Question

Which of the following is a high spin complex?

• A. $\left[Co\right(N{H}_3{)}_6{]}^{3+}$
• B. $\left[Fe\right(CN{)}_6{]}^{4-}$
• C. $\left[Ni\right(CN{)}_4{]}^{2-}$
• D. $[Fe{F}_6{]}^{4-}$

Answer 1

The correct option is D $[Fe{F}_6{]}^{4-}$

The complex $[Fe{F}_6{]}^{4-}$is paramagnetic and uses outer orbital (4d) in hybridisation $\left(s{p}^3{d}^2\right)$; it is thus called as outer orbital or high spin or spin free complex. So :
$F{e}^{2+},\left[Ar\right]3d^6$
$s{p}^3{d}^2$ hybrid orbitals
Six pairs of electrons from six $F^{-}$ ions.

(A) $\left[Co\right(N{H}_3{)}_6{]}^{3+}$(strong field ligand), it pair up againt Hund's rule $\to C{o}^{3+}\to 3d^6$
$d^{2}s{p}^3\to$ low spin complex

(B) $\left[Fe\right(CN{)}_6{]}^{4-}$(strong field ligand), it pair up againt Hund's rule$\to F{e}^{2+}\to 3d^6\to$
$d^{2}s{p}^3⟶$ low spin complex

(C) $\left[Ni\right(CN{)}_4{]}^{2-}$(strong field ligand), it pair up againt Hund's rule$\to d^8$ $ds{p}^2\to$ low spin complex