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In which of the following sets central atom of each member involves $sp^{3}$ hybridisation?
Option: 1
$IO_{4}^{-},ICI_{4}^{-},IF_{4}^{+}$

Option: 2
$XeO_{3},XeO_{4},XeF_{4}$

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

Option: 4
$PCl_{4}^{+},BF_{4}^{-},ClO_{4}^{-}$

Answers (1)

best_answer

Hybridisation is calculated by steric number rule.

Steric no. = no. of atoms attached to central atom + no. of lone pairs present on central atom.

Steric no.for $IO_{4}^{-}=4+0=4\; \; \; \; \; sp^{3}$

Steric no.for $ICI_{4}^{-}=4+2=6\; \; \; \; \; sp^{3}d^{2}$

Steric no.for $IF_{4}^{+}=4+1=5\; \; \; \; \: \; sp^{3}d$

Steric no.for $SO_{3}=3+0=3\; \; \; \; \; sp^{2}$

Steric no.for $SO{_{4}}^{2-}= 4 + 0 = 4 \; \; \; sp^{3}$

Steric no.for $SO{_{3}}^{2-}= 3 + 1 = 4 \; \; \; \; \; \; sp^{3}$

Steric no. for $PCl_{4}^{+} = 4 + 0 = 4\; \; \; \; sp^{3}$

Steric no. for $BF_{4}^{-} = 4 + 0 = 4\; \; \; \; sp^{3}$

Steric no. for $XeO_{3} = 3 + 1 = 4\; \; \; \; sp^{3}$

Steric no. for $XeO_{4} = 4 + 0 = 4\; \; \; \; sp^{3}$

Steric no.for $XeF_{4} = 4 + 2 = 6\; \; \; \; sp^{3}d^{2}$

Steric no.for $ClO_{4}^{-} = 4 + 0 = 4\; \; \; \; sp^{3}$

Therefore, Option(4) is correct.

Posted by

manish painkra

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