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Chapter 9: Problem 87

Consider the following \(\mathrm{XF}_{4}\) ions: \(\mathrm{PF}_{4}^{-}, \mathrm{BrF}_{4}^{-}, \mathrm{ClF}_{4}^{+},\) and \(\mathrm{AlF}_{4}^{-}\) (a) Which of the ions have more than an octet of electrons around the central atom? (b) For which of the ions will the electron-domain and molecular geometries be the same? (c) Which of the ions will have an octahedral electron-domain geometry (d) Which of the ions will exhibit a see-saw molecular geometry?

Short Answer

Expert verified
a) \(\mathrm{BrF}_{4}^{-}\) and \(\mathrm{ClF}_{4}^{+}\) have more than an octet of electrons around the central atom. b) All ions have the same electron-domain and molecular geometries. c) Only \(\mathrm{BrF}_{4}^{-}\) has an octahedral electron-domain geometry. d) None of the ions exhibit a see-saw molecular geometry.

Step by step solution

01

Valence electrons

For each ion, count the total number of valence electrons: - \(\mathrm{PF}_{4}^{-}\): P has 5 valence electrons and each F has 7. Add 1 electron for the negative charge: 5 + (4 × 7) + 1 = 33 electrons. - \(\mathrm{BrF}_{4}^{-}\): Br has 7 valence electrons and each F has 7. Add 1 electron for the negative charge: 7 + (4 × 7) + 1 = 36 electrons. - \(\mathrm{ClF}_{4}^{+}\): Cl has 7 valence electrons and each F has 7. Subtract 1 electron for the positive charge: 7 + (4 × 7) - 1 = 34 electrons. - \(\mathrm{AlF}_{4}^{-}\): Al has 3 valence electrons and each F has 7. Add 1 electron for the negative charge: 3 + (4 × 7) + 1 = 32 electrons.
02

Electron-domain geometry

Determine electron-domain geometry for each ion using VSEPR Theory: - \(\mathrm{PF}_{4}^{-}\): With 33 electrons, the electron-domain geometry is trigonal bipyramidal. - \(\mathrm{BrF}_{4}^{-}\): With 36 electrons, the electron-domain geometry is octahedral. - \(\mathrm{ClF}_{4}^{+}\): With 34 electrons, the electron-domain geometry is square planar. - \(\mathrm{AlF}_{4}^{-}\): With 32 electrons, the electron-domain geometry is tetrahedral.
03

Matching electron-domain and molecular geometries

Identify ions with matching electron-domain and molecular geometries: - For \(\mathrm{PF}_{4}^{-}\), \(\mathrm{BrF}_{4}^{-}\), \(\mathrm{ClF}_{4}^{+}\), and \(\mathrm{AlF}_{4}^{-}\), the molecular geometries match the electron-domain geometries, so the answer is all ions.
04

Octahedral electron-domain geometry

Identify ions with an octahedral electron-domain geometry: - Only \(\mathrm{BrF}_{4}^{-}\) has an octahedral electron-domain geometry.
05

See-saw molecular geometry

Identify ions with a see-saw molecular geometry: - None of the given ions exhibit a see-saw molecular geometry. #Results#: a) \(\mathrm{BrF}_{4}^{-}\) and \(\mathrm{ClF}_{4}^{+}\) have more than an octet of electrons around the central atom. b) All ions have the same electron-domain and molecular geometries. c) Only \(\mathrm{BrF}_{4}^{-}\) has an octahedral electron-domain geometry. d) None of the ions exhibit a see-saw molecular geometry.

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Most popular questions from this chapter

The Lewis structure for allene is Make a sketch of the structure of this molecule that is analogous to Figure \(9.25 .\) In addition, answer the following three questions: (a) Is the molecule planar? (b) Does it have a nonzero dipole moment? (c) Would the bonding in allene be described as delocalized? Explain.

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

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