Introduction

Filling atomic orbitals requires a set number of electrons. The s-block is composed of elements of Groups I and II, the alkali and alkaline earth metals (sodium and calcium belong to this block). Groups XIII through XVIII comprise of the p-block, which contains the nonmetals, halogens, and noble gases (carbon, nitrogen, oxygen, fluorine, and chlorine are common members). Transition metals reside in the d-block, between Groups III and XII. If the following table appears strange, or if the orientations are unclear, please review the section on atomic orbitals.

The key thing to remember about electronic configuration is that the most stable noble gas configuration is ideal for any atom. Forming bonds are a way to approach that configuration. In particular, the transition metals form more lenient bonds with anions, cations, and neutral complexes in comparision to other elements. This is because the d orbital is rather diffused (the f orbital of the lanthanide and actinide series more so).

Neutral-Atom Electron Configuration

Counting through the periodic table is an easy way to determine which electrons exist in which orbitals. As mentioned before, by counting protons (atomic number), you can tell the number of electrons in a neutral atom. Organizing by block quickens this process. If you do not feel confident about this counting system and how electron orbitals are filled, please see the section on electron configuration.

For example, if we were interested in determining the electronic organization of Vanadium (atomic number 23), we would start from hydrogen and make our way down (refer to the Periodic Table).

***1s (H, He), 2s (Li, Be), 2p (B, C, N, O, F, Ne), 3s (Na, Mg), 3p (Al, Si, P, S, Cl, Ar), 4s (K, Ca), 3d (Sc, Ti, V).
 

Referring to the periodic table below confirms this organization. We have 3 elements in the 3d orbital. Therefore, we write in the order the orbitals were filled.

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d³, or
[Ar] 4s² 3d³.

The neutral atom configurations of the fourth period transition metals are:

What may appear anomalous is the case that takes advantage of the degeneracy.

Take a brief look at where the element Chromium (atomic number 24) lies on the Periodic Table (found below). The electronic configuration for chromium is not
***4s²*******************([Ar] 4s²3d⁴)
***3d⁴_{x²-y² z² xy yz xz}
but instead it is
***4s¹*******************([Ar] 4s¹3d⁵)
***3d⁵ _{x²-y² z² xy yz xz}
especially because of the s and d orbitals' degeneracy. There is no error in assuming that a s-orbital electron will be displaced to fill the place of a d-orbital electron because their associated energies are equal. After all, the Aufbau Principle states that the lowest energy configuration is of unpaired electrons in the most space possible. Electrostatic force is inversely proportional to distance according to Coulomb's Law; this unnecessarily paired s-orbital electron can be relieved of its excess energy.

Lastly, for the two above energy diagrams to be true in nature, the distance between the 4s and the 3d orbitals would be neglected. There is a slight separation for transition metals on the right of the block, but for the purpose of discussing ionization, the order indicated is true. More energetic orbitals are labeled above lesser ones.

Example 1. Write the electronic configuration of neutral iron, iron(II) ion and iron(III) ion. The atomic number of iron is 26.

Example 2. Determine the more stable configuration between the following pair:
Example 2. A. [Kr] 5s² 4d⁶ vs. [Kr] 5s¹ 4d⁷
Example 2. B. Ag¹⁺ vs. Ag²⁺

Solution 1. Fe: [Ar] 4s² 3d⁶; Fe²⁺: [Ar] 3d⁶; Fe³⁺: [Ar] 3d⁵. Note that the s-orbital electrons are lost first, then the d-orbital electrons.

Solution 2. A. This describes Ruthenium. There is only one 5s electron.
Solution 2. B. Once-oxidized silver ([Kr] 4d¹⁰) is more stable than twice- ([Kr] 4d⁹).

Oxidation States of Transition Metals

When considering ions, we add or subtract negative charges from an atom. Keeping the atomic orbitals when assigning oxidation numbers in mind helps in recognizing that transition metals pose a special case but not an exception to this convenient method. An atom that accepts an electron to achieve a more stable configuration is assigned an oxidation number of -1. The donation of an electron is then +1.  Remember that when a transition metal loses electrons, it tends to lose it's s orbital electrons before any of its d orbital electrons. For more discussion of these compounds form, see formation of coordination complexes.

Physical Properties

Transition metals have high boiling points. The d orbitals allow electrons to become diffused and enables them to be delocalized within solid metal. This increases the attractive forces between the atoms and requires more energy to dissociate them in order to change phases. This attraction reaches a maximum in Group IV for manganese (boiling point of 2061 °C), which has 5 unpaired electrons. The potential for manganese to form strong and numerous bonds is greater than its neighbors. For this same reason, zinc has a low boiling point (907 °C): it does not have much attractive force between like atoms. Also in the 12^th period, mercury has a low melting point (-39 °C), which allows it to be liquid at standard conditions.

The lanthanide contraction is a term that describes two different periodic trends. The first is that the Group VI transition metals are separated by 15 additional elements which are displaced to the bottom of the table. The lanthanides introduce the f orbital, which are very diffused and do not shield well. Since additional protons are now more visible to these electrons, the atomic radius of a Group VI transition metal is contracted enough to have approximately equal atomic radii to Group V transition metals. The second definition explains the general decrease in ionic radii and atomic radii as one looks at transition metals from left to right. This is due to the addition of electrons to the same diffused f orbital while protons are added. This results in greater attraction between protons and neutrons. The term refers to the same idea that f orbitals do not shield electrons efficiently, but refer to comparisons between elements horizontally and vertically.

Varying Oxidation States

The following chart describes the most common oxidation states of the period 3 elements.

This diagram brings up a few concepts illustrating the stable states for specific elements.
Example 1. What makes zinc stable as Zn²⁺? What makes scandium stable as Sc³⁺?
Example 2. Why is iron almost always Fe²⁺ or Fe³⁺?
Example 3. Write manganese oxides in a few different oxidation states. Which ones are possible and/or reasonable?

Solution 1. Zinc has the neutral configuration [Ar]4s²3d¹⁰. Losing 2 electrons does not alter the complete d orbital. Neutral scandium is written as [Ar]4s²3d¹. Losing 3 electrons brings the configuration to the noble state with valence 3p⁶.
Solution 2. Iron is written as [Ar]4s²3d⁶. Losing 2 electrons from the s-orbital (3d⁶) or 2 s- and 1 d-orbital (3d⁵) electron are fairly stable oxidation states.
Solution 3. Although Mn⁺² is the most stable ion for manganese, the d-orbital can be made to remove 0 to 7 electrons. Compounds of manganese therefore range from Mn(0) as Mn_(s), Mn(II) as MnO, Mn(II,III) as Mn₃O₄, Mn(IV) as MnO₂, or manganese dioxide, Mn(VII) in the permanganate ion MnO₄^-, and so on.

Manganese: A Case Study

Manganese is widely studied because it is an important reducing agent in chemical analysis. it is also studied in biochemistry for catalysis, as well as in fortifying alloys. In plants, manganese is required in trace amounts; stronger doses begin to react with enzymes and inhibit some cellular function. Due to manganese's flexibility in accepting many oxidation states, it becomes a good example to describe general trends and concepts behind electron configurations.

Identification and Spectroscopy

Regular lattice structures can be calculated using the Schrödinger equation to determine specific solutions called Bloch waves. These waves describe electron distribution in solids, which are further interpreted to describe valence bands and conduction bands. The conduction band is where the electrons from a valence orbital may be excited to pass the charge elsewhere, this is a trait which influences intermolecular attractions. An alternative to these calculations are the more complicated cluster structure calculations, which tend to be more congruous with experimental data. By calculating and measuring respective peaks representing the binding energy for a given complex, the oxidation number can be determined. The intensity of these energies, measured in counts, help identify the bonds of interest by defining the shape and position of these peaks. These diagrams are known as energy spectra. Manganese is interesting to analyze in this way because it has many possible oxidation states.

Crystallography techniques, in particular x-ray diffraction spectroscopy, are usually employed in these analyses. For more information about how shooting electrons at regular crystal structures tells us about physical composition.

Magnetic Properties

Electron configurations of unpaired electrons are said to be paramagnetic and respond to the proximity of magnets. Fully paired electrons are diamagnetic and do not feel this influence. Manganese, in particular, has paramagnetic and diamagnetic orientations depending on what its oxidation state is.

Mn₂O₃ is manganese(III) oxide, where manganese is in the +3 state. 4 unpaired electrons means this complex is paramagnetic.

4s⁰
3d⁴ _{x²-y² z² xy yz xz}

MnO₂ is manganese(IV) oxide, where manganese is in the +4 state. 3 unpaired electrons means this complex is less paramagnetic than Mn³⁺.

4s⁰
3d³ _{x²-y² z² xy yz xz}

KMnO₄ is potassium permanganate, where manganese is in the +7 state. No electrons exist in the 4s and 3d orbitals. The 3p orbitals have no unpaired electrons, so this complex is diamagnetic.

4s⁰
3d⁰ _{x²-y² z² xy yz xz}

Summary

Oxidation states of transition metals follow the general rules for most other ions, except for the fact that the d orbital is degenerated with the s orbital of the higher quantum number. Transition metals achieve stability by arranging their electrons accordingly and are oxidized, or they lose electrons to other atoms and ions. These resulting cations participate in the formation of coordination complexes or synthesis of other compounds.

References

1. Oxtoby D, Gillis H P, Campion, A. Principles of Modern Chemistry, 6^th ed. Thomson Brooks/Cole, Belmont. 2008; 313-318.
2. Audi A, Sherwood, P. Valence-band x-ray photoelectron spectroscopic studies of manganese and its oxides interpreted by cluster and band structure calculations; Surf. Interface Anal.; 2002; 33; 274-282.
3. Reaney S, Kwik-Uribe C, Smith D. Manganese Oxidation State and Its Implications for Toxicity. Chem. Res. Toxicol.; 2002; 15; 1119-1126.
4. CRC Handbook, 88^th ed. Sct. 1, Prt. 1 Electron Configuration and Ionization Energy of Neutral Atoms in the Ground State; 13-14.
5. CRC Handbook, 88^th ed. Sct. 4, Prt. 1 Melting, Boiling, Triple, and Critical Point Temperatures of the Elements; 133-134.

Outside Links

• Bare, William D.; Resto, Wilfredo. "Vanadium lons as Visible Electron Carriers in a Redox System (TD)." J. Chem. Educ. 1994, 71, 692.
• For more help in writing these states, all neutral and +1 cations are listed at the NIST website: http://physics.nist.gov/PhysRefData/DFTdata/configuration.html

Contributors

• Margaux Kreitman (UCD)

This page viewed 23005 times
The ChemWiki has 9279 Modules.

   UC Davis ChemWiki by University of California, Davis is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License.  Tweet

Permissions beyond the scope of this license may be available at copyright@ucdavis.edu. Terms of Use

Powered by Mindtouch Core 2010