Question

which of the following lewis diagrams best represents the bonding in the n₂o molecule, considering formal charges? a :n=n=o: b :n≡n - o: c :n - n=o: d :n - o=n:

Explanation:

Step1: Recall formal - charge formula

The formal - charge formula is $FC = V - N - \frac{B}{2}$, where $V$ is the number of valence electrons, $N$ is the number of non - bonding electrons, and $B$ is the number of bonding electrons.

Step2: Analyze Option A

For the left - hand $N$: $V = 5$, $N = 2$, $B = 4$, so $FC=5 - 2-\frac{4}{2}=1$. For the middle $N$: $V = 5$, $N = 0$, $B = 8$, so $FC = 5-0 - \frac{8}{2}=1$. For $O$: $V = 6$, $N = 4$, $B = 4$, so $FC=6 - 4-\frac{4}{2}=0$.

Step3: Analyze Option B

For the left - hand $N$: $V = 5$, $N = 2$, $B = 6$, so $FC=5 - 2-\frac{6}{2}=0$. For the middle $N$: $V = 5$, $N = 0$, $B = 8$, so $FC = 5-0 - \frac{8}{2}=1$. For $O$: $V = 6$, $N = 6$, $B = 2$, so $FC=6 - 6-\frac{2}{2}=-1$.

Step4: Analyze Option C

For the left - hand $N$: $V = 5$, $N = 4$, $B = 2$, so $FC=5 - 4-\frac{2}{2}=0$. For the middle $N$: $V = 5$, $N = 2$, $B = 4$, so $FC = 5-2 - \frac{4}{2}=1$. For $O$: $V = 6$, $N = 4$, $B = 4$, so $FC=6 - 4-\frac{4}{2}=0$.

Step5: Analyze Option D

For the left - hand $N$: $V = 5$, $N = 4$, $B = 2$, so $FC=5 - 4-\frac{2}{2}=0$. For $O$: $V = 6$, $N = 4$, $B = 4$, so $FC=6 - 4-\frac{4}{2}=0$. For the right - hand $N$: $V = 5$, $N = 2$, $B = 4$, so $FC = 5-2 - \frac{4}{2}=1$.
The best Lewis structure has the lowest formal charges. In Option D, the sum of the formal charges is $0 + 0+1=1$, in Option C the sum is $0 + 1+0 = 1$, in Option B the sum is $0 + 1+( - 1)=0$, and in Option A the sum is $1 + 1+0 = 2$. Option B has the most stable distribution of formal charges.

Answer:

B. $:N\equiv N - \ddot{O}:$