The formal charge of the O-atoms in the ion :N·  · = O·  ·:  is

# The formal charge of the O-atoms in the ion $\left[:\stackrel{·\text{\hspace{0.17em}\hspace{0.17em}}·}{N}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\stackrel{·\text{\hspace{0.17em}\hspace{0.17em}}·}{O}:\right]\text{\hspace{0.17em}\hspace{0.17em}}is$

1. A

$-2$

2. B

+ 1

3. C

$-1$

4. D

0

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### Solution:

$\left[Formal\text{\hspace{0.17em}\hspace{0.17em}}ch\mathrm{arg}e\text{\hspace{0.17em}\hspace{0.17em}}on\text{\hspace{0.17em}\hspace{0.17em}}O-atom\right]\text{\hspace{0.17em}}=\frac{Total\text{\hspace{0.17em}\hspace{0.17em}}ch\mathrm{arg}e}{No.\text{\hspace{0.17em}\hspace{0.17em}}of\text{\hspace{0.17em}\hspace{0.17em}}O-atoms}=0$

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