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(B) To determine the number of species that do not follow the octet rule (i.e.,do not have $8$ electrons around the central atom),we analyze the valen

Vedclass · Accepted Answer

$3$

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(B) To determine the number of species that do not follow the octet rule (i.e.,do not have $8$ electrons around the central atom),we analyze the valence electrons of each molecule:$1$. $NH_3$: Nitrogen has $5$ valence electrons and forms $3$ bonds with Hydrogen,plus $1$ lone pair. Total electrons = $3 	imes 2 + 2 = 8$ electrons.$2$. $AlCl_3$: Aluminum has $3$ valence electrons and forms $3$ bonds with Chlorine. Total electrons = $3 	imes 2 = 6$ electrons (Incomplete octet).$3$. $BeCl_2$: Beryllium has $2$ valence electrons and forms $2$ bonds with Chlorine. Total electrons = $2 	imes 2 = 4$ electrons (Incomplete octet).$4$. $CCl_4$: Carbon has $4$ valence electrons and forms $4$ bonds with Chlorine. Total electrons = $4 	imes 2 = 8$ electrons.$5$. $PCl_5$: Phosphorus has $5$ valence electrons and forms $5$ bonds with Chlorine. Total electrons = $5 	imes 2 = 10$ electrons (Expanded octet).Species with $NOT$ $8$ electrons are $AlCl_3$,$BeCl_2$,and $PCl_5$.Therefore,the total number of such species is $3$.

Amongst the following,the total number of species $NOT$ having $8$ electrons around the central atom in its outermost shell is $NH_3$,$AlCl_3$,$BeCl_2$,$CCl_4$,$PCl_5$:

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