Liquid BrF3 autoionises as 2BrF3 BrF2+ + BrF4-. The shapes (hybridisation of Br) of the cation and the anion are

BrF2+ bent (sp^3); BrF4- square planar (sp^3d^2)

BrF2+ linear (sp^3d); BrF4- tetrahedral (sp^3)

BrF2+ bent (sp^3); BrF4- see-saw (sp^3d)

BrF2+ linear (sp); BrF4- square planar (sp^3d^2)

BrF2+ bent (sp^3); BrF4- square planar (sp^3d^2)

JEE Advanced levelSingle correct optionDifficulty 3 of 3ph203

Liquid $BrF X_{3}$ autoionises as $2 BrF X_{3} BrF X_{2} X^{+} + BrF X_{4} X^{-}$ . The shapes (hybridisation of Br) of the cation and the anion are

A $BrF X_{2} X^{+}$ linear ( $s p^{3} d$ ); $BrF X_{4} X^{-}$ tetrahedral ( $s p^{3}$ )
B $BrF X_{2} X^{+}$ bent ( $s p^{3}$ ); $BrF X_{4} X^{-}$ see-saw ( $s p^{3} d$ )
C $BrF X_{2} X^{+}$ linear ( $s p$ ); $BrF X_{4} X^{-}$ square planar ( $s p^{3} d^{2}$ )
D $BrF X_{2} X^{+}$ bent ( $s p^{3}$ ); $BrF X_{4} X^{-}$ square planar ( $s p^{3} d^{2}$ )

BrF X_{2} X^{+}

: Br has

7 - 1 = 6

valence electrons, 2 used in bonds, leaving 2 lone pairs. So 4 pairs,

s p^{3}

, bent.

BrF X_{4} X^{-}

7 + 1 = 8

electrons, 4 used in bonds, leaving 2 lone pairs. So 6 pairs,

s p^{3} d^{2}

, square planar. The anion is thus isostructural with

XeF X_{4}