Question

Draw three resonance structures for the molecule ${\mathrm{N}}_2\mathrm{O}$ in which the atoms are arranged in the order NNO. Indicate formal charges and arrange the resonance structures in order of increasing relative importance.

Short answer

N=N=O is the most stable; N-N≡O is less stable; N≡N-O is the least stable.

Step-by-step solution

Assign Valence Electrons

Identify the total number of valence electrons for the molecule. N has 5 valence electrons, and O has 6. Since there are 2 N atoms and 1 O atom, the total is: $$2(5)+6=16$$ valence electrons.

Draw the First Resonance Structure (N≡N-O)

Start by drawing a structure with a triple bond between the nitrogen atoms and a single bond to oxygen: N≡N-O. Assign lone pairs to fulfill the octet rule: - Each N in the triple bond takes 2 electrons each, leaving one lone pair on N. - O gets three lone pairs to complete its octet. - Assign formal charges: - Left N: 5 - (2+3) = 0 - Middle N: 5 - (6) = -1 - O: 6 - (1+6) = +1

Draw the Second Resonance Structure (N=N=O)

Draw a structure with a double bond between each of the N atoms and O: N=N=O. Assign lone pairs properly: - Each N gets one lone pair. - O receives two lone pairs. Calculate formal charges: - Left N: 5 - (4) = +1 - Middle N: 5 - (2+4) = 0 - O: 6 - (2+4) = 0

Draw the Third Resonance Structure (N-N≡O)

Create a structure with a triple bond to O and a single bond between the N atoms: N-N≡O. Add lone pairs to achieve full octets: - Left N gets three lone pairs. - Middle N shares three pairs with O. - O gets one lone pair. Calculate the formal charges: - Left N: 5 - (6) = -1 - Middle N: 5 - (3) = +1 - O: 6 - (3+2) = 0

Rank Resonance Structures by Stability

To order the resonance structures by stability, consider formal charges and complete octets: 1. Second structure (N=N=O) is the most stable as it has fewer formal charges, and the negative charge is on N. 2. Third structure (N-N≡O) is less stable due to unfavorable positive charge on N. 3. First structure (N≡N-O) is least stable because it places a positive formal charge on O, which is less electronegative.

Key concepts

Valence Electrons

Valence electrons are the outermost electrons of an atom that participate in bonding. Understanding valence electrons is crucial for drawing Lewis structures and resonance forms. In you count the valence electrons as follows: For nitrogen (N), there are 5 valence electrons, and for oxygen (O), there are 6. For the molecule ${\mathrm{N}}_2\mathrm{O}$:

• 2 nitrogen atoms contribute $2\times 5=10$ electrons.
• 1 oxygen atom contributes 6 electrons.

Thus, ${\mathrm{N}}_2\mathrm{O}$ has a total of 16 valence electrons."Keeping track of all valence electrons helps ensure that all resonance structures maintain the molecule's electron count throughout the drawing process.

Formal Charge

Formal charge helps in determining the most stable resonance structure by indicating the distribution of electrons in a molecule. To calculate a formal charge, follow this formula:$$\text{Formal Charge}=\text{Valence Electrons}-\text{Non-bonding Electrons}-\frac{1}{2}(\text{Bonding Electrons})$$Let's apply this to resonance structures of ${\mathrm{N}}_2\mathrm{O}$:

• First structure $(\mathrm{N}\equiv \mathrm{N}-\mathrm{O})$: The formal charge balances out with a +1 on oxygen and -1 on the middle nitrogen.
• Second structure $(\mathrm{N}=\mathrm{N}=\mathrm{O})$: There are minimal formal charges, showing a more stable distribution.
• Third structure $(\mathrm{N}-\mathrm{N}\equiv \mathrm{O})$: The presence of a positive formal charge on the first nitrogen shows decreased stability.

Using formal charges helps select the most favorable resonance structures for the stability of the molecule.

Octet Rule

The octet rule states that atoms tend to bond in a way that each atom has eight electrons in its valence shell, achieving a noble gas configuration. This rule is essential in drawing resonance structures for molecules like ${\mathrm{N}}_2\mathrm{O}$.When creating resonance structures, ensure each atom satisfies the octet rule:

• In $\mathrm{N}\equiv \mathrm{N}-\mathrm{O}$: Both nitrogens share six electrons, while oxygen gets three lone pairs to complete its octet.
• In $\mathrm{N}=\mathrm{N}=\mathrm{O}$: Each nitrogen and the oxygen atom have fulfilled the octet rule with lone pairs and double bonds.
• In $\mathrm{N}-\mathrm{N}\equiv \mathrm{O}$: Though the octet rule seems fulfilled, the distribution of electrons affects stability adversely due to formal charge imbalance.

Remember, while the octet rule guides the stability of structures, exceptions do exist varying with electronegativity and size of the elements involved.

Molecular Stability

Molecular stability in the context of resonance structures refers to the likelihood of a particular structure based on the distribution of electrons and alignment of charges. Key factors affecting stability include:

• Fewer formal charges lead to greater stability.
• Negative charges located on more electronegative atoms (like O rather than N) increase stability.
• A complete octet for each atom significantly contributes to stability.

For the ${\mathrm{N}}_2\mathrm{O}$ molecule:- The $\mathrm{N}=\mathrm{N}=\mathrm{O}$ structure is prioritized due to minimal and balanced formal charges. The oxygen carries a neutral charge where it should be.- Conversely, $\mathrm{N}-\mathrm{N}\equiv \mathrm{O}$ is less stable because of the unfavorable placement of a formal positive charge on nitrogen.- The $\mathrm{N}\equiv \mathrm{N}-\mathrm{O}$ structure, although fulfilling octet rules, places a positive formal charge on oxygen, further reducing stability.Effectively evaluating molecular stability can predict which resonance form contributes most to the actual molecule's properties.