In NO_3^{-} ion,the number of bond pairs and | vedclass

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(D) The nitrogen atom in the $NO_3^{-}$ ion has a valence shell configuration of $2s^2 2p^3$. In the nitrate ion,nitrogen forms three sigma bonds (one

Vedclass · Accepted Answer

$4,0$

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(D) The nitrogen atom in the $NO_3^{-}$ ion has a valence shell configuration of $2s^2 2p^3$. In the nitrate ion,nitrogen forms three sigma bonds (one with each oxygen atom) and one pi bond (delocalized over the three oxygen atoms). This means nitrogen uses all its $5$ valence electrons to form $4$ bonds ($3$ single bonds and $1$ double bond equivalent in resonance). Therefore,the number of bond pairs of electrons on the nitrogen atom is $4$. Since all valence electrons are involved in bonding,there are $0$ lone pairs of electrons on the nitrogen atom. Thus,the correct answer is $4,0$.

List-$I$	List-$II$
$(P)$ $XeF_2$	$(5)$ Trigonal bipyramidal and three lone pairs of electrons
$(Q)$ $XeF_4$	$(3)$ Square planar and two lone pairs of electrons
$(R)$ $XeO_3$	$(2)$ Pyramidal and one lone pair of electrons
$(S)$ $XeO_3F_2$	$(4)$ Trigonal bipyramidal and no lone pairs of electrons

In $NO_3^{-}$ ion,the number of bond pairs and lone pairs of electrons on the nitrogen atom are

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