# In $N{{O}_{3}}^{-}$ ion, number of bond pair and lone pair of electrons on nitrogen are:(A) 2,2(B) 3,1(C) 1,3(D) 4,0

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Hint: Draw the structure of $N{{O}_{3}}^{-}$ representing all the bonds. Nitrogen cannot have 5 bonds although it shows oxidation state of +5, hence there is a possibility of coordinate bond. The bond pair for every coordinate bond is taken as 1.

Let us draw the structure of $N{{O}_{3}}^{-}$ to understand the types of bond present in the ion:
We see that there are 2$\sigma$ bonds, 1$\pi$bond and 1 coordinate bond. Theoretically the coordinate bond is equivalent to a double bond but differs in terms of reactivity towards an incoming ion.
Note: Nitrogen shows the oxidation state of +5 in $N{{O}_{3}}^{-}$ ion. However, nitrogen cannot have 5 bonds attached to itself. This is because nitrogen does not have vacant orbitals to expand to its octet. From the subsequent period, the elements have vacant d orbitals and can expand their octet. For e.g. phosphorus can have 5 bonds attached to itself as seen in the molecule $PC{{l}_{5}}$,but $NC{{l}_{5}}$cannot exist.