There are several oxides of nitrogen; among the most common are N2O, NO, and NO2.
1.Write the Lewis structures for each of these molecules. In each case the oxygens are terminal atoms.
2. Which of these molecules "violate" the octet rule?
3. Draw resonance structures for N2O.
4. How would you expect the nitrogen to oxygen bond distance in NO compare to the average bond distance in NO2?
5. What experimental evidence leads us to believe that there is a double bond in the oxygen molecule? What experimental evidence leads us to believe that there are unpaired electrons in an oxygen molecule? Can you write a reasonable Lewis structure for O2? Explain!!
IM SOO STUCK!!
There are rules to follow to draw the electron dot structure of molecules. Read your text and your notes then try your hand at it. Here is a site that should help you.
http://dbhs.wvusd.k12.ca.us/webdocs/Bonding/Lewis-Structure-Rules.html
Post specific questions about the above which you don't understand.
It is almost impossible to draw these structures on the computer.
I understand that you are stuck on drawing Lewis structures for the oxides of nitrogen and have questions about them. Drawing Lewis structures can be a bit tricky, but I can certainly guide you through the process.
1. Lewis structures for N2O, NO, and NO2:
To draw a Lewis structure, you need to determine the total number of valence electrons for each molecule. For N2O, N has 5 valence electrons and O has 6 valence electrons. Since there are two N atoms and one O atom, the total number of valence electrons is (2 * 5) + 6 = 16.
N2O:
N≡N=O
NO:
O=N
NO2:
O−N=O
2. Violation of the octet rule:
The octet rule states that atoms tend to gain, lose, or share electrons to have a complete octet (8 valence electrons) in their outer shell, except for hydrogen, which aims for a duet (2 valence electrons). To determine if a molecule violates the octet rule, you need to count the number of valence electrons contributed by the atoms in the molecule and compare it to the number of valence electrons required for an octet.
In N2O, N contributes 5 electrons, O contributes 6 electrons, and there are two N atoms and one O atom. So, the total number of valence electrons is (2 * 5) + 6 = 16. This means every atom in N2O satisfies the octet rule.
In NO, N contributes 5 electrons, and O contributes 6 electrons. So, the total number of valence electrons is 5 + 6 = 11. Since there is an odd number of electrons, NO violates the octet rule but it is still a stable molecule due to the presence of an unpaired electron.
In NO2, O contributes 6 electrons, and N contributes 5 electrons. So, the total number of valence electrons is 5 + (2 * 6) = 17. Since there is one extra electron, NO2 violates the octet rule.
3. Resonance structures for N2O:
To draw resonance structures, you need to move electrons around while keeping the atoms in the same positions. Here are the resonance structures for N2O:
N≡N=O ⇌ N=N=O
4. Nitrogen-oxygen bond distance in NO compared to NO2:
In general, the nitrogen-oxygen bond distance is shorter in NO compared to the average bond distance in NO2. This is because NO has a triple bond between nitrogen and oxygen, which is shorter and stronger than the double bond in NO2. The presence of more shared electrons in a bond leads to a higher bond order and shorter bond distance.
5. Experimental evidence for a double bond in the oxygen molecule:
Experimental evidence for a double bond in the oxygen molecule (O2) can be observed through spectroscopy. Oxygen molecules absorb wavelengths of light that correspond to the breaking and formation of double bonds. This absorption occurs in the ultraviolet region, indicating the presence of a double bond.
Experimental evidence for unpaired electrons in an oxygen molecule can be observed through electron paramagnetic resonance (EPR) spectroscopy. EPR spectroscopy measures the interaction between unpaired electrons and an external magnetic field. Oxygen molecules exhibit a characteristic EPR spectrum, indicating the presence of unpaired electrons.
A reasonable Lewis structure for O2 is:
O=O
In this structure, both oxygen atoms share a double bond, contributing 2 valence electrons each, making a total of 4 shared electrons. This satisfies the octet rule for each oxygen atom, as they each have a total of 8 valence electrons.