Consider the following reaction:2Se_2Cl_2 to | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $

Vedclass · Accepted Answer

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

Vedclass · Answer

(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.

List-$I$	List-$II$
$A$. $XeF_4$	$1$. Pyramidal
$B$. $XeF_6$	$2$. $T$-shape
$C$. $XeO_3$	$3$. Distorted octahedral
$D$. $XeOF_2$	$4$. Square planar

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).

Similar Questions

The shape of a molecule of $NH_3$,in which the central atom contains a lone pair of electrons,is

Bond angle in $PH_{4}^{+}$ is more than that of $PH_{3}$. This is because

What are the formal charges on the terminal oxygen atoms of the ozone $(O_3)$ molecule?

Which of the following has the smallest bond angle?

Select the correct matching:

List-$I$ List-$II$

$A$. $XeF_4$ $1$. Pyramidal

$B$. $XeF_6$ $2$. $T$-shape

$C$. $XeO_3$ $3$. Distorted octahedral

$D$. $XeOF_2$ $4$. Square planar

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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).