If `AB_(4)^(n)` types species are tetrahedral, then which of the following is /are correctly match ?

Options

A. `{:(A,B,n),(Xe,O,0):}`
B. `{:(A,B,n),(Se,F,0):}`
C. `{:(A,B,n),(P,O,-3):}`
D. `{:(A,B,n),(N,H,+1):}`

Answer

To determine which of the given species of the form AB₄ⁿ are tetrahedral, we will calculate the steric number using the formula:

**Steric Number (SN) = 1/2 [V + M + (Charge)]**

Where:
- V = number of valence electrons of the central atom (A)
- M = number of monovalent atoms (B) attached to the central atom
- Charge = the overall charge of the species (subtract if it's a cation)

### Step-by-Step Solution:

1. **Identify the first species: XeO₄**
- A = Xe (Xenon)
- B = O (Oxygen)
- Number of O atoms (M) = 4
- Charge = 0 (neutral species)
- Valence electrons of Xe = 8
- Calculation:
\[ SN = \frac{1}{2} [8 + 4 + 0] = \frac{1}{2} [12] = 6 \]
- Hybridization: sp³ (since SN = 4, it is tetrahedral)

2. **Identify the second species: SF₄**
- A = S (Sulfur)
- B = F (Fluorine)
- Number of F atoms (M) = 4
- Charge = 0 (neutral species)
- Valence electrons of S = 6
- Calculation:
\[ SN = \frac{1}{2} [6 + 4 + 0] = \frac{1}{2} [10] = 5 \]
- Hybridization: sp³d (since SN = 5, it is trigonal bipyramidal, not tetrahedral)

3. **Identify the third species: PO₄³⁻**
- A = P (Phosphorus)
- B = O (Oxygen)
- Number of O atoms (M) = 4
- Charge = -3
- Valence electrons of P = 5
- Calculation:
\[ SN = \frac{1}{2} [5 + 4 - 3] = \frac{1}{2} [6] = 3 \]
- Hybridization: sp³ (since SN = 4, it is tetrahedral)

4. **Identify the fourth species: NH₄⁺**
- A = N (Nitrogen)
- B = H (Hydrogen)
- Number of H atoms (M) = 4
- Charge = +1
- Valence electrons of N = 5
- Calculation:
\[ SN = \frac{1}{2} [5 + 4 - 1] = \frac{1}{2} [8] = 4 \]
- Hybridization: sp³ (since SN = 4, it is tetrahedral)

### Conclusion:
The species that are tetrahedral are:
- XeO₄ (1)
- PO₄³⁻ (3)
- NH₄⁺ (4)

Thus, the correctly matched options are **1, 3, and 4**.