In the electronic structure of H_2SO_4,the to | vedclass

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(B) The total number of valence electrons in $H_2SO_4$ is calculated as: $2 	imes (H) + 1 	imes (S) + 4 	imes (O) = 2 	imes 1 + 6 + 4 	imes 6 = 3

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$16$

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(B) The total number of valence electrons in $H_2SO_4$ is calculated as: $2 	imes (H) + 1 	imes (S) + 4 	imes (O) = 2 	imes 1 + 6 + 4 	imes 6 = 32$ electrons. In the Lewis structure of $H_2SO_4$,there are $8$ covalent bonds (two $S-OH$ bonds,two $S=O$ bonds). Each covalent bond consists of $2$ shared electrons,so the total number of shared electrons is $8 	imes 2 = 16$. The number of unshared electrons (lone pair electrons) is the total valence electrons minus the shared electrons: $32 - 16 = 16$. Thus,the total number of unshared electrons is $16$. Hence,the correct option is $B$.

List-$I$	List-$II$
$(A)$ $C_{2}H_{4}$	$(I)$ $3\sigma$ bonds, $2\pi$ bonds
$(B)$ $C_{2}H_{2}$	$(II)$ $3\sigma$ bonds, $1$ lone pair
$(C)$ $CH_{4}$	$(III)$ $4\sigma$ bonds
$(D)$ $NH_{3}$	$(IV)$ $5\sigma$ bonds, $1\pi$ bond

In the electronic structure of $H_2SO_4$,the total number of unshared electrons is

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