QJEEM02Ch8f58: JEE Main 2002 Chemistry - Class 11 JEEM02Ch8f58 Solution: Hybridisation of underline atom changes in

(a) $\,\,\,$ AlH₃ + H ^$-$ $\to$ AlH $_4^ -$

Steric number of AlH₃ is = ${1 \over 2}$ [3 + 3] = 3

$\therefore\,\,\,$ AlH₃ is sp² hybridized.

Steric number of AlH $_4^ -$ = ${1 \over 2}$ [3 + 4 +1] = 4

$\therefore\,\,\,$ AlH $_4^ -$ is sp³ hybridized.

(b) $\,\,\,$ H₂O + H⁺ $\to$ H₃O⁺

Steric no of H₂O = ${1 \over 2}$ (6+ 2) = 4

$\therefore\,\,\,\,$ H₂O s sp³ hybridized.

Steric no of H₃O⁺ = ${1 \over 2}$ [ 8 + 3 $-$ 1] = 4

$\therefore\,\,\,$ H₃O⁺ s also sp³ hybridized.

(c) $\,\,\,$ NH₃ + H⁺ $\to$ NH₄⁺

Steric no of NH₃ = ${1 \over 2}$ [5 + 3] = 4

$\therefore\,\,\,$ hybridization of NH₃ is sp³

Steric number of NH₄⁺ = ${1 \over 2}$ [5 + 4 $-$ ] = 4

$\therefore\,\,\,$ Hybridization of NH₄⁺ is sp³