Consider the following reaction:2Se_2Cl_2 to | vedclass

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(A) The given reaction is $2Se_2Cl_2 	o 3Se + SeCl_4$.In $Se_2Cl_2$,the oxidation state of $Se$ is $+1$. In $SeCl_4$,the oxidation state of $Se$ is $

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$A$ has a see-saw shape with $sp^3d$ hybridization.

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(A) The given reaction is $2Se_2Cl_2 	o 3Se + SeCl_4$.In $Se_2Cl_2$,the oxidation state of $Se$ is $+1$. In $SeCl_4$,the oxidation state of $Se$ is $+4$,and in $Se$ (elemental),it is $0$. Since the oxidation state changes,it is a redox reaction.For $SeCl_4$,the central atom $Se$ has $6$ valence electrons. It forms $4$ bonds with $Cl$ atoms and has $1$ lone pair.Steric number $= 4 	ext{ (bond pairs)} + 1 	ext{ (lone pair)} = 5$.This corresponds to $sp^3d$ hybridization.Due to the presence of one lone pair in the equatorial position of a trigonal bipyramidal geometry,the shape of $SeCl_4$ is see-saw.

Consider the following reaction:
$2Se_2Cl_2 \to 3Se + SeCl_4$
Choose the $CORRECT$ option regarding the product $SeCl_4$ (referred to as $A$ in the options).

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Consider the following reaction:$2Se_2Cl_2 \to 3Se + SeCl_4$Choose the $CORRECT$ option regarding the product $SeCl_4$ (referred to as $A$ in the options).