Question

In which of the following species, \(\text{sp}^{3}\text{d}^{2}\) hybridisation is not associated?

• A. $\text{XeF}_6$
• B. $\text{BrF}_5^+$
• C. $\text{IF}_5$
• D. $\text{XeF}_4$

Hint:

To quickly determine hybridization, count the number of lone pairs and sigma bonds on the central atom:
- $\text{XeF}_6$ has 6 sigma bonds and 1 lone pair, total = 7 $\rightarrow \text{sp}^3\text{d}^3$.
- $\text{XeF}_4$ has 4 sigma bonds and 2 lone pairs, total = 6 $\rightarrow \text{sp}^3\text{d}^2$.
- $\text{IF}_5$ has 5 sigma bonds and 1 lone pair, total = 6 $\rightarrow \text{sp}^3\text{d}^2$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to determine which of the given chemical species does not have a central atom with $\text{sp}^3\text{d}^2$ hybridization.


Step 2: Key Formula or Approach:
The steric number ($\text{S.N.}$) of the central atom is calculated using the formula:
\[ \text{S.N.} = \frac{1}{2} [V + M - C + A] \]
Where:
- $V = \text{number of valence electrons of the central atom}$
- $M = \text{number of monovalent atoms surrounding it}$
- $C = \text{cationic charge}$
- $A = \text{anionic charge}$


Step 3: Detailed Explanation:
Let us calculate the steric number and hybridization for each given species:
- (A) $\text{XeF}_6$:
Central atom is Xenon ($\text{Xe}$), which has 8 valence electrons ($V = 8$).
Number of monovalent fluorine atoms = 6 ($M = 6$).
\[ \text{S.N.} = \frac{1}{2} [8 + 6] = 7 \]
A steric number of 7 corresponds to $\text{sp}^3\text{d}^3$ hybridization (distorted octahedral geometry).
- (B) $\text{BrF}_5^+$:
Central atom is Bromine ($\text{Br}$), which has 7 valence electrons ($V = 7$).
Monovalent fluorine atoms = 5 ($M = 5$).
Cationic charge = +1 ($C = 1$).
\[ \text{S.N.} = \frac{1}{2} [7 + 5 - 1] = 6 \]
A steric number of 6 corresponds to $\text{sp}^3\text{d}^2$ hybridization.
- (C) $\text{IF}_5$:
Central atom is Iodine ($\text{I}$), which has 7 valence electrons ($V = 7$).
Monovalent fluorine atoms = 5 ($M = 5$).
\[ \text{S.N.} = \frac{1}{2} [7 + 5] = 6 \]
A steric number of 6 corresponds to $\text{sp}^3\text{d}^2$ hybridization.
- (D) $\text{XeF}_4$:
Central atom is Xenon ($\text{Xe}$), which has 8 valence electrons ($V = 8$).
Monovalent fluorine atoms = 4 ($M = 4$).
\[ \text{S.N.} = \frac{1}{2} [8 + 4] = 6 \]
A steric number of 6 corresponds to $\text{sp}^3\text{d}^2$ hybridization.
Thus, $\text{sp}^3\text{d}^2$ hybridization is not associated with $\text{XeF}_6$.


Step 4: Final Answer:
The correct option is (A).