Does N2O violates the octet rule?
No. https://answers.yahoo.com/question/index?qid=20091111034406AAqpzhB
To determine whether N2O violates the octet rule, let's first review what the octet rule is. The octet rule states that atoms tend to gain, lose, or share electrons in order to achieve a stable configuration with eight valence electrons.
In N2O (nitrous oxide), we have one nitrogen atom (N) and two oxygen atoms (O). Nitrogen has five valence electrons, while oxygen has six valence electrons.
To check if N2O follows the octet rule, we count the total number of valence electrons in the molecule. In this case, we have:
1 nitrogen atom (N) x 5 valence electrons = 5 valence electrons
2 oxygen atoms (O) x 6 valence electrons = 12 valence electrons
So the total number of valence electrons in N2O is 5 + 12 = 17.
Next, we place the atoms in the molecule, connecting them with single bonds. In this case, the nitrogen atom will form a double bond with one of the oxygen atoms, and a single bond with the other oxygen atom.
The single bond consists of 2 electrons, and the double bond consists of 4 electrons.
Now let's distribute the remaining electrons to satisfy the octet rule.
Nitrogen already has a total of 8 electrons surrounding it (from the double bond and its three lone pairs). Each oxygen atom has a total of 8 electrons surrounding it as well (from the single bond and its three lone pairs).
When we add up the valence electrons, we have a total of:
Nitrogen: 5 (valence electrons) + 2 (from the single bond) + 6 (from the double bond) = 13 electrons
Oxygen: 6 (valence electrons) + 2 (from the single bond) = 8 electrons
Therefore, both nitrogen and each oxygen atom have achieved the octet rule, with 8 valence electrons. Hence, N2O does not violate the octet rule.