In the previous post, we talked about electron configurations, what they show, how to read, and how to write them based on energy levels and sublevels.

As a reminder, electron configurations show the **number** and **location of electrons** in an atom or an ion. For example, the electron configuration of **fluorine **is1*s*^{2}2*s*^{2}2*p*^{5}.

The number before each letter indicates the **main** **energy level**, and this is the principal quantum number, *n*. The letters (*s*, *p*, *d*, *f*) tell us the type of the orbital, and the exponent is for the number of electrons.

The 1*s*^{2 }means there are 2 electrons in the s orbital that is in the first row. Next, we have the orbitals in the second energy level because fluorine is in the second row of the periodic table. In this level, there are 2 electrons in an s orbital, and 5 electrons in the p orbital:

This is the very least you should know in addition to the order of filling the energy levels and sublevels so please go over the detailed post on the electron configurations before getting to ions.

**Electron Configuration of Ions**

The first thing you need to remember here is that cations are formed by losing an electron(s), and anions are formed by gaining an electron(s).

The **charge of the ion is a result of an imbalance between the number of protons** and electrons. If it is a **cation**, then the positive indicates how many **more protons** it has compared to the number of electrons. For **anions**, the charge tells how many **extra electrons **there are compared to the number of protons.

Recall, this pattern for the formation of anions and cations: Metals tend to lose electron(s) and become **cations** (positively charged ions).

Nonmetals tend to gain an electron(s) and become **anions** (negatively charged ions).

Notice that the number of protons is not changed, and the **ions** are charged because, unlike atoms, their number of **protons and electrons is not equal**.

**Electron Configuration of Main Group Cations**

Now, how do we determine the electron configuration of an ion? If it is, for example, a **+1 charged** cation, that means the atom has **lost one electron**. This electron is **going to be from the outermost** **valence shell** as these are the electrons farthest away from the nuclei and thus not as strongly attracted to it.

Let’s see an **example on Na**. It is in the first group, so it **loses one electron** to become **Na ^{+}**. The electron configuration of sodium is 1

*s*

^{²}2

*s*

^{²}2

*p*

^{⁶}3

*s*

^{¹}, and the electron is removed from the energy level with the greatest n value – 3

*s*. Therefore, the electron configuration of the

**Na**ion will be

^{+}**1**:

*s*^{²}2*s*^{²}2*p*^{⁶}

Notice that the ion has a configuration with a **complete shell of p orbitals** which is characteristic of **noble gases**. In fact, this is the electron configuration of Ne, and we say that Na^{+} and Ne are **isoelectronic** (same electronic structure). The reason for this is that, remember, noble gases are very stable because of the low energy level of complete orbitals.

Na (1*s*^{2}2*s*^{2}2*p*^{6}3*s*^{1}) ⟶ e^{−} + **Na ^{+}**([He]2

*s*

^{2}2

*p*

^{6}) [

**isoelectronic with Ne**([He] 2

*s*

^{2}2

*p*

^{6})]

Let’s now determine the electron configuration of the Ca^{2+} ion. It is formed when an atom of Ca (1*s*^{2}2*s*^{2}2*p*^{6}3*s*^{2}3*p*^{6}**4 s^{2 }**or [Ar]

**4**) loses two electrons from the outer shell 4s.

*s*^{2}

Ca ([Ar]**4 s^{2}**) ⟶ 2e

^{−}+ Ca

^{2+}([Ar]) [

**isoelectronic with Ar**]

This pattern explains why the **metals in the first group become +1**, the ones in the **second group** become +2, and **Al**, for example, becomes **+3**. It takes removing one electron from a metal in the first group to obtain the electron configuration of the previous noble gas, it takes two for the group two metals, etc.

**Electron Configuration of Anions**

Anions are formed when the atom **gains as many electrons** as necessary to **attain the electron configuration of the next noble gas** in the periodic table. For example, **oxygen** is in group 6, and therefore, it will need **two electrons** to attain the electron configuration of Ne:

Notice again that the two electrons go to the outermost valence orbital. Oxygen has 6 valence electrons – those in the 2*s* and 2*p* orbitals, however, since p sublevels are higher in energy, and they are the only ones capable of accepting additional electrons, the two electrons go to the 2*p* orbitals. The electron configuration of the oxide ion (**O ^{2-}**) is therefore,

**1**.

*s*^{²}2*s*^{²}2*p*^{⁶}

O (1*s*^{2}2*s*^{2}2*p*^{4}) + 2e^{−}⟶ + O** ^{2- }**([He]2

*s*

^{2}2

*p*

^{6}) [

**isoelectronic with Ne**([He] 2

*s*

^{2}2

*p*

^{6})]

Another common type of monoatomic anions are the **halides**. Halogens are in **group 7**, and therefore, they **only need one electron** to attain the electron configuration of the noble gas following them in the periodic table. **For example**, bromine takes one electron and becomes isoelectronic to t Kr:

Br([Ar] 4*s*^{2}3*d*^{10}4*p*^{5}) + e^{− }⟶ **Br ^{−}**([Ar] 4

*s*

^{2}3

*d*

^{10}4

*p*

^{6}) [

**isoelectronic with Kr**([Ar] 4

*s*

^{2}3

*d*

^{10}4

*p*

^{6})]

**Electron Configuration of Transition Metal Cations**

In contrast to main-group ions, transition metal ions **do not usually attain a noble gas configuration**. This is because the ns level is the outermost level, and the (n-1)d is considered an inner level therefore, it will take too much energy to remove those electrons and achieve a noble gas configuration. Therefore, the cation of a transition metal is formed by removing **first the electrons from the n s** (highest principal quantum number) orbital and

**then from the (n -1)**orbitals.

*d***For example**, the **electron configuration of Zn** is 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}**4 s^{2}3d^{10}** or [Ar]

**4**, and it loses the two electrons from the 4s orbital to become Zn

*s*^{2}3*d*^{10}^{2+}[Ar]

**3**. This is “not a bad” electron configuration considering the filled d orbitals.

*d*^{10}** **

**Check Also**

- Atomic Orbitals
- Electron Configurations
- Orbital Diagrams
- Aufbau’s Principle, Hund’s Rule, and Pauli’s Exclusion Principle
- Hund’s Rule
- Pauli Exclusion Principle
- Quantum Numbers (
*n*,*l*,*m*_{l},*m*_{s}) - Bohr Model of the Hydrogen Atom
- Rydberg Formula
- The Photoelectric Effect
- Calculating The Energy of a Photon
- Ionization Energy
- Electron Affinity
**Energy, Wavelength, and Frequency Practice Problems**

#### Practice

What is the maximum number of electrons that can occupy one p orbital?

a.14

b. 2

c. 10

d.1

e. 6

The n = 2 shell can accommodate a maximum of ____ electrons.

A. 10

B. 8

C. 4

D. 6

E. 2

Which of the following is an incorrect orbital occupancy representation?

- 4d
^{3} - 3d
^{5} - 2s
^{3} - 1s
^{1} - 2p
^{2}

The maximum number of electrons that can be accommodated in 3s subshell is

- 10
- 2
- 6
- 1
- 8

The following electron configuration is incorrect:

- 1s
^{2}2s^{2} - 1s
^{2}2s^{3}2p^{3} - 1s
^{2}2s^{2}2p^{5} - 1s
^{2}2s^{2}2p^{4} - 1s
^{2}2s^{2}2p^{6}3s^{2}

Which electron configuration represents an excited state of the indicated atom?

**He**: 1s^{2}**Na**: 1s^{2}2s^{2 }2p^{6}3s^{2}3p^{2}3s^{1}**P**: 1s^{2}2s^{2}2p^{6}3s^{2 }3p^{3}**N**: 1s^{2}2s^{2 }2p^{2}3s^{1}**Ne**: 1s^{2}2s^{2}2p^{6}

Which of the following represents the ground state electron configuration of a transition element?

- 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{5} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{1} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{4} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2} - 1s
^{2}2s^{2}2p^{3}

Which of the following is the correct ground-state electron configuration of V is?

- 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{3} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5} - 1s
^{2}2s^{2}2p^{6}3s^{2}4s^{2}3d^{3} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{5} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{3}

Which of the following represents a possible excited-state electron configuration for an iron atom?

- [Ar]3d
^{7}4s^{2} - [Ar]3d
^{6}4s^{2} - [Ar]3d
^{7}4s^{1} - [Ar]3d
^{6}4s^{1} - [Kr]3d
^{7}4s^{1}

Each of the following is an accurate representation of ground-state electron configuration except:

**Fe**: [Ar]4s^{2}3d^{6}.**Ca**: [Ar]4s^{2}.**Se**: [Ar] 4s^{2}3d^{10}4p^{4}.**Ag**: [Kr] 5s^{2}4d^{9}.**Ti**: [Ar] 4s^{2}3d^{2}

How many electrons, in total, are present in p orbitals in a ground-state nickel atom?

- 8
- 12
- 6
- 24
- 3

Which one of the following represents electron configuration of an excited carbon atom?

- 1s
^{2}2s^{2}2p^{3} - 1s
^{2}2s^{2}2p^{1} - 1s
^{2}2s^{2}2p^{1}3s^{1} - 1s
^{2}2s^{2}3s^{1} - 1s
^{2}2s^{2}2p^{2}

Which of the following ground-state electron configurations is incorrect?

**Cl**: [Ne]3s^{2}3p^{5}**Ge**: [Ar] 4s^{2 }3d^{10}3p^{2}**Co**: [Ar]4s^{2}3d^{7}**Rb**: [Kr]5s^{1}**Mn**: [Ar] 4s^{2}3d^{5}

What is the ground-state electron configuration of chromium (Cr)?

- 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^{4} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{5} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10} - 1s
^{2}2s^{2 }2p^{6}3s^{2}3p^{6}4s^{2}3d^{4} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}

What is the ground-state electron configuration of bromine (Br)?

- 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{10 }4p^{5} - [Kr] 4s
^{2}3s^{10}4p^{5} - [Ne] 3s
^{2}3p^{5} - [Ar] 4s
^{2}3d^{10}4p^{5} - 1s
^{2}2s^{2}2p^{5}3s^{2}3p^{6}4s^{2}3d^{10}4p^{5}

The ground state electron configuration of As is ________.

- 1s
^{2}2s^{2}3s^{2}3p^{6}4s^{2}3d^{10}4p^{2} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{1} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{3} - 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4d^{3} - [Kr]4s
^{2}3d^{10}4d^{3}

Which of the following elements has the same valence-shell electron configuration as lithium?

- calcium.
- sulfur.
- potassium.
- magnesium.
- argon.

How many unpaired electrons are found in the d orbital of nickel.

- 4 electrons
- 3 electrons
- 2 electrons
- 8 electrons
- none of these

The ground-state electron configuration of the element ________ is [Kr]5s^{2}4d^{10}

- Fe
- Cd
- Cr
- Mn
- Zn

The ground-state electron configuration of ________ is [Ar]4s^{2}3d^{7}.

- Ti
- Co
- Fe
- Cr
- Cs

Chlorine has the following ground-state configuration:

- [He]3s
^{2}3p^{2} - [Ne]3s
^{2}3p^{5} - [He]2s
^{2}2p^{5} - [Ne]2s
^{2}2p^{3} - [He]3s
^{2}3p^{5}

What is the maximum number of electrons that can be accommodated in each d-subshell?

- 10
- 5
- 6
- 3
- 2

What noble gas core should be used when writing the condensed electron configuration of Strontium (Sr)?

- [Xe]
- [Kr]
- [Ne]
- [Ar]
- [He]

# Electron Configuration of Ions

Which of the following is the ground state electron configuration of S^{2-}?

A) [Ne]

B) [Ne]3s^{2}3p^{6 }C) [Ne]3s^{2}3p^{4 }D) [Ne]3s^{2}3p^{2 }E) [Ne]4s^{2}

A cation of +3 indicates that an element has

A) lost three neutrons.

B) lost three protons.

C) lost three electrons.

D) gained three protons.

E) gained three electrons.

Indicate the ground state electron configuration for Cl⁻.

A) 1s^{2}2s^{2}2p^{6}3s^{2}3p^{4 }B) 1s^{2}2s^{2}2p^{6}3s^{2}3p^{5 }C) 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6 }D) 1s^{2}2s^{2}2p^{6}3s^{1}3p^{6 }E) 1s^{2}2s^{2}2p^{6}3s^{3}3p^{5}

Which of the following is the ground state electron configuration of Ba^{2+ }?

A) [Kr]5s^{2}4d^{10}5p^{6}6s^{2 }B) [Kr]5s^{2}4d^{10}5p^{6 }C) [Kr]5s^{2}4d^{10}5p^{6}5d^{2 }D) [Kr]5s^{2}5p^{6 }E) [Kr]5s^{2}4d^{10}5p^{6}6s^{2}

Which of the following is the ground state electron configuration of Ti^{2⁺}.

A) [Ar]3d^{4 }B) [Ar]3d^{2 }C) [Ar]4s^{2 }D) [Ar]4s^{2}3d^{4 }E) [Ar]4s^{2}3d^{2}

Which of the following is the ground state electron configuration of Zn^{2⁺}?

A) [Ar]4s^{2}3d^{6 }B) [Ar]3d^{10 }C) [Ar]4s^{2}3d^{8 }D) [Ar]3d^{8 }E) [Ar]

Which of the following is the ground state electron configuration of Cr^{3⁺}?

A) [Ar]

B) [Ar]4s^{1}3d^{2 }C) [Ar]3d^{3 }D) [Ar]4s^{2}3d^{1 }E) [Ar]4s^{2}3d^{7}

How many valence electrons are present in the azide (N^{3-}) ion?

A) 3

B) 5

C) 8

D) 7

E) 2

Ca^{2+} has the following ground state electron configuration:

A) 1s^{2}2s^{2}2p^{6 }B) 1s^{2}2s^{2}2p^{6}3s^{2 }C) 1s^{2}2s^{2}2p^{6}3s^{2}3p^{2 }D) 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{2 }E) 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}

Rb+ has the following ground state electron configuration:

A) [Kr]5s^{2}4d^{6 }B) [Kr]4s^{2 }C) [Ar]4s^{2}4p^{4 }D) [Kr]5s^{1 }E) [Ar]4s^{2}3d^{10}4p^{6}

Which of the following is the ground state electron configuration of Fe^{3⁺}?

A) [Ar]3d^{5}** ^{ }**B) [Ar]4s

^{1}3d

^{3 }C) [Ar]

D) [Ar]4s

^{2}3d

^{9 }E) [Ar]4s

^{2}3d

^{1}

The correct ground state electron configuration for Br^{–} is:

A) [Ar]4s^{2}3d^{10}4p^{6 }B) [Ar]4s^{2}3d^{10}4p^{3 }C) [Ar]4s^{2}3d^{8}4p^{6 }D) [Ar]4s^{2}3d^{10}4p^{5 }E) [Ar]4s^{2}4p^{6}

How many electrons are present in the ground state of Mg^{2+} ion?

A) 2

B) 6

C) 10

D) 8

E) 12