The number of species below that have two lon | vedclass

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

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$2$

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(C) To determine the number of lone pairs $(l.p.)$ on the central atom,we use the formula: $l.p. = \frac{1}{2} (V - N)$,where $V$ is the number of valence electrons of the central atom and $N$ is the number of monovalent atoms bonded to it.$1$. $SF_{4}$: $S$ has $6$ valence electrons,$4$ bonded atoms. $l.p. = \frac{1}{2}(6 - 4) = 1$.$2$. $BF_{4}^{-}$: $B$ has $3$ valence electrons,$4$ bonded atoms,plus $1$ negative charge. $l.p. = \frac{1}{2}(3 + 1 - 4) = 0$.$3$. $ClF_{3}$: $Cl$ has $7$ valence electrons,$3$ bonded atoms. $l.p. = \frac{1}{2}(7 - 3) = 2$.$4$. $AsF_{3}$: $As$ has $5$ valence electrons,$3$ bonded atoms. $l.p. = \frac{1}{2}(5 - 3) = 1$.$5$. $PCl_{5}$: $P$ has $5$ valence electrons,$5$ bonded atoms. $l.p. = \frac{1}{2}(5 - 5) = 0$.$6$. $BrF_{5}$: $Br$ has $7$ valence electrons,$5$ bonded atoms. $l.p. = \frac{1}{2}(7 - 5) = 1$.$7$. $XeF_{4}$: $Xe$ has $8$ valence electrons,$4$ bonded atoms. $l.p. = \frac{1}{2}(8 - 4) = 2$.$8$. $SF_{6}$: $S$ has $6$ valence electrons,$6$ bonded atoms. $l.p. = \frac{1}{2}(6 - 6) = 0$.The species with two lone pairs are $ClF_{3}$ and $XeF_{4}$.Therefore,the total number of such species is $2$.

The number of species below that have two lone pairs of electrons on their central atom is ..... . (Round off to the nearest integer)
$SF_{4}, BF_{4}^{-}, ClF_{3}, AsF_{3}, PCl_{5}, BrF_{5}, XeF_{4}, SF_{6}$

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The number of species below that have two lone pairs of electrons on their central atom is ..... . (Round off to the nearest integer)$SF_{4}, BF_{4}^{-}, ClF_{3}, AsF_{3}, PCl_{5}, BrF_{5}, XeF_{4}, SF_{6}$