Question

Out of $[Co{F}_6{]}^{3-}$ and $\left[Co\right(en{)}_3{]}^{3+}$, which one complex is:
(i) paramagnetic
(ii) more stable
(iii) inner orbital complex and
(iv) high spin complex (Atomic no. of Co = 27)

Answer 1

$C{o}^{3+}=\left[Ne\right]3s^{2}3p^{6}3d^6$
$F$ is weak field ligand for $C{o}^{3+}$ thus pairing of d-orbitals electrons does not happen, so it forms:
1. outer orbital complex of hybridization $s{p}^3{d}^2$
2. high spin complex means that there are more number of unpaired electrons
$en=ethylenediamine$ is a strong field ligand for $C{o}^{3+}$ thus the pairing of d-orbitals happening for formation of complex of $C{o}^{3+}$ with ligand $ethylenediamine$ so it forms:
1. inner-orbital complex of hyridization $d^{2}s{p}^3$
2. low spin complex
3. Stable complex form beacuse of ethylenediamine is a chelating legand.
(i) Unpaired of electrons present in $[Co{F}_6{]}^{3-}$ thus it is paramagnetic complex.
(ii) $\left[Co\right(en{)}_3{]}^{3+}$ is more stable beacuse of chelation.
(iii) $\left[Co\right(en{)}_3{]}^{3+}$
(iv) $[Co{F}_6{]}^{3-}$