Question

Correct electronic configuration of d-orbital of iron in complex anion $[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6{]}^{4-}$is (atomic no. of iron is $26$)?

Answer 1

Electronic configuration of d-orbital of iron in complex anion $[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6{]}^{4-}$:

• In $[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6{]}^{4-}$complex the ligand $\mathrm{CN}$is known as a strong field ligand, therefore, will form a low spin complex by using inner $3\mathrm{d}$-orbitals.
• Since it has an octahedral geometry, having 6 ligands, so its hybridization will be ${\mathrm{d}}^2{\mathrm{sp}}^3.$
• The central metal has an oxidation number $+2$ .
• $\mathrm{Fe}$ has the ground state given as :$\left[\mathrm{Ar}\right]3{\mathrm{d}}^{6}4{\mathrm{s}}^2$
• Hence for the complex $[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6{]}^{4-}$, the electronic configuration can be shown as $:3{\mathrm{d}}^{6}4{\mathrm{s}}^0$.

Hence, Correct electronic configuration of the d-orbital of iron in complex anion $[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6{]}^{4-}$is $:3{\mathrm{d}}^{6}4{\mathrm{s}}^0$.