The hybridization of the central atom in NH_3 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

Similar Questions

Chromium hexacarbonyl,$Cr(CO)_6$,is an octahedral compound involving which type of hybridization?

Which set of $d-$ orbitals is used by the central metal during the formation of $MnO_4^-$?

Which of the following complexes is tetrahedral as well as diamagnetic?

What is the coordination geometry around the manganese atom in decacarbonyldimanganese$(0)$?

Among the following,the species having square planar geometry for the central atom are:
$(i)$ $XeF_4$
$(ii)$ $SF_4$
$(iii)$ $[NiCl_4]^{2-}$
$(iv)$ $[PtCl_4]^{2-}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module