How many electrons are in the Lewis structure of a nitrite ion (NO2-)?
The Correct Answer and Explanation is:
To determine how many electrons are in the Lewis structure of a nitrite ion (NO₂⁻), we need to count the total number of valence electrons from each atom involved, accounting for the negative charge.
Step 1: Determine the number of valence electrons for each atom.
- Nitrogen (N): Nitrogen is in group 15 of the periodic table, so it has 5 valence electrons.
- Oxygen (O): Oxygen is in group 16 of the periodic table, so each oxygen atom has 6 valence electrons. Since there are two oxygen atoms in the nitrite ion, this gives us a total of ( 6 \times 2 = 12 ) valence electrons from oxygen.
- Charge: The nitrite ion has a negative charge of -1. This means that we must add 1 additional electron to the total to account for the extra negative charge.
Step 2: Add up the total number of valence electrons.
- Valence electrons from nitrogen: 5
- Valence electrons from oxygen: ( 6 \times 2 = 12 )
- Extra electron due to the negative charge: 1
Total electrons = ( 5 + 12 + 1 = 18 ) electrons.
Step 3: Draw the Lewis structure.
- In the Lewis structure of NO₂⁻, nitrogen will be the central atom. Oxygen atoms will form single bonds with nitrogen.
- Each oxygen will have lone pairs of electrons to complete their octet.
- One of the oxygen atoms will form a double bond with nitrogen to minimize formal charges and to ensure nitrogen has a full octet.
- The extra electron from the negative charge will be placed as a lone pair on one of the oxygen atoms.
Thus, the total number of electrons involved in the Lewis structure of the nitrite ion (NO₂⁻) is 18 electrons.
This count includes bonding electrons and lone pairs on each atom, as well as the additional electron from the negative charge, which is essential to achieving the correct overall electron count for the ion.