What is the molecular geometry of IBr3 according to the VSEPR theory? Seesaw B. T-shaped Bent or V-shaped D, Trigonal planar Tetrahedral
The Correct Answer and Explanation is:
The correct answer is: B. T-shaped
Explanation:
To determine the molecular geometry of IBr₃ using VSEPR (Valence Shell Electron Pair Repulsion) theory, we follow these steps:
Step 1: Determine the central atom and valence electrons
Iodine (I) is the central atom because it’s less electronegative than bromine (Br).
- Iodine is in Group 17 (halogens), so it has 7 valence electrons.
- Each bromine atom forms a single bond with iodine, using 1 electron from iodine per bond.
- 3 Br atoms form 3 bonds, using 3 of iodine’s 7 electrons.
- That leaves 4 electrons (2 lone pairs) on iodine.
Step 2: Count the regions of electron density around iodine
- There are 3 bonding pairs (one for each I–Br bond).
- There are 2 lone pairs of electrons on iodine.
So, iodine has 5 regions of electron density. According to VSEPR theory, 5 regions correspond to a trigonal bipyramidal electron geometry.
Step 3: Determine molecular geometry
In a trigonal bipyramidal electron geometry:
- Lone pairs prefer the equatorial positions (to minimize repulsion).
- If two lone pairs occupy equatorial positions, the three Br atoms occupy two axial and one equatorial position.
This leads to a T-shaped molecular geometry, where:
- Two Br atoms are opposite each other (axial).
- One Br atom is in the equatorial plane.
- The lone pairs occupy the other two equatorial positions.
Final Result:
- Electron Geometry: Trigonal bipyramidal
- Molecular Geometry: T-shaped
Therefore, the correct answer is B. T-shaped, due to the presence of two lone pairs and three bonding pairs on the central iodine atom
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