Question

Determine the shape of $ClF_3$ using VSEPR theory.

Hint:

For 5 electron pairs (trigonal bipyramidal arrangement), if there are 2 lone pairs, the molecular shape becomes T-shaped.

Answer 1

Step 1: Find total valence electrons.
Chlorine has 7 valence electrons.
Each fluorine has 7 valence electrons.
Total valence electrons:
\[ = 7 + 3(7) = 28. \]
Step 2: Determine number of electron pairs around central atom.
Chlorine forms three bonds with fluorine atoms.
Thus, 3 bonding pairs are present.
Remaining electrons on chlorine:
\[ 7 - 3 = 4 \text{ electrons} \] which form 2 lone pairs.
Hence total electron pairs around chlorine:
\[ 3 \text{ bonding pairs} + 2 \text{ lone pairs} = 5. \]
Step 3: Determine electron geometry.
Five electron pairs correspond to:
\[ \text{Trigonal Bipyramidal electron geometry}. \]
Step 4: Arrange lone pairs.
According to VSEPR, lone pairs occupy equatorial positions to minimize repulsion.
Two equatorial positions are occupied by lone pairs.
Remaining three positions are occupied by fluorine atoms.
Step 5: Molecular shape.
After placing lone pairs, the molecular geometry becomes:
\[ \boxed{\text{T-shaped}}. \]