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Chapter 8: Problem 34

Using Lewis symbols and Lewis structures, diagram the formation of \(\mathrm{BF}_{3}\) from \(\mathrm{B}\) and \(\mathrm{F}\) atoms, showing valence- shell electrons. (a) How many valence electrons does B have initially? (b) How many bonds F has to make in order to achieve an octet? (c) How many valence electrons surround the \(\mathrm{B}\) in the \(\mathrm{BF}_{3}\) molecule? (d) How many valence electrons surround each \(\mathrm{F}\) in the \(\mathrm{BF}_{3}\) molecule? (e) Does \(\mathrm{BF}_{3}\) obey the octet rule?

Short Answer

Expert verified
(a) B has 3 valence electrons initially. (b) F has to make 1 bond to achieve an octet. (c) In the \(\mathrm{BF}_3\) molecule, B is surrounded by 6 valence electrons. (d) Each F in the \(\mathrm{BF}_3\) molecule is surrounded by 8 valence electrons. (e) \(\mathrm{BF}_3\) does not obey the octet rule, as B has an incomplete octet.

Step by step solution

01

(a) Determine the number of valence electrons for B

To find the number of valence electrons of B, we need to look at its position in the periodic table. Boron (B) belongs to Group 13, which means it has 3 valence electrons.
02

(b) Determine how many bonds F has to make to achieve an octet

Fluorine (F) belongs to Group 17 in the periodic table, meaning it has 7 valence electrons. Since each atom wants to achieve a stable octet (8 valence electrons), F needs to form 1 bond (with 1 more electron) to achieve an octet.
03

(c) Create the Lewis structure for \(\mathrm{BF}_3\) and count valence electrons around B

To create the Lewis structure for \(\mathrm{BF}_3\), we'll follow these steps: 1. Place the least electronegative atom (B) in the center. 2. Surround B with F atoms, connected by single bonds 3. Distribute the remaining valence electrons as lone pairs on the F atoms The Lewis structure will look like this: ``` F | B-F-B | F ``` In the \(\mathrm{BF}_3\) molecule, there are 3 single bonds (each containing 2 valence electrons) attached to B and no lone pair electrons. Thus, B is surrounded by 6 valence electrons.
04

(d) Count valence electrons around each F in \(\mathrm{BF}_3\)

In the \(\mathrm{BF}_3\) molecule, each F is connected to B by a single bond and has 3 lone pairs of electrons. Thus, each F has 2 (from the bond) + 6 (from the lone pairs) = 8 valence electrons, achieving an octet.
05

(e) Determine if \(\mathrm{BF}_3\) obeys the octet rule

The octet rule states that an atom seeks to have a total of 8 valence electrons, either through sharing (covalent bonds) or transferring (ionic bonds). In the \(\mathrm{BF}_3\) molecule, each F atom has achieved an octet with 8 valence electrons. However, the central B atom is surrounded by only 6 valence electrons. Therefore, \(\mathrm{BF}_3\) does not obey the octet rule, as B has an incomplete octet.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Valence Electrons
Valence electrons are the outermost electrons found in an atom and are crucial in determining how the element interacts with others. Every atom seeks stability, which is often achieved when it has a full outer shell of electrons. This behavior can be understood through the concept of valence electrons, which are involved in bond formation.
For example, boron (B), found in Group 13 of the periodic table, has 3 valence electrons. Its chemical characteristics and reactivity are dictated by these electrons.
  • It can form covalent bonds by sharing its valence electrons with other atoms.
  • In a compound like \(\mathrm{BF}_3\), boron shares its 3 valence electrons with three fluorine (F) atoms, each contributing one valence electron to the bond formation.
Fluorine atoms, with 7 valence electrons (as they belong to Group 17), only need one more to achieve a full octet, making them highly reactive and eager to form bonds.
Octet Rule
The octet rule is a principle used to predict the formation of many molecules by suggesting that atoms seek to complete eight electrons in their valence shell. This rule stems from the observation that elements become chemically stable when they have a full set of eight electrons similar to that of noble gases.
This is crucial for understanding molecular structures such as \(\mathrm{BF}_3\). When discussing the octet rule:
  • Fluorine atoms in \(\mathrm{BF}_3\) achieve a full set of eight valence electrons by forming one covalent bond each with boron.
  • Boron in \(\mathrm{BF}_3\), however, is an exception to the octet rule. It only manages to surround itself with 6 electrons after bonding, thus completing fewer than 8 valence electrons.
This misalignment with the octet rule illustrates a situation called an 'incomplete octet'. It is noteworthy in chemical bonding, especially for elements like boron, aluminum, and other elements in Groups 3A, 3B, and some transition metals.
Covalent Bonds
Covalent bonds are the connectors in a molecule formed from the sharing of valence electrons between atoms. They allow each atom in the molecule to achieve greater stability by reaching a noble gas electron configuration. The formation of covalent bonds can be demonstrated with Lewis structures, which visually represent the sharing of electrons.
In the case of \(\mathrm{BF}_3\):
  • A covalent bond forms between each boron and fluorine atom by sharing one valence electron from each, resulting in 3 single covalent bonds formed.
  • Through these bonds, each fluorine atom in \(\mathrm{BF}_3\) achieves an octet, making them stable.
  • The central boron atom, though stable enough to exist, remains without an octet, having 6 valence electrons.
This type of bonding showcases how atoms can reach an agreeable electron configuration through sharing. Understanding covalent bonds highlights how molecules are constructed and the stability they aim to achieve through electron sharing.

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Most popular questions from this chapter

We can define average bond enthalpies and bond lengths for ionic bonds, just like we have for covalent bonds. Which ionic bond is predicted to have the smaller bond enthalpy, \(\mathrm{Li}-\mathrm{F}\) or \(\mathrm{Cs}-\mathrm{F}\) ?

Calculate the formal charge on the indicated atom in each of the following molecules or ions: (a) the central oxygen atom in \(\mathrm{O}_{3},(\mathbf{b})\) phosphorus in \(\mathrm{PF}_{6}^{-},(\mathbf{c})\) nitrogen in \(\mathrm{NO}_{2}\), (d) iodine in ICl \(_{3}\), (e) chlorine in \(\mathrm{HClO}_{4}\) (hydrogen is bonded to \(\mathrm{O}\) ).

A common form of elemental phosphorus is the white phosphorus, where four \(\mathrm{P}\) atoms are arranged in a tetrahedron. All four phosphorus atoms are equivalent. White phosphorus reacts spontaneously with the oxygen in air to form \(\mathrm{P}_{4} \mathrm{O}_{6} .\) (a) How many valance electron pairs are in the \(\mathrm{P}_{4} \mathrm{O}_{6}\) molecule? (b) When \(\mathrm{P}_{4} \mathrm{O}_{6}\) is dissolved in water, it produces a \(\mathrm{H}_{3} \mathrm{PO}_{3}\), molecule. \(\mathrm{H}_{3} \mathrm{PO}_{3}\) has two forms, \(\mathrm{P}\) forms 3 covalent bonds in the first form and \(\mathrm{P}\) forms 5 covalent bonds in the second form. Draw two possible Lewis structures of \(\mathrm{H}_{3} \mathrm{PO}_{3}\). (c) Which structure obeys the octet rule?

(a) True or false: The hydrogen atom is most stable when it has a full octet of electrons. (b) How many electrons must a sulfur atom gain to achieve an octet in its valence shell? (c) If an atom has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{3},\) how many electrons must it gain to achieve an octet?

You and a partner are asked to complete a lab entitled "Carbonates of Group 2 metal" that is scheduled to extend over two lab periods. The first lab, which is to be completed by your partner, is devoted to carrying out compositional analysis and determine the identity of the Group 2 metal (M). In the second lab, you are to determine the melting point of this compound. Upon going to lab you find two unlabeled vials containing white powder. You also find the following notes in your partner's notebook-Compound 1: \(40.04 \% \mathrm{M}\) and \(12.00 \%\) C, \(47.96 \%\) O (by mass), Compound \(2: 69.59 \% \mathrm{M}\), \(6.09 \% \mathrm{C},\) and \(24.32 \% \mathrm{O}\) (by mass). (a) What is the empirical formula for Compound 1 and the identity of \(\mathrm{M} ?(\mathbf{b})\) What is the empirical formula for Compound 2 and the identity of \(\mathrm{M}\) ? Upon determining the melting points of these two compounds, you find that both compounds do not melt up to the maximum temperature of your apparatus, instead, the compounds decompose and liberate colorless gas. (c) What is the identity of the colorless gas? (d) Write the chemical equation for the decomposition reactions of compound 1 and 2. (e) Are compounds 1 and 2 ionic or molecular?
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