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(D) To determine the number of lone pairs,we draw the Lewis structures:$1$. $CO$: The structure is $:C \equiv O:$. It has $1$ lone pair on $C$ and $1$

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$A < B < D < C$

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(D) To determine the number of lone pairs,we draw the Lewis structures:$1$. $CO$: The structure is $:C \equiv O:$. It has $1$ lone pair on $C$ and $1$ lone pair on $O$,total = $2$ lone pairs.$2$. $NO_2^-$: The structure is $[:O-N=O:]^-$. $N$ has $1$ lone pair,one $O$ has $3$ lone pairs,and the other $O$ has $2$ lone pairs,total = $6$ lone pairs.$3$. $CO_3^{2-}$: The structure has $3$ oxygen atoms each with $3$ lone pairs (for single-bonded $O$) or $2$ lone pairs (for double-bonded $O$). Total lone pairs = $8$.$4$. $NF_3$: $N$ has $1$ lone pair and each of the $3$ $F$ atoms has $3$ lone pairs,total = $1 + 3 	imes 3 = 10$ lone pairs.Comparing the counts: $CO (2) Thus,the increasing order is $A < B < D < C$. The correct option is $(D)$.

Arrange the following species in the increasing order of lone pairs of electrons:
$(A)$ $CO$
$(B)$ $NO_2^-$
$(C)$ $NF_3$
$(D)$ $CO_3^{2-}$

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Arrange the following species in the increasing order of lone pairs of electrons:$(A)$ $CO$$(B)$ $NO_2^-$$(C)$ $NF_3$$(D)$ $CO_3^{2-}$