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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 $$ \text {a} C F_{4} \quad \text { e. BeH }_{2} \quad \text { i. } \operatorname{KrF}_{4} $$ $$ \text {b} \mathrm{NF}_{3} \quad \text { f. } \mathrm{TeF}_{4} \quad \text { j. SeF }_{6} $$ $$ \text {c} \mathrm{OF}_{2} \quad \text { g. AsF_ } \quad \text { k. } \mathrm{IF}_{5} $$ $$ \text {d} \mathrm{BF}_{3} \quad \text { h. } \mathrm{KrF}_{2} \quad \text { L. } \mathrm{IF}_{3} $$

Short Answer

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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. It is determined by the number of bonding electron pairs and lone pairs around a central atom.
Understanding molecular geometry is essential as it helps predict the shape and the potential reactivity of the molecule. Various molecular shapes arise, such as:
  • Linear: Seen in molecules like BeH2, where two bonding pairs lead to a straight configuration with bond angles of 180°.
  • Tetrahedral: Highlighted by CF4, characterized by four bonding pairs creating a 3D shape with bond angles of 109.5°.
  • Trigonal Planar: Found in BF3, with three bonding pairs situated in a flat plane, forming angles of 120°.
  • Trigonal Pyramidal: As seen in NF3, where three bonding pairs and a lone pair result in angles around 107°.
  • Bent or Angular: For OF2, featuring two bonding pairs and two lone pairs, creating a bent shape with angles around 104.5°.
This geometric classification enables scientists and students alike to understand molecular behavior and interactions.
Hybridization
Hybridization is a concept in chemistry where atomic orbitals mix to form new hybrid orbitals, which dictate molecular bonding and geometry. Each type of hybridization corresponds to specific geometries and bond angles based on the molecule's configuration.
Here are some common types:
  • sp Hybridization: In linear molecules like BeH2, the central atom uses sp hybrid orbitals, combining an s and a p orbital.
  • sp2 Hybridization: In trigonal planar molecules like BF3, one s orbital mixes with two p orbitals.
  • sp3 Hybridization: Seen in tetrahedral (CF4) and trigonal pyramidal (NF3) geometries, involving one s and three p orbitals.
This rearrangement of orbitals allows for optimal overlap and formation of stable bonds, giving rise to the molecule's observed shape and properties.
Bond Angles
Bond angles in molecules are the angles between adjacent lines representing bonds. They are directly influenced by the number of electron pairs—both bonded and lone pairs—around the central atom. The spatial configuration of these pairs determines the molecule's overall shape.
  • Linear Geometry: As observed in BeH2, bond angles are 180° due to two electron pairs.
  • Tetrahedral Geometry: In CF4, the four bonding pairs create uniform 109.5° angles.
  • Trigonal Planar Geometry: BF3 displays bond angles of 120° from three equally spaced bonds.
  • Bent/Angular Geometry: OF2 with two lone pairs leads to bond angles of approximately 104.5°.
  • Trigonal Pyramidal Geometry: NF3 shows bond angles slightly compressed to around 107° by the lone pair.
Bond angles help determine the reactivity and properties of molecules, and small changes in these angles can lead to significantly different chemical behaviors.
Molecular Polarity
Molecular polarity results from the distribution of electron density across a molecule, affecting how that molecule interacts with others. Polarity is primarily influenced by the difference in electronegativity between bonded atoms and the molecule's shape and symmetry.
Here's how it applies to our molecules:
  • Nonpolar Molecules: Exhibited by CF4 and BF3, where symmetrical shapes lead to a balanced charge distribution, nullifying any dipole moment.
  • Polar Molecules: Found in NF3 and OF2, where asymmetrical shapes and lone pairs lead to an imbalance. For NF3, the lone pair on nitrogen makes the molecule polar, while in OF2, the two lone pairs on oxygen drive polarity.
Understanding molecular polarity is crucial as it dictates properties like solubility, boiling points, and interaction with other molecules, playing a key role in chemical reactivity and application.

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Most popular questions from this chapter

As the head engineer of your starship in charge of the warp drive, you notice that the supply of dilithium is critically low. While searching for a replacement fuel, you discover some diboron, B. a. What is the bond order in \(\mathrm{Li}_{2}\) and \(\mathrm{B}_{2} ?\) b. How many electrons must be removed from \(\mathrm{B}_{2}\) to make it isoelectronic with \(\mathrm{Li}_{2}\) so that it might be used in the warp drive? c. The reaction to make \(\mathrm{B}_{2}\) isoelectronic with \(\mathrm{Li}_{2}\) is generalized (where \(n=\) number of electrons determined in part \(\mathrm{b}\) ) as follows: $$ \mathrm{B}_{2} \longrightarrow \mathrm{B}_{2}^{n+}+n \mathrm{e}^{-} \quad \Delta H=6455 \mathrm{kJ} / \mathrm{mol} $$ How much energy is needed to ionize 1.5 \(\mathrm{kg} \mathrm{B}_{2}\) to the desired isoelectronic species?

Many important compounds in the chemical industry are derivatives of ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right) .\) Two of them are acrylonitrile and methyl methacrylate. Complete the Lewis structures, showing all lone pairs. Give approximate values for bond angles \(a\) through \(f\) . Give the hybridization of all carbon atoms. In acrylonitrile, how many of the atoms in the molecule must lie in the same plane? How many \(\sigma\) bonds and how many \(\pi\) bonds are there in methyl methacrylate and acrylonitrile?

Use the MO model to determine which of the following has the smallest ionization energy: \(\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{N}_{2}^{2-}, \mathrm{N}_{2}^{-}, \mathrm{O}_{2}^{+} .\) Explain your answer.

Describe the bonding in the \(\mathrm{CO}_{3}^{2-}\) ion using the localized electron model. How would the molecular orbital model describe the \(\pi\) bonding in this species?

The transport of \(\mathrm{O}_{2}\) in the blood is carried out by hemoglobin. Carbon monoxide (CO) can interfere with \(\mathrm{O}_{2}\) transport because hemoglobin has a stronger affinity for CO than for \(\mathrm{O}_{2}\) If \(\mathrm{CO}\) is present, normal uptake of \(\mathrm{O}_{2}\) is prevented, depriving the body of needed \(\mathrm{O}_{2} .\) Using the molecular orbital model, write the electron configurations for \(\mathrm{CO}\) and for \(\mathrm{O}_{2} .\) From your configurations, give two property differences between \(\mathrm{CO}\) and \(\mathrm{O}_{2}\)

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