Question

2nd attempt
part 1 (1 point)
see periodic table see hint
complete both resonance structures of nitrous acid, hno₂ by adding the missing lone pair electrons, multiple bonds, and non - zero formal charges. do not add curved arrows, reaction arrows, or plus signs between the structures.

Explanation:

Step1: Determine the total valence electrons

Nitrogen (N) has 5 valence - electrons, oxygen (O) has 6 valence - electrons, and hydrogen (H) has 1 valence - electron. So for HNO₂, the total number of valence electrons is \(1 + 5+2\times6=18\).

Step2: Draw the first resonance structure

• Place the N atom in the center. Connect H to O, and O to N, and another O to N.
• Distribute the remaining electrons as lone - pairs to satisfy the octet rule.
• Calculate formal charges. The formal charge on an atom \(FC = V - N_{l.p.}- \frac{N_{b.p.}}{2}\), where \(V\) is the number of valence electrons of the atom in isolation, \(N_{l.p.}\) is the number of non - bonding (lone - pair) electrons, and \(N_{b.p.}\) is the number of bonding electrons.
• For the first resonance structure, one O has a single bond to N and 3 lone pairs (\(FC = 6 - 6-\frac{2}{2}=- 1\)), the other O has a double bond to N and 2 lone pairs (\(FC = 6 - 4-\frac{4}{2}=0\)), N has \(FC = 5 - 2-\frac{6}{2}=0\), and H has \(FC = 1 - 0-\frac{2}{2}=0\).

Step3: Draw the second resonance structure

• Move a lone pair from the singly - bonded O to form a double bond with N, and move the double - bond electrons between the other O and N to become a lone pair on that O.
• Recalculate the formal charges. Now the first O (new double - bonded) has \(FC = 6 - 4-\frac{4}{2}=0\), the second O (new single - bonded) has \(FC = 6 - 6-\frac{2}{2}=-1\), N has \(FC = 5 - 2-\frac{6}{2}=0\), and H has \(FC = 1 - 0-\frac{2}{2}=0\).

Since this is a visual - drawing task, it's not possible to directly provide the final answer in text format here. But the general steps for constructing the resonance structures are as above. You would draw two Lewis structures for HNO₂ with the appropriate lone - pair electrons, double bonds, and formal charges as calculated.

Answer:

Step1: Determine the total valence electrons

Nitrogen (N) has 5 valence - electrons, oxygen (O) has 6 valence - electrons, and hydrogen (H) has 1 valence - electron. So for HNO₂, the total number of valence electrons is \(1 + 5+2\times6=18\).

Step2: Draw the first resonance structure

• Place the N atom in the center. Connect H to O, and O to N, and another O to N.
• Distribute the remaining electrons as lone - pairs to satisfy the octet rule.
• Calculate formal charges. The formal charge on an atom \(FC = V - N_{l.p.}- \frac{N_{b.p.}}{2}\), where \(V\) is the number of valence electrons of the atom in isolation, \(N_{l.p.}\) is the number of non - bonding (lone - pair) electrons, and \(N_{b.p.}\) is the number of bonding electrons.
• For the first resonance structure, one O has a single bond to N and 3 lone pairs (\(FC = 6 - 6-\frac{2}{2}=- 1\)), the other O has a double bond to N and 2 lone pairs (\(FC = 6 - 4-\frac{4}{2}=0\)), N has \(FC = 5 - 2-\frac{6}{2}=0\), and H has \(FC = 1 - 0-\frac{2}{2}=0\).

Step3: Draw the second resonance structure

• Move a lone pair from the singly - bonded O to form a double bond with N, and move the double - bond electrons between the other O and N to become a lone pair on that O.
• Recalculate the formal charges. Now the first O (new double - bonded) has \(FC = 6 - 4-\frac{4}{2}=0\), the second O (new single - bonded) has \(FC = 6 - 6-\frac{2}{2}=-1\), N has \(FC = 5 - 2-\frac{6}{2}=0\), and H has \(FC = 1 - 0-\frac{2}{2}=0\).

Since this is a visual - drawing task, it's not possible to directly provide the final answer in text format here. But the general steps for constructing the resonance structures are as above. You would draw two Lewis structures for HNO₂ with the appropriate lone - pair electrons, double bonds, and formal charges as calculated.