Species like `SbCl_(6)^(-), SnCl_(6)^(2-), XeF_(5)^(+) and IO_(6)^(5-)` has hybridization as `sp^(3)d^(x - 1)`. The value of "x" is

Answer

To determine the value of "x" in the hybridization formula \( sp^3d^{(x-1)} \) for the species \( SbCl_6^-, SnCl_6^{2-}, XeF_5^+, \) and \( IO_6^{5-} \), we will calculate the hybridization for each species step by step.

### Step 1: Determine the hybridization of \( SbCl_6^- \)

1. **Identify the central atom and its valence electrons**:
- Antimony (Sb) is in Group 15, so it has 5 valence electrons.

2. **Count the number of single bonded atoms**:
- There are 6 chlorine (Cl) atoms bonded to Sb.

3. **Account for the negative charge**:
- The negative charge (+1) means we add 1 electron.

4. **Apply the hybridization formula**:
\[ \text{Hybridization} = \frac{1}{2} \left( \text{Number of valence electrons} + \text{Number of single bonded atoms} + \text{Charge} \right) \]
\[ = \frac{1}{2} \left( 5 + 6 + 1 \right) = \frac{12}{2} = 6 \]
- Thus, the hybridization of \( SbCl_6^- \) is \( sp^3d^2 \).

### Step 2: Determine the hybridization of \( SnCl_6^{2-} \)

1. **Identify the central atom and its valence electrons**:
- Tin (Sn) is in Group 14, so it has 4 valence electrons.

2. **Count the number of single bonded atoms**:
- There are 6 chlorine (Cl) atoms bonded to Sn.

3. **Account for the negative charge**:
- The charge is -2, so we add 2 electrons.

4. **Apply the hybridization formula**:
\[ \text{Hybridization} = \frac{1}{2} \left( 4 + 6 + 2 \right) = \frac{12}{2} = 6 \]
- Thus, the hybridization of \( SnCl_6^{2-} \) is \( sp^3d^2 \).

### Step 3: Determine the hybridization of \( XeF_5^+ \)

1. **Identify the central atom and its valence electrons**:
- Xenon (Xe) is in Group 18, so it has 8 valence electrons.

2. **Count the number of single bonded atoms**:
- There are 5 fluorine (F) atoms bonded to Xe.

3. **Account for the positive charge**:
- The charge is +1, so we subtract 1 electron.

4. **Apply the hybridization formula**:
\[ \text{Hybridization} = \frac{1}{2} \left( 8 + 5 - 1 \right) = \frac{12}{2} = 6 \]
- Thus, the hybridization of \( XeF_5^+ \) is \( sp^3d^2 \).

### Step 4: Determine the hybridization of \( IO_6^{5-} \)

1. **Identify the central atom and its valence electrons**:
- Iodine (I) is in Group 17, so it has 7 valence electrons.

2. **Count the number of single bonded atoms**:
- There are 6 oxygen (O) atoms bonded to I. Note that oxygen typically forms double bonds, but in this case, we will consider them as single bonds for the purpose of this calculation.

3. **Account for the negative charge**:
- The charge is -5, so we add 5 electrons.

4. **Apply the hybridization formula**:
\[ \text{Hybridization} = \frac{1}{2} \left( 7 + 6 + 5 \right) = \frac{18}{2} = 9 \]
- Thus, the hybridization of \( IO_6^{5-} \) is \( sp^3d^3 \).

### Conclusion

From the calculations, we found that the hybridization for all species except \( IO_6^{5-} \) is \( sp^3d^2 \), which corresponds to 6 hybrid orbitals.

Now, since the hybridization is given as \( sp^3d^{(x-1)} \), we can equate:
\[ x - 1 = 2 \implies x = 3 \]

### Final Answer
The value of "x" is **3**.