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Ch.21 - Transition Elements and Coordination Chemistry
McMurry - Chemistry 8th Edition
McMurry8th EditionChemistryISBN: 9781292336145Not the one you use?Change textbook
All textbooksMcMurry 8th EditionCh.21 - Transition Elements and Coordination ChemistryProblem 21.127b
Chapter 21, Problem 21.127b

For each of the following complexes, describe the bonding using valence bond theory. Include orbital diagrams for the free metal ion and the metal ion in the complex. Indicate which hybrid orbitals the metal ion uses for bonding, and specify the number of unpaired electrons. 
(b) [Ag(NH3)2]+

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1
Identify the central metal ion in the complex. In this case, it is Ag^+.
Determine the electron configuration of the free metal ion, Ag^+. Silver (Ag) has an atomic number of 47, so its electron configuration is [Kr] 4d^10 5s^1. As Ag^+ loses one electron, the configuration becomes [Kr] 4d^10.
Draw the orbital diagram for the free metal ion, Ag^+. Since Ag^+ has a filled 4d subshell, all 4d orbitals are paired, and there are no unpaired electrons.
Consider the ligands and their effect on the metal ion. NH3 is a neutral ligand and acts as a Lewis base, donating a pair of electrons to the metal ion. In [Ag(NH3)2]^+, the Ag^+ ion forms two coordinate covalent bonds with NH3 ligands.
Determine the hybridization of the metal ion in the complex. For [Ag(NH3)2]^+, the Ag^+ ion uses sp hybrid orbitals to accommodate the two pairs of electrons from the NH3 ligands, resulting in a linear geometry around the metal ion.

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Key Concepts

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

Valence Bond Theory

Valence Bond Theory (VBT) explains how atoms bond by overlapping their atomic orbitals to form covalent bonds. It emphasizes the role of hybridization, where atomic orbitals mix to create new hybrid orbitals that can accommodate electron pairs. This theory helps in understanding the geometry and bonding characteristics of complexes, such as how the shape of [Ag(NH3)2]+ is determined by the hybridization of the silver ion.
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Hybridization

Hybridization is the process of combining atomic orbitals to form new hybrid orbitals that are suitable for the pairing of electrons to form chemical bonds. In the case of [Ag(NH3)2]+, the silver ion undergoes hybridization to form dsp2 hybrid orbitals, which allows it to bond with two ammonia ligands. Understanding hybridization is crucial for predicting the geometry and bond angles in coordination complexes.
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Unpaired Electrons

Unpaired electrons are electrons that are alone in an orbital and are crucial for determining the magnetic properties of a complex. In [Ag(NH3)2]+, the silver ion has a specific electron configuration that can lead to the presence of unpaired electrons, affecting its reactivity and interaction with ligands. Identifying the number of unpaired electrons is essential for understanding the electronic structure and bonding behavior of the metal ion in the complex.
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Nickel(II) complexes with the formula NiX2L2, where X is Cl- or N-bonded NCS- and L is the monodentate triphenylphosphine ligand P(C6H5)3, can be square planar or tetrahedral.

(b) If NiCl2L2 is paramagnetic and Ni(NCS)2L2 is diamagnetic, which of the two complexes is tetrahedral and which is square planar?

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Textbook Question

What is the systematic name for each of the following ions? 

(c) [Co(CO3)3]3-

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Textbook Question

For each of the following complexes, draw a crystal field energy-level diagram, assign the electrons to orbitals, and predict the number of unpaired electrons.

(b) [MnCl4]2- (tetrahedral)

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Textbook Question

The percent iron in iron ore can be determined by dissolving the ore in acid, then reducing the iron to Fe2+, and finally titrating the Fe2+ with aqueous KMnO4. The reaction products are Fe2+ and Mn2+.

(c) Draw a crystal field energy-level diagram for the reactants and products, MnO4-, 3Fe1H2O2642+, 3Fe1H2O2643+, and 3Mn1H2O2642+, and predict the number of unpaired electrons for each.

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For each of the following complexes, draw a crystal field energy-level diagram, assign the electrons to orbitals, and predict the number of unpaired electrons.

(c) [Co(NCS)4]2- (tetrahedral)

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Spinach contains a lot of iron but is not a good source of dietary iron because nearly all the iron is tied up in the oxalate complex [Fe(C2O4)3]3-.

(c) Draw a crystal field energy-level diagram for [Fe(C2O4)3]3-, and predict the number of unpaired electrons. (C2O42- is a weak-field bidentate ligand.)

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