Question

Using CFT, write electronic configuration for: (i) \( [CoF_6]^{3-} \) (ii) \( [Co(NH_3)_6]^{3+} \). [At. No. Co=27]

Hint:

Spectrochemical Series: $I^-<Br^-<S^{2-}<SCN^-<Cl^-<F^-<OH^-<C_2O_4^{2-}<H_2O<NCS^-<edta^{4-}<NH_3<en<CN^-<CO$.

Answer 1

(i) For \( [CoF_6]^{3-} \):
- The atomic number of Co is 27, so its electronic configuration is: \[ \text{Co} = [Ar] 3d^7 4s^2 \] - In \( [CoF_6]^{3-} \), Co is in the +3 oxidation state. To find the electronic configuration of \( Co^{3+} \), we remove 3 electrons from the 4s and 3d orbitals: \[ \text{Co}^{3+} = [Ar] 3d^6 \] - In the octahedral crystal field of \( [CoF_6]^{3-} \), the d-orbitals split into two sets: \( t_{2g} \) (lower energy) and \( e_g \) (higher energy). The 6 electrons will occupy the orbitals as follows: \[ \text{Electronic configuration of } [CoF_6]^{3-} \text{:} \quad t_{2g}^6 e_g^0 \] This means all six electrons are in the lower-energy \( t_{2g} \) orbitals, resulting in a low-spin configuration due to the weak field ligand \( F^- \) (fluoride).

(ii) For \( [Co(NH_3)_6]^{3+} \):
- In \( [Co(NH_3)_6]^{3+} \), Co is again in the +3 oxidation state, so the electronic configuration of \( Co^{3+} \) is: \[ \text{Co}^{3+} = [Ar] 3d^6 \] - In an octahedral field created by the stronger ligand \( NH_3 \) (ammonia), which is a weak field ligand, the d-orbitals will split, but since \( NH_3 \) induces a weaker splitting, the electrons will occupy the lower and higher energy \( t_{2g} \) and \( e_g \) orbitals according to Hund's rule. \[ \text{Electronic configuration of } [Co(NH_3)_6]^{3+} \text{:} \quad t_{2g}^6 e_g^0 \] This means that all six electrons occupy the lower-energy \( t_{2g} \) orbitals, with no electrons in the higher-energy \( e_g \) orbitals, forming a low-spin configuration.