# Complete Electron Configuration for Iodine (I, I– ion)

Iodine is the 53rd element in the periodic table and its symbol is ‘I’. Iodide is a classified halogen element. In this article, I have discussed in detail how to easily write the complete electron configuration of iodine.

## What is the electron configuration of iodine?

The total number of electrons in iodine is fifty-three. These electrons are arranged according to specific rules in different orbitals.

The arrangement of electrons in iodine in specific rules in different orbits and orbitals is called the electron configuration of iodine.

The electron configuration of iodine is [Kr] 4d^{10} 5s^{2} 5p^{5}, if the electron arrangement is through orbitals. Electron configuration can be done in two ways.

- Electron configuration through orbit (Bohr principle)
- Electron configuration through orbital (Aufbau principle)

Electron configuration through orbitals follows different principles. For example Aufbau principle, Hund’s principle, and Pauli’s exclusion principle.

## Iodine atom electron configuration through orbit

Scientist Niels Bohr was the first to give an idea of the atom’s orbit. He provided a model of the atom in 1913. The complete idea of the orbit is given there.

The electrons of the atom revolve around the nucleus in a certain circular path. These circular paths are called orbit(shell). These orbits are expressed by n. [n = 1,2,3,4 . . . The serial number of the orbit]

K is the name of the first orbit, L is the second, M is the third, and N is the name of the fourth orbit. The electron holding capacity of each orbit is 2n^{2}.

Shell Number (n) | Shell Name | Electrons Holding Capacity (2n^{2}) |

1 | K | 2 |

2 | L | 8 |

3 | M | 18 |

4 | N | 32 |

For example,

- n = 1 for K orbit.

The maximum electron holding capacity in K orbit is 2n^{2}= 2 × 1^{2}= 2. - For L orbit, n = 2.

The maximum electron holding capacity in L orbit is 2n^{2}= 2 × 2^{2}= 8. - n=3 for M orbit.

The maximum electron holding capacity in M orbit is 2n^{2}= 2 × 3^{2 }= 18. - n=4 for N orbit.

The maximum electron holding capacity in N orbit is 2n^{2}= 2 × 4^{2}= 32.

Therefore, the maximum electron holding capacity in the first shell is two, the second shell is eight and the 3rd shell can have a maximum of eighteen electrons. The atomic number is the number of electrons in that element.

The atomic number of iodine is 53. That is, the number of electrons in iodine is fifty-three. Therefore, an iodine atom will have two electrons in the first shell, eight in the 2nd orbit, and eighteen electrons in the 3rd shell.

According to Bohr’s formula, the fourth shell will have twenty-five electrons but the fourth shell of iodine will have eighteen electrons and the remaining seven electrons will be in the fifth shell.

Therefore, the order of the number of electrons in each shell of the iodine(I) atom is 2, 8, 18, 18, 7. Electrons can be arranged correctly through orbits from elements 1 to 18.

The electron configuration of an element with an atomic number greater than 18 cannot be properly determined according to the Bohr atomic model. The electron configuration of all the elements can be done through the orbital diagram.

## Electron configuration of iodine through orbital

Atomic energy shells are subdivided into sub-energy levels. These sub-energy levels are also called orbital. The most probable region of electron rotation around the nucleus is called the orbital.

The sub-energy levels depend on the azimuthal quantum number. It is expressed by ‘l’. The value of ‘l’ is from 0 to (n – 1). The sub-energy levels are known as s, p, d, and f.

Orbit Number | Value of ‘l’ | Number of subshells | Number of orbital | Subshell name | Electrons holding capacity | Electron configuration |

1 | 0 | 1 | 1 | 1s | 2 | 1s^{2} |

2 | 0 1 | 2 | 1 3 | 2s 2p | 2 6 | 2s^{2} 2p^{6} |

3 | 0 1 2 | 3 | 1 3 5 | 3s 3p 3d | 2 6 10 | 3s^{2} 3p^{6} 3d^{10} |

4 | 0 1 2 3 | 4 | 1 3 5 7 | 4s 4p 4d 4f | 2 6 10 14 | 4s^{2} 4p^{6} 4d^{10} 4f^{14} |

For example,

- If n = 1,

(n – 1) = (1–1) = 0

Therefore, the value of ‘l’ is 0. So, the sub-energy level is 1s. - If n = 2,

(n – 1) = (2–1) = 1.

Therefore, the value of ‘l’ is 0, 1. So, the sub-energy levels are 2s, and 2p. - If n = 3,

(n – 1) = (3–1) = 2.

Therefore, the value of ‘l’ is 0, 1, 2. So, the sub-energy levels are 3s, 3p, and 3d. - If n = 4,

(n – 1) = (4–1) = 3

Therefore, the value of ‘l’ is 0, 1, 2, 3. So, the sub-energy levels are 4s, 4p, 4d, and 4f. - If n = 5,

(n – 1) = (n – 5) = 4.

Therefore, l = 0,1,2,3,4. The number of sub-shells will be 5 but 4s, 4p, 4d, and 4f in these four subshells it is possible to arrange the electrons of all the elements of the periodic table.

Subshell name | Name source | Value of ‘l’ | Value of ‘m’(0 to ± l) | Number of orbital (2l+1) | Electrons holding capacity2(2l+1) |

s | Sharp | 0 | 0 | 1 | 2 |

p | Principal | 1 | −1, 0, +1 | 3 | 6 |

d | Diffuse | 2 | −2, −1, 0, +1, +2 | 5 | 10 |

f | Fundamental | 3 | −3, −2, −1, 0, +1, +2, +3 | 7 | 14 |

The orbital number of the s-subshell is one, three in the p-subshell, five in the d-subshell and seven in the f-subshell. Each orbital can have a maximum of two electrons.

The sub-energy level ‘s’ can hold a maximum of two electrons, ‘p’ can hold a maximum of six electrons, ‘d’ can hold a maximum of ten electrons, and ‘f’ can hold a maximum of fourteen electrons.

Aufbau is a German word, which means building up. The main proponents of this principle are scientists Niels Bohr and Pauli. The Aufbau method is to do electron configuration through the sub-energy level.

The Aufbau principle is that the electrons present in the atom will first complete the lowest energy orbital and then gradually continue to complete the higher energy orbital.

The energy of an orbital is calculated from the value of the principal quantum number ‘n’ and the azimuthal quantum number ‘l’. The orbital for which the value of (n + l) is lower is the low energy orbital and the electron will enter that orbital first.

Orbital | Orbit (n) | Azimuthal quantum number (l) | Orbital energy (n + l) |

3d | 3 | 2 | 5 |

4s | 4 | 0 | 4 |

Here, the energy of 4s orbital is less than that of 3d. So, the electron will enter the 4s orbital first and enter the 3d orbital when the 4s orbital is full. The method of entering electrons into orbitals through the Aufbau principle is 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d.

The first two electrons of iodine enter the 1s orbital. The s-orbital can have a maximum of two electrons. Therefore, the next two electrons enter the 2s orbital.

The p-orbital can have a maximum of six electrons. So, the next six electrons enter the 2p orbital. The second orbit is now full. So, the remaining electrons will enter the third orbit.

Then two electrons will enter the 3s orbital and the next six electrons will be in the 3p orbital of the third orbit. The 3p orbital is now full of electrons. So, the next two electrons will enter the 4s orbital and ten electrons will enter the 3d orbital.

Then the next six electrons enter the 4p orbital. The 4p orbital is now full. So, the next two electrons will enter the 5s orbital and the next ten electrons will enter the 4d orbital.

The 4d orbital is now full. So, the remaining five electrons enter the 5p orbital. Therefore, the iodine complete electron configuration will be 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p^{5}.

Note:The unabbreviated electron configuration of iodine is [Kr] 4d^{10}5s^{2}5p^{5}. When writing an electron configuration, you have to write serially.

## Video for how to write the electron configuration for Iodine (I)

## Iodine excited state electron configuration

Atoms can jump from one orbital to another orbital in an excited state. This is called quantum jump. The ground state electron configuration of iodine is 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p^{5}.

In the iodine ground-state electron configuration, the last electrons of the 5p orbital are located in the 5p_{x}(2), 5p_{y}(2) and 5p_{z} orbitals. We already know that the p-subshell has three orbitals. The orbitals are p_{x}, p_{y}, and p_{z} and each orbital can have a maximum of two electrons.

Then the correct electron configuration of iodine in ground state will be 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p_{x}^{2} 5p_{y}^{2} 5p_{z}^{1}. This electron configuration shows that the last shell of the iodine atom has an unpaired electron. So the valency of iodine is 1.

When iodine atoms are excited, then iodine atoms absorb energy. As a result, an electron in the 5p_{y} orbital jumps to the 5d_{xy} orbital. We already know that the d-subshell has five orbitals. The orbitals are d_{xy}, d_{yz}, d_{zx}, d_{x2-y2} and d_{z2} and each orbital can have a maximum of two electrons.

Therefore, the electron configuration of iodine(I*) in an excited state will be 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p_{x}^{2} 5p_{y}^{1} 5p_{z}^{1} 5d_{xy}^{1}. Here, iodine has three unpaired electrons. In this case, the valency of iodine is 3.

When iodine is further excited, then an electron in the 5p_{x} orbital jumps to the 5d_{yz} orbital. Therefore, the electron configuration of iodine(I**) in an excited state will be 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p_{x}^{1} 5p_{y}^{1} 5p_{z}^{1} 5d_{xy}^{1} 5d_{yz}^{1}.

Here, iodine has five unpaired electrons. So in this case, the valency of iodine is 5. When iodine is further excited, then an electron in the 5s orbital jumps to the 5d_{zx} orbital.

Therefore, the electron configuration of iodine(I***) in an excited state will be 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{1} 5p_{x}^{1} 5p_{y}^{1} 5p_{z}^{1} 5d_{xy}^{1} 5d_{yz}^{1} 5d_{zx}^{1}. Here, iodine has seven unpaired electrons. In this case, the valency of iodine is 7.

From the above information, we can say that iodine exhibits variable valency. Due to this, the oxidation states of iodine are 1, 3, 5, 7. Iodine atom exhibits -1, +1, +3, +5, +7 oxidation states. The oxidation state of the element changes depending on the bond formation.

## Iodide ion(I^{–}) electron configuration

After arranging the electrons, it is seen that the last shell of the iodine atom has seven electrons. Therefore, the valence electrons of iodine are seven. The elements that have 5, 6, or 7 electrons in the last shell receive the electrons in the last shell during bond formation.

The elements that receive electrons and form bonds are called anion. During the formation of a bond, the last shell of iodine receives an electron and turns into an iodide ion(I^{–}). Therefore, iodide is an anion element.

I + e^{–} → I^{–}

Here, the electron configuration of iodide ion(I^{–}) is 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p^{6}.

This electron configuration shows that iodide ion(I^{–}) has five shells and the last shell has eight electrons. The electron configuration shows that the iodide ion(I^{–}) has acquired the electron configuration of xenon and it achieves a stable electron configuration.

## FAQs

### What is the symbol for iodine?

**Ans:**The symbol for iodine is ‘I’.### How many electrons does iodine have?

**Ans:**53 electrons.### How do you write the full electron configuration for iodine?

**Ans:**1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^{5}.### How many valence electrons does iodine have?

**Ans:**Seven valence electrons.### What is the valency of iodine?

**Ans:**The valency of iodine is 1, 3, 5, and 7.