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## Q. 7.2

Draw the Lewis structures of ${NO _{2}}^{- }$ and $N _{2}$.

STRATEGY

Follow the steps outlined in Figure 7.3. ## Verified Solution

Skeleton for ${NO _{2}}^{- }$        $[O–N–O]^{-}$

VE            2(6 (for O)) + 5 (for N) + 1(-1 charge) = 18

AE          AE = VE – 2(bonds) = 18 – 2(2 bonds) = 14

NE              2(6 (for each O)) + 4 (for N) = 16

AE = NE ?              No; 2 electrons short

Convert a single bond to a double bond.<\p>

Lewis structure             $\left[:\overset{..}{\underset{..}{\overset{..}{O} }-\overset{..}{N}= \overset{..}{O}: } \right] ^{-}$
Skeleton for $N _{2}$          N–N

VE           2(5 (for each N)) = 10

AE            AE = VE – 2(bonds) = 10 – 2(1 bond) = 8

NE             2 × 6 (for each N to have an octet) = 12
AE= NE ?           No; 4 electrons short

Convert a single bond to a triple bond.

Lewis structure      :N$\equiv$ N:

END POINT

For the Lewis structure of $NO^{- } _{2}$, it does not matter which single bond you convert to a double bond. We will talk about this in more detail when we discuss resonance forms.