Question

The hybridised state of bromine in bromine pentafluoride is

• A. sp$^3$d
• B. dsp$^3$
• C. d$^2$sp$^3$
• D. sp$^2$d
• E. sp$^3$d$^2$

Hint:

Steric number 6 → always sp$^3$d$^2$ (octahedral arrangement).

Answer 1

Answer: (E)

Understanding the 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. \[ \boxed{sp^3d^2} \]