If lone pairs of electrons occupy axial posit | vedclass

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Question

(C) The central atom $Cl$ in $ClF_3$ has $7$ valence electrons. It forms $3$ $Cl-F$ bonds and has $2$ lone pairs,resulting in $sp^3d$ hybridization.Ac

Vedclass · Accepted Answer

All bond lengths are equal

Vedclass · Answer

(C) The central atom $Cl$ in $ClF_3$ has $7$ valence electrons. It forms $3$ $Cl-F$ bonds and has $2$ lone pairs,resulting in $sp^3d$ hybridization.According to $VSEPR$ theory,lone pairs occupy equatorial positions to minimize $lp-lp$ and $lp-bp$ repulsions.If the lone pairs were to occupy axial positions,the $lp-lp$ repulsion would be higher,and the molecule would not be stable in that configuration.In the actual $T$-shaped structure,the two axial $Cl-F$ bonds are longer than the equatorial $Cl-F$ bond due to the repulsion from the lone pairs.Therefore,the statement that 'All bond lengths are equal' is incorrect.

List-$I$	List-$II$
$A$. $XeO_{3}$	$I$. $BrF_{5}$
$B$. $XeF_{2}$	$II$. $NH_{3}$
$C$. $XeO_{2}F_{2}$	$III$. $[I_{3}]^{-}$
$D$. $XeOF_{4}$	$IV$. $SF_{4}$

If lone pairs of electrons occupy axial positions,then which of the following statements about $ClF_3$ is incorrect?

Similar Questions

Which of the following has a pyramidal shape?

In which of the following conversions does a planar compound change into a non-planar compound?
$(a)$ $ClF_3 \rightarrow ClF_4^{-}$
$(b)$ $NH_2^{-} \rightarrow NH_4^{+}$
$(c)$ $I_3^{+} \rightarrow I_3^{-}$
$(d)$ $SO_2 \rightarrow SO_3$

Provide the shapes of $H_2O$,$NH_3$,and $SO_2$ including their lone pairs of electrons.

Match List-$I$ with List-$II$ according to shape.

List-$I$ List-$II$

$A$. $XeO_{3}$ $I$. $BrF_{5}$

$B$. $XeF_{2}$ $II$. $NH_{3}$

$C$. $XeO_{2}F_{2}$ $III$. $[I_{3}]^{-}$

$D$. $XeOF_{4}$ $IV$. $SF_{4}$

Choose the correct answer from the options given below :

Which species is planar?

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