The hybrid state of sulphur in SO_3 molecule | vedclass

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(D) The hybridization of the central atom can be calculated using the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence ele

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$sp^2$

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(D) The hybridization of the central atom can be calculated using the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.For $SO_3$,the central atom is $S$ (valence electrons $V = 6$). Oxygen is a divalent atom,so $M = 0$.$H = \frac{1}{2}(6 + 0 - 0 + 0) = 3$.$A$ value of $H = 3$ corresponds to $sp^2$ hybridization.Thus,the $S$ atom in $SO_3$ is $sp^2$ hybridized,resulting in a trigonal planar structure.

Molecule	Hybridisation of central atom	Shape
$(a)$ $PCl_5$	$dsp^3$	square pyramidal
$(b)$ $[Ni(CN)_4]^{2-}$	$sp^3$	tetrahedral
$(c)$ $SF_6$	$sp^3d^2$	octahedral
$(d)$ $IF_3$	$dsp^3$	pyramidal

The hybrid state of sulphur in $SO_3$ molecule is

Similar Questions

Identify the correct set from the following table:
Molecule Hybridisation of central atom Shape
$(a)$ $PCl_5$ $dsp^3$ square pyramidal
$(b)$ $[Ni(CN)_4]^{2-}$ $sp^3$ tetrahedral
$(c)$ $SF_6$ $sp^3d^2$ octahedral
$(d)$ $IF_3$ $dsp^3$ pyramidal

What is the $CORRECT$ geometry and hybridisation for $SF_4$?

Percentage of $s-$character in $sp^3$ hybrid orbital is (in $\%$)

Consider the following reaction $BCl_3 + NH_3 \longrightarrow BCl_3 \cdot NH_3$. The geometries of $BCl_3$ and $BCl_3 \cdot NH_3$ respectively are

$sp^3$ hybridisation is found in

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