JEE Main 2026 — Chemistry Inorganic Chemistry

2026mcqmedium

Bromine trifluoride autoionizes to form $\text{BrF}_2^{\oplus}$ and $\text{BrF}_4^{\ominus}$ . The shapes of the cation and anion are respectively ________, and ________.

Official previous-year question

Held on 8 Apr 2026 · Verified 6 Jul 2026.

Options

A
bent, square planar
B
linear, square planar
C
bent, see-saw
D
linear, tetrahedral

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Solution

The autoionization of bromine trifluoride is given by the reaction:

$2\text{BrF}_3 \rightleftharpoons \text{BrF}_2^{\oplus} + \text{BrF}_4^{\ominus}$

For the cation $\text{BrF}_2^{\oplus}$ :

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 $\text{BrF}_4^{\ominus}$ :

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

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Solution

The autoionization of bromine trifluoride is given by the reaction:

$2\text{BrF}_3 \rightleftharpoons \text{BrF}_2^{\oplus} + \text{BrF}_4^{\ominus}$

For the cation $\text{BrF}_2^{\oplus}$ :

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 $\text{BrF}_4^{\ominus}$ :

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

JEE Main 2026 — Chemistry Inorganic Chemistry

Options

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Solution

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JEE Main 2026 — Chemistry Inorganic Chemistry

Options

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Solution

Your note

JEE Main Inorganic Chemistry in other years

More JEE Main Inorganic Chemistry questions

Other Chemistry topics for JEE Main

Physical Chemistry

Organic Chemistry

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