(A) In $SO_3$,the central sulfur atom is bonded to three oxygen atoms with no lone pairs.Using the formula: $\text{Hybridization} = \frac{1}{2} (V + M

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(A) In $SO_3$,the central sulfur atom is bonded to three oxygen atoms with no lone pairs.Using the formula: $\text{Hybridization} = \frac{1}{2} (V + M - C + A)$,where $V = 6$ (valence electrons of $S$),$M = 0$ (monovalent atoms),$C = 0$,and $A = 0$.$\text{Hybridization} = \frac{1}{2} (6 + 0) = 3$,which corresponds to $sp^2$ hybridization.

Class 11 Chemistry Chemical Bonding and Molecular Structure Hybridisation

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What is the type of hybridization in $SO_3$?

A
$sp^2$
B
$sp^3$
C
$sp^2d$
D
$sp^2d^2$

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Class 11 Questions

Class 11

Chemistry Questions

Chapter

Chemical Bonding and Molecular Structure

Subtopic

Hybridisation

Match the compounds given in $Column\, I$ with the hybridisation and shape given in $Column\, II$ and mark the correct option.
$Column\, I$ $Column\, II$
$(A)\, XeF_6$ $(i)\, \text{Distorted octahedral}$
$(B)\, XeO_3$ $(ii)\, \text{Square planar}$
$(C)\, XeOF_4$ $(iii)\, \text{Pyramidal}$
$(D)\, XeF_4$ $(iv)\, \text{Square pyramidal}$

$Column\, I$	$Column\, II$
$(A)\, XeF_6$	$(i)\, \text{Distorted octahedral}$
$(B)\, XeO_3$	$(ii)\, \text{Square planar}$
$(C)\, XeOF_4$	$(iii)\, \text{Pyramidal}$
$(D)\, XeF_4$	$(iv)\, \text{Square pyramidal}$

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