Number of compounds with one lone pair of ele | vedclass

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(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 . . . . . . $O_3, H_2O, SF_4, ClF_3, NH_3, BrF_5, XeF_4$

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