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Ch.9 - Molecular Geometry and Bonding Theories
All textbooksBrown 14th EditionCh.9 - Molecular Geometry and Bonding TheoriesProblem 96b
Chapter 9, Problem 96b

(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-.

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Identify the central atom in the IO6^5- ion, which is iodine (I).
Determine the total number of valence electrons available. Iodine has 7 valence electrons, and each oxygen has 6 valence electrons. Since there are 6 oxygen atoms, calculate the total valence electrons from oxygen and iodine, and account for the 5- charge by adding 5 more electrons.
Use the VSEPR (Valence Shell Electron Pair Repulsion) theory to predict the molecular geometry. Count the number of electron pairs around the central iodine atom, including both bonding pairs and lone pairs.
Assign the electron pairs to positions around the iodine atom to minimize repulsion, considering that the IO6^5- ion has 6 bonding pairs and no lone pairs on the central atom.
Based on the arrangement of electron pairs, determine the molecular geometry. For 6 bonding pairs and no lone pairs, the geometry is expected to be octahedral.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. It is determined by the number of bonding pairs and lone pairs of electrons around the central atom, which influences the shape according to VSEPR (Valence Shell Electron Pair Repulsion) theory. Understanding molecular geometry is crucial for predicting the behavior and reactivity of molecules.
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VSEPR Theory

VSEPR theory posits that electron pairs around a central atom will arrange themselves to minimize repulsion, leading to specific molecular shapes. This theory helps predict the geometry of molecules based on the number of bonding and non-bonding electron pairs. For example, a tetrahedral arrangement occurs when there are four regions of electron density around a central atom.
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Oxidation States

Oxidation states indicate the degree of oxidation of an atom in a compound, reflecting the number of electrons lost or gained. In the case of IO<sub>6</sub><sup>5-</sup>, understanding the oxidation states of iodine and oxygen is essential for determining the overall charge and the bonding characteristics of the molecule. This knowledge aids in predicting the molecular structure and stability.
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Related Practice
Textbook Question

The O¬H bond lengths in the water molecule 1H2O2 are 96 pm, and the H¬O¬H angle is 104.5°. The dipole moment of the water molecule is 1.85 D. (b) Calculate the magnitude of the bond dipole of the O¬H bonds. (Note: You will need to use vector addition to do this.)

Textbook Question

a) Predict the electron-domain geometry around the central S atom in SF2, SF4, and SF6.

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Textbook Question

Which of the following statements about hybrid orbitals is or are true? (i) After an atom undergoes sp hybridization, there is one unhybridized p orbital on the atom, (ii) Under sp2 hybridization, the large lobes point to the vertices of an equilateral triangle, and (iii) The angle between the large lobes of sp3 hybrids is 109.5°.

Textbook Question

Sodium azide is a shock-sensitive compound that releases N2 upon physical impact. The compound is used in automobile airbags. The azide ion is N3-. (a) Draw the Lewis structure of the azide ion that minimizes formal charge (it does not form a triangle). Is it linear or bent?

Textbook Question

Sodium azide is a shock-sensitive compound that releases N2 upon physical impact. The compound is used in automobile airbags. The azide ion is N3-. (b) State the hybridization of the central N atom in the azide ion.