Return to Answer

Here are the resonance structures we can draw for the azide anion.

• Resonance structure A has an octet on each nitrogen and just one unit of formal charge on each nitrogen
• Resonance structures B and C also have an octet around each nitrogen, but there is a rather large formal charge of -2 on one of the nitrogens. Such a large charge on a small atom like nitrogen makes structures B and C less favorable and therefore we would expect them to count less to the "true" representation of the azide anion, than resonance structure A
• Resonance structures D and E do not have an octet around the singly-bound, terminal nitrogen, these resonance structures will count even less. In fact, since the azide ion is linear and these two resonance structures are bent, we can say that they don't really contribute to the true picture of the azide ion.

So if we consider only resonance structures A and B (C is equivalent to B) we can see that the central nitrogen is $\ce{sp}$ hybridized in both cases (like in allene for A and like in acetylene for B). The $\ce{sp}$ hybridization of the central nitrogen is consistent with the linear structure of the azide ion.

In A, the terminal nitrogens may both be $\ce{sp^2}$ hybridized; after all the drawing looks similar to the double bond in ethylene. However, there are other options. For example, both nitrogens could be $\ce{sp}$ hybridized, with an $\ce{sp}$ single bond, one lone pair in a $\ce{sp}$ orbital and the other lone pair in a $\ce{p}$ orbital (just like the 2 lone pairs on oxygen in water). Without more data there is no way to tell for sure, but in an introductory class $\ce{sp^2}$ hybridization would be a reasonable answer.

In B, the terminal nitrogen with the triple bond appears to be $\ce{sp}$ hybridized just like in acetylene. The hybridization of the other terminal nitrogen in resonance structure B is anybody's guess; there are many possibilities and since there is only one ligand attached (the central nitrogen), we don't know what direction the 2 electron lone pairs are pointing in. This nitrogen might be unhybridized, $\ce{sp}$ hybridized, $\ce{sp^2}$ hybridized - again, we just don't know. Since the true hybridization of the terminal nitrogen is an equal mixture of these two hybridizations (e.g. a blend of B and C), we really don't know how the terminal nitrogens in B and c are hybridized.

In an introductory class, the answer would be that the central nitrogen is $\ce{sp}$ hybridized, and, since resonance structure A is the major resonance contributor, we might say that the terminal nitrogen is $\ce{sp^2}$ hybridized (but, from the above discussion, we know that is an oversimplification).

Here are the resonance structures we can draw for the azide anion.

• Resonance structure A has an octet on each nitrogen and just one unit of formal charge on each nitrogen
• Resonance structures B and C also have an octet around each nitrogen, but there is a rather large formal charge of -2 on one of the nitrogens. Such a large charge on a small atom like nitrogen makes structures B and C less favorable and therefore we would expect them to count less to the "true" representation of the azide anion, than resonance structure A
• Resonance structures D and E do not have an octet around the singly-bound, terminal nitrogen, these resonance structures will count even less. In fact, since the azide ion is linear and these two resonance structures are bent, we can say that they don't really contribute to the true picture of the azide ion.

So if we consider only resonance structures A and B (C is equivalent to B) we can see that the central nitrogen is $\ce{sp}$ hybridized in both cases (like in allene for A and like in acetylene for B). The $\ce{sp}$ hybridization of the central nitrogen is consistent with the linear structure of the azide ion.

In A, the terminal nitrogens may both be $\ce{sp^2}$ hybridized; after all the drawing looks similar to the double bond in ethylene. However, there are other options. For example, both nitrogens could be $\ce{sp}$ hybridized, with an $\ce{sp}$ single bond, one lone pair in a $\ce{sp}$ orbital and the other lone pair in a $\ce{p}$ orbital (just like the 2 lone pairs on oxygen in water). Without more data there is no way to tell for sure, but in an introductory class $\ce{sp^2}$ hybridization would be a reasonable answer.

In B, the terminal nitrogen with the triple bond appears to be $\ce{sp}$ hybridized just like in acetylene. The hybridization of the other terminal nitrogen in resonance structure B is anybody's guess; there are many possibilities and since there is only one ligand attached (the central nitrogen), we don't know what direction the 2 electron lone pairs are pointing in. This nitrogen might be unhybridized, $\ce{sp}$ hybridized, $\ce{sp^2}$ hybridized - again, we just don't know. Since the true hybridization of the terminal nitrogen is an equal mixture of these two hybridizations (e.g. a blend of B and C), we really don't know how the terminal nitrogens in B and C are hybridized.

In an introductory class, the answer would be that the central nitrogen is $\ce{sp}$ hybridized, and, since resonance structure A is the major resonance contributor, we might say that the terminal nitrogen is $\ce{sp^2}$ hybridized (but, from the above discussion, we know that is an oversimplification).

Here are the resonance structures we can draw for the azide anion.

• Resonance structure A has an octet on each oxygen and just one unit of formal charge on each nitrogen
• Resonance structures B and C also have an octet around each nitrogen, but there is a rather large formal charge of -2 on one of the nitrogens. Such a large charge on a small atom like nitrogen makes structures B and C less favorable and therefore we would expect them to count less to the "true" representation of the azide anion, than resonance structure A
• Resonance structures D and E do not have an octet around the singly-bound, terminal nitrogen, these resonance structures will count even less. In fact, since the azide ion is linear and these two resonance structures are bent, we can say that they don't really contribute to the true picture of the azide ion.

So if we consider only resonance structures A and B (C is equivalent to B) we can see that the central nitrogen is $\ce{sp}$ hybridized in both cases (like in allene for A and like in acetylene for B). The $\ce{sp}$ hybridization of the central nitrogen is consistent with the linear structure of the azide ion.

In A, the terminal nitrogens may both be $\ce{sp^2}$ hybridized; after all the drawing looks similar to the double bond in ethylene. However, there are other options. For example, both nitrogens could be $\ce{sp}$ hybridized, with an $\ce{sp}$ single bond, one lone pair in a $\ce{sp}$ orbital and the other lone pair in a $\ce{p}$ orbital (just like the 2 lone pairs on oxygen in water). Without more data there is no way to tell for sure, but in an introductory class $\ce{sp^2}$ hybridization would be a reasonable answer.

In B, the terminal nitrogen with the triple bond appears to be $\ce{sp}$ hybridized just like in acetylene. The hybridization of the other terminal nitrogen in resonance structure B is anybody's guess; there are many possibilities and since there is only one ligand attached (the central nitrogen), we don't know what direction the 2 electron lone pairs are pointing in. This nitrogen might be unhybridized, $\ce{sp}$ hybridized, $\ce{sp^2}$ hybridized - again, we just don't know. Since the true hybridization of the terminal nitrogen is an equal mixture of these two hybridizations (e.g. a blend of B and C), we really don't know how the terminal nitrogens in B and c are hybridized.

In an introductory class, the answer would be that the central nitrogen is $\ce{sp}$ hybridized, and, since resonance structure A is the major resonance contributor, we might say that the terminal nitrogen is $\ce{sp^2}$ hybridized (but, from the above discussion, we know that is an oversimplification).

Here are the resonance structures we can draw for the azide anion.

• Resonance structure A has an octet on each nitrogen and just one unit of formal charge on each nitrogen
• Resonance structures B and C also have an octet around each nitrogen, but there is a rather large formal charge of -2 on one of the nitrogens. Such a large charge on a small atom like nitrogen makes structures B and C less favorable and therefore we would expect them to count less to the "true" representation of the azide anion, than resonance structure A
• Resonance structures D and E do not have an octet around the singly-bound, terminal nitrogen, these resonance structures will count even less. In fact, since the azide ion is linear and these two resonance structures are bent, we can say that they don't really contribute to the true picture of the azide ion.

So if we consider only resonance structures A and B (C is equivalent to B) we can see that the central nitrogen is $\ce{sp}$ hybridized in both cases (like in allene for A and like in acetylene for B). The $\ce{sp}$ hybridization of the central nitrogen is consistent with the linear structure of the azide ion.

In A, the terminal nitrogens may both be $\ce{sp^2}$ hybridized; after all the drawing looks similar to the double bond in ethylene. However, there are other options. For example, both nitrogens could be $\ce{sp}$ hybridized, with an $\ce{sp}$ single bond, one lone pair in a $\ce{sp}$ orbital and the other lone pair in a $\ce{p}$ orbital (just like the 2 lone pairs on oxygen in water). Without more data there is no way to tell for sure, but in an introductory class $\ce{sp^2}$ hybridization would be a reasonable answer.

In B, the terminal nitrogen with the triple bond appears to be $\ce{sp}$ hybridized just like in acetylene. The hybridization of the other terminal nitrogen in resonance structure B is anybody's guess; there are many possibilities and since there is only one ligand attached (the central nitrogen), we don't know what direction the 2 electron lone pairs are pointing in. This nitrogen might be unhybridized, $\ce{sp}$ hybridized, $\ce{sp^2}$ hybridized - again, we just don't know. Since the true hybridization of the terminal nitrogen is an equal mixture of these two hybridizations (e.g. a blend of B and C), we really don't know how the terminal nitrogens in B and c are hybridized.

In an introductory class, the answer would be that the central nitrogen is $\ce{sp}$ hybridized, and, since resonance structure A is the major resonance contributor, we might say that the terminal nitrogen is $\ce{sp^2}$ hybridized (but, from the above discussion, we know that is an oversimplification).

Here are the resonance structures we can draw for the azide anion.

• Resonance structure A has an octet on each oxygen and just one unit of formal charge on each nitrogen
• Resonance structures B and C also have an octet around each nitrogen, but there is a rather large formal charge of -2 on one of the nitrogens. Such a large charge on a small atom like nitrogen makes structures B and C less favorable and therefore we would expect them to count less to the "true" representation of the azide anion, than resonance structure A
• Resonance structures D and E do not have an octet around the singly-bound, terminal nitrogen, these resonance structures will count even less. In fact, since the azide ion is linear and these two resonance structures are bent, we can say that they don't really contribute to the true picture of the azide ion.

So if we consider only resonance structures A and B (C is equivalent to B) we can see that the central nitrogen is $\ce{sp}$ hybridized in both cases (like in allene for A and like in acetylene for B). The $\ce{sp}$ hybridization of the central nitrogen is consistent with the linear structure of the azide ion.

In A, the terminal nitrogens are both $\ce{sp^2}$ hybridized, just like the double bond in ethylene. In B, the terminal nitrogen with the triple bond is $\ce{sp}$ hybridized just like in acetylene. The hybridization of the other terminal nitrogen in resonance structure B is anybody's guess; there are many possibilities and since there is only one ligand attached (the central nitrogen), we don't know what direction the 2 electron lone pairs are pointing in. This nitrogen might be unhybridized, $\ce{sp}$ hybridized (like in water where one of the oxygen lone pairs resides in an $\ce{sp}$ orbital, and the other in a $\ce{p}$ orbital), $\ce{sp^2}$ hybridized - again, we just don't know.

In an introductory class, the answer would be that the central nitrogen is $\ce{sp}$ hybridized, and, since resonance structure A is the major resonance contributor, we might say that the terminal nitrogen is $\ce{sp^2}$ hybridized (but, from the above discussion, we know that is an oversimplification).

Here are the resonance structures we can draw for the azide anion.

• Resonance structure A has an octet on each oxygen and just one unit of formal charge on each nitrogen
• Resonance structures B and C also have an octet around each nitrogen, but there is a rather large formal charge of -2 on one of the nitrogens. Such a large charge on a small atom like nitrogen makes structures B and C less favorable and therefore we would expect them to count less to the "true" representation of the azide anion, than resonance structure A
• Resonance structures D and E do not have an octet around the singly-bound, terminal nitrogen, these resonance structures will count even less. In fact, since the azide ion is linear and these two resonance structures are bent, we can say that they don't really contribute to the true picture of the azide ion.

So if we consider only resonance structures A and B (C is equivalent to B) we can see that the central nitrogen is $\ce{sp}$ hybridized in both cases (like in allene for A and like in acetylene for B). The $\ce{sp}$ hybridization of the central nitrogen is consistent with the linear structure of the azide ion.

In A, the terminal nitrogens may both be $\ce{sp^2}$ hybridized; after all the drawing looks similar to the double bond in ethylene. However, there are other options. For example, both nitrogens could be $\ce{sp}$ hybridized, with an $\ce{sp}$ single bond, one lone pair in a $\ce{sp}$ orbital and the other lone pair in a $\ce{p}$ orbital (just like the 2 lone pairs on oxygen in water). Without more data there is no way to tell for sure, but in an introductory class $\ce{sp^2}$ hybridization would be a reasonable answer.

In B, the terminal nitrogen with the triple bond appears to be $\ce{sp}$ hybridized just like in acetylene. The hybridization of the other terminal nitrogen in resonance structure B is anybody's guess; there are many possibilities and since there is only one ligand attached (the central nitrogen), we don't know what direction the 2 electron lone pairs are pointing in. This nitrogen might be unhybridized, $\ce{sp}$ hybridized, $\ce{sp^2}$ hybridized - again, we just don't know. Since the true hybridization of the terminal nitrogen is an equal mixture of these two hybridizations (e.g. a blend of B and C), we really don't know how the terminal nitrogens in B and c are hybridized.

In an introductory class, the answer would be that the central nitrogen is $\ce{sp}$ hybridized, and, since resonance structure A is the major resonance contributor, we might say that the terminal nitrogen is $\ce{sp^2}$ hybridized (but, from the above discussion, we know that is an oversimplification).