0.1.2.2: Predicting configurations of transition metal atoms and ions

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A note from Dr. Haas: There's a problem with the way that electron configurations are taught in Introductory Chemistry; the main issue is that you weren't told the whole story. Details were left out on purpose for introductory courses to make it simple for you to arrive at the correct electron configuration most of the time. But, in advanced courses like Bioinorganic Chemistry, we're all about the details and the exceptions to the general rules! Read on to learn what they didn't tell you in Intro Chemistry.

First, I want to point out an inconsistency in what you learned before. There are two things that you learned that don't quite jive with one another:

You learned that energy of orbitals increases with the shell number (\(n\)).
You learned that the filling of orbitals always follows a pattern where 4s fills before 3d and 5s fills before 4d, etc. (See the table on the previous page.)

If you apply the Aufbau principle to the two "facts" listed above, you would arrive at opposite conclusions about the relative energies of some orbitals. Take the 3d and 4s orbitals as an example. #1 would lead you to believe that 3d is lower in energy than 4s. While #2 would leave you expecting the opposite; that 4s is lower in energy than 3d. In the description below, we will get to the bottom of this conundrum, and we will focus on the 3d and 4s orbitals, since that's where we will spend much of our time this semester.

The true story on relative energies of 3d and 4s orbitals

Hydrogen: Hydrogen is an outlier in this discussion; for reasons we will discuss later, all subshells within a shell are degenerate in a hydrogen atom. For example, all of the orbitals within shell 2, including 2s and 2p orbitals, are degenerate in a hydrogen atom.

The first eighteen multi-electron atoms, from He to Ar: The general "rules" about relative energy levels of orbitals in multi-electron atoms are that energy level increases with shell number (1<2<3<4...), and within a shell the relative energy levels of subshells are s<p<d<f. These "rules" are perfectly true for the light elements in the first three periods of the periodic table, from atomic number 2 (helium, He) to 18 (argon, Ar).

Everything is straightforward up to element 18, argon; but the \(n=3\) shell is not full because the 3d subshell is still unoccupied. As we go from element 18 to 19, where will the 19th electron go?

Elements K and Ca: At this point, we have arrived at the fourth period where the relative energy levels of 3d and 4s "swap". The 3d and 4s orbitals were getting closer and closer to one another up to this point, and now they have finally crossed. K and Ca and exceptions to rule #1 stated above, and 4s is lowered in energy than 3d even though 4s has a higher shell number. Coincidentally, this change in relative energies of 4s and 3d is exactly at the point where these subshells become relevant for the ground states of K and Ca. However, K and Ca are the only two elements that have 4s lower in energy than 3d.

**Table \(\PageIndex{1}\).** Electron configurations of K and Ca. Note that the orbitals are listed in the order of increasing energy! 4s is LOWER in energy than 3d in these two elements.
Element	Electron configuration
K	1s²2s²2p⁶3s²3p⁶4s¹
Ca	1s²2s²2p⁶3s²3p⁶4s²

Elements Sc to Zn (the d-block elements) and beyond: Now we're back to the expected trend in relative energy levels. The 3d is lower in energy than 4s. However, across this row of transition metals, the difference between 3d and 4s is small; from Sc to Zn, the 3d and 4s are so close in energy that they are almost degenerate in some cases. When the orbitals are close enough in energy that they are nearly degenerate, other minor energetic factors like pairing energy and electron-electron repulsion within a subshell determine whether electrons will fill the slightly lower-energy 3d orbital or the slightly higher energy 4s orbital.

In the case of Sc, Ti, V, Mn, Fe, Co, and Ni, it may seem counterintuitive for the 4s orbital to be full while the lower-energy 3d orbital is only partially filled. This is explained by electron-electron repulsion in the smaller-sized 3d orbital making it more favorable for electrons to occupy, and even pair in, the larger-sized 4s orbital. In the case of Cr, the difference in energy between 3d and 4s is significant enough that it is more favorable for electrons to occupy 3d before pairing in 4s. In Cu, 3d has dropped quite a bit below the energy of 4s, and so even pairing in 3d is more favorable than occupation of 4s.

Figure 2: Periodic table of periods 2-4.

The electronic structures of the d-block elements from Sc to Zn are shown in the table below. Each additional electron usually goes into a 3d orbital. For convenience, [Ar] is used to represent 1s²2s²2p⁶3s²3p⁶.

**Table \(\PageIndex{1}\).** Electron configurations of d-block atoms across the first-row of the transition metals (or fourth period) from Sc to Zn. Orbitals are listed in order of increasing energy.
Element	Electron Configuration
Sc	[Ar] 3d¹4s²
Ti	[Ar] 3d²4s²
V	[Ar] 3d³4s²
Cr	[Ar] 3d⁵4s¹
Mn	[Ar] 3d⁵4s²
Fe	[Ar] 3d⁶ 4s²
Co	[Ar] 3d⁷4s²
Ni	[Ar] 3d⁸4s²
Cu	[Ar] 3d¹⁰4s¹
Zn	[Ar] 3d¹⁰4s²

Remember that, in reality, for Sc through to Zn the 3d orbitals have the lower energy - not the 4s.

d-block ions

In the case of Sc to Zn, a 4s electron is lost first in the formation of an ion because 4s is the higher-energy orbital. The removal of any one electron lowers the energy of all the orbitals, but the \(n=3\) orbitals are affected more than the \(n=4\) orbitals. The result is that 3d drops much lower relative to 4s. Thus, all valence electrons of d-block ions from Sc to Zn are in the 3d shell, and none are in 4s. This trend holds true for all elements in the d-block; ions contain all valence electrons in the appropriate d shell, and never the higher-energy s subshell.

When d-block (first row) elements form ions, the 4s electrons are lost first.

Exercises \(\PageIndex{1}\): Iron

Write out the electron configuration for atomic Fe and Fe³⁺. Make sure you write orbitals in order of increasing energy.

Answer

Fe: [Ar] 1s²2s²2p⁶3s²3p⁶3d⁶4s²

When we are careful to write the orbitals in order of increasing energy, it's clear that the electrons would be removed first from the higher-energy 4s and then from the 3d.

Fe²⁺: [Ar] 1s²2s²2p⁶3s²3p⁶3d⁵

Exercise \(\PageIndex{2}\): Scandium

A) Write out the electron configurations for Sc, Sc⁺, and Sc⁺².

B) Explain why the electron configuration of Sc is not [Ar] 3d³.

C) Sc and Ti⁺ are isoelectronic. Explain why their electron configurations are different even though they have the same number of valence electrons.

Answer A

Sc: [Ar] 3d¹4s²

Sc⁺: [Ar] 3d² *when a d-block ion forms, it pulls the 3d orbital much lower than the 4s so that all electrons that were in 4s would move to 3d.

Sc⁺²: [Ar] 3d¹ *when a d-block ion forms, it pulls the 3d orbital much lower than the 4s so that all electrons that were in 4s would move to 3d.

Answer B

In Sc, the 3d orbitals are lower than the 4s orbital, but they are still very close in energy. If we imagine a situation where we are "adding the last three electrons to Sc one at a time", we can describe it as follows: The first of the three valence electrons would go into a 3d orbital because it is lower-energy than 4s. The energy cost of electron-electron repulsions of two electrons in the small 3d subshell is greater than the difference in the 3d and 4s energy levels, so it is more energetically favorable to put the second valence electron in 4s. The third electron goes into 4s because the pairing energy within the large 4s orbital is still less of an energetic cost than the electron-electron repulsions in the 3d subshell.

Answer C

Even though Sc and Ti are isoelectronic, formation of the Ti⁺ ion lowers the 3d energy level much more than the 4s.

Sc: [Ar] 3d¹4s²--> The 4s is full in Sc because the 3d and 4s are very close in energy, even though 3d is slightly lower (see answer to B).
Ti⁺: [Ar] 3d³ --> Formation of the ion lowers the 3d orbital more than the 4s, and so all of the valence electrons are in the lower-energy 3d subshells.

Exercise \(\PageIndex{3}\): Chromium

Why is the electronic structure of chromium (Cr) [Ar]3d⁵4s¹ instead of [Ar]3d⁴4s²?

Answer

Because that is the structure in which the balance of repulsions and the size of the energy gap between the 3d and 4s orbitals happens to produce the lowest energy for the system.

WARNING: Many chemistry textbooks, websites, and teachers try to explain this by saying that the half-filled orbitals minimize repulsions, but that is a flawed, incomplete argument. You are not taking into account the size of the energy gap between the lower energy 3d orbitals and the higher energy 4s orbital.

Two rows directly underneath chromium in the Periodic Table is tungsten. Tungsten has exactly the same number of outer electrons as chromium, but its outer structure is 5d⁴6s², NOT 5d⁵6s¹.

In this case, the most energetically stable structure is not the one where the orbitals are half-full. You cannot make generalizations like this!

Jim Clark (Chemguide.co.uk)

Contributions

Jim Clark (Chemguide.co.uk)
Curated or created by Kathryn Haas