Question

Write reasonable Lewis structures for the following species, none of which follow the octet rule. (a) \(\mathrm{BeH}_{2}\) (b) \(\mathrm{CO}^{-}\) (c) \(\mathrm{SO}_{2}^{-}\) (d) \(\mathrm{CH}_{3}\)

Short answer

Question: Draw the Lewis structures for the following chemical species that do not follow the octet rule: (a) BeH2, (b) CO-, (c) SO2-, and (d) CH3. Explain the formation of each structure and the presence of any resonance structures. Answer: (a) BeH2 Lewis structure: ``` H | Be - H ``` Beryllium has only 4 electrons around it, fulfilling its valence requirement, which is less than the octet rule. Both hydrogens have filled their valence requirement (2 electrons each). (b) CO- Lewis structure has two resonance forms: ``` Resonance Form1: C ≡ O Resonance Form2: C = O: :O ``` The actual structure is an average of the two resonance structures, with a single negative charge on carbon. (c) SO2- Lewis structure: ``` O=S=O .S ``` The structure has a total of 19 valence electrons, with the central sulfur atom forming a radical species. (d) CH3 Lewis structure: ``` H | H-C-H . ``` Carbon has only 7 electrons around it, breaking the octet rule.

Step-by-step solution

(a) BeH2 Lewis structure

To draw the Lewis structure for \(\mathrm{BeH}_{2}\), first, we count the total number of valence electrons. Beryllium (Be) has 2 valence electrons and each hydrogen (H) has 1 valence electron. The total number of electrons is 4. Place Be in the center and connect it with two hydrogens as follows: ``` H | Be - H ``` Now, Be has only 4 electrons around it and has fulfilled its valence requirement which is less than the octet rule. Both hydrogens have also filled their valence requirement (2 electrons each).

(b) CO- Lewis structure

To draw the Lewis structure for \(\mathrm{CO}^{-}\), firstly we should count the total number of valence electrons. Carbon has 4 valence electrons, oxygen has 6, and due to the negative charge, we add one more electron. The total number of valence electrons is 11. Carbon is the least electronegative atom, so it becomes the central atom. Now we can connect carbon (C) and oxygen (O) with a single bond. ``` C - O ``` Each hydrogen has fulfilled its valence requirement. However, the central atom now has only two electrons. Now we can add a double and then a triple bond between C and O to fulfill the valence requirements. ``` Resonance Form1: C ≡ O Resonance Form2: C = O: :O ``` There are two resonance structures for \(\mathrm{CO}^{-}\) with a single negative charge on carbon. The actual structure is an average of the two resonance structures.

(c) SO2- Lewis structure

To draw the Lewis structure for \(\mathrm{SO}_{2}^{-}\), let's count the total number of valence electrons. Sulfur has 6 valence electrons, each oxygen has 6, and due to the negative charge, we add one additional electron. The total number of valence electrons is 19. Sulfur is the central atom. Connect it with each oxygen using a double bond, and then place the remaining electron on the sulfur, forming a radical species: ``` O=S=O .S ``` This is the Lewis structure for \(\mathrm{SO}_{2}^{-}\) with a total of 19 valence electrons.

(d) CH3 Lewis structure

To draw the Lewis structure for \(\mathrm{CH}_{3}\), count the total number of valence electrons. Carbon has 4 valence electrons and each hydrogen has 1 valence electron. The total number of electrons is 7. Putting carbon in the center, connect it to each hydrogen atom with a single bond. You'll end up with a central carbon atom with a single unpaired electron: ``` H | H-C-H . ``` This is the Lewis structure for \(\mathrm{CH}_{3}\) with a total of 7 valence electrons, in which carbon has only 7 electrons around it, breaking the octet rule.

Key concepts

Octet Rule

The octet rule is a fundamental concept in chemistry that describes how atoms tend to bond in ways that achieve a full outer shell of eight valence electrons. This rule is guided by the principle that atoms are more stable when they have eight electrons in their outermost shell, resembling the electronic configuration of noble gases.

However, there are exceptions to this rule, especially with elements like hydrogen, beryllium, and boron. For example:

• Hydrogen is stable with 2 valence electrons.
• Beryllium, as in \( \text{BeH}_2 \), holds only 4 valence electrons.
• Boron compounds often form with only 6 valence electrons around Boron.

These exceptions do not follow the octet rule but still form stable compounds. Understanding these exceptions helps in accurately drawing Lewis structures for such species.

Valence Electrons

Valence electrons are the outermost electrons of an atom and play a crucial role in chemical bonding. These electrons are involved in forming bonds between atoms and determining the atom's bonding capabilities.

• To determine the number of valence electrons, look at the element's group number in the periodic table. For instance, carbon, in group 14, has 4 valence electrons.
• Valence electrons dictate how atoms will bond and interact. For example: \( \text{CO}^- \) has a total of 11 valence electrons, contributing to the formation of multiple bonds.

A fundamental skill in chemistry is counting these electrons accurately to understand the bond formation and structure of molecules. Bach structure or ion will often require a different count of valence electrons due to electron sharing or loss/gain.

Resonance Structures

Resonance structures are different ways to represent a molecule, showing the same arrangement of atoms but with a different distribution of electrons between bonds.
These structures illustrate a concept where bonds can shift without changing the overall geometry of the molecule. Let's consider:

• In \( \text{CO}^- \), resonance is showcased through forms:
  • One with a triple bond between C and O.
  • Another with a double bond, sharing a different set of lone electron pairs.
• The structure of \( \text{SO}_2^- \) also displays resonance because of varying electron placement between sulfur and oxygen.

These structures help chemists understand how molecules like \( \text{CO}^- \) have a more stable configuration, as the actual molecule is a hybrid of all resonance forms. Recognizing and drawing these structures are key in visualizing electron delocalization within a molecule.