2.3.3: Electron Configuration of Transition Metals

Electron configuration describes the distribution of electrons among different orbitals (including shells and subshells) within atoms and molecules.

Introduction

There are four principle orbitals (s, p, d, and f) which are filled according to the energy level and valence electrons of the element. All four orbitals can hold different number of electrons. The s-orbital can hold 2 electrons, and the other three orbitals can hold up to 6, 10, and 14 electrons, respectively. The s-orbital primarily denotes group 1 or group 2 elements, the p-orbital denotes group 13, 14, 15, 16, 17, or 18 elements, and the f-orbital denotes the Lanthanides and Actinides group. The main focus of this module however will be on the electron configuration of transition metals, which are found in the d-orbitals (d-block).

The electron configuration of transition metals is special in the sense that they can be found in numerous oxidation states. Although the elements can display many different oxidation states, they usually exhibit a common oxidation state depending on what makes that element most stable. For this module, we will work only with the first row of transition metals; however the other rows of transition metals generally follow the same patterns as the first row.

The s, p, d, and f-orbitals are identified on the periodic table below:

First Row Transition Metals

In the first row of the transition metals, the ten elements that can be found are: Scandium (Sc), Titanium (Ti), Vanadium (V), Chromium (Cr), Manganese (Mn), Iron (Fe), Cobalt (Co), Nickel (Ni), Copper (Cu), and Zinc (Zn).

Below is a table of the oxidation states that the transition metals can or cannot form. As stated in the boxes, the “No” indicates that the elements are not found with that oxidation state. The “Rare” signifies the oxidation states that the elements are rarely found in. Lastly, the “Common” identifies the oxidation states that the elements readily found in.

Oxidation States of the First Row Transition Metals

First Row Transition Metals		Oxidation States
Element Symbol	Atomic Number	+1	+2	+3	+4	+5	+6	+7
Sc	21	No	Rare	Common	No	No	No	No
Ti	22	No	Rare	Rare	Common	No	No	No
V	23	Rare	Common (lilac)	Common (green)	Common (blue)	Common (yellow)	No	No
Cr	24	Rare	Common	Common (Most stable)	Rare	Rare	Common	No
Mn	25	Rare	Common (Most stable) (pink/red)	Common (purple/red)	Common	Rare (blue)	Common (green)	Common (purple)
Fe	26	Rare	Common (ferrous)	Common (ferric)	Rare	Rare	Rare	No
Co	27	Rare	Common	Common	Rare	Rare	Rare	No
Ni	28	Rare	Common	Rare	Rare	No	No	No
Cu	29	Rare	Common (blue/green)	No	No	No	No	No
Zn	30	No	Common	No	No	No	No	No

Electron Configuration of Transition Metal Atoms

The electron configuration for the first row transition metals consists of 4s and 3d subshells with an argon (noble gas) core. This only applies to the first row transition metals, adjustments will be necessary when writing the electron configuration for the other rows of transition metals. The noble gas before the first row of transition metals would be the core written with brackets around the element symbol (i.e. [Ar] would be used for the first row transition metals), and the electron configuration would follow a [Ar] ns^xnd^x format. In the case of first row transition metals, the electron configuration would simply be [Ar] 4s^x3d^x. The energy level, "n", can be determined based on the periodic table, simply by looking at the row number in which the element is in. However, there is an exception for the d-block and f-block, in which the energy level, "n" for the d block is "n-1" ("n" minus 1) and for the f block is "n-2"

Example of Determining Energy Levels (n)

For example, if we want to determine the ground state electron configuration for cobalt (Co), we would first look at the row number, which is 4 according to the periodic table below; meaning n = 4 for the s-orbital. In addition, since we know that the energy level for the d orbital is "n-1", therefore n = 3 for the d-orbital in this case. Thus, the electron configuration for cobalt would simply be Co: [Ar] 4s²3d⁷. The reason why it is 3d⁷ can be explained using the periodic table. As stated, you could simply count the boxes on the periodic table, and since Cobalt is the 7th element of the first row transition metals, we get Co: [Ar] 4s²3d⁷.

Exceptions in Transition Metals

Some subtle variations are encountered across the transition metals. Filling the d orbitals is not as straightforward as the s and p orbitals. For example, in the first row of the transition metals, all but two elements have configurations [Ar]4s²3d^x. However, chromium has configuration [Ar]4s¹3d⁵, and copper has configuration [Ar]4s¹3d¹⁰. Similarly, two elements in the third row of the transition metals do not have the configuration [Xe]6s²5d^x. Platinum has configuration [Xe]6s¹5d⁹, and gold has configuration [Xe]6s¹5d¹⁰. Things are even worse in the second row of transition metals, in which half of the elements do not follow the "correct" order of filling. Niobium is [Kr]5s¹4d⁴, molybdenum is [Kr]5s¹4d⁵, ruthenium is [Kr]5s¹4d⁷, silver is [Kr]5s¹4d¹⁰, and palladium doesn't have any s electrons at all in its outer shell: it is [Kr]4d¹⁰.

What's going on? The main reason things are complicated here is that the 4s and 3d levels are quite close to each other in energy (as are 5s and 4d, and 6s and 5d). As a result, slight changes are causing the electron configuration to vary from one element to the next. Pairing energy is certainly a culprit; that's the amount of energy it costs to put two electrons in the same orbital. If the electron configuration is [Ar]4s²3d^x, then two electrons are always being forced to occupy the same space, the 4s orbital. That costs energy, because electrons repel each other. Pairing energy changes from one element to another, but by the time we reach chromium, pairing energy is evidently high enough (or the difference in energy between the 4s and 3d levels is low enough) that the energy is lower if the electrons just spread themselves out.

So, a balance has to be struck between pairing energy and orbital energy. Both are changing as we move from one element to the next. Sometimes the pairing energy of the s orbital is small compared to the energy jump to the d orbital, so two electrons go into an s orbital. Sometimes the pairing energy of the s orbital is large compared to the energy jump to the d orbital, so the electron goes in a d orbital. Of course, the pairing energy of the d orbitals also plays a role in some cases, and it also varies from one element to another.

Well, what are you supposed to do with that information? Usually, you are expected to know the most general rule (such as filling like [Ar]4s²3d^x). Twenty-one out of thirty transition metals have two s electrons and some number of d electrons. Sometimes, you are expected to know the most common exceptions; those are chromium and copper (there's a lot more chromium and copper in the world than there is niobium), and they are relatively easy to remember because one has a half-filled d shell and the other has a completely filled d shell.

• Most transition metals have two s electrons and some d electrons.
• Copper and chromium have only one s electron; the other one is "promoted" into a d orbital.
• You can keep track those two exceptions if you remember that copper has one electron in each d orbital and copper has an electron pair in each d orbital ("half-filled d" and "filled d" is a good rule to remember).

One more important thing to know is that these cases describe only the transition metals in their elemental state. They do not apply to transition metal ions or compounds, in which the transition metal is found bound with atoms of different types to form salts or other materials. The atoms in a chunk of silver metal have the electronic configuration [Kr]5s¹4d¹⁰, but the atoms in a silver ion, which have one less electron, have configuration [Kr]4d¹⁰. The missing electron is lost from the s orbital, not from the d. In general, compounds and ions of the transition metals do not have s electrons in the valence shells.

Transition Metal Ions

Why would metal ions and compounds be different from neutral atoms? The difference is easiest to see in the case of ions, in which the metal loses one or more electrons. Because it not longer has the same amount of electrons as protons, it becomes positively charged (it has more positive protons than it has negative electrons). The positive charge causes the electrons to become more attracted to the nucleus; that atom contracts or shrinks.

Figure \(\PageIndex{1}\): A 4s orbital responds to an increase in positive charge when an electron is lost to form a cation. (Chris Schaller; CC-NC-BY)

Figure \(\PageIndex{2}\): A 3d orbital responds to an increase in positive charge when an electron is lost to form a cation. (Chris Schaller; CC-NC-BY)

As a result of the charge when a transition metal atom becomes an ion, the 3d level falls below the 4s level. Remember, these two orbitals are very close in energy to begin with, so small changes can reverse their order. Similarly, the 3d level generally falls below the 4s level anytime a transition metal joins other atoms to become part of a compound.

Figure \(\PageIndex{3}\): The 3d orbital energy level drops below the 4s orbital energy level in a cation. (Chris Schaller; CC-NC-BY)

The way the 3d electrons fall in energy with increasing charge is one of the factors making the electron configurations of transition metals complicated.

• In ions and compounds, the d orbital is lower in energy than the s orbital of the next level, not the other way around.

• Examples

  It is helpful to first write down the electron configuration of an element at its ground state before attempting to determine the electron configuration of an element with an oxidation state. See examples below.

  Example \(\PageIndex{1}\)

  What is the valence electron configuration of V⁴⁺

  Solution

  Vanadium Atom:

  V: 5 valence electrons = [Ar] 4s²3d³

  Vanadium with an Oxidation State of +4:

  V⁴⁺: [Ar] 4s⁰3d¹

  Or you can also write it as V⁴⁺: [Ar] 3d¹

  Example \(\PageIndex{2}\)

  What is the valence electron configuration of Ni²⁺

  Solution

  Nickel Atom:

  Ni: 10 valence electrons = [Ar] 4s²3d⁸

  Nickel with an Oxidation State of +2:

  Ni²⁺: [Ar] 4s⁰3d⁸

  Or simply Ni²⁺: [Ar] 3d⁸

• Example \(\PageIndex{3}\)

  What is the valence electron configuration of Os²⁺ and Os³⁺

  Solution

  Osmium Atom:

  For third row transition metals the valence electrons also include 4f electrons (although these are generally inert)

  Os: 22 valence electrons = [Xe] 6s²4f¹⁴5d⁶

  Osmium with an Oxidation State of +2:

  Os: [Xe] 4f¹⁴5d⁶

  Osmium with an Oxidation State of +3:

  Os: [Xe] 4f¹⁴5d⁵

Spin Multiplicity and Degenerate Orbitals

Earlier, we saw that electrons tend to go into unfilled orbitals before pairing up in the same one, provided other orbitals are available at the same energy level.

This idea is part of Hund's rule. Hund's rule says, in part, that if you have two electrons, and there are two degenerate orbitals, then one electron will go into each orbital. It's partly about avoiding electron-electron repulsion that would occur if you put two electrons into the same orbital -- that is, into the exact same region of space.

It's partly something else though, and that is a quantum mechanical bias toward high multiplicity. Multiplicity refers to the number of unpaired electrons there are in an atom or molecule. By paired, we mean two electrons that have opposite spin. Remember, spin is a fundamental quantum mechanical property of an electron. It can only have two values, and the numerical values seem arbitrary but it's important to know that the two possible options are opposite numbers: they can be either +1/2 or -1/2. Unpaired electrons would be those that don't have a partner somewhere with an opposite spin value.

To illustrate that idea, consider the following drawing. It shows three different ways a set of five electrons might fill in a group of five orbital (maybe the 3d level; suppose these are the valence electrons on a vanadium atom). Often, electrons in orbital diagrams are indicated by arrows, with the direction of the arrow indicating the spin. An up arrow means spin = 1/2; a down arrow means spin = -1/2.

In two cases, some of the electrons are paired; they have a partner somewhere with opposite spin. The multiplicity is basically a tally of how many unpaired electrons are left over; you get the multiplicity by adding up the spin value of all the electrons. One case has all the electrons unpaired. All of the electrons have the same spin. Multiplicity is maximized in this case. Hund's rule says this case has the lowest energy.

What about if two electrons occupy the same orbital? We could still maximize multiplicity by keeping their spins "parallel"; that is, they could both have spin = 1/2 or both spin = -1/2. That doesn't happen, though. Remember the quantum rule that no two electrons on the same atom can be described by the same set of quantum numbers. In other words, each electron on the atom must have a unique identity. This rule has a name, too: the Pauli exclusion principle.

• Electrons always occupy the lowest energy orbital available.
• Multiplicity is maximized; electrons are given the same spins when possible.
• However, when they are found in the same orbital, two electrons must have opposite spin (Pauli Exclusion Principle).

Exercise \(\PageIndex{1}\)

Determine the valence electron configuration for the following metal atoms and ions.

a. Cu

b. Co⁺²

c. Fe⁺³

d. Ti⁺⁴

e. V

f. Co

g. Mn⁺²

h. Zn⁺²

i. Zn

Answer

a. [Ar]4s¹3d¹⁰

b. [Ar]3d⁷

c. [Ar]3d⁵

d. [Ar]3d⁰

e. [Ar]4s²3d³

f. [Ar]4s²3d⁷

g. [Ar]3d⁵

h. [Ar]3d¹⁰

i. [Ar]4s²3d¹⁰

References

1. Petrucci, Ralph. Harwood, William. Herring, Geoffrey. Madura, Jeffery. General Chemistry: Principles and Modern Applications, 9th Edition. Pearson Education, Inc. New Jersey, 2007.
2. Hein, et al. Introduction to General, Organic, and Biochemistry. 9th ed. Hoboken, NJ 2009.

Problems

1. What is the maximum number of electrons each orbital (s, p, d, f) can hold, respectively?
2. Write the electron configuration for Sc³⁺.
3. Write the electron configuration for Ti²⁺.
4. Which first row transition metal is the only element that forms an oxidation state of +7?
5. For additional practice, try to write the electron configuration of all the first row transition metals with their common oxidation states:

To see an example of an element from the second row or third row transition metals, see "Electron Configuration of a Second Row Transition Metal (Rhodium)" and "Electron Configuration of a Third Row Transition Metal (Osmium)".)

A) V²⁺ B) V³⁺ C) V⁵⁺ D) Cr²⁺ E) Cr³⁺ F) Cr⁶⁺ G) Mn²⁺ H) Mn³⁺

I) Mn⁴⁺ J) Mn⁶⁺ K) Mn⁷⁺ L) Fe²⁺ M) Fe³⁺ N) Co²⁺ O) Co³⁺ P) Cu²⁺ Q) Zn²⁺

Contributors

• Liza Chu (UCD)
• Modified by Catherine McCusker (ETSU)