Question

Calculate CFSE of the following complex:
${[Fe{(CN)_6}]^{4 - }}$
(A) $- 0.4{\Delta _t}$
(B) $- 2.4{\Delta _{\text{o}}}$
(C) $0.4{\Delta _{\text{o}}}$
(D) $0.6{\Delta _{\text{o}}}$

Answer 1

To find the CFSE, we need to fill the ${t_{2g}}$ and ${e_g}$ orbitals according to CFT ( crystal field theory). In the complex ${[Fe{(CN)_6}]^{4 - }}$, Iron ( $Fe$ ) is in $+ 2$ oxidation state and thus it is a $3{d^6}$ system. $CN$ is a strong field ligand and thus the complex will be a low spin complex and all the six electrons will be filled in ${t_{2g}}$ orbitals.

Complete step by step solution:
-Iron ($Fe$) has electronic configuration $3{d^6}4{s^2}$ in the ground state. But in the complex ${[Fe{(CN)_6}]^{4 - }}$, iron is in $+ 2$ oxidation state. Therefore, it will have configuration $3{d^6}4{s^0}$ . Also, the coordination number of iron is six, therefore the complex will have octahedral geometry.
-According to CFT( crystal field theory), five degenerate $d$ orbitals split into three ${t_{2g}}$ and two ${e_g}$ orbitals in the presence of ligands. This splitting of the degenerate levels due to the presence of ligands in a definite geometry is termed as crystal field splitting and the energy separation is denoted by ${\Delta _{\text{o}}}$ (the subscript o is for octahedral the energy of the two ${e_g}$ orbitals will increase by $\left( {\dfrac{3}{5}} \right){\Delta _{\text{o}}}$ and that of the three ${t_{2g}}$ will decrease by $\left( {\dfrac{2}{5}} \right){\Delta _{\text{o}}}$ . Thus, from here we get the formula for crystal field splitting energy i.e. CFSE and it is:
${\Delta _{\text{o}}} = {\text{no}}{\text{. of electrons in }}{{\text{t}}_{{\text{2g}}}} \times \left( {\dfrac{2}{5}} \right){\Delta _{\text{o}}} + {\text{no}}{\text{. of electrons in }}{{\text{e}}_g} \times \left( {\dfrac{3}{5}} \right){\Delta _{\text{o}}}$
Or, ${\Delta _{\text{o}}} = {\text{no}}{\text{. of electrons in }}{{\text{t}}_{{\text{2g}}}} \times ( - 0.4{\Delta _{\text{o}}}) + {\text{no}}{\text{. of electrons in }}{{\text{e}}_g} \times (0.6{\Delta _{\text{o}}})$
-In the complex, ${[Fe{(CN)_6}]^{4 - }}$, we have $CN$ as a ligand which is a strong field ligand. For strong field ligands, ${\Delta _{\text{o}}}$(CFSE) is greater than the pairing energy,P i.e. ${\Delta _{\text{o}}} > P$ and they form low spin complexes. Therefore, ${[Fe{(CN)_6}]^{4 - }}$will be a low spin complex and all the six electrons (since, $F{e^{ + 2}}$ is $3{d^6}$ system) will enter in ${t_{2g}}$ orbital. Thus, configuration of $F{e^{ + 2}}$ in the complex will be $t_{2g}^6e_g^0$ .
Now, CFSE of the complex:
Since, ${\Delta _{\text{o}}} = {\text{no}}{\text{. of electrons in }}{{\text{t}}_{{\text{2g}}}} \times ( - 0.4{\Delta _{\text{o}}}) + {\text{no}}{\text{. of electrons in }}{{\text{e}}_g} \times (0.6{\Delta _{\text{o}}})$
Therefore, ${\Delta _{\text{o}}} = 6 \times ( - 0.4{\Delta _{\text{o}}}) + 0 \times (0.6{\Delta _{\text{o}}}) = - 2.4{\Delta _{\text{o}}}$

Thus, option (B) is the correct answer.

Note: The crystal field splitting ${\Delta _{\text{o}}}$, depends upon the field produced by the ligand and charge on the metal ion. Some ligands produce strong field and are called strong field ligands while some produce weak field and are called weak field ligands. Ligands are generally arranged in a series called spectrochemical series, in the order of increasing field strength as given below:
${I^ - } < B{r^ - } < SC{N^ - } < C{l^ - } < {S^{2 - }} < {F^ - } < O{H^ - } < {C_2}{O_4}^{2 - } < {H_2}O < NC{S^ - } < edt{a^{4 - }} < N{H_3} < en < C{N^ - } < CO$