JEE Main 2026 — Chemistry Inorganic Chemistry

2026integermedium

The crystal field splitting energy of $\left[\mathrm{Co}(\text { oxalate })_{3}\right]^{3-}$ complex is ' $\mathrm{n}^{\prime}$ times that of the $\left[\mathrm{Cr}(\text { oxalate })_{3}\right]^{3-}$ complex. Here ' $\mathrm{n}^{\prime}$ is $\_\_\_\_$ . (Assume $\Delta_{0} \gg \mathrm{P}$ )

Official previous-year question

Held on 23 Jan 2026 · Verified 6 Jul 2026.

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Solution

The Crystal Field Stabilization Energy (CFSE) is calculated based on the electronic configuration of the central metal ion in the octahedral field.

For $[Co(ox)_3]^{3-}$ :

Cobalt is in $+3$ oxidation state. $Co^{3+}$ has a $d^6$ configuration.

Given $\Delta_0 \gg P$ , it is a low spin complex. The electrons fill the $t_{2g}$ level first.

Configuration: $t_{2g}^6 e_g^0$ .

CFSE for $Co^{3+} = 6 \times (-0.4 \Delta_0) + 0 \times (0.6 \Delta_0) = -2.4 \Delta_0$ .

For $[Cr(ox)_3]^{3-}$ :

Chromium is in $+3$ oxidation state. $Cr^{3+}$ has a $d^3$ configuration.

Configuration: $t_{2g}^3 e_g^0$ .

CFSE for $Cr^{3+} = 3 \times (-0.4 \Delta_0) + 0 \times (0.6 \Delta_0) = -1.2 \Delta_0$ .

Comparing the magnitudes of the stabilization energies:

Magnitude of CFSE for $Co^{3+} = 2.4 \Delta_0$ .

Magnitude of CFSE for $Cr^{3+} = 1.2 \Delta_0$ .

The ratio $n = \frac{2.4 \Delta_0}{1.2 \Delta_0} = 2$ .

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Solution

The Crystal Field Stabilization Energy (CFSE) is calculated based on the electronic configuration of the central metal ion in the octahedral field.

For $[Co(ox)_3]^{3-}$ :

Cobalt is in $+3$ oxidation state. $Co^{3+}$ has a $d^6$ configuration.

Given $\Delta_0 \gg P$ , it is a low spin complex. The electrons fill the $t_{2g}$ level first.

Configuration: $t_{2g}^6 e_g^0$ .

CFSE for $Co^{3+} = 6 \times (-0.4 \Delta_0) + 0 \times (0.6 \Delta_0) = -2.4 \Delta_0$ .

For $[Cr(ox)_3]^{3-}$ :

Chromium is in $+3$ oxidation state. $Cr^{3+}$ has a $d^3$ configuration.

Configuration: $t_{2g}^3 e_g^0$ .

CFSE for $Cr^{3+} = 3 \times (-0.4 \Delta_0) + 0 \times (0.6 \Delta_0) = -1.2 \Delta_0$ .

Comparing the magnitudes of the stabilization energies:

Magnitude of CFSE for $Co^{3+} = 2.4 \Delta_0$ .

Magnitude of CFSE for $Cr^{3+} = 1.2 \Delta_0$ .

The ratio $n = \frac{2.4 \Delta_0}{1.2 \Delta_0} = 2$ .

JEE Main 2026 — Chemistry Inorganic Chemistry

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JEE Main 2026 — Chemistry Inorganic Chemistry

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JEE Main Inorganic Chemistry in other years

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