Sure, let's go through a detailed, step-by-step solution for determining the best Lewis diagram for the [tex]\( \mathrm{NO_2^-} \)[/tex] ion.
1. Determine Total Number of Valence Electrons:
- Nitrogen (N) has 5 valence electrons.
- Oxygen (O) has 6 valence electrons.
- We have two oxygens, so that's [tex]\( 2 \times 6 = 12 \)[/tex] valence electrons from oxygen.
- There is an extra electron because of the negative charge.
So, the total number of valence electrons is:
[tex]\[
5 \, (\text{from N}) + 12 \, (\text{from 2 O atoms}) + 1 \, (\text{additional e}^-) = 18 \, \text{valence electrons}
\][/tex]
2. Lewis Structure Considerations:
- Central Atom: The less electronegative atom is typically the central atom, so nitrogen will be the central atom.
- Bonding Framework: Connect nitrogen to two oxygen atoms.
- Distribute Remaining Electrons: First, satisfy the octet for oxygen atoms, then place any remaining electrons on the central atom.
3. Constructing the Lewis Diagram:
- Every bond (single line) represents 2 electrons.
- Start by forming single bonds between nitrogen and each oxygen [tex]\((\text{N}-\text{O})\)[/tex], which uses [tex]\(2 \times 2 = 4\)[/tex] electrons.
- This leaves [tex]\(18 - 4 = 14\)[/tex] electrons for lone pairs.
4. Placing Lone Pairs:
- First, place lone pairs on oxygen atoms to satisfy their octet. Each oxygen will need 6 more electrons to complete their octet.
- Placing 3 lone pairs on each oxygen uses [tex]\(2 \times 6 = 12\)[/tex] electrons.
- This leaves [tex]\(14 - 12 = 2\)[/tex] electrons.
5. Placing Remaining Electrons:
- Place the remaining 2 electrons on nitrogen, giving nitrogen a total of 5 shared and unshared electrons.
6. Final Lewis Structure:
- Double bonds can help nitrogen achieve an octet while minimizing formal charge.
- One oxygen might form a double bond with nitrogen, while the other oxygen remains with a single bond and bearing a negative charge.
- The most stable structure satisfies octet rule and minimizes formal charge distribution.
Therefore, the best Lewis structure for [tex]\( \mathrm{NO_2^-} \)[/tex] can be represented as:
[tex]\[
\begin{array}{c}
\ddot{O} = \ddot{N} - \ddot{O}
\end{array}
\][/tex]
Among the choices given:
a.
[tex]\[
r: \ddot{O}-\ddot{N}-\ddot{O}=
\][/tex]
b. [tex]\( N=5 e \)[/tex]
c.
[tex]\[
[\ddot{N}-\ddot{O}=\ddot{O}]
\][/tex]
d.
[tex]\[
20=2 \times b e-
\][/tex]
e.
[tex]\[
\left[\begin{array}{c}
\ddot{N} \\
: O: O:
\end{array}\right]
\][/tex]
The best representation corresponds to the structure exhibited in choice (a). Here is the correctly formatted Lewis diagram:
[tex]\[
\ddot{O}-\ddot{N}=\ddot{O} \quad \text{with lone pairs appropriately on oxygens and nitrogen.}
\][/tex]
Therefore, the correct answer is a.
