Step 1: Understanding the Concept:
Hybridization of a central atom is determined by the steric number (SN), which is the sum of the number of sigma bonds and lone pairs on the central atom.
According to VSEPR theory and hybridization rules:
- $SN = 4 \rightarrow sp^3$
- $SN = 5 \rightarrow sp^3d$
- $SN = 6 \rightarrow sp^3d^2$
- $SN = 7 \rightarrow sp^3d^3$
Step 2: Key Formula or Approach:
The steric number is calculated using: $SN = \frac{1}{2} [V + M - C + A]$
Where:
- $V = $ Valence electrons of the central atom.
- $M = $ Number of monovalent atoms (H, F, Cl, etc.).
- $C = $ Positive charge (cationic).
- $A = $ Negative charge (anionic).
Step 3: Detailed Explanation:
Let's evaluate each species:
- (A) $XeF_6$:
$V = 8$ (Xenon), $M = 6$ (Fluorines).
$SN = \frac{1}{2} [8 + 6] = 7$.
A steric number of 7 corresponds to $sp^3d^3$ hybridization. The geometry is distorted octahedral or pentagonal bipyramidal.
- (B) $BrF_5^+$:
$V = 7$ (Bromine), $M = 5$, $C = +1$.
$SN = \frac{1}{2} [7 + 5 - 1] = 6$.
$SN = 6$ corresponds to $sp^3d^2$ hybridization.
- (C) $IF_5$:
$V = 7$ (Iodine), $M = 5$.
$SN = \frac{1}{2} [7 + 5] = 6$.
$SN = 6$ corresponds to $sp^3d^2$ hybridization.
- (D) $XeF_4$:
$V = 8$ (Xenon), $M = 4$.
$SN = \frac{1}{2} [8 + 4] = 6$.
$SN = 6$ corresponds to $sp^3d^2$ hybridization.
Step 4: Final Answer:
The only species that does not have $sp^3d^2$ hybridization is $XeF_6$ (it has $sp^3d^3$). Thus, the correct option is (A).