Question
Consider the following species: BrF_5, XeF_5^-, BF_4^-, ICl_4^-, XeF_4, SF_4, NH_4^+, ClF_3, XeF_2, ICl_2^- Number of species having sp^3d hybridized central atom is __________.
JEE Main
Chemistry
Chemical Bonding and Molecular Structure
2026
JEE Main 2026 (05 April Shift 1)
JEE Main Chemistry Question (2026) — Solution
Options
- A. A
- B. B
- C. C
- D. D
Answer
A. A
Step-by-step solution
The steric number H is calculated as H = 1 2 (V + M - C + A), where V is the number of valence electrons on the central atom, M is the number of monovalent atoms, C is the cationic charge, and A is the anionic charge. For sp^3d hybridization, H = 5. BrF_5: H = 1 2 (7 + 5) = 6 (sp^3d^2) XeF_5^-: H = 1 2 (8 + 5 + 1) = 7 (sp^3d^3) BF_4^-: H = 1 2 (3 + 4 + 1) = 4 (sp^3) ICl_4^-: H = 1 2 (7 + 4 + 1) = 6 (sp^3d^2) XeF_4: H = 1 2 (8 + 4) = 6 (sp^3d^2) SF_4: H = 1 2 (6 + 4) = 5 (sp^3d) NH_4^+: H = 1 2 (5 + 4 - 1) = 4 (sp^3) ClF_3: H = 1 2 (7 + 3) = 5 (sp^3d) XeF_2: H = 1 2 (8 + 2) = 5 (sp^3d) ICl_2^-: H = 1 2 (7 + 2 + 1) = 5 (sp^3d) The species with sp^3d hybridization are SF_4, ClF_3, XeF_2, and ICl_2^-. The number of such species is 4. Answer: 4
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