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Electron Configurations

>> Parts of this equation/concept include:

A -	Atoms
B -	Ions
C -	Paired and Unpaired Electrons

Electron configurations are a simple method of representing the arrangement of electrons within an atom. If the arrangement is such that each electron occupies the lowest energy orbital available to it, the electron configuration represents the ground state. If electrons occupy higher energy orbitals, the electron configuration represents an excited state. Excited states do not produce "impossible" energy levels, like 1p, but instead move electrons to an empty, higher-energy orbital.

Electron configurations provide a list of all the electrons in an atom, with the principal quantum number, then the angular momentum quantum number (as a letter), then the number of electrons with those quantum numbers as a superscript.

The periodic table is very useful for determining electron configurations. The position of the element in the periodic table reflects the energy of the highest-energy electron. The position of each element previous to that one reflects another electron.

The period is used to determine the value of n. The periodic table can be divided into sections that represent values of l. The first (leftmost) two columns represent s electrons. The six columns on the right side of the periodic table represent the p electrons. The 10 elements in the middle are d electrons. The two rows that are offset at the bottom (lanthanides and actinides) represent f electrons.

The tricky part is with the d and f electrons. The d electrons have a principal quantum number one less than the period. The f electrons are n = period – 2. (It might be easier just to remember the first row, lanthanides, as 4f and the second, actinides, as 5f.) The first two electrons are always 1s, regardless of their position.

Some people write the electron configurations in slightly different order (e.g., the 3d before the 4s). The order is not important, provided the electrons are associated with the correct orbital.

>> Example 1

What is the electron configuration of Ru?

Solution:

1s² (The first two electrons in the positions of H and He.)

1s²2s² (Li and Be positions.)

1s²2s²2p⁶ (B through Ne.)

1s²2s²2p⁶3s² (Moved to the next period, positions of Na and Mg.)

1s²2s²2p⁶3s²3p⁶ (Al through Ar positions.)

1s²2s²2p⁶3s²3p⁶4s² (K and Ca are in the fourth period.)

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰ (For the first time the d group is included. Remember the n = period – 1.)

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶ (The p group has n = period.)

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s² (Next period, s group.)

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d⁶ (The d group has the lower n, and the last electron of ruthenium is in the d group; there are six in that group to get to Ru.)

Because the electron configuration of each element is the same until the last electron (i.e., it just ends at different places), an abbreviated version of the electron configuration is often used. With this method the electron configuration starts with chemical symbol, in square brackets, of the noble gas (last column) of the row previous to the element.

>> Example 2

What is the abbreviated electron configuration of Sn?

Solution:

Sn is in period 5; the noble gas in period 4 is Kr. The rest of the electron configuration starts with the s section of period 5, then the d section (where n = period – 1 = 4), then two electrons in the p section, so

[Kr]5s²4d¹⁰5p²

>> Example 3

What is the abbreviated electron configuration of platinum (Pt)?

Solution:

[Xe]6s²4f¹⁴5d⁸

The lanthanides, elements 58-71, are part of the sixth period, even though it is traditional (as a space-saving device) to put them at the bottom of the table. You need to follow the numbers. Also remember that the f electrons have principal quantum numbers that are two lower than the period.

Because the ns and (n – 1)d energy levels are very close, an exception can happen; occasionally an electron will move from the s to the d orbital. This occurs when it will create a full or half-full d orbital (d⁵ and d¹⁰). The other exceptions are not predictable enough to worry about.

>> Example 4

What are the electron configurations of Mo and Au?

Solution:

Mo = [Kr]5s²4d⁴, but the s electron can move to make a half-full orbital, so the actual configuration would be

[Kr]5s¹4d⁵

Au = [Xe]6s²4f¹⁴5d⁹, but one more electron would make the d orbital full, so

[Xe]6s¹4f¹⁴5d¹⁰

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Cations have a positive charge because the element has lost one or more electrons. Metals form cations. When electrons are lost, they are lost from the orbital with the highest n value. If there are two orbitals with the same n value, then electrons will be lost from the highest l value.

Elements in the s group will lose enough electrons to have a noble gas configuration

Elements in the d group often have a +2 charge as they lose the s electrons. However, they often have other charges that are less predictable.

Metals in the p group lose their p electrons first, then the s electrons. These metals normally have two charges, one to reflect the loss of p electrons, the other reflect the loss of both the s and p electrons.

>> Example 5

What is the electron configuration of Na⁺?

Solution:

The atom Na has the electron configuration of [Ne]3s¹.

When it loses one electron, the electron configuration is [Ne].

>> Example 6

What are the electron configurations of Fe²⁺ and Fe³⁺?

Solution:

The electron configuration of the atom Fe is [Ar]4s²3d⁶.

The ion Fe²⁺ loses two electrons from the highest n value of 4, so

Fe²⁺ = [Ar]3d⁶

The ion Fe³⁺ loses three electrons, the two from the 4s orbital and one more, so

Fe³⁺ = [Ar]3d⁵

>> Example 7

What are the electron configurations of Pb²⁺ and Pb⁴⁺?

Solution:

The electron configuration of the atom Pb is [Xe]6s²4f¹⁴5d¹⁰6p²

For Pb²⁺ the first two electrons lost are from the 6p orbital.

Pb²⁺ = [Xe]6s²4f¹⁴5d¹⁰

For Pb⁴⁺ the s electrons are also lost.

Pb⁴⁺ = [Xe]4f¹⁴5d¹⁰

Anions are formed by the addition of electrons. Nonmetals normally form anions. Enough electrons will be added to reach the noble gas configuration.

>> Example 8

What are the electron configurations of I^– and N^3–?

Solution:

The electron configuration of the atom I is [Xe]6s²4f¹⁴5d¹⁰6p⁵.

To make I^–, one electron is added.

I^– = [Xe] 6s²4f¹⁴5d¹⁰6p⁶

The electron configuration of the atom N is [He]2s²2p³.

N^3– means that three electrons have been added to the atom.

N^3– = [He]2s²2p⁶

If you consider that metals form cations and nonmetals make anions, and both gain or lose electrons so that they have the electron configuration of a noble gas, the charge an atom is likely to acquire can be predicted for atoms in the s and p groups.

>> Example 9

What ions will be produced from the following atoms?

1. Ca
2. Li
3. O
4. Bi
5. Al

Solution:

1. Ca is a metal. Losing two electrons will give it the same number of electrons as argon. Ca²⁺.
2. Li is a metal. Losing one electron will give it the same number of electrons as He. Li⁺. (When writing charges of ions do not include a 1 in the formula!)
3. O is a nonmetal, so it will gain electrons. Two will give it the same number of electrons as Ne. O^2–.
4. Bi is a metal, so it will lose electrons. Its electron configuration is [Xe]6s²4f¹⁴5d¹⁰6p³. Recall that these atoms lose their p, then their s electrons. Bi might have a charge of Bi³⁺ or Bi⁵⁺.
5. Al is a metal, so it will lose electrons. Losing three will give it the same number of electrons as Ne. Al³⁺.

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To determine whether or not an atom has unpaired electrons, only the electrons in the highest-energy orbital need to be considered (unless it is one of the exceptions, then the s and d electrons should be considered).

Hund's rule says that each orbital (orientation) will have one electron before any pair-up. It is traditional to use a line or a box to represent each orientation and arrows to represent electrons. So distribute the electrons available (see electron configuration), one in each orientation (with arrows in the same direction) before doubling up.

>> Example 10

How many unpaired electrons are in the following atoms or ions?

1. Co
2. Cl
3. Sr
4. Mn²⁺
5. Si

Solution:

1. The electron configuration of Co is [Ar]4s²3d⁷. 3d is the highest-energy orbital. There are five orientations of d orbitals and seven electrons to distribute in these orbitals.

       There are three unpaired electrons.

2. The electron configuration of Cl is [Ne]3s²3p⁵. 3p is the highest-energy orbital. There are three orientations of p orbitals and five electrons to distribute.

       There is one unpaired electron.

3. Sr has the electron configuration of [Kr]5s². 5s is the highest-energy orbital. It has one orientation and two electrons , zero unpaired electrons.
4. Mn²⁺ has the electron configuration of [Ar]3d⁵. (The s electrons are lost in forming the ion.) There are five orientations of the d orbital, with one electron in each; there are five unpaired electrons.
5. Si has the electron configuration of [Ne]3s²3p². There are three orientations / p orbitals. Therefore the two p electrons can each have their own orbital, and there are two unpaired electrons.

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