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Chapter 8: Problem 122

Predict the molecular structure (including bond angles) for each of the following. (See Exercises 115 and 116.) a. \(\mathrm{ICl}_{5}\) b. \(\mathrm{XeCl}_{4}\) c. \(\mathrm{SeCl}_{6}\)

Short Answer

Expert verified
The molecular structures and bond angles for the given molecules are as follows: - \(\mathrm{ICl}_{5}\): Trigonal bipyramidal structure with axial bond angles of 180° and equatorial bond angles of 120°. - \(\mathrm{XeCl}_{4}\): Square planar structure with bond angles of 90°. - \(\mathrm{SeCl}_{6}\): Octahedral structure with bond angles of 90°.

Step by step solution

01

Find the number of valence electrons

Count the valence electrons in the molecule. Iodine (I) has 7 valence electrons, and each chlorine (Cl) has 7 valence electrons. As there are 5 chlorine atoms, the total number of valence electrons is: \(7 + 5\times7 = 42\)
02

Determine the electron-pair geometry

Place the central atom (I) in the center and distribute the valence electrons among the surrounding atoms (Cl) to satisfy the octet rule. Since there are 5 surrounding atoms and no lone pairs on the central atom, the steric number is 5. This corresponds to a trigonal bipyramidal electron-pair geometry.
03

Predict the molecular structure and bond angles

The molecular structure will match the electron-pair geometry since there are no lone pairs on the central atom. Thus, the molecular structure of \(\mathrm{ICl}_{5}\) is trigonal bipyramidal. In this structure, the axial bond angle is 180°, and the equatorial bond angles are 120°. For molecule \(\mathrm{XeCl}_{4}\):
04

Find the number of valence electrons

Count the valence electrons in the molecule. Xenon (Xe) has 8 valence electrons, and each chlorine (Cl) has 7 valence electrons. As there are 4 chlorine atoms, the total number of valence electrons is: \(8 + 4\times7 = 36\)
05

Determine the electron-pair geometry

Place the central atom (Xe) in the center and distribute the valence electrons among the surrounding atoms (Cl) and the central atom. Since there are 4 surrounding atoms and 2 lone pairs on the central atom, the steric number is 6. This corresponds to an octahedral electron-pair geometry.
06

Predict the molecular structure and bond angles

The molecular structure will be a square planar based on the octahedral electron-pair geometry with two lone pairs on opposite positions. The bond angle between any two chlorine atoms surrounding the central Xe atom is 90°. Thus, the molecular structure of \(\mathrm{XeCl}_{4}\) is square planar with 90° bond angles. For molecule \(\mathrm{SeCl}_{6}\):
07

Find the number of valence electrons

Count the valence electrons in the molecule. Selenium (Se) has 6 valence electrons, and each chlorine (Cl) has 7 valence electrons. As there are 6 chlorine atoms, the total number of valence electrons is: \(6 + 6\times7 = 48\)
08

Determine the electron-pair geometry

Place the central atom (Se) in the center and distribute the valence electrons among the surrounding atoms (Cl). Since there are 6 surrounding atoms and no lone pairs on the central atom, the steric number is 6. This corresponds to an octahedral electron-pair geometry.
09

Predict the molecular structure and bond angles

The molecular structure will match the electron-pair geometry since there are no lone pairs on the central atom. Thus, the molecular structure of \(\mathrm{SeCl}_{6}\) is octahedral. In this structure, the bond angles between any two chlorine atoms surrounding the central Se atom are 90°. Summary: - \(\mathrm{ICl}_{5}\) has a trigonal bipyramidal molecular structure with axial bond angles of 180° and equatorial bond angles of 120°. - \(\mathrm{XeCl}_{4}\) has a square planar molecular structure with bond angles of 90°. - \(\mathrm{SeCl}_{6}\) has an octahedral molecular structure with bond angles of 90°.

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Most popular questions from this chapter

Use the following data to estimate \(\Delta H_{\mathrm{f}}^{\circ}\) for magnesium fluoride. $$\mathrm{Mg}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{MgF}_{2}(s)$$ $\begin{array}{l}{\text { Lattice energy }} & {-22913 . \mathrm{kJ} / \mathrm{mol}} \\ {\text { First ionization energy of } \mathrm{Mg}} & \quad{735 \mathrm{kJ} / \mathrm{mol}} \\ {\text {Second ionization energy of } \mathrm{Mg}} & \quad {1445 \mathrm{kJ} / \mathrm{mol}}\\\\{\text { Electron affinity of } \mathrm{F}} & {-328 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Bond energy of } \mathrm{F}_{2}} & \quad {154 \mathrm{kJ} / \mathrm{mol}} \\\ {\text { Enthalpy of sublimation for } \mathrm{Mg}} & \quad {150 . \mathrm{kJ} / \mathrm{mol}} \end{array}$

Use the following data to estimate \(\Delta H_{\mathrm{f}}^{\circ}\) for barium bromide. $$\mathrm{Ba}(s)+\mathrm{Br}_{2}(g) \longrightarrow \mathrm{BaBr}_{2}(s)$$ $\begin{array}{ll}{\text { Lattice energy }} & {-1985 \mathrm{kJ} / \mathrm{mol}} \\ {\text { First ionization energy of Ba }} & \quad {503 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Second ionization energy of } \mathrm{Ba}} & \quad {965 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Electron affinity of } \mathrm{Br}} & {-325 \mathrm{kJ} / \mathrm{mol}}\\\\{\text { Bond energy of } \mathrm{Br}_{2}} & \quad {193 \mathrm{kJ} / \mathrm{mol}} \\\ {\text { Enthalpy of sublimation of } \mathrm{Ba}} & \quad {178 \mathrm{kJ} / \mathrm{mol}}\end{array}$

Which of the following molecules have net dipole moments? For the molecules that are polar, indicate the polarity of each bond and the direction of the net dipole moment of the molecule. a. \(\mathrm{CH}_{2} \mathrm{Cl}_{2}, \mathrm{CHCl}_{3}, \mathrm{CCl}_{4}\) b. \(\mathrm{CO}_{2}, \mathrm{N}_{2} \mathrm{O}\) c. \(\mathrm{PH}_{3}, \mathrm{NH}_{3}\)

Which compound in each of the following pairs of ionic substances has the most exothermic lattice energy? Justify your answers a. \(\mathrm{LiF}, \mathrm{CsF} \quad\) d. $\mathrm{Na}_{2} \mathrm{SO}_{4}, \mathrm{CaSO}_{4}$ b. \(\mathrm{NaBr}, \mathrm{Nal} \quad\) e. $\mathrm{KF}, \mathrm{K}_{2} \mathrm{O}$ c. \(\mathrm{BaCl}_{2}, \mathrm{BaO} \quad\) f. $\mathrm{Li}_{2} \mathrm{O}, \mathrm{Na}_{2} \mathrm{S}$

Consider the following bond lengths: $$\mathrm{C}-\mathrm{O} \quad 143 \mathrm{pm} \quad \mathrm{C}=\mathrm{O} \quad 123 \mathrm{pm} \quad \mathrm{C} \equiv \mathrm{O} \quad 109 \mathrm{pm}$$ In the \(\mathrm{CO}_{3}^{2-}\) ion, all three \(\mathrm{C}-\mathrm{O}\) bonds have identical bond lengths of 136 \(\mathrm{pm} .\) Why?
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