Download Coordination Compounds

Example 24.1 Writing Electron Configurations for
Transition Metals
Write the ground state electron configuration for Zr.
Procedure For…
Writing Electron Configurations
Solution
Identify the noble gas that precedes the element and write it in square brackets.
[Kr]
Count down the periods to determine the outer principal quantum level—this is the quantum level for the s orbital.
Subtract one to obtain the quantum level for the d orbital. If the element is in the third or fourth transition series,
include (n − 2) f 14 electrons in the configuration.
Zr is in the fifth period so the orbitals used are
[Kr] 5s4d
Count across the row to see how many electrons are in the neutral atom and fill the orbitals accordingly.
Zr has four more electrons than Kr.
[Kr] 5s24d2
For an ion, remove the required number of electrons, first from the s and then from the d orbitals.
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 24.1 Writing Electron Configurations for
Transition Metals
Continued
For Practice 24.1
Write the ground state electron configuration for Os.
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 24.2 Writing Electron Configurations for
Transition Metals
Write the ground state electron configuration for Co3+.
Procedure For…
Writing Electron Configurations
Solution
Identify the noble gas that precedes the element and write it in square brackets.
[Ar]
Count down the periods to determine the outer principal quantum level—this is the quantum level for the s orbital.
Subtract one to obtain the quantum level for the d orbital. If the element is in the third or fourth transition series,
include (n − 2)f 14 electrons in the configuration.
Co is in the fourth period so the orbitals used are
[Ar] 4s3d
Count across the row to see how many electrons are in the neutral atom and fill the orbitals accordingly.
Co has nine more electrons than Ar.
[Ar] 4s23d7
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 24.2 Writing Electron Configurations for
Transition Metals
Continued
For an ion, remove the required number of electrons, first from the s and then from the d orbitals.
Co3+ has lost three electrons relative to the Co atom.
[Ar] 4s03d 6 or [Ar] 3d 6
For Practice 24.2
Write the ground state electron configuration for Nb2+.
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 24.3 Naming Coordination Compounds
Name the following compound: [Cr(H2O)5Cl]Cl2.
Procedure For…
Naming Coordination Compounds
Solution
Identify the cation and anion and first name the simple ion (i.e., not the complex one).
[Cr(H2O)5Cl]2+ is a complex cation. Cl− is chloride.
Give each ligand a name and list them in alphabetical order.
H2O is aqua.
Cl− is chloro.
Name the metal ion.
Cr3+ is chromium(III).
Name the complex ion by adding prefixes to indicate the number of each ligand followed by the name of each ligand
followed by the name of the metal ion.
[Cr(H2O)5Cl]2+ is pentaaquachlorochromium(III).
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 24.3 Naming Coordination Compounds
Continued
Name the compound by writing the name of the cation before the anion. The only space is between ion names.
[Cr(H2O)5Cl]Cl2 is pentaaquachlorochromium(III) chloride.
For Practice 24.3
Name the following compound: [Mn(CO)(NH3)5]SO4.
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 24.4 Naming Coordination Compounds
Name the following compound: K3[Fe(CN)6].
Procedure For…
Naming Coordination Compounds
Solution
Identify the cation and anion and first name the simple ion (i.e., not the complex one).
K+ is potassium. [Fe(CN)6]3− is a complex anion.
Give each ligand a name and list them in alphabetical order.
CN− is cyano.
Name the metal ion.
Fe3+ is ferrate(III) because the complex is anionic.
Name the complex ion by adding prefixes to indicate the number of each ligand followed by the name of each ligand
followed by the name of the metal ion.
[Fe(CN)6]3− is hexacyanoferrate(III).
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 24.4 Naming Coordination Compounds
Continued
Name the compound by writing the name of the cation before the anion. The only space is between ion names.
K3[Fe(CN)6] is potassium hexacyano ferrate(III).
For Practice 24.4
Name the following compound: Na2[PtCl4].
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 24.5 Identifying and Drawing Geometric Isomers
Draw the structures and label the type of all the isomers of [Co(en) 2Cl2]+.
Procedure For…
Identifying and Drawing Geometric Isomers
Solution
Identify the coordination number and the geometry around the metal.
The ethylenediamine (en) ligand is bidentate so each occupies two coordination sites. Each Cl − is monodentate,
occupying one site. The total coordination number is 6, so this must be an octahedral complex.
Identify if this is cis–trans or fac–mer isomerism.
With ethylenediamine occupying four sites and Cl − occupying two sites, it fits the general formula MA4B2, leading
to cis–trans isomers.
Draw and label the two isomers.
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 24.5 Identifying and Drawing Geometric Isomers
Continued
For Practice 24.5
Draw the structures and label the type of all the isomers of [Cr(H2O)3Cl3]+.
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 24.6 Identifying and Drawing Geometric Isomers
Draw the structures and label the type for of the isomers of [Ni(CN) 2Cl2]2−.
Procedure For…
Identifying and Drawing Geometric Isomers
Solution
Identify the coordination number and the geometry around the metal.
All the ligands are monodentate, so the total coordination number is 4. Ni 2+ is a d8 electronic configuration, so we
expect a square planar complex.
Identify if this is cis–trans or fac–mer isomerism.
Square planar complexes can only have cis–trans isomers.
Draw and label the two isomers.
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 24.6 Identifying and Drawing Geometric Isomers
Continued
For Practice 24.6
Draw the structures and label the type of all the isomers of [Co(NH3)2Cl2(ox)]−.
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 24.7 Recognizing and Drawing Optical Isomers
Determine whether the cis or trans isomers in Example 24.5 are optically active (demonstrate optical isomerism).
Solution
Draw the trans isomer of [Co(en)2Cl2]+ and its mirror image. Check to see if they are superimposable by
rotating one isomer 180°.
In this case the two are identical, so there is no optical activity.
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 24.7 Recognizing and Drawing Optical Isomers
Continued
Draw the cis isomer and its mirror image. Check to see if they are superimposable by rotating one isomer 180°.
In this case the two structures are not superimposable, so the cis isomer does exhibit optical activity.
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 24.7 Recognizing and Drawing Optical Isomers
Continued
For Practice 24.7
Determine whether the fac or mer isomers of [Cr(H2O)3Cl3]+ are optically active.
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 24.8 Crystal Field Splitting Energy
The complex ion [Cu(NH3)6]2+ is blue in aqueous solution. Estimate the crystal field splitting energy
(in kJ/mol) for this ion.
Solution
Begin by consulting the color wheel to determine approximately what wavelength is being absorbed.
Since the solution is blue, you can deduce that orange light is absorbed since orange is the complementary color
to blue.
Estimate the absorbed wavelength.
The color orange ranges from 580 to 650 nm, so you can estimate the average wavelength as 615 nm.
Calculate the energy corresponding to this wavelength, using E = hc/λ. This energy corresponds to Δ.
Convert J/ion into kJ/mol.
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 24.8 Crystal Field Splitting Energy
Continued
For Practice 24.8
The complex ion [Co(NH3)5NO2]2+ is yellow. Estimate the crystal field splitting energy (in kJ/mol) for this ion.
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 24.9 High- and Low-Spin Octahedral Complexes
How many unpaired electrons are there in the complex ion [CoF 6]3−?
Procedure For…
Determining the Number of Unpaired Electrons in Octahedral Complexes
Solution
Begin by determining the charge and number of d electrons on the metal.
The metal is Co3+ and has a d6 electronic configuration.
Look at the spectrochemical series to determine whether the ligand is a strong-field or a weak-field ligand.
F− is a weak-field ligand, so Δ is relatively small.
Decide if the complex is high- or low-spin and draw the electron configuration.
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 24.9 High- and Low-Spin Octahedral Complexes
Continued
Weak-field ligands yield high-spin configurations.
Count the unpaired electrons.
This configuration has four unpaired electrons.
For Practice 24.9
How many unpaired electrons are there in the complex ion [FeCl 6]3−?
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 24.10 High- and Low-Spin Octahedral Complexes
How many unpaired electrons are there in the complex ion [Co(NH3)5NO2]2+?
Procedure For…
Determining the Number of Unpaired Electrons in Octahedral Complexes
Solution
Begin by determining the charge and number of d electrons on the metal.
The metal is Co3+ and has a d6 electronic configuration.
Look at the spectrochemical series to determine whether the ligand is a strong-field or a weak-field ligand.
NH3 and NO2− are both strong-field ligands, so Δ is relatively large.
Decide if the complex is high- or low-spin and draw the electron configuration.
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.
Example 24.10 High- and Low-Spin Octahedral Complexes
Continued
Strong-field ligands yield low-spin configurations.
Count the unpaired electrons.
This configuration has no unpaired electrons.
For Practice 24.10
How many unpaired electrons are there in the complex ion [Co(CN) 6]4−?
Chemistry: A Molecular Approach, 3rd Edition
Nivaldo J. Tro
© 2014 Pearson Education, Inc.