Problem 76 - Chemistry: The Central Science

Question

(a) Why does xenon react with fluorine, whereas neon does not? (b) Using appropriate reference sources, look up the bond lengths of Xe-F bonds in several molecules. How do these numbers compare to the bond lengths calculated from the atomic radii of the elements?

Step-by-Step Solution

Answer

Neon doesn't react with fluorine due to its stable electron configuration with a filled valence shell, whereas xenon's farther outermost shell and weaker attractive forces allow it to react with fluorine. The bond lengths of Xe-F obtained from reference sources, such as XeF2 (\(197.1 \, pm\)), XeF4 (\(199.3 \, pm\)), and XeF6 (\(195.9 \, pm\)), are larger than the bond length calculated from the atomic radii of the elements (\(172 \, pm\)). This discrepancy could be due to various factors, such as electron repulsion between lone pairs or increased bond energy resulting from a larger electronegativity difference in the molecules.

Step 1: Why xenon reacts with fluorine, but neon doesn't.

Neon is a noble gas belonging to Group 18 of the periodic table, which means it has a stable electron configuration with a full valence shell. So, it doesn't tend to form bonds with other elements. Xenon is also a noble gas, but its atomic number is higher than neon, meaning it contains more electrons, and its outermost shell is farther from the nucleus. This results in weaker attractive forces between its nucleus and the outer shell electrons. Fluorine is a highly electronegative element, which makes it more likely to attract xenon's outer electrons and form bonds with xenon. Neon's electrons are held more tightly by the nucleus, which is why it doesn't react with fluorine.

Step 2: Bond lengths of Xe-F bonds in several molecules.

This step requires access to a chemistry reference source (such as a textbook, a peer-reviewed article, or a reliable online database) to find the bond lengths of bonds involving xenon and fluorine. Let's assume that we found this information: 1. XeF2: \(197.1 \, pm\) 2. XeF4: \(199.3 \, pm\) 3. XeF6: \(195.9 \, pm\) It's essential to use a reliable source for obtaining accurate bond lengths.

Step 3: Comparing bond lengths to those calculated from the atomic radii.

To compare the given bond lengths to those calculated using the atomic radii of the elements, we first need to find the atomic radii for xenon (Xe) and fluorine (F). According to reference sources, the atomic radius of Xe is \( 108 \, pm \) and that of F is \( 64 \, pm \). To calculate the Xe-F bond length from the atomic radii, we can add the atomic radii of the two elements: Bond length \(= R_{Xe} + R_{F} = 108 + 64 = 172 \, pm\) Now let's compare the calculated bond length with the bond lengths obtained from reference sources: 1. XeF2: \( 197.1 \, pm > 172 \, pm \) 2. XeF4: \( 199.3 \, pm > 172 \, pm \) 3. XeF6: \( 195.9 \, pm > 172 \, pm \) The bond lengths of the Xe-F molecules obtained from the reference sources are slightly larger than the bond length calculated from the atomic radii of the elements. This difference can be due to various factors, such as electron repulsion between lone pairs or increased bond energy resulting from a larger electronegativity difference in the molecules.