The increasing order of the number of lone pa | vedclass

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

Question

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

Vedclass · Accepted Answer

$IV < III < I < II$

Vedclass · Answer

(D) To determine the number of lone pairs on the central atom,we use the formula: $	ext{Lone pairs} = \frac{1}{2} 	imes (V - N)$,where $V$ is the number of valence electrons of the central atom and $N$ is the number of bonding electrons (monovalent atoms attached).$I) \ ClF_3$: Central atom $Cl$ ($7$ valence electrons). $3$ $F$ atoms attached. $	ext{Lone pairs} = \frac{1}{2} 	imes (7 - 3) = 2$.$II) \ XeF_2$: Central atom $Xe$ ($8$ valence electrons). $2$ $F$ atoms attached. $	ext{Lone pairs} = \frac{1}{2} 	imes (8 - 2) = 3$.$III) \ SF_4$: Central atom $S$ ($6$ valence electrons). $4$ $F$ atoms attached. $	ext{Lone pairs} = \frac{1}{2} 	imes (6 - 4) = 1$.$IV) \ SiH_4$: Central atom $Si$ ($4$ valence electrons). $4$ $H$ atoms attached. $	ext{Lone pairs} = \frac{1}{2} 	imes (4 - 4) = 0$.Comparing the values: $IV (0) Therefore,the increasing order is $IV < III < I < II$.

The increasing order of the number of lone pairs of electrons on the central atom of the following molecules is:
$I) \ ClF_3$
$II) \ XeF_2$
$III) \ SF_4$
$IV) \ SiH_4$

Similar Questions

Of the following species,the one having a square planar structure is:

The geometrical arrangement of orbitals and shape of $I_3^-$ are respectively

Out of $CHCl_3$,$CH_4$,and $SF_4$,the molecules that do not have regular geometry are

Explain: The shape of $H_2O$ is $V$ (angular).

According to $VSEPR$ theory,

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The increasing order of the number of lone pairs of electrons on the central atom of the following molecules is:$I) \ ClF_3$$II) \ XeF_2$$III) \ SF_4$$IV) \ SiH_4$