no3- lewis structure
What is the Lewis structure of NO₃⁻?
Answer:
To determine the Lewis structure of the nitrate ion (\text{NO}_3^-), follow these steps:
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Determine the total number of valence electrons:
- Nitrogen (\text{N}) is in Group 15 and has 5 valence electrons.
- Each oxygen (\text{O}) is in Group 16 and has 6 valence electrons. Since there are three oxygen atoms, the total number of valence electrons from oxygen is 3 \times 6 = 18.
- The \text{NO}_3^- ion as a whole carries a -1 charge, which means an additional electron.
\text{Total valence electrons} = 5 (\text{from N}) + 18 (\text{from 3 O atoms}) + 1 (\text{negative charge}) = 24 \text{electrons} -
Draw the skeletal structure:
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Place the nitrogen atom in the center because nitrogen is less electronegative than oxygen.
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Arrange the three oxygen atoms around the nitrogen atom.
O | O-N-O
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Distribute the electrons to form bonds and satisfy the octet rule:
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Initially, place a single bond (2 electrons) between nitrogen and each oxygen atom.
O | O–N–O
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This uses up 6 of the 24 valence electrons, leaving us with 18 electrons.
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Place remaining electrons around the oxygen atoms to complete their octets:
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Each oxygen atom needs 8 valence electrons. Each single bond already contributes 2 electrons, so each oxygen atom needs 6 more electrons to complete its octet.
.. .. ..
–N–
| |
… …- This configuration uses 18 more electrons (6 electrons per oxygen atom, totaling 18 electrons), and we have now used all 24 valence electrons.
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Check if all atoms satisfy the octet rule:
- Each oxygen atom has 8 valence electrons (2 from the bond with nitrogen and 6 lone pair electrons).
- Nitrogen has 6 valence electrons only and needs 2 more to have a full octet.
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Use double bonds if necessary and minimize formal charges:
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To minimize formal charges, we can form a double bond between nitrogen and one of the oxygen atoms. This means we move a lone pair of electrons from one of the oxygen atoms into a bonding position with nitrogen.
O || O==N–O
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Now, we adjust the lone pairs to ensure all atoms satisfy the octet rule:
.. ..
==N–
| |
… -
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Calculate formal charges to ensure the stability and resonance:
- Formal charge formula: \text{Formal charge} = \text{valence electrons} - \text{(non-bonding electrons} + \frac{1}{2} \times \text{bonding electrons})
For the oxygen with a double bond:
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Valence electrons = 6
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Non-bonding electrons = 4
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Bonding electrons = 4 (since it shares 2 bonds with nitrogen)
\text{Formal charge} = 6 - (4 + \frac{4}{2}) = 6 - 6 = 0
For the oxygen with single bonds:
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Valence electrons = 6
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Non-bonding electrons = 6
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Bonding electrons = 2
\text{Formal charge} = 6 - (6 + \frac{2}{2}) = 6 - 7 = -1
For nitrogen:
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Valence electrons = 5
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Non-bonding electrons = 0
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Bonding electrons = 8 (since it shares 4 bonds in total)
\text{Formal charge} = 5 - (0 + \frac{8}{2}) = 5 - 4 = +1
To equalize the charges, we have resonance structures, where the double bond can shift among the oxygen atoms. The resonance-stabilized structure is:
\begin{array}{c} .. .. .. :O==N–O–:O– \leftrightarrow :O:=N–O–:O– \leftrightarrow :O–N:=O–:O: \\ .. \end{array}
Final Answer:
\text{NO}_3^- has a resonance structure with a delocalized double bond, resulting in equivalent bonding situations for all three oxygen atoms, portrayed as a combination of all possible resonance structures. The most accurate representation would acknowledge the equal distribution of electrons among the oxygen atoms surrounding the nitrogen atom.