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One member of cach of the following sets of compounds is not isoelectronic with the others. Which one in each set is the odd one out? (a) \(\left[\mathrm{NO}_{2}\right]^{+}, \mathrm{CO}_{2},\left[\mathrm{NO}_{2}\right]^{-}\) and \(\left[\mathrm{N}_{3}\right]^{-}\) (b) \([\mathrm{CN}]^{-}, \mathrm{N}_{2}, \mathrm{CO},[\mathrm{NO}]^{+}\) and \(\left[\mathrm{O}_{2}\right]^{2-}\) (c) \(\left[\mathrm{SiF}_{6}\right]^{2-},\left[\mathrm{PF}_{6}\right]^{-},\left[\mathrm{AlF}_{6}\right]^{3-}\) and \(\left[\mathrm{BrF}_{6}\right]^{-}\)

Short Answer

Expert verified
(a) \\([\mathrm{N}_3]^-\\), (b) \\([\mathrm{O}_2]^{2-}\\), (c) \\([\mathrm{BrF}_6]^{-}\\)

Step by step solution

01

Understanding Isoelectronic

Isoelectronic species have the same number of electrons. To determine which compound is not isoelectronic, we need to calculate the total number of electrons for each compound in the set.
02

Analyze Set (a)

- \([\mathrm{NO}_2]^+\): Nitrogen (7 electrons) + 2*Oxygen (16 electrons each) - 1 (charge) = 7 + 32 - 1 = 38 electrons. - \(\mathrm{CO}_2\): Carbon (6 electrons) + 2*Oxygen (16 electrons each) = 6 + 32 = 38 electrons. - \([\mathrm{NO}_2]^-\): Nitrogen (7 electrons) + 2*Oxygen (16 electrons each) + 1 (charge) = 7 + 32 + 1 = 40 electrons.- \([\mathrm{N}_3]^-\): 3*Nitrogen (7 electrons) + 1 (charge) = 21 + 1 = 22 electrons.Since \([\mathrm{N}_3]^-\) has 22 electrons, it is the odd one out.
03

Analyze Set (b)

- \([\mathrm{CN}]^-\): Carbon (6 electrons) + Nitrogen (7 electrons) + 1 (charge) = 14 electrons.- \(\mathrm{N}_2\): 2*Nitrogen (7 electrons each) = 14 electrons.- \(\mathrm{CO}\): Carbon (6 electrons) + Oxygen (16 electrons) = 6 + 16 = 22 electrons.- \([\mathrm{NO}]^+\): Nitrogen (7 electrons) + Oxygen (16 electrons) - 1 (charge) = 7 + 16 - 1 = 22 electrons.- \([\mathrm{O}_2]^{2-}\): 2*Oxygen (16 electrons each) + 2 (charge) = 34 electrons.Given \(\mathrm{O}_2\) and \([\mathrm{O}_2]^{2-}\) have drastically more, while \(\mathrm{CO}\) and \([\mathrm{NO}]^+\) are greater but still similar, \(\mathrm{N}_2\) and \([\mathrm{CN}]^-\) are unique, thus \([\mathrm{O}_2]^{2-}\) is the odd one.
04

Analyze Set (c)

- \([\mathrm{SiF}_6]^{2-}\): Silicon (14 electrons) + 6*Fluorine (9 electrons each) + 2 (charge) = 14 + 54 + 2 = 70 electrons.- \([\mathrm{PF}_6]^{-}\): Phosphorus (15 electrons) + 6*Fluorine (9 electrons each) + 1 (charge) = 15 + 54 + 1 = 70 electrons.- \([\mathrm{AlF}_6]^{3-}\): Aluminum (13 electrons) + 6*Fluorine (9 electrons each) + 3 (charge) = 13 + 54 + 3 = 70 electrons.- \([\mathrm{BrF}_6]^{-}\): Bromine (35 electrons) + 6*Fluorine (9 electrons each) + 1 (charge) = 35 + 54 + 1 = 90 electrons.Thus, \([\mathrm{BrF}_6]^{-}\) is the odd one out due to having 90 electrons versus the 70 in others.

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Key Concepts

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

Electron counting
Electron counting involves determining the total number of electrons in a molecule or ion. It's essential for understanding whether compounds are isoelectronic, meaning they possess the same number of electrons. To perform electron counting, follow these steps:

  • Identify the number of valence electrons for each atom in the molecule, which can be found on the periodic table.
  • Add together the valence electrons for all atoms in the compound.
  • Consider the charge of the ion, adding electrons for negative charges and subtracting for positive charges.
This process allows us to calculate a total electron count, helping to ascertain whether molecules in a set are isoelectronic or have varied electron contributions. Understanding electron counting aids in predicting chemical bonds and the geometry of a molecule.
Chemical bonding
Chemical bonding is the process where atoms or ions are held together in a molecule or crystal. This packaging can occur through different types of bonds:

  • Covalent bonds: These occur when atoms share electrons, typically between non-metals, aiming to fill their outer electron shell.
  • Ionic bonds: Here, electrons are transferred between atoms, usually from a metal to a non-metal, creating charged ions.
  • Metallic bonds: In metals, electrons are shared across a lattice of atoms.
The nature and strength of these bonds influence molecular properties and reactivity. Bonds affect electron distribution, which in turn affects molecular behavior, such as polarity and interaction with other molecules.
Charge of ions
Ions are atoms or molecules that have gained or lost one or more electrons, resulting in a net electrical charge. The charge of an ion is crucial because it affects how ions combine in chemical reactions. For example, in the compound \([\text{NO}_2]^+\), "+" indicates a positive charge from one less electron. In set \([\text{SiF}_6]^{2-}\), "2-" illustrates the gain of two electrons, tailoring how it interacts with other ions.

  • Cations: Positively charged ions formed by losing electrons. Typically metals.
  • Anions: Negatively charged ions formed by gaining electrons. Typically non-metals.
Understanding the charge on ions is important for predicting how different ions will interact in compounds, influencing properties like solubility and crystal structure.
Molecular geometry
Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. It impacts physical properties such as polarity, reactivity, and interactions with other molecules. The electronic structure determined by electron counting influences molecular shapes through:

  • VSEPR theory: Valence Shell Electron Pair Repulsion theory predicts molecular structure based on electron pair repulsion. Lone pairs and bond pairs want to minimize repulsive forces, thus arranging themselves as far apart as possible.
  • Hybridization: The mixing of atomic orbitals into new hybrid orbitals that influence molecular geometry, such as \(sp^3\) hybridization leading to tetrahedral shapes.
Different molecular geometries, like linear, trigonal planar, or tetrahedral, result from these interactions and determine how molecules stack or interact with each other.

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Most popular questions from this chapter

Using VB theory and the Lewis structure model, determine the bond order in (a) \(\mathrm{H}_{2},(\mathrm{b}) \mathrm{Na}_{2},(\mathrm{c}) \mathrm{S}_{2},(\mathrm{d}) \mathrm{N}_{2}\) and (e) \(\mathrm{Cl}_{2}\) Is there any ambiguity with finding the bond orders by this method?

(a) Write down the ions that are present in the compound \(\left[\mathrm{PCl}_{4}\right]\left[\mathrm{PCl}_{3} \mathrm{F}_{3}\right] .\) What shape do you expect each ion to adopt? In theory, does cither ion possess stereoisomers? (b) Use the VSEPR model to rationalize why \(\mathrm{BCl}_{3}\) and \(\mathrm{NCl}_{3}\) do not adopt similar structures. Is either molecule expected to be polar? Rationalize your answer.

Use the VSEPR model to predict the structures of (a) \(\mathrm{H}_{2} \mathrm{Se},\) (b) \(\left[\mathrm{BH}_{4}\right]^{-},(\mathrm{c}) \mathrm{NF}_{3},(\mathrm{d}) \mathrm{SbF}_{5}\) \((\mathrm{e})\left[\mathrm{H}_{3} \mathrm{O}\right]^{+}\) (f) \(\mathrm{IF}_{7},(\mathrm{g})\left[\mathrm{I}_{3}\right]^{-},(\mathrm{h})\left[\mathrm{I}_{3}\right]^{+},\) (i) \(\mathrm{SO}_{3}\)

(a) Construct an MO diagram for the formation of \(\mathrm{O}_{2}\) show only the participation of the valence orbitals of the oxygen atoms. (b) Use the diagram to rationalize the following trend in \(\mathrm{O}-\mathrm{O}\) bond distances: \(\mathrm{O}_{2}, 121 \mathrm{pm}\) \(\left[\mathrm{O}_{2}\right]^{+}, 112 \mathrm{pm} ;\left[\mathrm{O}_{2}\right]^{-}, 134 \mathrm{pm} ;\left[\mathrm{O}_{2}\right]^{2-}, 149 \mathrm{pm} .\) (c) Which of these species are paramagnetic?

State whether you expect the following species to possess stereoisomers and, if so, draw their structures and give them distinguishing labels: (a) \(\mathrm{BF}_{2} \mathrm{Cl}\) (b) \(\mathrm{POCl}_{3}\) (c) MePF \(_{4} ;(\mathrm{d})\left[\mathrm{PF}_{2} \mathrm{Cl}_{4}\right]^{-}\)

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