The hybridisation of the central atom of BF_3 | vedclass

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

Question

(A) For $BF_3$: The central atom $B$ has $3$ valence electrons and forms $3$ bonds with $F$ atoms. Steric number $= 3 + 0 = 3$,which corresponds to $s

Vedclass · Accepted Answer

$sp^2, sp^2, sp$

Vedclass · Answer

(A) For $BF_3$: The central atom $B$ has $3$ valence electrons and forms $3$ bonds with $F$ atoms. Steric number $= 3 + 0 = 3$,which corresponds to $sp^2$ hybridisation.For $SnCl_2$: The central atom $Sn$ has $4$ valence electrons. It forms $2$ bonds with $Cl$ atoms and has $1$ lone pair. Steric number $= 2 + 1 = 3$,which corresponds to $sp^2$ hybridisation.For $HgCl_2$: The central atom $Hg$ has $2$ valence electrons and forms $2$ bonds with $Cl$ atoms. Steric number $= 2 + 0 = 2$,which corresponds to $sp$ hybridisation.Therefore,the correct sequence is $sp^2, sp^2, sp$.

The hybridisation of the central atom of $BF_3$,$SnCl_2$,and $HgCl_2$ respectively is:

Similar Questions

Consider the following species: $BrF_5, XeF_5^-, BF_4^-, ICl_4^-, XeF_4, SF_4, NH_4^+, ClF_3, XeF_2, ICl_2^-$. Number of species having $sp^3d$ hybridized central atom is . . . . . . .

In which of the following does the central atom $NOT$ use $sp^3$ hybrid orbitals in its bonding?

Write the state of hybridisation of carbon in the following compounds and the shapes of each of the molecules:
$(a)$ $H_2C=O$
$(b)$ $CH_3F$
$(c)$ $HC \equiv N$

The nature of hybridization in the $NH_3$ molecule is

The geometry and the type of hybrid orbital present about the central atom in $BF_3$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module