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(D) To determine the total number of lone pairs in $N_2O_3$,we look at its Lewis structure.In the structure of $N_2O_3$ $(O=N-O-N=O)$,the distribution

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$8$

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(D) To determine the total number of lone pairs in $N_2O_3$,we look at its Lewis structure.In the structure of $N_2O_3$ $(O=N-O-N=O)$,the distribution of lone pairs is as follows:$1$. Each terminal oxygen atom $(=O)$ has $2$ lone pairs.$2$. The central oxygen atom $(-O-)$ has $2$ lone pairs.$3$. Each nitrogen atom has $1$ lone pair.Total lone pairs = $(2 	imes 2) + 2 + (2 	imes 1) = 4 + 2 + 2 = 8$.Thus,the total number of lone pairs is $8$.

List-$I$	List-$II$
$(P). \ IF_7$	$(a). \ \text{Planar} \ \& \ \text{Polar}$
$(Q). \ SO_2$	$(b). \ \text{Non-planar} \ \& \ \text{Polar}$
$(R). \ SF_4$	$(c). \ \text{Planar} \ \& \ \text{Non-polar}$
$(S). \ CS_2$	$(d). \ \text{Non-planar} \ \& \ \text{Non-polar}$

The total number of lone pairs of electrons in $N_2O_3$ is

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