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(C) Let us analyze the geometry of each species:$(a)$ $ClF_3$ is $T$-shaped (planar) and $ClF_4^{-}$ is square planar (planar). No change in planarity

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only $b$

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(C) Let us analyze the geometry of each species:$(a)$ $ClF_3$ is $T$-shaped (planar) and $ClF_4^{-}$ is square planar (planar). No change in planarity.$(b)$ $NH_2^{-}$ is $V$-shaped (planar) and $NH_4^{+}$ is tetrahedral (non-planar). Thus,it changes from planar to non-planar.$(c)$ $I_3^{+}$ is angular (planar) and $I_3^{-}$ is linear (planar). No change in planarity.$(d)$ $SO_2$ is angular (planar) and $SO_3$ is trigonal planar (planar). No change in planarity.Therefore,only $(b)$ involves a change from a planar to a non-planar compound.

List-$I$	List-$II$
$a$. See-Saw Shape	$i$. $XeF_4$
$b$. Square Pyramidal	$ii$. $ClF_3$
$c$. $T$-Shape	$iii$. $PbCl_2$
$d$. Bent Shape	$iv$. $SF_4$
	$v$. $BrF_5$

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

Similar Questions

Match the following:
List-$I$ List-$II$
$a$. See-Saw Shape $i$. $XeF_4$
$b$. Square Pyramidal $ii$. $ClF_3$
$c$. $T$-Shape $iii$. $PbCl_2$
$d$. Bent Shape $iv$. $SF_4$
$v$. $BrF_5$

The correct answer is

Assertion : $SeCl_4$ does not have a tetrahedral structure.
Reason : $Se$ in $SeCl_4$ has two lone pairs.

The number of bent-shaped molecule/s from the following is $.....$. $N_3^{-}$,$NO_2^{-}$,$I_3^{-}$,$O_3$,$SO_2$

$XeF_4$ is square planar while $XeF_6$ has a distorted octahedral structure. What is the correct explanation for this observation?

The shape of $O_2F_2$ is similar to which of the following?

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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)$ $SO_2 \rightarrow SO_3$