Question

Find the number of unpaired electrons calculated in ${\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$ and ${\left[ {Co{F_6}} \right]^{3 - }}$
A.4 unpaired electrons in both complexes
B. 4 unpaired electrons in ${\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$ 7 and no unpaired electrons in ${\left[ {Co{F_6}} \right]^{3 - }}$
C. 4 unpaired electrons in ${\left[ {Co{F_6}} \right]^{3 - }}$ and no unpaired electrons in ${\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$
D. no unpaired electrons in both complexes

Answer 1

The transition elements are found in the group 3, 4, 5, 6, 7, 8, 9 ,10, 11 and 12 of the periodic table. These are also known as transition metals. The d orbital is filled with an electronic shell n-1. There are a total of 40 d block elements.

Complete step by step answer:
-The elements which lie in the middle of the group II-A and group IIB elements in the periodic table are d block elements. They are known as transition elements as they are the elements that lie between the metals and non-metals of the periodic table.
-In the case of transition elements due to the presence of electrons at d orbitals, which is closer to the outermost shell of the metal. They show a variable oxidation state. the electrons of the d orbitals.
-Transition metals form a complex with a different ligand using its d-orbital electrons. They can combine with a different number of ligands and form different kinds of geometry like octahedral, square planar, tetrahedral, etc.
-In the complex, the number of the unpaired electron depends upon the ligand field strength of the ligand.
-In the case of, ${\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$ the number of d electron is 6. $N{H_3}$ can act as both a Strong Field Ligand as well as a Weak Field Ligand. In this case, because of the presence of cobalt, ammonia behaves as a strong field ligand. -The electronic configuration of d orbital in the complex is ${t_{2g}}^6e{g^0}$ . Therefore, all the d electrons will be paired. On the other hand, in the case of ${\left[ {Co{F_6}} \right]^{3 - }}$ fluoride ligand is a weal flew ligand. Therefore, here the electronic configuration of d orbital in the complex is ${t_{2g}}^4e{g^2}$ . The number of unpaired electrons is, 4 . The diagrams are shown below,

So, the correct answer is C.

Note:
These transition metals usually have high melting and boiling points. This is mainly because they have filled d orbitals because of which no unpaired electron is available. Because of the unavailability of unpaired electrons, these metals do not undergo covalent bonding.