Which compound of sulphur is formed in the se | vedclass

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(D) The ground state electronic configuration of sulphur $(S)$ is $[Ne] 3s^2 3p^4$.In the first excited state,one electron from the $3p$ orbital is pr

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$SF_6$

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(D) The ground state electronic configuration of sulphur $(S)$ is $[Ne] 3s^2 3p^4$.In the first excited state,one electron from the $3p$ orbital is promoted to the $3d$ orbital,resulting in $3s^2 3p^3 3d^1$. This state allows for the formation of $SF_4$ through $sp^3d$ hybridization.In the second excited state,one electron from the $3s$ orbital is also promoted to the $3d$ orbital,resulting in $3s^1 3p^3 3d^2$. This provides six unpaired electrons for bonding.These six orbitals undergo $sp^3d^2$ hybridization to form $SF_6$ with octahedral geometry.Therefore,$SF_6$ is the compound formed in the second excited state.

List-$I$ (Hybridisation)	List-$II$ (Shape)
$A$. $dsp^2$	$I$. Square planar
$B$. $sp^3$	$II$. Tetrahedral
$C$. $d^2sp^3$	$III$. Octahedral
$D$. $sp^3d$	$IV$. Trigonal bipyramidal

Which compound of sulphur is formed in the second excited state?

Similar Questions

Among $[I_3]^{+}$,$[SiO_4]^{4-}$,$SO_2Cl_2$,$XeF_2$,$SF_4$,$ClF_3$,$Ni(CO)_4$,$XeO_2F_2$,$[PtCl_4]^{2-}$,$XeF_4$,and $SOCl_2$,the total number of species having $sp^3$ hybridised central atom is. . . . .

In the process $XH_3 + H^{+} \to XH_4^+ (X = N, P, As, Sb)$,the maximum increase in bond angle occurs in the case of:

Match the following List-$I$ (Hybridisation) with List-$II$ (Shape):
List-$I$ (Hybridisation) List-$II$ (Shape)
$A$. $dsp^2$ $I$. Square planar
$B$. $sp^3$ $II$. Tetrahedral
$C$. $d^2sp^3$ $III$. Octahedral
$D$. $sp^3d$ $IV$. Trigonal bipyramidal

The compound formed by $sp^3d$ hybridization will have which structure?

The shape of ethane is

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