Observe the following reactions:P_4 + 8SOCl_2 | vedclass

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(A) The reactions are:$1) P_4 + 8SOCl_2 \longrightarrow 4PCl_3 + 4SO_2 + 2S_2Cl_2$$2) P_4 + 10SO_2Cl_2 \longrightarrow 4PCl_5 + 10SO_2$In both reactio

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(A) The reactions are:$1) P_4 + 8SOCl_2 \longrightarrow 4PCl_3 + 4SO_2 + 2S_2Cl_2$$2) P_4 + 10SO_2Cl_2 \longrightarrow 4PCl_5 + 10SO_2$In both reactions,the common product '$x$' is sulfur dioxide $(SO_2)$.In $SO_2$,the central atom is sulfur $(S)$.The electronic configuration of sulfur is $[Ne] 3s^2 3p^4$.In $SO_2$,sulfur forms two double bonds with two oxygen atoms.Sulfur has $6$ valence electrons. It uses $4$ electrons for bonding with oxygen atoms (two double bonds).Therefore,$6 - 4 = 2$ electrons remain,which form $1$ lone pair on the sulfur atom.Thus,the number of lone pair of electrons on the central atom of $SO_2$ is $1$.

Observe the following reactions:
$P_4 + 8SOCl_2 \longrightarrow 4PCl_3 + 4SO_2 + 2S_2Cl_2$
$P_4 + 10SO_2Cl_2 \longrightarrow 4PCl_5 + 10SO_2$
In both the reactions,a common product '$x$' is obtained. The number of lone pair of electrons on the central atom of '$x$' is:

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Observe the following reactions:$P_4 + 8SOCl_2 \longrightarrow 4PCl_3 + 4SO_2 + 2S_2Cl_2$$P_4 + 10SO_2Cl_2 \longrightarrow 4PCl_5 + 10SO_2$In both the reactions,a common product '$x$' is obtained. The number of lone pair of electrons on the central atom of '$x$' is: