How would you expect the extent of overlap of the bonding atomic orbitals to vary in the series IF, ICl, IBr, and \(I_{2}\) ? Explain your answer.

Short Answer

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The extent of overlap of bonding atomic orbitals in the series IF, ICl, IBr, and \(I_{2}\) varies as IF < ICl ≈ IBr < \(I_{2}\). This is due to the differences in atomic size and electronegativity between the involved atoms, which affect the efficiency of orbital overlap and the strength of the bond.

Step by step solution

01

Understanding atomic orbitals and overlap

Atomic orbitals are regions in an atom where electrons with a specific energy level are most likely to be found. When two atomic orbitals belonging to different atoms overlap, they can share electrons and form a chemical bond. The extent of orbital overlap has a significant effect on the bond strength; stronger bonds form when there is a greater overlap of atomic orbitals.
02

Comparing the size of halogen atoms

For analyzing the extent of overlap, we need to first understand the size of the halogen atoms involved in the series (F, Cl, Br, and I). As we move down the periodic table from F to I, the atomic size increases, as they have more electron shells. Here is the order of atomic sizes (from smallest to largest): F < Cl < Br < I.
03

The role of the nature of the bond in orbital overlap

In a covalent bond between similar atoms (like \(I_{2}\)), atomic orbitals overlap directly, resulting in maximum overlap and a strong bond. However, in a bond between different atoms (like IF, ICl, IBr), the difference in electronegativity of the atoms affects the extent of the orbital overlap. Due to the difference in electronegativity, the shared electron pair is attracted more by the more electronegative atom (for example, F is more electronegative than I). This results in partial charge separation and thus, reduced overlap of atomic orbitals as compared to the bond between similar atoms.
04

Analyzing the extent of orbital overlap in the given series

As we have seen earlier, the atomic size plays an important role in determining the overlap of atomic orbitals. In IF, the atomic size difference between I and F is the greatest, making it challenging for the atomic orbitals to overlap efficiently. As a result, the extent of overlap in IF is expected to be the smallest. Moving on to ICl and IBr, the atomic size difference between the involved atoms (I and Cl or I and Br) is smaller compared to IF. Consequently, the extent of overlap in ICl and IBr will be more significant compared to IF. However, the exact extent of overlap between ICl and IBr will depend on other factors like electronegativity difference and may not vary vastly. Finally, in \(I_{2}\), we have two similar atoms (iodine) involved in the bond, which results in direct orbital overlap and maximum overlap. Therefore, we can expect the extent of overlap in \(I_{2}\) to be the highest among the given series of compounds. In conclusion, the extent of overlap of bonding atomic orbitals in the given series varies in the following order: IF < ICl ≈ IBr < \(I_{2}\). This is due to the differences in atomic size and electronegativity between the involved atoms, which affect the efficiency of orbital overlap and the strength of the bond.

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Most popular questions from this chapter

(a) Using only the valence atomic orbitals of a hydrogen atom and a fluorine atom, and following the model of Figure 9.46 , how many MOs would you expect for the HF molecule? (b) How many of the MOs from part (a) would be occupied by electrons? (c) It turns out that the difference in energies between the valence atomic orbitals of \(\mathrm{H}\) and \(\mathrm{F}\) are sufficiently different that we can neglect the interaction of the \(1 s\) orbital of hydrogen with the \(2 s\) orbital of fluorine. The \(1 s\) orbital of hydrogen will mix only with one \(2 p\) orbital of fluorine. Draw pictures showing the proper orientation of all three \(2 p\) orbitals on \(\mathrm{F}\) interacting with a \(1 s\) orbital on \(\mathrm{H}\). Which of the \(2 p\) orbitals can actually make a bond with a \(1 s\) orbital, assuming that the atoms lie on the \(z\) -axis? (d) In the most accepted picture of HF, all the other atomic orbitals on fluorine move over at the same energy into the molecular orbital energy-level diagram for HF. These are called "nonbonding orbitals." Sketch the energy-level diagram for HF using this information and calculate the bond order. (Nonbonding electrons do not contribute to bond order.) (e) Look at the Lewis structure for HF. Where are the nonbonding electrons?

In which of the following \(\mathrm{AF}_{n}\) molecules or ions is there more than one \(\mathrm{F}-\mathrm{A}-\mathrm{F}\) bond angle: \(\mathrm{SiF}_{4}, \mathrm{PF}_{5}, \mathrm{SF}_{4}, \mathrm{AsF}_{3} ?\)

(a) Sketch the molecular orbitals of the \(\mathrm{H}_{2}^{-}\) ion and draw its energy-level diagram. (b) Write the electron configuration of the ion in terms of its MOs. (c) Calculate the bond order in \(\mathrm{H}_{2}^{-}\). (d) Suppose that the ion is excited by light, so that an electron moves from a lower-energy to a higher- energy molecular orbital. Would you expect the excited-state \(\mathrm{H}_{2}^{-}\) ion to be stable? Explain.

Butadiene, \(\mathrm{C}_{4} \mathrm{H}_{6}\) is a planar molecule that has the following carbocarbon bond lengths: (a) Predict the bond angles around each of the carbon atoms and sketch the molecule. (b) From left to right, what is the hybridization of each carbon atom in butadiene? (c) The middle \(C-\) bond length in butadiene \((1.48\) A) is a little shorter than the average \(\mathrm{C}-\mathrm{C}\) single bond length \((1.54 \hat{\mathrm{A}}) .\) Does this imply that the middle \(\mathrm{C}-\mathrm{Cbond}\) in butadiene is weaker or stronger than the average \(\mathrm{C}-\mathrm{C}\)? (\mathbf{d} ) Based on your answer for part ( c ),discuss what additional aspects of bonding in butadiene might support the shorter middle \(\mathrm{C}-\) C bond.

In which of these molecules or ions does the presence of nonbonding electron pairs produce an effect on molecular shape, assuming they are all in the gaseous state? (a) \(\mathrm{SiH}_{4}\) (b) \(\mathrm{PF}_{3},\) (c) \(\mathrm{HBr}\), (d) \(\mathrm{HCN},\) (e) \(\mathrm{SO}_{2}\)

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