Number of compounds with one lone pair of ele | vedclass

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

Question

(A) To determine the number of lone pairs on the central atom,we calculate the number of lone pairs using the formula: $	ext{Lone pair} = \frac{V - B

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) To determine the number of lone pairs on the central atom,we calculate the number of lone pairs using the formula: $	ext{Lone pair} = \frac{V - B}{2}$,where $V$ is the number of valence electrons of the central atom and $B$ is the number of bonding electrons (shared with surrounding atoms).$1$. $O_3$: Central $O$ has $1$ lone pair.$2$. $H_2O$: Central $O$ has $2$ lone pairs.$3$. $SF_4$: Central $S$ has $1$ lone pair.$4$. $ClF_3$: Central $Cl$ has $2$ lone pairs.$5$. $NH_3$: Central $N$ has $1$ lone pair.$6$. $BrF_5$: Central $Br$ has $1$ lone pair.$7$. $XeF_4$: Central $Xe$ has $2$ lone pairs.The compounds with exactly one lone pair on the central atom are $O_3$,$SF_4$,$NH_3$,and $BrF_5$.Thus,the total count is $4$.

Number of compounds with one lone pair of electrons on the central atom amongst the following is . . . . . . $O_3, H_2O, SF_4, ClF_3, NH_3, BrF_5, XeF_4$

Similar Questions

Are all the five bonds in $PCl_{5}$ molecule equivalent? Justify your answer.

What is the meaning of $VSEPR$?

The compound$(s)$ with $2$ lone pairs of electrons on the central atom is(are):
$(A)$ $BrF_5$
$(B)$ $ClF_3$
$(C)$ $XeF_4$
$(D)$ $SF_4$

When iodine is dissolved in aqueous potassium iodide,the shape of the species formed is

In which of the following covalent compounds is the bond angle maximum?

Vedclass Test Series

Exam Paper Generator

Online Exam Module