For each of the following molecules, write the Lewis structure(s), predict the molecular structure (including bond angles), give the expected hybrid orbitals on the central atom, and predict the overall polarity $$ \text {a} C F_{4} \quad \text { e. BeH }_{2} \quad \text { i. } \operatorname{KrF}_{4} $$ $$ \text {b} \mathrm{NF}_{3} \quad \text { f. } \mathrm{TeF}_{4} \quad \text { j. SeF }_{6} $$ $$ \text {c} \mathrm{OF}_{2} \quad \text { g. AsF_ } \quad \text { k. } \mathrm{IF}_{5} $$ $$ \text {d} \mathrm{BF}_{3} \quad \text { h. } \mathrm{KrF}_{2} \quad \text { L. } \mathrm{IF}_{3} $$

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#tag_title# f. TeF4#tag_content# Lewis structure: Draw a tellurium atom in the center. Attach four fluorine atoms to the tellurium atom with single bonds. There are two lone pairs of electrons on the tellurium atom. Molecular structure: See-saw (six electron pairs) Bond angles: Approximately 90° and 120° Hybrid orbitals on central atom (Tellurium): sp3d2 hybridized Overall polarity: Polar (due to unsymmetric distribution of polar Te-F bonds and the presence of lone pairs) #tag_title# g. AsF3#tag_content# Lewis structure: Draw an arsenic atom in the center. Attach three fluorine atoms to the arsenic atom with single bonds. There is a lone pair of electrons on the arsenic atom. Molecular structure: Trigonal pyramidal (four electron pairs) Bond angles: 107° (approx.) Hybrid orbitals on central atom (Arsenic): sp3 hybridized Overall polarity: Polar (due to the presence of a lone pair, causing unsymmetric distribution of polar As-F bonds) #tag_title# h. KrF2#tag_content# Lewis structure: Draw a krypton atom in the center. Attach two fluorine atoms to the krypton atom with single bonds. There are three lone pairs of electrons on the krypton atom. Molecular structure: Linear (five electron pairs) Bond angles: 180° Hybrid orbitals on central atom (Krypton): sp3d hybridized Overall polarity: Nonpolar (due to the symmetric distribution of polar Kr-F bonds) #tag_title# i. KrF4#tag_content# Lewis structure: Draw a krypton atom in the center. Attach four fluorine atoms to the krypton atom with single bonds. There are two lone pairs of electrons on the krypton atom. Molecular structure: Square planar (six electron pairs) Bond angles: 90° Hybrid orbitals on central atom (Krypton): sp3d2 hybridized Overall polarity: Nonpolar (due to symmetric distribution of polar Kr-F bonds) #tag_title# j. SeF6#tag_content# Lewis structure: Draw a selenium atom in the center. Attach six fluorine atoms to the selenium atom with single bonds. Molecular structure: Octahedral (six electron pairs) Bond angles: 90° Hybrid orbitals on central atom (Selenium): sp3d2 hybridized Overall polarity: Nonpolar (due to symmetric distribution of polar Se-F bonds) #tag_title# k. IF5#tag_content# Lewis structure: Draw an iodine atom in the center. Attach five fluorine atoms to the iodine atom with single bonds. There is a lone pair of electrons on the iodine atom. Molecular structure: Square pyramidal (six electron pairs) Bond angles: 90° (approx.) Hybrid orbitals on central atom (Iodine): sp3d2 hybridized Overall polarity: Polar (due to the presence of a lone pair, causing unsymmetric distribution of polar I-F bonds) #tag_title# l. IF3#tag_content# Lewis structure: Draw an iodine atom in the center. Attach three fluorine atoms to the iodine atom with single bonds. There are two lone pairs of electrons on the iodine atom. Molecular structure: T-shaped (five electron pairs) Bond angles: Approximately 90° and 180° Hybrid orbitals on central atom (Iodine): sp3d hybridized Overall polarity: Polar (due to unsymmetric distribution of polar I-F bonds and the presence of lone pairs)

Step by step solution

01

a. CF4

Lewis structure: Draw a carbon atom in the center. Attach four fluorine atoms to the carbon atom with single bonds. Molecular structure: Tetrahedral (four electron pairs) Bond angles: 109.5° Hybrid orbitals on central atom (Carbon): sp3 hybridized Overall polarity: Nonpolar (due to symmetric distribution of polar C-F bonds)
02

b. NF3

Lewis structure: Draw a nitrogen atom in the center. Attach three fluorine atoms to the nitrogen atom with single bonds. There is a lone pair of electrons on the nitrogen atom. Molecular structure: Trigonal pyramidal (four electron pairs) Bond angles: 107° (approx.) Hybrid orbitals on central atom (Nitrogen): sp3 hybridized Overall polarity: Polar (due to the presence of a lone pair, causing unsymmetric distribution of polar N-F bonds)
03

c. OF2

Lewis structure: Draw an oxygen atom in the center. Attach two fluorine atoms to the oxygen atom with single bonds. There are two lone pairs of electrons on the oxygen atom. Molecular structure: Bent/angular (four electron pairs) Bond angles: 104.5° (approx.) Hybrid orbitals on central atom (Oxygen): sp3 hybridized Overall polarity: Polar (due to the presence of two lone pairs, causing unsymmetric distribution of polar O-F bonds)
04

d. BF3

Lewis structure: Draw a boron atom in the center. Attach three fluorine atoms to the boron atom with single bonds. Molecular structure: Trigonal planar (three electron pairs) Bond angles: 120° Hybrid orbitals on central atom (Boron): sp2 hybridized Overall polarity: Nonpolar (due to symmetric distribution of polar B-F bonds) Proceed similarly for the rest of the molecules.
05

e. BeH2

Lewis structure: Draw a beryllium atom in the center. Attach two hydrogen atoms to the beryllium atom with single bonds. Molecular structure: Linear (two electron pairs) Bond angles: 180° Hybrid orbitals on central atom (Beryllium): sp hybridized Overall polarity: Nonpolar (due to symmetric distribution of polar Be-H bonds) Continue in the same fashion for the remaining molecules.

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Most popular questions from this chapter

For each of the following molecules or ions that contain sulfur, write the Lewis structure(s), predict the molecular structure (including bond angles), and give the expected hybrid orbitals for sulfur. $$ \begin{array}{l}{\text { a. } \mathrm{SO}_{2}} \\ {\text { b. } \mathrm{SO}_{3}}\end{array} $$ $$ \text {c} \mathrm{s}_{2} \mathrm{O}_{3}^{2-}\left[\begin{array}{c}{\mathrm{o}} \\\ {\mathrm{s}-\mathrm{s}-\mathrm{o}} \\ {\mathrm{o}} \\\ {\mathrm{o}}\end{array}\right]^{2-} $$ e. \(\mathrm{SO}_{3}^{2-}\) f. \(\mathrm{SO}_{4}^{2-}\) g. \(\mathrm{SF}_{2}\) h. \(\mathrm{SF}_{4}\) i. \(\mathrm{SF}_{6}\) j. \(\mathrm{F}_{3} \mathrm{S}-\mathrm{SF}\) k. \(\mathrm{SF}_{5}+\)

Explain the difference between the \(\sigma\) and \(\pi\) MOs for homo- nuclear diatomic molecules. How are bonding and antibonding orbitals different? Why are there two \(\pi\) MOs and one \(\sigma\) MO? Why are the \(\pi\) MOs degenerate?

Draw the Lewis structure for HCN. Indicate the hybrid orbitals, and draw a picture showing all the bonds between the atoms, labeling each bond as \(\sigma\) or \(\pi .\)

A flask containing gaseous \(\mathrm{N}_{2}\) is irradiated with 25 -nm light. a. Using the following information, indicate what species can form in the flask during irradiation. $$ \begin{array}{ll}{\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{N}(g)} & {\Delta H=941 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{N}_{2}(g) \longrightarrow \mathrm{N}_{2}^{+}(g)+\mathrm{e}^{-}} & {\Delta H=1501 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{N}(g) \longrightarrow \mathrm{N}^{+}(g)+\mathrm{e}^{-}} & {\Delta H=1402 \mathrm{kJ} / \mathrm{mol}}\end{array} $$ b. What range of wavelengths will produce atomic nitrogen in the flask but will not produce any ions? c. Explain why the first ionization energy of $\mathrm{N}_{2}(1501 \mathrm{kJ} /\( mol) is greater than the first ionization energy of atomic nitrogen \)(1402 \mathrm{kJ} / \mathrm{mol})$

Use the localized electron model to describe the bonding in $\mathrm{H}_{2} \mathrm{O}$ .

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