The complex ion \[{\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_6}} \right]^{4 - }}\] contains:
A.\[{\text{s}}{{\text{p}}^3}{{\text{d}}^2}\] hybrid orbital with octahedral structure
B.Total 104 electrons on \[{\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_6}} \right]^{4 - }}\]
C.6 sigma bonds
D.Total 36 electrons on \[{\text{F}}{{\text{e}}^{2 + }}\] ion.

Hint: The given complex follows the EAN rule and hence the number of electrons can be calculated from EAN rule. It forms inner orbital complexes. Sigma bonds are the single bonds that are formed between any two elements.

Complete step by step answer:
The complex is known as Ferrocyanide. Since cyanide is a very strong field ligand it always prefers pairing of electrons in d orbital and hence the inner orbital complex forms instead of outer orbital complex. So the hybridization will be \[{{\text{d}}^2}{\text{s}}{{\text{p}}^3}\]instead of \[{\text{s}}{{\text{p}}^3}{{\text{d}}^2}\]to so the given option ‘a’ is incorrect.
The total electrons can be calculated by using the number of electrons in iron that is 26. 4 electrons are added due to charge on the complex. 6 electrons are in 6 bonds of iron with cyanide. Hence the electron can be calculated as:
\[26 + 6 + 4 + 12 \times 6 = 108\]
The above complex contains 12 sigma bonds and not six Sigma bonds. 6 sigma bonds are contained in between iron and cyanide and another six Sigma bonds are contained in between carbon and nitrogen in cyanide. Hence option c is incorrect.
The electrons on \[{\text{F}}{{\text{e}}^{2 + }}\]will be calculated as.
\[26 - 2 + 2 \times 6 = 36\]
The oxidation state of metal is \[ + 2\] and hence we subtracted 2 electrons. Electronic contribution of cyanide is 2.

Hence the correct option is D.

Note:
\[{\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_6}} \right]^{4 - }}\] is anionic complex. Usually found as potassium and sodium salts of this complex in the laboratory with the yellow colour solutions. Unlike Other cyanide complexes the ferrocyanide and ferricyanide are less toxic because they do not really free cyanide.