Shape of BrF2+ and BrF4- | Hybridization Solved
Welcome to another important session on Chemca. In this post, we are tackling a critical JEE Main 2026 PYQ (Previous Year Question) from the Chemical Bonding chapter. We will determine the hybridization, geometry, and exact molecular shape of the BrF2+ and BrF4- ions using the VSEPR (Valence Shell Electron Pair Repulsion) theory.
1. Shape and Hybridization of BrF2+ (Bromine difluoride cation)
To find the shape of the BrF2+ ion, we first need to determine the steric number of the central Bromine (Br) atom.
- Central Atom: Bromine (Br) belongs to Group 17, so it has 7 valence electrons.
- Positive Charge: The +1 charge means it has lost one electron. Remaining valence electrons = 7 - 1 = 6 electrons.
- Bond Pairs (bp): It forms 2 single bonds with 2 Fluorine atoms, using up 2 electrons.
- Lone Pairs (lp): The remaining 4 electrons form 2 lone pairs.
Steric Number (Z) = Bond Pairs + Lone Pairs = 2 + 2 = 4.
Since the steric number is 4, the hybridization is sp3.
The standard electron geometry for sp3 is Tetrahedral. However, due to the presence of 2 lone pairs repelling the bond pairs (lp-lp > lp-bp > bp-bp repulsion), the molecular shape becomes Bent or V-shaped (similar to the water molecule, H2O).
2. Shape and Hybridization of BrF4- (Tetrafluorobromate anion)
Now, let's analyze the BrF4- ion. The process is similar, but the extra electron changes the geometry entirely.
- Central Atom: Bromine (Br) with 7 valence electrons.
- Negative Charge: The -1 charge means it has gained one electron. Total valence electrons = 7 + 1 = 8 electrons.
- Bond Pairs (bp): It forms 4 single bonds with 4 Fluorine atoms, using up 4 electrons.
- Lone Pairs (lp): The remaining 4 electrons form 2 lone pairs.
Steric Number (Z) = Bond Pairs + Lone Pairs = 4 + 2 = 6.
A steric number of 6 corresponds to sp3d2 hybridization.
The electron geometry for sp3d2 is Octahedral. To minimize lone-pair to lone-pair repulsion at 90 degrees, the two lone pairs occupy the axial positions (180 degrees apart). This leaves the four Fluorine atoms in the equatorial plane, resulting in a Square Planar molecular shape.
Summary for JEE Main Exams
BrF2+: sp3 hybridized, Bent (V-shape).
BrF4-: sp3d2 hybridized, Square Planar.
Make sure to watch the full video lecture embedded above for a more visual explanation and tips on solving hybridization questions quickly in the exam hall. For more chemistry lectures, visit the Chemca Lectures page.
