Understanding $MnCr_2O_4$ Spinel Structure

The compound $MnCr_2O_4$ adopts a spinel crystal structure, generally represented as $AB_2O_4$. In a normal spinel structure, the divalent cation (A) occupies the tetrahedral sites, and the trivalent cations (B) occupy the octahedral sites.

Ionic Distribution in $MnCr_2O_4$

$MnCr_2O_4$ typically exhibits a normal spinel structure. This means:

• Manganese ($Mn$) is present as the divalent ion, $Mn^{2+}$, and occupies the tetrahedral sites.
• Chromium ($Cr$) is present as the trivalent ion, $Cr^{3+}$, and occupies the octahedral sites.

The ionic distribution is represented as $(Mn^{2+})_T (Cr^{3+})_2O_4$.

Calculating Crystal Field Stabilization Energy (CFSE)

The total CFSE for the compound is the sum of the CFSE contributions from the cations in their respective sites.

CFSE for $Mn^{2+}$ (Tetrahedral)

• Electronic configuration of $Mn^{2+}$ is $d^5$.
• In a tetrahedral field, the CFSE is calculated using the formula: CFSE = $0.6 \times n(e) - 0.4 \times n(t_2)$.
• Assuming a high-spin configuration for $Mn^{2+}$ ($t_2^3 e^2$), the CFSE = $0.6 \times 2 - 0.4 \times 3 = 1.2 - 1.2 = 0$ Dq.

CFSE for $Cr^{3+}$ (Octahedral)

• Electronic configuration of $Cr^{3+}$ is $d^3$.
• In an octahedral field, the CFSE is calculated using the formula: CFSE = $-4.0 \times n(t_{2g}) + 6.0 \times n(e_g)$.
• For $d^3$ configuration ($t_{2g}^3 e_g^0$), the CFSE = $-4.0 \times 3 + 6.0 \times 0 = -12$ Dq.

Total CFSE for $MnCr_2O_4$

The total CFSE is the sum of CFSE from one $Mn^{2+}$ ion and two $Cr^{3+}$ ions.

Total CFSE = CFSE($Mn^{2+}_T$) + 2 x CFSE($Cr^{3+}_O$)

Total CFSE = $0$ Dq + 2 x ($-12$ Dq)

Total CFSE = $0 - 24$ Dq = $-24$ Dq.

Therefore, $MnCr_2O_4$ is a normal spinel with a total CFSE of $-24$ Dq.