Question

The correct increasing order of stability of the complexes based on $\Delta_o$ value is

• A. I < II < III < IV
• B. II < I < IV < III
• C. I < IV < II < III
• D. II < IV < I < III

Answer 1

Answer: (B)

II < I < IV < III

Answer 2

The stability of coordination complexes is related to their Crystal Field Stabilization Energy (CFSE). A more negative CFSE value indicates greater stability. The CFSE for octahedral complexes is calculated using the formula: $CFSE = (-0.4 n_t + 0.6 n_e) \Delta_o$, where $n_t$ is the number of electrons in $t_{2g}$ orbitals and $n_e$ is the number of electrons in $e_g$ orbitals.

I. $[Mn(CN)_6]^{3-}$: $Mn^{+3} \rightarrow 3d^4$. Low spin: $t_{2g}^4 e_g^0$. CFSE = $4(-0.4 \Delta_o) = -1.6 \Delta_o$. II. $[Co(CN)_6]^{4-}$: $Co^{+2} \rightarrow 3d^7$. High spin: $t_{2g}^5 e_g^2$. CFSE = $5(-0.4 \Delta_o) + 2(0.6 \Delta_o) = -2.0 \Delta_o + 1.2 \Delta_o = -0.8 \Delta_o$. III. $[Fe(CN)_6]^{4-}$: $Fe^{+2} \rightarrow 3d^6$. Low spin: $t_{2g}^6 e_g^0$. CFSE = $6(-0.4 \Delta_o) = -2.4 \Delta_o$. IV. $[Fe(CN)_6]^{3-}$: $Fe^{+3} \rightarrow 3d^5$. Low spin: $t_{2g}^5 e_g^0$. CFSE = $5(-0.4 \Delta_o) = -2.0 \Delta_o$.

Comparing the CFSE values: I: $-1.6 \Delta_o$ II: $-0.8 \Delta_o$ III: $-2.4 \Delta_o$ IV: $-2.0 \Delta_o$

The increasing order of stability (from least negative to most negative CFSE) is: $-0.8 > -1.6 > -2.0 > -2.4$ Thus, the order is II < I < IV < III.