Question

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

Answer 1

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.