Question

Give the Lewis structure of (a) \(\mathrm{NO}_{2}\) (b) \(\mathrm{NO}\) (c) \(\mathrm{SO}_{2}\) (d) \(\mathrm{SO}_{2}\)

Short answer

(a) NO₂: (b) NO: (c) SO₂: (d) SO₃:

Step-by-step solution

1. Count the total valence electrons for each molecule.

For each molecule, determine how many valence electrons are present based on the periodic table. (a) \(\mathrm{NO}_{2}\): Nitrogen has 5 valence electrons, and each oxygen atom has 6. So the total number of valence electrons is \(5 + 2\times6 = 17\). (b) \(\mathrm{NO}\): Nitrogen has 5 valence electrons and oxygen has 6. The total number of valence electrons is \(5 + 6 = 11\). (c) \(\mathrm{SO}_{2}\): Sulfur has 6 valence electrons, and each oxygen atom has 6. The total number of valence electrons is \(6 + 2\times6 = 18\). (d) \(\mathrm{SO}_{3}\): Sulfur has 6 valence electrons, and each oxygen atom has 6. The total number of valence electrons is \(6 + 3\times6 = 24\).

2. Arrange the atoms and place single bonds.

Place the least electronegative atom in the center and place the other atoms around it. For all molecules, nitrogen or sulfur is the central atom and oxygen atoms are surrounding the central atom. Put single bonds between the central atom and surrounding atoms. (a) \(\mathrm{NO}_{2}\): Each single bond uses 2 valence electrons, meaning 4 electrons are used in bonding. (b) \(\mathrm{NO}\): 2 valence electrons used in the single bond. (c) \(\mathrm{SO}_{2}\): 4 valence electrons used in two single bonds. (d) \(\mathrm{SO}_{3}\): 6 valence electrons used in three single bonds.

3. Distribute remaining electrons as lone pairs around the atoms.

Balance the remaining electrons around the atoms so that they follow the octet rule whenever possible. (a) \(\mathrm{NO}_{2}\): 17 - 4 = 13 electrons remaining. Distribute them as lone pairs: two lone pairs around each oxygen and a double bond to one of the oxygen atoms result in 1 electron remaining on nitrogen (unpaired). (b) \(\mathrm{NO}\): 11 - 2 = 9 electrons remaining. Distribute them as lone pairs: two lone pairs around oxygen and a double bond gives 7 electrons used, leaving 1 unpaired electron on both nitrogen and oxygen. (c) \(\mathrm{SO}_{2}\): 18 - 4 = 14 electrons remaining. Distribute them as lone pairs: two lone pairs around each oxygen and a double bond to one of the oxygen atoms result in 4 electrons remaining, which are placed as 2 lone pairs on sulfur. (d) \(\mathrm{SO}_{3}\): 24 - 6 = 18 electrons remaining. Distribute them as lone pairs: two lone pairs around each oxygen atom and a double bond to each oxygen atom result in no electrons remaining.

4. Final Lewis structures.

With the electrons distributed and the rules followed, the Lewis structures for the molecules are: (a) \(\mathrm{NO}_{2}\): (b) \(\mathrm{NO}\): (c) \(\mathrm{SO}_{2}\): (d) \(\mathrm{SO}_{3}\):