Step 1: Understanding the Topic
This question asks to identify which molecule has a T-shaped molecular geometry. This requires using the VSEPR (Valence Shell Electron Pair Repulsion) theory to predict the shape of each molecule based on the number of bonding and non-bonding electron pairs around the central atom. Step 2: Key Approach - Identifying the VSEPR Type for T-Shape
A T-shaped molecular geometry arises from a specific VSEPR classification: AX$_3$E$_2$.
'A' is the central atom.
'X$_3$' indicates three bonding pairs (three atoms bonded to the central atom).
'E$_2$' indicates two lone pairs of electrons on the central atom.
This combination (3 bonding + 2 lone pairs = 5 total electron domains) corresponds to a trigonal bipyramidal electron geometry. To minimize repulsion, the two lone pairs occupy equatorial positions, forcing the three bonded atoms into a T-shape. Step 3: Detailed Analysis of Each Option
We will determine the VSEPR type for the central atom in each molecule.
(A) XeF$_2$: Central atom is Xe (8 valence e$^-$). It forms 2 bonds with F.
Bonding pairs = 2.
Lone pairs = (8 - 2)/2 = 3.
Type: AX$_2$E$_3$. This results in a Linear shape.
(B) SO$_2$: Central atom is S (6 valence e$^-$). It forms two double bonds with O.
Bonding domains = 2.
Lone pairs = (6 - 4)/2 = 1.
Type: AX$_2$E$_1$. This results in a Bent shape.
(C) IO$_4^-$: Central atom is I (7 valence e$^-$ + 1 for charge = 8 e$^-$). It forms bonds with 4 O atoms.
Bonding pairs = 4.
Lone pairs = (8 - 8)/2 = 0.
Type: AX$_4$E$_0$. This results in a Tetrahedral shape.
(D) BrF$_3$: Central atom is Br (7 valence e$^-$). It forms 3 bonds with F.
Bonding pairs = 3.
Lone pairs = (7 - 3)/2 = 2.
Type: AX$_3$E$_2$. This results in a T-shaped molecular geometry.
Step 4: Final Answer
The molecule that fits the AX$_3$E$_2$ classification and exhibits a T-shaped geometry is BrF$_3$.
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