The compound(s) with 2 lone pairs of electron | vedclass

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

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$B, C$

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(B) To determine the number of lone pairs on the central atom,we use the formula: $	ext{Lone pairs} = \frac{1}{2} 	imes (V - N - C)$,where $V$ is the number of valence electrons of the central atom,$N$ is the number of monovalent atoms bonded to it,and $C$ is the charge on the species.$1$. For $BrF_5$: Central atom $Br$ has $7$ valence electrons. It is bonded to $5$ $F$ atoms. $	ext{Lone pairs} = \frac{1}{2} 	imes (7 - 5) = 1$.$2$. For $ClF_3$: Central atom $Cl$ has $7$ valence electrons. It is bonded to $3$ $F$ atoms. $	ext{Lone pairs} = \frac{1}{2} 	imes (7 - 3) = 2$.$3$. For $XeF_4$: Central atom $Xe$ has $8$ valence electrons. It is bonded to $4$ $F$ atoms. $	ext{Lone pairs} = \frac{1}{2} 	imes (8 - 4) = 2$.$4$. For $SF_4$: Central atom $S$ has $6$ valence electrons. It is bonded to $4$ $F$ atoms. $	ext{Lone pairs} = \frac{1}{2} 	imes (6 - 4) = 1$.Thus,$ClF_3$ and $XeF_4$ have $2$ lone pairs on their central atoms. The correct option is $B$.

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

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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)$ $SF_4$