Step 1: Understanding the Topic
This question requires the application of Valence Shell Electron Pair Repulsion (VSEPR) theory. The core principle of VSEPR theory is that electron pairs (both bonding and non-bonding/lone pairs) around a central atom arrange themselves in space to be as far apart as possible, thereby minimizing electrostatic repulsion. This arrangement determines the molecule's geometry.
Step 2: Key Approach - VSEPR Procedure
1. Find the central atom and count the total number of valence electrons.
2. Determine the number of bonding pairs and lone pairs around the central atom.
3. Use the total number of electron domains (bonding pairs + lone pairs) to determine the electron geometry.
4. Determine the final molecular shape by considering only the positions of the atoms, accounting for the greater repulsive effect of lone pairs.
Step 3: Detailed Explanation
A. Count Valence Electrons:
The central atom is Chlorine (Cl).
Valence electrons from Cl (Group 17) = 7
Valence electrons from 3 Fluorine atoms (Group 17) = $3 \times 7 = 21$
Total valence electrons = $7 + 21 = 28$.
B. Find Electron Pairs on Central Atom (Cl):
Chlorine forms three single bonds with the three fluorine atoms. This uses 3 of its 7 valence electrons.
Number of bonding pairs = 3
Electrons remaining on Chlorine = $7 - 3 = 4$ electrons.
These 4 electrons form lone pairs. Since each lone pair consists of 2 electrons:
Number of lone pairs = $4 / 2 = 2$
Total electron domains around Chlorine = 3 (bonding) + 2 (lone) = 5.
C. Determine Electron and Molecular Geometry:
With 5 electron domains, the electron geometry (the arrangement of all electron pairs) is trigonal bipyramidal.
In a trigonal bipyramidal arrangement, lone pairs occupy the equatorial positions to minimize repulsion (as they repel more strongly than bonding pairs). Both lone pairs will go to two of the three equatorial sites.
The three fluorine atoms will occupy the remaining two axial positions and one equatorial position.
The resulting shape formed by the atoms (Cl and 3 F) is a T-shape.
Step 4: Final Answer
The molecular shape of $ClF_3$ is:
\[
\boxed{\text{T-shaped}}
\]