If p_x and p_y orbitals participate in sp^2 h | vedclass

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(C) In $SO_3$,the sulphur atom undergoes $sp^2$ hybridisation. The orbitals involved in $sp^2$ hybridisation are $s$,$p_x$,and $p_y$. Since the molecu

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$d_{xz}, d_{yz}$

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(C) In $SO_3$,the sulphur atom undergoes $sp^2$ hybridisation. The orbitals involved in $sp^2$ hybridisation are $s$,$p_x$,and $p_y$. Since the molecule is planar (lying in the $xy$-plane),the $p_z$ orbital remains unhybridised and is involved in $p\pi-p\pi$ bonding. To form $p\pi-d\pi$ bonds,the $d$ orbitals must have the correct symmetry to overlap with the remaining $p$ orbitals of oxygen. Given the $sp^2$ hybridisation uses $p_x$ and $p_y$,the $d$ orbitals that can participate in $\pi$-bonding with the $p_z$ orbital of oxygen are $d_{xz}$ and $d_{yz}$.

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

If $p_x$ and $p_y$ orbitals participate in $sp^2$ hybridisation of the sulphur atom in $SO_3$,then the $d$ orbitals which are used in $p\pi-d\pi$ bonds are

Similar Questions

Is there any change in the hybridization of $B$ and $N$ atoms as a result of the following reaction? $BF_3 + NH_3 \to F_3B \cdot NH_3$

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

The species in which the $N$ atom is in a state of $sp$ hybridization is:

Which of the following compounds shows the hybridization mode $sp^2 - sp^2 - sp - sp$ from left to right?

What is the bond angle $\angle F-B-F$ in $BF_3$ (in $^{\circ}$)?

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