The molecule in which the central atom does N | vedclass

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

Question

(D) To determine the hybridisation of the central atom,we use the formula: $	ext{Steric Number} = \frac{1}{2} [V + M - C + A]$,where $V$ is the numbe

Vedclass · Accepted Answer

$BF_3$

Vedclass · Answer

(D) To determine the hybridisation of the central atom,we use the formula: $	ext{Steric Number} = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$A$) For $NH_3$: $	ext{Steric Number} = \frac{1}{2} [5 + 3] = 4$,which corresponds to $sp^3$ hybridisation.$B$) For $H_2O$: $	ext{Steric Number} = \frac{1}{2} [6 + 2] = 4$,which corresponds to $sp^3$ hybridisation.$C$) For $CH_4$: $	ext{Steric Number} = \frac{1}{2} [4 + 4] = 4$,which corresponds to $sp^3$ hybridisation.$D$) For $BF_3$: $	ext{Steric Number} = \frac{1}{2} [3 + 3] = 3$,which corresponds to $sp^2$ hybridisation.Thus,the central atom in $BF_3$ does not undergo $sp^3$ hybridisation.

Column-$I$	Column-$II$
$A$. Boron in $[B(OH)_4]^-$	$1$. $sp^2$
$B$. Aluminum in $[Al(H_2O)_6]^{3+}$	$2$. $sp^3$
$C$. Boron in $B_2H_6$	$3$. $sp^3d^2$
$D$. Carbon in Buckminsterfullerene
$E$. $Si$ in $SiO_4^{4-}$
$F$. $Ge$ in $[GeCl_6]^{2-}$

The molecule in which the central atom does $NOT$ undergo $sp^3$ hybridisation is

Similar Questions

Identify the pair of molecules which have same hybridisation as the hybridisation in $Xenon$ $(II)$ fluoride.

The hybridisations of $N$-orbitals in $NO_3^{-}$,$NO_2^{-}$ and $NH_4^{+}$ respectively are

In which of the following sets of molecules,do the central atoms have the same hybridisation?

As the $p-$character increases,the bond angle in hybrid orbitals formed by $s$ and $p$ atomic orbitals

Vedclass Test Series

Exam Paper Generator

Online Exam Module