Draw the Lewis structure for AsF

Draw the Lewis structure for AsF

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in the window below and then answer the questions that follow. – Do not include overall ion charges or formal charges in your drawing. Draw the Lewis structure for IBr

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in the window below and then answer the questions that follow. What is the electron-pair geometry for I in IBr

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? What is the the shape (molecular geometry) of IBr

The Correct Answer and Explanation is :

Lewis Structures of AsF₆⁻ and IBr₃

Step 1: Drawing the Lewis Structure for AsF₆⁻

  1. Count the valence electrons
  • Arsenic (As) has 5 valence electrons
  • Fluorine (F) has 7 valence electrons, and there are six fluorine atoms
  • The negative charge (-1) adds one extra electron
  • Total valence electrons = 5 + (7 × 6) + 1 = 48 electrons
  1. Determine the central atom and arrange the atoms
  • As is less electronegative than F, so As is the central atom
  • Distribute six fluorine atoms around arsenic
  1. Draw single bonds between As and F
  • Each single bond uses 2 electrons
  • 6 bonds × 2 electrons = 12 electrons used
  • Remaining electrons = 48 – 12 = 36 electrons
  1. Distribute remaining electrons to satisfy octets
  • Each fluorine needs 6 more electrons to complete its octet
  • 6 fluorine atoms × 6 electrons = 36 electrons (all remaining electrons are used)
  1. Check the octet rule and formal charge
  • As has 12 valence electrons (expanded octet, which is allowed for Group 15 elements in Period 4 and beyond)
  • Formal charges: As has a -1 formal charge, matching the given ion charge

Thus, the Lewis structure of AsF₆⁻ has As in the center with six F atoms surrounding it, each bonded by a single bond, and a negative charge on the ion.


Step 2: Drawing the Lewis Structure for IBr₃

  1. Count the valence electrons
  • Iodine (I) has 7 valence electrons
  • Bromine (Br) has 7 valence electrons, and there are three Br atoms
  • Total valence electrons = 7 + (7 × 3) = 28 electrons
  1. Determine the central atom and arrange the atoms
  • I is less electronegative than Br, so it is the central atom
  • Distribute three Br atoms around I
  1. Draw single bonds between I and Br
  • 3 bonds × 2 electrons = 6 electrons used
  • Remaining electrons = 28 – 6 = 22 electrons
  1. Distribute remaining electrons to satisfy octets
  • Each Br needs 6 more electrons to complete its octet
  • 3 Br × 6 electrons = 18 electrons used
  • Remaining electrons = 22 – 18 = 4 electrons
  1. Place remaining electrons on the central atom (I)
  • The 4 leftover electrons go on I as two lone pairs
  • Iodine now has 5 electron groups (3 bonds + 2 lone pairs)

Electron-Pair Geometry and Molecular Geometry of IBr₃

  1. Electron-Pair Geometry
  • The electron-domain (steric) number is 5 (three bonding pairs + two lone pairs).
  • The electron-pair geometry is trigonal bipyramidal.
  1. Molecular Geometry (Shape)
  • The two lone pairs occupy equatorial positions to minimize electron repulsions.
  • This results in a T-shaped molecular geometry.

Explanation (300 words)

The Lewis structures of AsF₆⁻ and IBr₃ help predict molecular geometry based on VSEPR (Valence Shell Electron Pair Repulsion) theory. For AsF₆⁻, the arsenic atom is surrounded by six fluorine atoms, forming an octahedral molecular shape with 90° bond angles. The molecule is nonpolar because the symmetrical distribution of fluorine atoms cancels out dipole moments.

For IBr₃, iodine is the central atom, forming three single bonds with bromine atoms and having two lone pairs. The presence of lone pairs affects the molecular shape. Although the electron-pair geometry is trigonal bipyramidal, the actual molecular geometry is T-shaped because lone pairs occupy equatorial positions to reduce repulsions. The bond angles are slightly less than 90° due to lone pair repulsions. The molecule is polar because of its asymmetric shape, leading to a net dipole moment.

These structures illustrate key concepts in molecular geometry:

  • Lone pairs cause deviations from ideal geometries
  • Symmetry determines molecular polarity
  • Expanded octets occur in elements beyond Period 2, such as arsenic and iodine

Understanding these structures helps predict chemical reactivity, intermolecular forces, and physical properties like boiling points and solubility.

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