Question

The Valence Bond Theory (VBT) explains the formation, magnetic behaviour and geometry of coordination compounds. The Crystal Field Theory (CFT) of coordination compounds is based on the effect of different crystal fields (provided by the ligands taken as point charges), on the degeneracy of d-orbital energies of the central metal atom/ion. The splitting of the d-orbitals provides different electronic arrangements in strong and weak crystal fields.

Answer the following question :

(b) Using crystal field theory, write the electronic configuration of central metal atom/ion of the following :

(i) \( \mathrm{[CoF_6]^{3-}} \)

(ii) \( \mathrm{[Co(NH_3)_6]^{3+}} \)

[At. No. : Co = 27]

Hint:

Both ions are Co(III), d6. F- is a weak ligand (t2g4 eg2) and NH3 is a strong ligand (t2g6 eg0).

Answer 1

Step 1: Find the oxidation state and d-count:
Cobalt has atomic number 27, so the atom is \( \mathrm{[Ar]\,3d^7\,4s^2} \).
In \( \mathrm{[CoF_6]^{3-}} \): x + 6(-1) = -3, so x = +3.
In \( \mathrm{[Co(NH_3)_6]^{3+}} \): x + 0 = +3, so x = +3.
In both cases the metal ion is \( \mathrm{Co^{3+}} \), which has the configuration \( \mathrm{3d^6} \).

Step 2: Recall the octahedral splitting:
In an octahedral field the five d orbitals split into a lower \( t_{2g} \) set (three orbitals) and a higher \( e_g \) set (two orbitals).
If the splitting \( \Delta_o \) is smaller than the pairing energy P, the electrons stay unpaired (weak field, high spin). If \( \Delta_o \) is larger than P, the electrons pair in \( t_{2g} \) first (strong field, low spin).

Step 3: Part (i), fluoride complex:
F- is a weak field ligand, so \( \Delta_o < P \).
The six d electrons fill \( t_{2g} \) singly, then \( e_g \) singly, and the sixth electron pairs in \( t_{2g} \).
\[ \mathrm{[CoF_6]^{3-}:\ t_{2g}^{4}\,e_g^{2}} \]
It has four unpaired electrons and is paramagnetic.

Step 4: Part (ii), ammine complex:
NH3 is a strong field ligand, so \( \Delta_o > P \).
All six electrons pair up in the lower \( t_{2g} \) set and \( e_g \) stays empty.
\[ \mathrm{[Co(NH_3)_6]^{3+}:\ t_{2g}^{6}\,e_g^{0}} \]
It has no unpaired electron and is diamagnetic.

Final Answer:
(i) \( t_{2g}^{4} e_g^{2} \) (ii) \( t_{2g}^{6} e_g^{0} \)
\[ \boxed{\mathrm{[CoF_6]^{3-}: t_{2g}^4 e_g^2;\quad [Co(NH_3)_6]^{3+}: t_{2g}^6 e_g^0}} \]