Problem 114 - Chemistry: The Central Science

Question

(a) The measured \(\mathrm{Bi}-\mathrm{Br}\) bond length in bismuth tribromide, \(\mathrm{BiBr}_{3}\), is \(2.63 \AA\). Based on this value and the data in Figure 7.8, predict the atomic radius of Bi. (b) Bismuth tribromide is soluble in acidic solution. It is formed by treating solid bismuth(III) oxide with aqueous hydrobromic acid. Write a balanced chemical equation for this reaction. (c) While bismuth(III) oxide is soluble in acidic solutions, it is insoluble in basic solutions such as \(\mathrm{NaOH}(a q)\). Based on these properties, is bismuth characterized as a metallic, metalloid, or nonmetallic element? (d) Treating bismuth with fluorine gas forms \(\mathrm{BiF}_{5}\). Use the electron configuration of \(\mathrm{Bi}\) to explain the formation of a compound with this formulation. (e) While it is possible to form \(\mathrm{BiF}_{5}\) in the manner just described, pentahalides of bismuth are not known for the other halogens. Explain why the pentahalide might form with fluorine but not with the other halogens. How does the behavior of bismuth relate to the fact that xenon reacts with fluorine to form compounds but not with the other halogens?

Step-by-Step Solution

Answer

The atomic radius of bismuth (Bi) is \(0.74 \,\text{Å}\). The balanced equation for the reaction between bismuth(III) oxide and hydrobromic acid is Bi2O3 + 6 HBr -> 2 BiBr3 + 3 H2O. Bismuth can be characterized as a metallic element due to its metallic properties and its behavior with acids and bases. The formation of BiF5 involves Bi5+ ions and five F- ions; the electron configuration of Bi5+ is [Xe] 4f14 5d10. Fluorine's high electronegativity allows it to stabilize the charge on Bi5+ and xenon, making the formation of pentahalide BiF5 and xenon compounds possible, while other halogens cannot.

Step 1: a) Finding the atomic radius of Bi

We are given the measured Bi-Br bond length in bismuth tribromide (BiBr3) as \(2.63 \,\text{Å}\). To find the atomic radius of bismuth (Bi), we first need the radius of a Br atom. In Figure 7.8, the atomic radius of bromine is given as \(1.15 \,\text{Å}\). Since a bond length is the sum of the two atomic radii involved, the radius of Bi can be calculated using the Bi-Br bond length and the Br atomic radius. Atomic radius of Bi = \( \frac{\text{Bi-Br bond length} - \text{Br atomic radius}}{2} \) = \( \frac{2.63 - 1.15}{2} \,\text{Å} \) = \(0.74 \,\text{Å}\)

Step 2: b) Balanced equation for the reaction between bismuth(III) oxide and hydrobromic acid

To find the balanced chemical equation for the reaction between bismuth(III) oxide (Bi2O3) and hydrobromic acid (HBr), we need to combine the reactants to form the products, bismuth tribromide (BiBr3) and water (H2O). Bi2O3 + 6 HBr -> 2 BiBr3 + 3 H2O

Step 3: c) Classifying bismuth based on its properties

Bismuth(III) oxide is soluble in acidic solutions but insoluble in basic solutions. Additionally, bismuth has metallic properties such as being a good conductor of electricity, malleable, and ductile. Due to its metallic properties and its behavior with acids and bases, bismuth can be characterized as a metallic element.

Step 4: d) Electron configuration of Bi and the formation of BiF5

The electron configuration of bismuth (Bi) is [Xe] 4f14 5d10 6s2 6p3. Bi can lose three electrons from the 6p orbital and form Bi3+ with an electron configuration of [Xe] 4f14 5d10 6s2. However, when reacting with highly electronegative fluorine, Bi can lose two more electrons from the 6s orbital forming Bi5+ with an electron configuration of [Xe] 4f14 5d10. The formation of BiF5 involves Bi5+ ions and five F- ions. So the reaction is as follows: Bi + 5 F2 -> BiF5

Step 5: e) Formation of pentahalides with fluorine and not with other halogens

Fluorine is the most electronegative halogen and has a stronger ability to attract and stabilize the charge on a cation such as Bi5+ making the formation of the pentahalide BiF5 possible. However, other halogens such as chlorine, bromine, and iodine are less electronegative and are less able to stabilize the charge on Bi5+. As a result, pentahalides of bismuth with other halogens are not known. The same principle applies to xenon, which reacts with fluorine to form compounds, but not with other halogens. The higher electronegativity of fluorine allows it to stabilize the charge on xenon, enabling it to form compounds.

Key Concepts

Atomic Radius Calculation

Understanding the atomic radius of an element is crucial for students who study chemistry, especially when dealing with molecular structures and bond lengths. As demonstrated in the textbook exercise, the atomic radius calculation of bismuth (Bi) requires knowledge of bond lengths and covalent radii of other atoms.

To calculate the atomic radius of Bi, we use the bond length of Bi-Br in bismuth tribromide (BiBr₃) and deduct the known radius of a bromine (Br) atom. By dividing the remainder by two, we obtain the radius of bismuth. This process not only highlights the additive nature of bond lengths but also emphasizes the importance of atomic radii in understanding molecular dimensions.

Chemical Reactions

Chemical reactions are the backbone of chemistry, signifying the transformation of substances through the breaking and making of bonds. Take the conversion of bismuth(III) oxide to bismouth tribromide, for example. This reaction is an illustration of how chemical reactions are represented by balanced equations, ensuring that the number of each type of atom is conserved. In an educational setting, students would be encouraged to balance chemical equations to understand the stoichiometric relationships between reactants and products, providing insight into conservation of mass during chemical changes.

Element Classification

Classification of elements is key to predicting their chemical behavior. Bismuth, in the provided exercise, displays properties that can be assessed to determine its element classification. Based on its reaction with other substances and its physical properties, we can categorize bismuth as a metallic element. Teaching this concept equips students with the knowledge to predict solubility, reaction tendencies, and other chemical properties based on an element's position on the periodic table and its characteristic properties.

Electron Configuration

To explain the formation of compounds and predict chemical behavior, it's essential to understand an atom's electron configuration. The electron configuration indicates the distribution of electrons in an atom's orbitals. For bismuth (Bi), the electron configuration helps us understand how it can form compounds like BiF₅ by losing electrons from specific orbitals, leading to its valency. Such foundational knowledge in electron distribution is instrumental in guiding students through the rationale behind the formation of various ionic and covalent bonds.

Halide Compounds

Exploring halide compounds opens up discussions about ionic and covalent bond strengths, as well as the role of electronegativity in compound stability. Bismuth forms halides with different halogens, but its ability to form pentahalides like BiF₅ is unique to its reaction with fluorine. This is due to fluorine's high electronegativity. The contrast between bismuth and xenon’s reaction with fluorine versus other halogens further exemplifies the concept of halide compound formation, an invaluable lesson in understanding chemical reactivity and molecular stability.