Electronic configuration is the electronic distribution among the atomic orbitals of element representing different energy level. Let us study Palladium’s electronic configuration.

**The atomic number of palladium is 46,and the electronic configuration of Pd is 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 4d^{10}. It belongs d- block, period 5 and group 10 of elements. A precious metal that was discovered in 1803 by the English has a silvery-white color and a beautiful appearance.**

Electronic configuration, nomenclature, and schematic diagram of Palladium will be discussed in this article, in addition to the ground and excited state of the element that is relevant to it.

**How to write palladium electron configuration**

**The correct configuration of Pd = 1s**^{2}** 2s**^{2}** 2p**^{6}** 3s**^{2}** 3p**^{6}** 4s**^{2}** 3d**^{10}** 4p**^{6}** 4d**^{10}**, and it can be determined using the steps mentioned below**.

**First, one needs to determine the element’s period. Here Pd, which is a member of the 5 periods. It tells us the total number of shells that were found.**

**The next thing that needs to be done is to figure out how many electrons an element has based on its atomic number.****Pd does in fact has a total of 46 electrons.**

**After that, the electron sequence model, also called the “yellow brick road” model, will be used to fill in the complete orbitals in accordance with the electrons that are accessible.****The order of rising energy level is 1s 2s 2p and so on, and since Pd has 46 electrons, the configuration of Pd is 1s**^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4d^{10}5s^{2}.

**Then calculate the remaining number of electrons, and fill a single electron in each subshell and again start to form the first subshell.****In this case we are left with 8 electrons and the next available sub-shell is d, as it has 10 subshells, so initially 5 electrons will be filled, which will leave 3 electrons. After filling the remaining 3 electrons in the first three subshells, the last two subshells will be left. This yield 5d**.^{8}

**This yields the predicted electronic configuration of Pd 1s**^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{8}.**However, the energy of 5d<5s, so for the formation of a stable compound 5d files first then 5s. The correct configuration of Pd = 1s**^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}4d^{10}.

**Palladium electron configuration diagram**

**The electron configuration is drawn as follows**** using the octet rule, where the first two electrons were filled followed by 8 and 18 and the remaining 18 were filled in the last orbital which yields 2, 8, 18, and 18 configurations.**

**Palladium electron configuration notation**

**The electronic configuration notation of Pd is written as [Kr] 4d^{10}, where Kr represents the electronic configuration of krypton as a noble gas which is 36, followed by the representation of the remaining 10 electrons in the 4d sub-shell.**

**Palladium unabbreviated electron configuration**

**The unabbreviated electronic configuration of Pd = 1s**^{2}** 2s**^{2}** 2p**^{6}** 3s**^{2}** 3p**^{6}** 4s**^{2}** 3d**^{10}** 4p**^{6}** 4d**^{10}**. A****s it contains 46 electrons so its unabbreviated electron configuration is the same as the electronic configuration**.

**Ground state Palladium electron configuration**

**The standard ground state configuration of Pd = 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 4d^{10}. **

**The excited state of Palladium electron configuration**

**Pd shows two excited states and is characterized by the shifting of electrons from 4d to 5s subshell. **

**First excited state confugration of Pd =****1s**^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{1}**4d**^{9}**The second excited state****confugration****of the element is Pd = 1s**^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{8}.

**Ground state Palladium orbital diagram**

**The atomic orbital diagram for the outermost shell of Pd in the ground state is shown below.**

**Conclusion**

We are now capable of computing the electrical configuration of Pd, which is denoted by the notation [Kr] 4d^{10}. While we have previously talked about the ground state and the excited state of the element, the configuration of the excited state shifts to [Kr] 4d^{8} which exhibits diamagnetic behavior in the ground state.