Q18 / 155 · WBJEE 2026 · Chemistry
In which of the following species, $sp^3d^2$ hybridisation is not associated?
- Correct answerA$XeF_6$
- B$BrF_6^+$
- C$IF_5$
- D$XeF_4$
Solution
The hybridization of a central atom is found using the steric number formula $H = \dfrac{1}{2}(V + M - C + A)$, where $V$ is the number of valence electrons, $M$ is the number of monovalent surrounding atoms, $C$ is the cationic charge, and $A$ is the anionic charge.
For $XeF_6$, $V = 8$ and $M = 6$. $H = \dfrac{1}{2}(8 + 6) = 7$, which corresponds to $sp^3d^3$ hybridization.
For $BrF_6^+$, $V = 7$, $M = 6$, and $C = 1$. $H = \dfrac{1}{2}(7 + 6 - 1) = 6$, which corresponds to $sp^3d^2$ hybridization.
For $IF_5$, $V = 7$ and $M = 5$. $H = \dfrac{1}{2}(7 + 5) = 6$, which corresponds to $sp^3d^2$ hybridization.
For $XeF_4$, $V = 8$ and $M = 4$. $H = \dfrac{1}{2}(8 + 4) = 6$, which corresponds to $sp^3d^2$ hybridization.
Thus, $XeF_6$ is not associated with $sp^3d^2$ hybridization.
Answer: $XeF_6$
For $XeF_6$, $V = 8$ and $M = 6$. $H = \dfrac{1}{2}(8 + 6) = 7$, which corresponds to $sp^3d^3$ hybridization.
For $BrF_6^+$, $V = 7$, $M = 6$, and $C = 1$. $H = \dfrac{1}{2}(7 + 6 - 1) = 6$, which corresponds to $sp^3d^2$ hybridization.
For $IF_5$, $V = 7$ and $M = 5$. $H = \dfrac{1}{2}(7 + 5) = 6$, which corresponds to $sp^3d^2$ hybridization.
For $XeF_4$, $V = 8$ and $M = 4$. $H = \dfrac{1}{2}(8 + 4) = 6$, which corresponds to $sp^3d^2$ hybridization.
Thus, $XeF_6$ is not associated with $sp^3d^2$ hybridization.
Answer: $XeF_6$
Hence, the correct answer is option: 'A'.