The number of lone pairs of electrons on the | vedclass

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

Question

(C) To determine the number of lone pairs on the central atom,we use the formula: $	ext{Lone pairs} = \frac{1}{2} (V - N)$,where $V$ is the number of

Vedclass · Accepted Answer

$2, 1, 1, 2$

Vedclass · Answer

(C) To determine the number of lone pairs on the central atom,we use the formula: $	ext{Lone pairs} = \frac{1}{2} (V - N)$,where $V$ is the number of valence electrons of the central atom and $N$ is the number of monovalent atoms bonded to it.$1$. For $ClF_3$: $V = 7, N = 3$. $	ext{Lone pairs} = \frac{1}{2} (7 - 3) = 2$.$2$. For $NF_3$: $V = 5, N = 3$. $	ext{Lone pairs} = \frac{1}{2} (5 - 3) = 1$.$3$. For $SF_4$: $V = 6, N = 4$. $	ext{Lone pairs} = \frac{1}{2} (6 - 4) = 1$.$4$. For $XeF_4$: $V = 8, N = 4$. $	ext{Lone pairs} = \frac{1}{2} (8 - 4) = 2$.CompoundLone pairs$ClF_3$$2$$NF_3$$1$$SF_4$$1$$XeF_4$$2$Therefore,the correct sequence is $2, 1, 1, 2$.

The number of lone pairs of electrons on the central atom of $ClF_3, NF_3, SF_4,$ and $XeF_4$ respectively are:

Similar Questions

$SF_6$ is a kinetically inert substance because

Molecular bromine on reaction with three moles of molecular fluorine produces an interhalogen compound. The total number of lone pairs at the central halogen atom is

Fluorine reacts with dilute $NaOH$ and forms a gaseous product $A$. The bond angle in the molecule of $A$ is

The numbers of lone pair$(s)$ on $Xe$ in $XeF_2$ and $XeF_4$ are,respectively:

In a $AX_2L_n$ compound,if $X$ is a bond pair and $n$ is the number of lone pairs of electrons,which of the following options is incorrect?

Vedclass Test Series

Exam Paper Generator

Online Exam Module