Question

Total number of unpaired electrons in ${Co^{3+}} $ ( $ Atomic\,number = 27 $ ):

Answer 1

Hint: To calculate the number of unpaired electrons of the given ion, we have to write and check the electronic configuration of this atom. Cobalt is a transition metal of the $ 3d - $ series. It is a silver-grey metal that is found in the earth’s crust.

Complete Step-by-step Answer:
Elements from $ Sc(scandium) $ to $ Zn(zinc) $ come in the $ 3d - $ series of the D block elements. These elements can be found from the $ {3^{rd}} $ group to the $ {12^{th}} $ group of the modern periodic table. Now we will write the electronic configuration of this element. The transition metals have a d-subshell that is partially filled with electrons. The compounds and ions of the d-block are colored due to the $ d - d $ transition shown by electrons. Many types of ligands bind themselves with these elements and form a coordination compound. These elements have great use as catalysts in several reactions and mechanisms. They also exhibit a wide range of oxidation states and due to this, they form stable complexes.
Now we will calculate the number of unpaired electrons in $C{o^{3 + }}$ ion with the help of electronic configuration of cobalt. Its atomic number is $ 27 $ . Its electronic configuration will be $ [Ar]\,3{d^7}\,4{s^2} $ . It has $ 7 $ electrons in its d-orbital. After forming $C{o^{3 + }}$ ion it will lose $ 3 $ electrons, two will be from $ s - orbital $ and $ $ 1 $ $ will be from the $ d $ orbital. Now, their arrangement will be like $ \matrix
{ \uparrow \downarrow } \& { \uparrow } \& {\uparrow } \& { \uparrow \,\,\,\, \uparrow } $ . Hence it will have $ 4 $ unpaired electrons. Therefore we can say that the total number of unpaired electrons in $ C{o^{3 + }} $ is $ 4 $ .

Note:
Elements of the $ 3d - $ series show great properties of metals like ductility, malleability, and lustre. They also show high melting and boiling points. Due to their small size, they have high ionization enthalpies.