# Iodine(I) electron configuration and orbital diagram

Iodine is the 53rd element in the periodic table and its symbol is ‘I’. Iodide is a classified halogen element. The total number of electrons in iodine is fifty-three. These electrons are arranged according to specific rules of different orbits. The arrangement of electrons in different orbits and orbitals of an atom in a certain order is called **electron configuration**. The electron configuration of iodine(I) atom 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. This article gives an idea about the electron configuration and orbital diagram of iodine, period and groups, valency and valence electrons of iodine, bond and compound formation, and application of different principles. Hopefully, after reading this article you will know in detail about this.

Table of Contents

## 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, N is the name of the fourth orbit. The electron holding capacity of each orbit is 2n^{2}.

For example,

- n = 1 for K orbit.

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

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

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

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

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 levels are subdivided into sub-energy levels. These sub-energy levels are called orbital. The sub energy levels are expressed by ‘l’. The value of ‘l’ is from 0 to (n – 1). The sub-energy levels are known as s, p, d, f. Determining the value of ‘l’ for different energy levels is-

- If n = 1,

(n – 1) = (1–1) = 0

Therefore, the orbital number of ‘l’ is 1; And the orbital is 1s. - If n = 2,

(n – 1) = (2–1) = 1.

Therefore, the orbital number of ‘l’ is 2; And the orbital is 2s, 2p. - If n = 3,

(n – 1) = (3–1) = 2.

Therefore, the orbital number of ‘l’ is 3; And the orbital is 3s, 3p, 3d. - If n = 4,

(n – 1) = (4–1) = 3

Therefore, the orbital number of ‘l’ is 4; And the orbital is 4s, 4p, 4d, 4f. - If n = 5,

(n – 1) = (n – 5) = 4.

Therefore, l = 0,1,2,3,4. The number of orbitals will be 5 but 4s, 4p, 4d, 4f in these four orbitals it is possible to arrange the electrons of all the elements of the periodic table. The electron holding capacity of these orbitals is s = 2, p = 6, d = 10 and f = 14. The German physicist Aufbau first proposed the idea of electron configuration through sub-orbits.

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. These orbitals are named s, p, d, f. The Aufbau electron configuration method 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 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(I) 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}.**

## How to write the orbital diagram for iodine?

To create an orbital diagram of an atom, you first need to know Hund’s principle and Pauli’s exclusion principle. Hund’s principle is that electrons in different orbitals with the same energy would be positioned in such a way that they could be in the unpaired state of maximum number and the spin of the unpaired electrons will be one-way.

And Pauli’s exclusion principle is that the value of four quantum numbers of two electrons in an atom cannot be the same. To write the orbital diagram of iodine(I), you have to do the electron configuration of iodine. Which has been discussed in detail above. 1s is the closest and lowest energy orbital to the nucleus. Therefore, the electron will first enter the 1s orbital.

According to Hund’s principle, the first electron will enter in the clockwise direction and the next electron will enter the 1s orbital in the anti-clockwise direction. The 1s orbital is now filled with two electrons. Then next two electrons will enter the 2s orbital just like the 1s orbital.

The next three electrons will enter the 2p orbital in the clockwise direction and the next three electrons will enter the 2p orbital in the anti-clockwise direction. The next two electrons will enter the 3s orbital just like the 1s orbital and the next six electrons will enter the 3p orbital just like the 2p orbital.

The 3p orbital is now full of electrons. So, the next two electrons will enter the 4s orbital just like the 1s orbital. Then next five electrons will enter the 3d orbital in the clockwise direction and the next five electrons will enter the 3d orbital in the anti-clockwise direction. The 3d orbital is now full. So, the next six electrons will enter the 4p orbital just like the 3p orbital.

**Then next two electrons will enter the 5s orbital just like the 1s orbital and the next ten electrons will enter the 4d orbital just like the 3d orbital. The 4d orbital is now full of electrons. Then next three electrons will enter the 5p orbital in the clockwise direction and the remaining two electrons will enter the 5p orbital in the anti-clockwise direction. This is clearly shown in the figure of the orbital diagram of iodine.**

## Iodine excited state electron configuration

Atoms can jump from one orbital to another orbital in the 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, five electrons of the 5p orbital are located in the 5p_{x}(2), 5p_{y}(2) and 5p_{z} sub-orbitals.

The p-orbital has three sub-orbitals. The sub-orbitals are p_{x}, p_{y}, and p_{z}. Each sub-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} sub-orbital jumps to the 5d_{xy} sub-orbital. The d-orbital has five sub-orbitals. The sub-orbitals are d_{xy}, d_{yz}, d_{zx}, d_{x2-y2} and d_{z2}. Each sub-orbital can have a maximum of two electrons. Therefore, the electron configuration of iodine(I*) in 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 5px sub-orbital jumps to the 5d_{yz} sub-orbital. Therefore, the electron configuration of iodine(I**) in 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} sub-orbital. Therefore, the electron configuration of iodine(I***) in 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(I) 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 bonds, 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.

## FAQ

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 electron configuration for iodine?**Ans:** Tellurium electron configuration is 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 1,3,5,7.

**Reference**