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.

List-$I$	List-$II$
$(a)$ $SF_6$	$(i)$ $sp^3d^2$
$(b)$ $PCl_5$	(ii) $sp^3d$
$(c)$ $XeF_4$	(iii) $sp^3d^3$
$(d)$ $IF_7$	(iv) $sp^3d^2$

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

Similar Questions

In which of the following pairs,both the species have the same hybridisation?
$I$. $SF_4, XeF_4$
$II$. $I_3^-, XeF_2$
$III$. $ICl_4^+, SiCl_4$
$IV$. $ClO_3^-, PO_4^{3-}$

Match the following:
List-$I$ List-$II$
$(a)$ $SF_6$ $(i)$ $sp^3d^2$
$(b)$ $PCl_5$ (ii) $sp^3d$
$(c)$ $XeF_4$ (iii) $sp^3d^3$
$(d)$ $IF_7$ (iv) $sp^3d^2$

The correct answer is:

On the transformation $PCl_5 \to PCl_6^-$,what changes will take place?

What is the hybridization state of the central atom in the conjugate base of $NH_4^{+}$ ion?

The carbon atoms in an acetylene molecule are in which state of hybridisation?

Vedclass Test Series

Exam Paper Generator

Online Exam Module