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Question: Write the electronic configuration of Cobalt and find the total no. of unpaired \[{{e}^{-}}\] and it...

Write the electronic configuration of Cobalt and find the total no. of unpaired e{{e}^{-}} and it's total spin.

Explanation

Solution

We know that we have to find the nearest noble gas element of the atom to write the abbreviated electron configuration of an atom in the periodic table. We should know the atomic number of an element to write the electronic configuration.

Complete answer:
The electronic configuration of any element tells us the total number of electrons present in that atom. It is identified through atomic number, as the atomic number of an atom tells us the total number of electrons in that atom. The electronic configuration requires filling of orbital or sub shells (s, p, d, f) \left( s,\text{ }p,\text{ }d,\text{ }f \right)~ The Electronic configuration of any atom tells us the total number of electrons which is equal to the total number protons in that atom that is equal to the atomic number of that element. Electronic configuration of any element consists of filling the orbital or the s, p, d, f sub – shells with electrons. The filling of electrons in various orbits is according to a principle of Aufbau that takes place from the lower energy level to the higher energy level. The s, p, d, f subshells are written along with the number of the shell, like 1,2,31,2,3 etc. We have been given cobalt which has atomic number z=27.z=27. The electronic configuration of cobalt will consist of filling 2727 electrons according to the Aufbau principle. The electronic configuration of Cobalt 1s22s22p63s23p64s23d71{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{7}} and the total number of electrons unpaired are 33 cause last shell has 77 electrons filled.
Number of spin=±12×number of unpaired electrons.Number\text{ }of\text{ }spin=\pm \dfrac{1}{2}\times number\text{ }of\text{ }unpaired\text{ }electrons.
Number of spin=±12×3=±1.5\Rightarrow Number\text{ }of\text{ }spin=\pm \dfrac{1}{2}\times 3=\pm 1.5

Note:
Remember that by using the abbreviated way we can decrease the length of the electronic configuration of the elements which have very lengthy electronic configuration. We can use the abbreviated electron configuration to write the electronic configuration of the d-block and f-block elements.