The correct order of hybridisations of the ce | vedclass

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(B) $1$. For $NH_3$: The central atom $N$ has $3$ bond pairs and $1$ lone pair,so the hybridisation is $sp^3$.$2$. For $[PtCl_4]^{2-}$: $Pt^{2+}$ is a

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$sp^3, dsp^2, sp^3d, sp^2$

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(B) $1$. For $NH_3$: The central atom $N$ has $3$ bond pairs and $1$ lone pair,so the hybridisation is $sp^3$.$2$. For $[PtCl_4]^{2-}$: $Pt^{2+}$ is a $d^8$ metal ion. With strong field ligands like $Cl^-$ (in square planar geometry),it undergoes $dsp^2$ hybridisation.$3$. For $PCl_5$: The central atom $P$ has $5$ bond pairs and $0$ lone pairs,resulting in $sp^3d$ hybridisation.$4$. For $BCl_3$: The central atom $B$ has $3$ bond pairs and $0$ lone pairs,resulting in $sp^2$ hybridisation.Therefore,the order is $sp^3, dsp^2, sp^3d, sp^2$.

The correct order of hybridisations of the central atom in $NH_3$,$[PtCl_4]^{2-}$,$PCl_5$,and $BCl_3$ is:

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