Question
Write the Lewis structure of nitrite ion, $\mathrm{NO}_2^{-}$.
✓
Answer
Step I: Count the total number of valence electrons of nitrogen atom, oxygen atom and one electron of additional negative charge.
Valence shell configuration of nitrogen and oxygen are:
$
N \Rightarrow\left(2 s^2 2 p^3\right), O \Rightarrow\left(2 s^2 2 p^4\right)
$
The total electrons available are:
$
5+(2 \times 6)+1=18 \text { electrons }
$
Step II: The skeletal structure of $\mathrm{NO}_2^{-}$is written as $\mathrm{O} \mathrm{N} \mathrm{O}$
Step III: Draw a single bond i.e., one shared electron pair between the nitrogen and each oxygen atoms. Then distribute the remaining electrons to achieve noble gas configuration for each atom. This does not complete the octet of nitrogen.
Hence, there is a multiple bond between nitrogen and one of the oxygen atoms (a double bond). The remaining two electrons constitute a lone pair on nitrogen. Following are Lewis dot structures of $\mathrm{NO}_2^{-}$.
Valence shell configuration of nitrogen and oxygen are:
$
N \Rightarrow\left(2 s^2 2 p^3\right), O \Rightarrow\left(2 s^2 2 p^4\right)
$
The total electrons available are:
$
5+(2 \times 6)+1=18 \text { electrons }
$
Step II: The skeletal structure of $\mathrm{NO}_2^{-}$is written as $\mathrm{O} \mathrm{N} \mathrm{O}$
Step III: Draw a single bond i.e., one shared electron pair between the nitrogen and each oxygen atoms. Then distribute the remaining electrons to achieve noble gas configuration for each atom. This does not complete the octet of nitrogen.
Hence, there is a multiple bond between nitrogen and one of the oxygen atoms (a double bond). The remaining two electrons constitute a lone pair on nitrogen. Following are Lewis dot structures of $\mathrm{NO}_2^{-}$.
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