# Tin(Sn) electron configuration and orbital diagram

Tin is the 50th element in the periodic table and its symbol is ‘Sn’. Tin is a post-transition metal element. The total number of electrons in tin is fifty. 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 a tin atom can be done in two ways.

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

Electron configuration through orbital 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 tin, period and groups, valency and valence electrons of tin, bond formation, and application of different principles. Hopefully, after reading this article you will know in detail about this.

Table of Contents

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

The maximum electrons 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 tin is 50. That is, the number of electrons in tin is fifty. Therefore, a tin 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-two electrons but the fourth shell of tin will have eighteen electrons and the remaining four electrons will be in the fifth shell. Therefore, the order of the number of electrons in each shell of the tin(Sn) atom is 2, 8, 18, 18, 4.

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 tin 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 tin 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 the 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 two electrons enter the 5p orbital. Therefore, the tin full 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^{2}.

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

## How to write the orbital diagram for tin?

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 tin(Sn), you have to do the electron configuration of tin. 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 the 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 the1s 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 the 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 the 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. So, the remaining two electrons will enter the 5p orbital in the clockwise direction. This is clearly shown in the figure of the orbital diagram of tin.

## Tin 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 tin is 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p^{2}. In the tin ground-state electron configuration, the last electrons of the 5p orbital are located in the 5p_{x} and 5p_{y} 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 tin in the 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}^{1} 5p_{y}^{1}.

This electron configuration shows that the last shell of the tin atom has two unpaired electrons. So in this case, the valency of tin is 2. When the tin atom is excited, then the tin atom absorbs energy. As a result, an electron in the 5s orbital jumps to the 5p_{z} orbital.

Therefore, the electron configuration of tin(Sn*) 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}. The valency of the element is determined by electron configuration in the excited state. Here, tin has four unpaired electrons. So, the valency of tin is 4.

## Tin ion(Sn^{4+}) electron configuration

The electron configuration shows that the last shell of tin has four electrons. Therefore, the valence electrons of tin are four. There are two types of tin ion. The tin atom exhibits Sn^{2+} and Sn^{4+} ions. The elements that form bonds by donating electrons are called cation. The tin atom donates two electrons in the 5p orbital to form a tin ion(Sn^{2+}). That is, tin is a cation element.

Sn – 2e^{–} → Sn^{2+}

Here, the electron configuration of tin ion(Sn^{2+}) is 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2}. On the other hand, the tin atom donates two electrons in the 5p orbital and two electrons in the 5s orbital to convert a tin ion(Sn^{4+}).

Sn – 4e^{–} → Sn^{4+}

The electron configuration of tin ion(Sn^{4+}) is 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10}. This electron configuration shows that the tin ion(Sn^{4+}) has four shells and the last shell has eighteen electrons and it achieves a stable electron configuration. Tin atom exhibit +2 and +4 oxidation states. The oxidation state of the element changes depending on the bond formation.

## FAQs

What is the symbol for tin?**Ans:** The symbol for tin is ‘Sn’.

How many electrons does tin have?**Ans:** 50 electrons.

How do you write the full electron configuration for tin?**Ans:** 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p^{2}.

How many valence electrons does tin have?**Ans:** Four valence electrons.

What is the valency of tin?**Ans:** The valency of tin is 2, 4.

**Reference**