Question:easy

What is the geometry of $\mathrm{SbF}_5$ molecule?

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Any neutral central atom from Group 15 (like $\mathrm{P}$, $\mathrm{As}$, $\mathrm{Sb}$) bonded to exactly five monovalent halogen atoms ($\mathrm{X}_5$) will have a steric number of 5 with zero lone pairs, guaranteeing a trigonal bipyramidal molecular shape!
Updated On: Jun 11, 2026
  • Trigonal pyramidal
  • Trigonal planar
  • Square pyramidal
  • Trigonal bipyramidal
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Identify the central atom and its group.
In $SbF_5$ the central atom is antimony, which sits in group 15, so it brings $5$ valence electrons.
Step 2: Count the things attached.
Five fluorine atoms are bonded to antimony, each forming one single bond, so there are $5$ bonding regions.
Step 3: Find lone pairs by electron bookkeeping.
Each $Sb-F$ bond uses one antimony electron, so $5$ electrons go into bonds. That uses all $5$ valence electrons, leaving $0$ lone pairs on antimony.
Step 4: Get the steric number.
Steric number $=$ bonding regions $+$ lone pairs $= 5 + 0 = 5$. A steric number of $5$ means $sp^3d$ hybridisation.
Step 5: Apply VSEPR for five clouds, zero lone pairs.
Five electron clouds with no lone pair always spread into the trigonal bipyramidal shape to keep repulsion lowest.
Step 6: Describe and conclude.
Three fluorines sit in the equatorial plane $120^\circ$ apart and two sit axially at $90^\circ$, giving a trigonal bipyramidal molecule.
\[ \boxed{\text{Trigonal bipyramidal (option D)}} \]
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