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Nitrogen forms largest number of oxides ...

Nitrogen forms largest number of oxides in which oxidation state of nitrogen varies from +1 to +5 `N_(2)O,NO,N_(2)O_(3),N_(2)O_(4) and N_(2)O_(5)` respectively. The largest number of oxides are formed due to capability of forming stable multiple bonds by nitrogen with oxygen.
Q. Which of the following is/are correct statement(s)?

A

In `N_(2)O_(4)` the N-N bond length is longer than the usual N-N single bond distance

B

`NO_(2)` molecule is angular and N-O bond length is intermediate between a single and a double bond

C

`N_(2)O` is linear molecule and has small dipole moment

D

All of these

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The correct Answer is:
To analyze the question regarding the nitrogen oxides and determine which statements are correct, we can follow these steps: ### Step 1: Understand the Oxides of Nitrogen Nitrogen can form several oxides with oxidation states ranging from +1 to +5. The oxides mentioned are: - N₂O (Nitrous oxide, oxidation state +1) - NO (Nitric oxide, oxidation state +2) - N₂O₃ (Dinitrogen trioxide, oxidation state +3) - N₂O₄ (Dinitrogen tetroxide, oxidation state +4) - N₂O₅ (Dinitrogen pentoxide, oxidation state +5) ### Step 2: Analyze Each Statement We need to evaluate the correctness of the statements provided. #### Statement A: "In N₂O₄, Nitrogen-Nitrogen bond length is longer than usual Nitrogen-Nitrogen single bond distance." - **Analysis**: The structure of N₂O₄ shows that there is resonance and delocalization of electrons, leading to partial double bond character between the nitrogen atoms. This results in a shorter bond length compared to a typical N-N single bond. - **Conclusion**: This statement is **incorrect**. #### Statement B: "NO₂ molecule is angular and NO bond length is intermediate between single and double bond." - **Analysis**: The structure of NO₂ is indeed angular (bent) due to the presence of a lone pair on nitrogen. The bond length of the N-O bond is intermediate because of resonance between the structures (N=O and N-O⁻). - **Conclusion**: This statement is **correct**. #### Statement C: "N₂O is a linear molecule and has a small dipole moment." - **Analysis**: The structure of N₂O is linear (N≡N-O) and the dipole moment is small due to the symmetry of the molecule. - **Conclusion**: This statement is **correct**. ### Final Conclusion Based on the analysis: - Statement A is incorrect. - Statement B is correct. - Statement C is correct. Thus, the correct statements are B and C.
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Nitrogen forms largest number of oxides in which oxidation state of nitrogen varies from +1 to +5 N_(2)O,NO,N_(2)O_(3),N_(2)O_(4) and N_(2)O_(5) respectively. The largest number of oxides are formed due to capability of forming stable multiple bonds by nitrogen with oxygen. Q. Which of the following statements is/are correct regarding oxides of nitrogen?

Nitrogen forms largest number of oxides in which oxidation state of nitrogen varies from +1 to +5 N_(2)O,NO,N_(2)O_(3),N_(2)O_(4) and N_(2)O_(5) respectively. The largest number of oxides are formed due to capability of forming stable multiple bonds by nitrogen with oxygen. Q. Which of the following oxides is known as anhydride of nitric acid?

Which of the following is true for N_(2)O_(5)

Calculate the oxidation number of nitrogen in the following oxides. N_(2)O,NO,N_(2)O_(3),NO_(2),N_(2)O_(5)

The correct order of the oxidation states of nitrogen in NO, N_(2)O, NO_(2) and N_(2)O_(3) is :

The correct order of the oxidation states of nitrogen in NO, N_(2)O, NO_(2) and N_(2)O_(3) is :

Arrange in the increasing order of oxidation state of nitrogen for following nitrogen oxides N_(2)O, NO_(2), NO, N_(2)O_(3)

Out of the following oxides of nitrogen how many of them are neutral oxides? N_(2)O, NO, N_(2)O_(3), NO_(2),N_(2)O_(4), N_(2)O_(5)

Arrange the oxides of nitrogen in increasing order of oxidation state of N from +1 to +5.

Valency and oxidation number of nitrogen in N_(2)O_(5)