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(a) Predict the electron-domain geometry around the central S atom in SF2,SF4, and SF6. ( b ) The anion IO4 has a tetrahedral structure: three oxygen atoms form double bonds with the central iodine atom and one oxygen atom which carries a negative charge forms a single bond. Predict the molecular geometry of IO65.

Short Answer

Expert verified
The electron-domain geometries of the molecules with Sulfur (S) atom are: - SF2: Tetrahedral (two bond pairs and two lone pairs) - SF4: Trigonal Bipyramidal (four bond pairs and one lone pair) - SF6: Octahedral (six bond pairs and no lone pairs) The molecular geometry of IO65 is Pentagonal Pyramidal, with its electron-domain geometry being pentagonal bipyramidal (six bond pairs and one lone pair).

Step by step solution

01

Determine the electron domain geometries of the molecules with Sulfur (S) atom

To find the electron domain geometries of SF2, SF4, and SF6, we need to consider the number of bonding pairs and lone pairs around the central sulfur atom. Sulfur (S) is in group 16 and has six valence electrons. In each of the compounds, bond pairs will form as Sulfur bonds with Fluorine (F). - In SF2, there are two bond pairs (S-F bonds) and two lone pairs (non-bonding electrons) on the Sulfur atom. - In SF4, there are four bond pairs (S-F bonds) and one lone pair on the Sulfur atom. - In SF6, there are six bond pairs (S-F bonds) and no lone pairs on the Sulfur atom. Now we can predict the electron-domain geometry for each compound using VSEPR theory: - In SF2, the electron-domain geometry is tetrahedral, as there are four electron domains (two bond pairs and two lone pairs). - In SF4, the electron-domain geometry is trigonal bipyramidal, as there are five electron domains (four bond pairs and one lone pair). - In SF6, the electron-domain geometry is octahedral, as there are six electron domains (six bond pairs).
02

Predict the molecular geometry of IO65

Given that IO4 has a tetrahedral structure, we can infer that the central Iodine (I) atom forms double bonds with three Oxygen (O) atoms and a single bond with a negatively charged Oxygen atom. In IO65, we need to predict the molecular geometry of the molecule considering additional Oxygen atoms. First, let's determine the electron-domain geometry: Iodine (I) is in group 17 and has seven valence electrons. In IO65, it forms six I-O single bonds, and the molecule has a 5- charge. The central iodine atom needs to gain an additional five electrons to have a 5- charge, so it will have six bond pairs (I-O bonds) and one lone pair. The electron-domain geometry of IO65 will be pentagonal bipyramidal, as there are seven electron domains (six bond pairs and one lone pair). Now, we can predict the molecular geometry: Since there is one lone pair in the electron-domain geometry, the molecular geometry of IO65 will be slightly distorted from a perfect pentagonal bipyramidal structure. The lone pair will repel the I-O bond pairs, giving the molecule a "Pentagonal Pyramidal" shape.

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Key Concepts

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

Electron-Domain Geometry
The concept of electron-domain geometry is crucial in understanding the shape of molecules. It examines how electron pairs (bonding and non-bonding) arrange themselves around a central atom to minimize repulsion. Using VSEPR (Valence Shell Electron Pair Repulsion) theory, we can predict these arrangements.

Consider a sulfur atom (S) which is central in compounds like SF2, SF4, and SF6. Sulfur belongs to group 16, having six valence electrons. Here's how the electron-domain geometry forms in each:

  • SF2: Contains two bond pairs and two lone pairs, leading to a tetrahedral electron-domain geometry.
  • SF4: Contains four bond pairs and one lone pair, resulting in a trigonal bipyramidal electron-domain geometry.
  • SF6: With six bond pairs and no lone pairs, it forms an octahedral electron-domain geometry.
Understanding these geometries helps us predict molecular shapes in a systematic way.
Molecular Geometry Prediction
Molecular geometry prediction involves determining the spatial arrangement of atoms in a molecule. This is slightly different from electron-domain geometry, which considers all electron pairs.

For example, in SF4, the electron-domain geometry is trigonal bipyramidal, but its molecular geometry is "see-saw" due to one lone pair. Lone pairs push bonding pairs closer together, causing deviations.

  • For SF2, the molecular shape is "bent" given the tetrahedral electron-domain geometry but with two lone pairs.
  • In SF4, despite the trigonal bipyramidal electron-domain structure, the lone pair adjusts it to a "see-saw" shape.
  • SF6's lack of lone pairs results in a molecular geometry that remains "octahedral," matching its electron-domain geometry.
This prediction guides understanding of individual molecular characteristics and interactions.
SF2, SF4, SF6 Structures
These sulfur-fluorine compounds illustrate how different electron and molecular geometries manifest in chemical structures.

Each compound showcases unique features based on the number of bonding and non-bonding pairs:

  • SF2: The presence of two lone pairs gives a bent shape. It resembles water, emphasizing contour differences due to lone electron pairs.
  • SF4: Its see-saw shape stems from one lone pair in a trigonal bipyramidal arrangement. The non-equivalent electron positions show how lone pairs skew geometry.
  • SF6: Lacking lone pairs leads to a perfect octahedral shape, indicating symmetrical distribution of six fluorine atoms around the sulfur.
These structures demonstrate the impact of lone pairs and bond pairs on shaping molecules, dictating interactions and properties crucial for their application.

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Most popular questions from this chapter

Sulfur tetrafluoride (SF4) reacts slowly with O2 to form sulfur tetrafluoride monoxide (OSF_4) according to the following unbalanced reaction: SF4(g)+O2(g)OSF4(g) The O atom and the four F atoms in OSF4 are bonded to a central S atom. (a) Balance the equation. (b) Write a Lewis structure of OSF4 in which the formal charges of all atoms are zero. (c) Use average bond enthalpies (Table 8.3) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electron-domain geometry of OSF 4, and write two possible molecular geometries for the molecule based on this electron-domain geometry. (e) For each of the molecules you drew in part (d), state how many fluorines are equatorial and how many are axial.

(a) What is the difference between a localized π bond and a delocalized one? (b) How can you determine whether a molecule or ion will exhibit delocalized π bonding? (c) Is the π bond in NO2 localized or delocalized?

Draw sketches illustrating the overlap between the following orbitals on two atoms: (a) the 2s orbital on each atom, (b) the 2pz orbital on each atom (assume both atoms are on the z -axis), (c) the 2 s orbital on one atom and the 2p2 orbital on the other atom.

Give the electron-domain and molecular geometries of a molecule that has the following electron domains on its central atom: (a) four bonding domains and no nonbonding domains, (b) three bonding domains and two nonbonding domains, (c) five bonding domains and one nonbonding domain, (d) four bonding domains and two nonbonding domains.

The molecule shown here is difluoromethane (CH_2F2), which is used as a refrigerant called R-32. (a) Based on the structure, how many electron domains surround the C atom in this molecule? (b) Would the molecule have a nonzero dipole moment? (c) If the molecule is polar, which of the following describes the direction of the overall dipole moment vector in the molecule: (i) from the carbon atom toward a fluorine atom, (ii) from the carbon atom to a point midway between the fluorine atoms, (iii) from the carbon atom to a point midway between the hydrogen atoms, or (iv) from the earbon atom toward a hydrogen atom? [Sections 9.2 and 9.3]

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