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

List - $I$ (Complex)	List – $II$ (Geometry)
$(a)$ $[Co(NH_3)_6]^{3+}$	$(i)$ Trigonal bipyramidal
$(b)$ $[NiCl_4]^{2-}$	(ii) Octahedral
$(c)$ $[Ni(CN)_4]^{2-}$	(iii) Tetrahedral
$(d)$ $[Fe(CO)_5]$	(iv) Square planar

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

Similar Questions

Among $[Co(CN)_4]^{4-}, [Co(CO)_3(NO)], XeF_4, [PCl_4]^{+}, [PdCl_4]^{2-}, [ICl_4]^{-}, [Cu(CN)_4]^{3-}$ and $P_4$,the total number of species with tetrahedral geometry is:

The geometry and magnetic behaviour of the complex $[Ni(CO)_4]$ are

Which of the following compounds has a different geometry from the others?

The hybridization states of the central atom in the complexes $[Fe(CN)_6]^{3-}$,$[Fe(CN)_6]^{4-}$ and $[Co(NO_2)_6]^{3-}$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module