Question

The hybridisation of Fe in K4[Fe(CN)6] is)

Answer 1

Answer:- Hybridization of Fe in $K_4[Fe(CN)_6]$ is $d^2sp^3$ .

Explanations:- Oxidation number of alkali metals in their compounds is +1, oxidation number of cyanide ion is -1 and there is no charge given for the complex. As per the oxidation number rules, sum of oxidation numbers of a compound equal to zero if the charge of the complex is zero. We could calculate the oxidation number of Fe from here:

$4(+1)+x+6(-1)=0$

$4+x-6=0$

$x-2=0$

$x=+2$

Electronic configuration for Fe is $1s^22s^22p^63s^23p^64s^23d^6$ . $Fe^2^+$ is formed when Fe loses two electrons from it's outer most shell which is 4s. So, electronic configuration of $Fe^2^+$ is $1s^22s^22p^63s^23p^63d^6$ .

From this electron configuration it is clear that the iron ion has 6 electrons in 3d orbital. Since, $CN^-$ is a strong field ligand, the pairing of Iron 3d orbitals takes place and two of the 3d orbitals are used by the ligands. Since there are total six ligands, one 4s and three 4p orbitals are also used by the ligands and in this way the hybridization of Fe is  $d^2sp^3$  and it's a low spin octahedral complex.

Answer 2

Answer:

d2sp3

Explanation:

Iron(Fe) is atomic no 26.In which 6 electrons are in 3d orbitals and 2 electrons are there in 4s orbital.

In this complex, it exists in +2 oxidation state.

When ligands CN approach to the central metal atom, the electrons remains in the inner orbital.And 6 electron pairs of 6CN's are filled in 2d,1s and 3p orbitals.

Hence theHybridisation is d 2sp3