Problem 140
Question
Draw plausible Lewis structures for the following species; use expanded valence shells where necessary. (a) \(\mathrm{Cl}_{2} \mathrm{O} ;\) (b) \(\mathrm{PF}_{3} ;\) (c) \(\mathrm{CO}_{3}^{2-} ;\) (d) \(\mathrm{BrF}_{5}\).
Step-by-Step Solution
Verified Answer
The Lewis representations for the given species are as follows: (a) Cl-O-Cl (b) F-P-F | F (c) -O=C=O- | O^- (d) F-Br-F | F / F | F
1Step 1: Identify Valence Electrons
First, identify the number of valence electrons for each atom. Chlorine has 7 valence electrons, Oxygen has 6, Phosphorus has 5, Fluorine has 7, Carbon has 4, Bromine has 7. Hydrogen, Beryllium, and Boron are the exceptions which do not follow the octet rule.
2Step 2: Draw Lewis Structures for Cl2O
For Cl2O, place the Oxygen in the center and a Chlorine atom on either side. Draw single bonds connecting each atom. Oxygen then has 4 non-bonding electrons, while each Chlorine atom has 6, fulfilling the octet rule.
3Step 3: Draw Lewis structures for PF3
For PF3, place the Phosphorus in the center with a Fluorine atom on all sides. Draw single bonds connecting each atom. The Phosphorus then has a lone pair of non-bonding electrons, while each Fluorine atom has 6, fulfilling the octet rule.
4Step 4: Draw Lewis structures for CO3^2-
For CO3^2-, place the Carbon in the center, surrounded by three Oxygen atoms. Create double bonds between Carbon and two Oxygen atoms, and a single bond to the third. The third Oxygen atom then carries a negative charge. The carbon and the double-bonded oxygens have no lone pairs, the singly bonded oxygen has 3 lone pairs.
5Step 5: Draw Lewis structures for BrF5
For BrF5, place the Bromine atom in the center, surrounded by five Fluorine atoms. Draw single bonds connecting each atom. The Bromine has a lone pair of non-bonding electrons, while each Fluorine atom has 6, fulfilling the octet rule. Here, Bromine exceeds the octet rule, emphasizing an expanded valence shell.
Key Concepts
Valence ElectronsOctet RuleExpanded Valence Shells
Valence Electrons
Valence electrons are the outermost electrons of an atom and play a critical role in chemical bonding and reactions. These electrons are situated in the highest energy level of an atom and are responsible for the atom's chemical properties.
Counting Valence Electrons
For most main group elements, the number of valence electrons corresponds to their group number in the periodic table. For example, Oxygen, from Group 16, has six valence electrons, while Chlorine, from Group 17, possesses seven valence electrons. These electrons can be represented as dots surrounding the atomic symbol in Lewis structures.Importance in Bond Formation
In chemical bonding, atoms strive to achieve a stable electron configuration, typically that of the nearest noble gas. Valence electrons are either shared with other atoms to form covalent bonds or transferred in the case of ionic bonds. In our exercise, for instance, Chlorine's seven valence electrons form bonds with Oxygen and the atoms share electrons to achieve stability.Octet Rule
The octet rule is a chemical rule of thumb that reflects the observation that atoms of main group elements tend to combine in such a way that each atom has eight electrons in its valence shell, giving it the same electronic configuration as a noble gas.
Understanding the Octet Rule
The rule is based on the principle that a full valence shell is energetically favorable and stable. For many elements, having eight electrons within the outer shell is seen as filling or having a complete octet. In drawing Lewis structures, such as for the molecule PF3 from our exercise, the aim is to ensure each Fluorine atom adheres to the octet rule, contributing to a stable molecule.Exceptions to the Rule
It's important to note that the octet rule isn't universally applicable. Elements like Hydrogen, Helium, Beryllium, and Boron often have fewer than eight electrons in their valence shells and are stable. On the other hand, some elements can have expanded valence shells, which lead us to the next concept.Expanded Valence Shells
Expanded valence shells occur for elements in period 3 (such as sulfur and chlorine) and beyond because these elements have d orbitals that can accommodate additional electrons beyond the usual octet.
Exceeding the Octet
Atoms with expanded valence shells can have more than eight valence electrons, which is observable in compounds such as BrF5. Here, Bromine forms five bonds, one with each Fluorine atom, totaling ten electrons around Bromine, thus exceeding the typical octet.Periodic Table Placement and d Orbitals
The capacity to expand the valence shell is related to the element's position on the periodic table and the availability of d orbitals. For many nonmetals after the second period, the presence of accessible d orbitals in their atomic structure allows these elements to hold on to more electrons and form additional bonds as manifested in our BrF5 example.Other exercises in this chapter
Problem 138
The highest bond-dissociation energy is found in (a) \(\mathrm{O}_{2} ;\) (b) \(\mathrm{N}_{2} ;\) (c) \(\mathrm{Cl}_{2} ;\) (d) \(\mathrm{I}_{2}\).
View solution Problem 139
The greatest bond length is found in (a) \(\mathrm{O}_{2} ;\) (b) \(\mathrm{N}_{2}\) (c) \(\mathrm{Br}_{2} ;\) (d) BrCl.
View solution Problem 141
Predict the shapes of the following sulfur-containing species. (a) \(\mathrm{SO}_{2} ;\) (b) \(\mathrm{SO}_{3}^{2-} ;\) (c) \(\mathrm{SO}_{4}^{2-}\).
View solution Problem 142
Without referring to tables or figures in the text other than the periodic table, indicate which of the following atoms, \(\mathrm{Bi}, \mathrm{S}, \mathrm{Ba},
View solution