The hybridization of the central atom in NH_3 | vedclass

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(B) $1$. For $NH_3$: The central atom $N$ has $5$ valence electrons. It forms $3$ bonds with $H$ and has $1$ lone pair. Steric number = $3 + 1 = 4$,so

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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 $5$ valence electrons. It forms $3$ bonds with $H$ and has $1$ lone pair. Steric number = $3 + 1 = 4$,so hybridization is $sp^3$.$2$. For $[PtCl_4]^{2-}$: $Pt$ is in $+2$ oxidation state ($d^8$ configuration). It forms a square planar complex,so hybridization is $dsp^2$.$3$. For $PCl_5$: $P$ has $5$ valence electrons and forms $5$ bonds with $Cl$. Steric number = $5$,so hybridization is $sp^3d$.$4$. For $BCl_3$: $B$ has $3$ valence electrons and forms $3$ bonds with $Cl$. Steric number = $3$,so hybridization is $sp^2$.Thus,the sequence is $sp^3, dsp^2, sp^3d, sp^2$.

List-$I$ (Molecules/ion)	List-$II$ (Hybridisation of central atom)
$A. PF_5$	$I. dsp^2$
$B. SF_6$	$II. sp^3d$
$C. Ni(CO)_4$	$III. sp^3d^2$
$D. [PtCl_4]^{2-}$	$IV. sp^3$

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

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Match the List-$I$ with List-$II$
List-$I$ (Molecules/ion) List-$II$ (Hybridisation of central atom)
$A. PF_5$ $I. dsp^2$
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Choose the correct answer from the options given below $:$

Which one of the following is square planar in structure and has diamagnetic property?

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