Question:

The hybridised state of bromine in bromine pentafluoride is

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Steric number 6 → always sp$^3$d$^2$ (octahedral arrangement).
Updated On: May 8, 2026
  • sp$^3$d
  • dsp$^3$
  • d$^2$sp$^3$
  • sp$^2$d
  • sp$^3$d$^2$
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Solution and Explanation

Concept: Hybridisation is determined using steric number: \[ \text{Steric number} = \text{bond pairs} + \text{lone pairs} \]

Step 1:
Find valence electrons.
Bromine (Br) has 7 valence electrons.

Step 2:
Bond formation.
In BrF$_5$:
• 5 fluorine atoms form 5 sigma bonds → 5 bond pairs
• Remaining electrons = 2 → 1 lone pair

Step 3:
Calculate steric number. \[ \text{Steric number} = 5 + 1 = 6 \]

Step 4:
Determine hybridisation.
Steric number 6 corresponds to: \[ sp^3d^2 \]

Step 5:
Geometry.

• Electron geometry → Octahedral
• Molecular shape → Square pyramidal (due to one lone pair)

Step 6:
Final answer. \[ \boxed{sp^3d^2} \]
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