Among the following transformations,the hybri | vedclass

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(C) To determine the change in hybridisation,we calculate the steric number for the central atom in each case:$1$. $CO_2 \rightarrow HCOOH$: In $CO_2$

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$NH_3 \rightarrow NH_4^+$

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(C) To determine the change in hybridisation,we calculate the steric number for the central atom in each case:$1$. $CO_2 \rightarrow HCOOH$: In $CO_2$,$C$ is $sp$ hybridised (steric number = $2$). In $HCOOH$,$C$ is $sp^2$ hybridised (steric number = $3$).$2$. $BF_3 \rightarrow BF_4^-$: In $BF_3$,$B$ is $sp^2$ hybridised (steric number = $3$). In $BF_4^-$,$B$ is $sp^3$ hybridised (steric number = $4$).$3$. $NH_3 \rightarrow NH_4^+$: In $NH_3$,$N$ has $3$ bond pairs and $1$ lone pair,so steric number = $4$ ($sp^3$ hybridised). In $NH_4^+$,$N$ has $4$ bond pairs and $0$ lone pairs,so steric number = $4$ ($sp^3$ hybridised).$4$. $PCl_3 \rightarrow PCl_5$: In $PCl_3$,$P$ is $sp^3$ hybridised (steric number = $4$). In $PCl_5$,$P$ is $sp^3d$ hybridised (steric number = $5$).Thus,the hybridisation of the central atom remains unchanged in $NH_3 \rightarrow NH_4^+$.

Among the following transformations,the hybridisation of the central atom remains unchanged in

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