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Exercise · Q7

Q.For the nitrate ion, NO3−\text{NO}_3^-, draw one Lewis structure and calculate the formal charge on nitrogen and on each type of oxygen atom present. Show that the formal charges add up to the overall charge on the ion.

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NO3−\text{NO}_3^- has 5+3(6)+1=245 + 3(6) + 1 = 24 valence electrons — the same total as CO32−\text{CO}_3^{2-}, since nitrogen's one extra valence electron relative to carbon is exactly offset by the ion's smaller (1-, rather than 2-) charge. The Lewis structure again places one oxygen doubly bonded to the central atom and the other two singly bonded, each singly bonded oxygen carrying three lone pairs and the doubly bonded oxygen carrying two lone pairs. Applying FC=(valence electrons)−(lone-pair electrons)−12(bonding electrons)FC = (\text{valence electrons}) - (\text{lone-pair electrons}) - \tfrac{1}{2}(\text{bonding electrons}): nitrogen has 44 bonding pairs and no lone pairs, so FCN=5−0−12(8)=5−4=+1FC_N = 5 - 0 - \tfrac{1}{2}(8) = 5 - 4 = +1. The doubly bonded oxygen has two lone pairs and 44 bonding electrons, so FC=6−4−2=0FC = 6 - 4 - 2 = 0. Each singly bonded oxygen has three lone pairs and 22 bonding electrons, so FC=6−6−1=−1FC = 6 - 6 - 1 = -1. Summing: (+1)+0+(−1)+(−1)=−1(+1) + 0 + (-1) + (-1) = -1, exactly matching the ion's actual charge of 1−1-, confirming the structure is correctly drawn — and showing that, unlike neutral CO2\text{CO}_2, the central atom itself (nitrogen) carries a nonzero formal charge in this ion. [!ANSWER] Nitrogen carries formal charge +1+1, the double-bonded oxygen 00, and each single-bonded oxygen −1-1, summing to the ion's overall −1-1 charge.

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