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 solve the problem, we need to determine the value of "x" in the hybridization formula \( sp^3d^{(x-1)} \) for the given species: \( SbCl_6^{-} \), \( SnCl_6^{2-} \), \( XeF_5^{+} \), and \( IO_6^{5-} \).

### Step-by-Step Solution:

1. **Identify the Hybridization of Each Species:**
- The species mentioned are known to have an octahedral geometry. In octahedral geometry, the central atom is surrounded by six bonding pairs of electrons.

2. **Understanding the Hybridization Types:**
- The two common types of hybridization for octahedral structures are:
- \( d^2sp^3 \)
- \( sp^3d^2 \)
- The hybridization \( sp^3d^{(x-1)} \) suggests that there are 4 \( sp^3 \) hybridized orbitals and \( (x-1) \) \( d \) orbitals involved in bonding.

3. **Determine the Number of d Orbitals Used:**
- For octahedral geometry, we typically need 6 orbitals for bonding (which can be a combination of s, p, and d orbitals).
- In the case of \( sp^3d^{(x-1)} \), we can equate the total number of orbitals:
\[ 4 (from \, sp^3) + (x-1) = 6 \]
- This simplifies to:
\[ x - 1 = 2 \]

4. **Solve for x:**
- Adding 1 to both sides gives:
\[ x = 3 \]

5. **Conclusion:**
- The value of \( x \) is 3.

### Final Answer:
The value of "x" is \( 3 \).