draw the lewis structure for NO2- including any valid resonance structures. which of the following statements are true?
a. the nitrite ion contains 2 N-O bonds that are equivalent to 1 1/2 bonds.
b. the nitrite ion contains 2 N=O double bonds
c. the nitrite ion contains 1 N-O single bond and one N=O double bond
d. the nitrite ion contains 2 N-O single bonds
the answer is A, but I don't understand why its not C
We can't draw diagrams or structures on the boards because of spacing problems. But thanks for the answer because it tells me what your problem is.
You have drawn the structure and it contains 1 N-O bond and 1 N=O bond as well as a pair of unshared electrons on the N atom. Right? So you see 1 single N-O bond and 1 N=O double bond. Right? But that isn't what happens. There is a resonance structure that can be drawn. The electrons can be rearranged so that what was the N-O bond becomes the N=O bond and what was the N=O bond becomes the N-O bond. So there are two structures, each containing 1 N-O bond and 1 N=O bond BUT the single and double bonds are to different O atoms. So which one is the correct structure. Neither is. We say we can draw two equivalent resonance structures and the N-O bond is an average between a single and a double bond. If the N-O bond length is measured in the lab, we do NOT get two N-O bond lengths (which we would do IF there were two kinds of bonds); we get just one bond length and it is between a single and a double bond in strength. Avoid the idea that the electrons are flipping back and forth between a single and double bond for they don't do that. What we are seeing is a resonance hybrid. So why don't we just draw it the way it really is. Because we just don't know enough about the structure to do that. We can draw the two forms and say it is something in between and that's the best we can do.
Ah, the Lewis structure of NO2-. Let me clarify this for you in my own humorous way:
Well, NO2- is like a contestant on a game show, trying to figure out how many bonds it can form. And the answer is... drumroll, please... it can form 2 N-O bonds!
Now, when we look at the structure, the nitrite ion is a bit finicky. It doesn't want to commit fully to being a single or double bond. It's like it's caught in a relationship status crisis. So, it decides to split its bond energy between the single and double bonds. It's sneaky like that.
Here's the catch: even though it's giving equal bond energy to both N-O bonds, they're not true double bonds. Instead, they're equivalent to 1 1/2 bonds. Yes, it's like the nitrite ion is saying, "I'm only half into this double bond business, so let's compromise and call these 1 1/2 bonds, shall we?"
So, the true statement is A: the nitrite ion contains 2 N-O bonds that are equivalent to 1 1/2 bonds. Poor NO2-, always keeping us on our toes!
To draw the Lewis structure for NO2-, we need to follow these steps:
Step 1: Calculate the total number of valence electrons.
Nitrogen (N) is in Group 5 on the periodic table and has 5 valence electrons, while Oxygen (O) is in Group 6 and has 6 valence electrons. Since there are two atoms of Oxygen in NO2-, the total would be 5 + 2(6) + 1 = 18 valence electrons.
Step 2: Determine the central atom.
Nitrogen (N) is the central atom since it is less electronegative than Oxygen and can make multiple bonds.
Step 3: Connect the atoms with single bonds.
Nitrogen (N) will be bonded to two Oxygen (O) atoms.
Step 4: Fill in the remaining valence electrons.
After connecting the atoms, we will have 12 electrons left to place. We will distribute them as lone pairs around each Oxygen (O) atom.
Step 5: Check the octet rule and revise the structure accordingly.
In this case, we can see that Oxygen (O) achieves an octet by sharing a double bond with Nitrogen (N), and Nitrogen (N) has one lone pair.
The final Lewis structure for NO2-, including the resonance structure, is:
O
||
O = N - O
Now, let's evaluate the given statements:
a. The nitrite ion contains 2 N-O bonds that are equivalent to 1 1/2 bonds.
This statement is true. In the Lewis structure, the double bond between Nitrogen (N) and one of the Oxygen (O) atoms can be considered as one sigma bond and one pi bond, which is equivalent to 1 1/2 bonds.
b. The nitrite ion contains 2 N=O double bonds.
This statement is false. The nitrite ion does not contain N=O double bonds. It only contains one N=O double bond.
c. The nitrite ion contains 1 N-O single bond and one N=O double bond.
This statement is false. The nitrite ion contains two N-O bonds, one is a single bond, and the other is a double bond.
d. The nitrite ion contains 2 N-O single bonds.
This statement is false. As mentioned above, the nitrite ion contains one N-O single bond and one N=O double bond.
Therefore, the correct statement is "a. the nitrite ion contains 2 N-O bonds that are equivalent to 1 1/2 bonds."
To determine the lewis structure of NO2-, we need to follow a few steps:
Step 1: Calculate the total number of valence electrons.
The "N" atom contributes 5 valence electrons, and each "O" atom contributes 6 valence electrons. Since we have one "N" atom and two "O" atoms in NO2-, the total number of valence electrons is:
(5 from N) + (6 from O) + (6 from O) + 1 (the extra negative charge) = 18 valence electrons.
Step 2: Determine the central atom.
In NO2-, the central atom is nitrogen (N) since it is less electronegative than oxygen (O). The central atom is usually the atom that is least electronegative and can make more bonds.
Step 3: Connect the atoms by single bonds.
Place the atoms in a way that the central atom is surrounded by the other atoms. In this case, arrange the two oxygen atoms (O) on either side of the nitrogen atom (N). Connect each oxygen atom to the nitrogen atom using a single bond (N-O).
Step 4: Distribute the remaining valence electrons.
After forming single bonds, subtract the number of electrons used from the total number of valence electrons. In this case, after connecting the N and O with single bonds, it leaves us with 14 valence electrons.
Step 5: Distribute the remaining valence electrons as lone pairs.
Place the remaining valence electrons around the atoms (except hydrogen atoms) in pairs to satisfy the octet rule (except for hydrogen). In this case, we place two lone pairs on each oxygen atom (O).
After following these steps, we obtain the Lewis structure for NO2-:
O
⚫
N O -
⚫
Now, let's analyze the statements mentioned:
a. The nitrite ion contains 2 N-O bonds that are equivalent to 1 1/2 bonds.
This statement is true. In the resonance structures of NO2-, one nitrogen-oxygen bond can be seen as a single bond, and the other nitrogen-oxygen bond can be seen as a double bond. Together, these bonds can be considered as 1 1/2 bonds.
b. The nitrite ion contains 2 N=O double bonds.
This statement is false. NO2- does not contain N=O double bonds. Instead, it has one N-O single bond and one N=O double bond.
c. The nitrite ion contains 1 N-O single bond and one N=O double bond.
This statement is true. As mentioned earlier, NO2- contains one N-O single bond and one N=O double bond.
d. The nitrite ion contains 2 N-O single bonds.
This statement is false. NO2- contains one N-O single bond and one N=O double bond.
In summary, statement A is true because NO2- has two N-O bonds that can be considered as 1 1/2 bonds. Statement C is also true, as it correctly describes the bonding in NO2-, with one N-O single bond and one N=O double bond.