sp):}`

Answer

To solve the problem of matching the species in Column I with the type of hybrid orbitals in Column II, we will follow a systematic approach using the formula for determining the hybridization based on the steric number (H). The formula is:

\[ H = \frac{1}{2} \left( \text{Number of valence electrons on central atom} + \text{Number of monovalent atoms} + \text{Charge} \right) \]

We will apply this formula to each species in Column I.

### Step 1: Analyze SF₄
1. **Identify the central atom**: Sulfur (S).
2. **Count the valence electrons**: Sulfur has 6 valence electrons.
3. **Count the monovalent atoms**: There are 4 fluorine (F) atoms.
4. **Consider the charge**: There is no charge (0).
5. **Calculate H**:
\[ H = \frac{1}{2} (6 + 4 + 0) = \frac{1}{2} (10) = 5 \]
6. **Determine hybridization**: A steric number of 5 corresponds to \( sp^3d \).

### Step 2: Analyze IF₅
1. **Identify the central atom**: Iodine (I).
2. **Count the valence electrons**: Iodine has 7 valence electrons.
3. **Count the monovalent atoms**: There are 5 fluorine (F) atoms.
4. **Consider the charge**: There is no charge (0).
5. **Calculate H**:
\[ H = \frac{1}{2} (7 + 5 + 0) = \frac{1}{2} (12) = 6 \]
6. **Determine hybridization**: A steric number of 6 corresponds to \( sp^3d^2 \).

### Step 3: Analyze NO₂⁺
1. **Identify the central atom**: Nitrogen (N).
2. **Count the valence electrons**: Nitrogen has 5 valence electrons.
3. **Count the monovalent atoms**: There are 2 oxygen (O) atoms.
4. **Consider the charge**: There is a positive charge (+1), so we subtract 1.
5. **Calculate H**:
\[ H = \frac{1}{2} (5 + 2 - 1) = \frac{1}{2} (6) = 3 \]
6. **Determine hybridization**: A steric number of 3 corresponds to \( sp \).

### Step 4: Analyze NH₄⁺
1. **Identify the central atom**: Nitrogen (N).
2. **Count the valence electrons**: Nitrogen has 5 valence electrons.
3. **Count the monovalent atoms**: There are 4 hydrogen (H) atoms.
4. **Consider the charge**: There is a positive charge (+1), so we subtract 1.
5. **Calculate H**:
\[ H = \frac{1}{2} (5 + 4 - 1) = \frac{1}{2} (8) = 4 \]
6. **Determine hybridization**: A steric number of 4 corresponds to \( sp^3 \).

### Final Matching:
- (a) SF₄ → (iii) \( sp^3d \)
- (b) IF₅ → (i) \( sp^3d^2 \)
- (c) NO₂⁺ → (v) \( sp \)
- (d) NH₄⁺ → (iv) \( sp^3 \)

### Summary of Matches:
- A matches with (iii)
- B matches with (i)
- C matches with (v)
- D matches with (iv)