Question

How many unpaired electrons are present in $ d $ -orbitals of central metal ions in $ {[Fe(C{N_6})]^{3 - }} $ complex ion?

Answer 1

Hint: To find the number of unpaired electrons simply find the oxidation state of the central metal ion first. Then corresponding to it which $ d $ orbital it possesses. Then pair the electrons of the surrounding metals. Number of electrons left unpaired will be your answer.

Complete answer:
Complex ions have a metal ion at its centre with a number of molecules or ions surrounding it by a coordinate or dative covalent bond.
We will start by first finding the oxidation number of the central metal atom, which in this case is $ Fe $ . We are aware that the atomic number of $ Fe $ is $ 26 $ . Let us write the electronic configuration of $ Fe $
 $ Fe $ ( $ 26 $ ) = $ [Ar]4{s^2}3{d^6} $
Electronic configuration for $ F{e^{3 + }} $
 $ F{e^{3 + }} $ = $ [Ar]4{s^0}3{d^5} $
Thus it is clear that the electrons of these $ 3d $ orbitals will participate in bonding with the $ CN $ metal. Also one must know that $ CN $ is a powerful field ligand therefore will favour electron pairing and uses inner $ 3d $ orbitals to form a low spin complex. As we can see in the complex given that there are six ligands present, during bonding the electrons of $ CN $ pairs with the four electrons of $ 3d $ orbitals of $ Fe $ . Since there are five unpaired electrons out of which four are used for bonding and only one electron is left which is unpaired, therefore the central metal atom $ Fe $ will have one unpaired electron.

Note:
The geometry of the complex given, which is $ {[Fe(C{N_6})]^{3 - }} $ is octahedral in shape. Also the hybridisation of the complex $ {[Fe(C{N_6})]^{3 - }} $ is $ {d^2}s{p^3} $ . The oxidation of the central metal atom $ Fe $ is $ - 3 $ oxidation state. Due to one unpaired electron it is a paramagnetic complex.