JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

The autoionization of bromine trifluoride is given by the reaction: 2 BrF _3 BrF _2^ + BrF _4^ For the cation BrF _2^ : The central atom Br has 7 valence electrons. Due to the positive charge, it has 7 - 1 = 6 electrons available. It forms 2 single bonds with F atoms, leaving 4 electrons, which constitute 2 lone pairs. Steric number = 2 (bond pairs) + 2 (lone pairs) = 4. The hybridization is sp^3. With 2 bond pairs and 2 lone pairs, the shape is bent (or V-shaped). For the anion BrF _4^ : The central atom Br has 7 valence electrons. Due to the negative charge, it has 7 + 1 = 8 electrons available. It forms 4 single bonds with F atoms, leaving 4 electrons, which constitute 2 lone pairs. Steric number = 4 (bond pairs) + 2 (lone pairs) = 6. The hybridization is sp^3d^2. To minimize repulsion, the 2 lone pairs occupy opposite (trans) positions in the octahedral geometry, resulting in a square planar shape. Thus, the shapes of the cation and anion are bent and square planar, respectively. Answer: bent, square planar