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There is no change in the type of hybridisation when
Option: 1
$NH_{3}$ combines with $H^{+}$

Option: 2
$AlH_{3}$ combines with $H^{-}$

Option: 3
$SO_{3}$ forms $SO_{4}^{2-}$

Option: 4
$SiF_{4}$ forms $SiF_{6}^{2-}$

Answers (1)

best_answer

How to find Hybridisation

The hybridisation depends upon sigma bonds and lone pair of electrons.

Thus,

Hybridisation = Number of sigma bonds + Number of lone pairs present on central atom

For example, hybridisation for NH₃ is sp³ and its molecular geometry is tetrahedral.

NH₃ has 3 sigma bonds and 1 lone pair, thus hybridisation for NH₃:

3 sigma bonds + 1 lone pair = 4

Thus hybridisation for NH₃ is sp³ and its geometry is tetrahedral.

As we have learnt,

$\mathrm{NH_{3}\overset{H^{+}}{\rightarrow}NH_{4}^{+}}$

$sp^{3}$ $sp^{3}$

$\Rightarrow$ No change in hybridisation

$\mathrm{AlH_{3}+H^{-}\rightarrow AlH_{4}^{-}}$

$sp^{2}$ $sp^{3}$

$\Rightarrow$ Hybridisation changes

$\mathrm{SO_{3}+H_{2}O\rightarrow H_{2}SO_{4}}$

$\Rightarrow$ $sp^{2}$ Hybridisation of Sulphur in $SO_{4}^{2-}$ is $sp^{3}$

Hence, option number (1) is the correct answer

Posted by

jitender.kumar

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