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JEE Main 202623 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Coordination Compounds

JEE Main 2026 · 23 January · Morning Shift · Chemistry · Coordination Compounds

The crystal field splitting energy of [Co( oxalate )_ 3 ]^ 3-...

The crystal field splitting energy of [Co( oxalate )3]3−\left[\mathrm{Co}(\text{ oxalate })_{3}\right]^{3-} complex is ' n′\mathrm{n}^{\prime} times that of the [Cr( oxalate )3]3−\left[\mathrm{Cr}(\text{ oxalate })_{3}\right]^{3-} complex. Here ' n′\mathrm{n}^{\prime} is ____\_\_\_\_. (Assume Δ0≫P)\Delta_{0} \gg \mathrm{P} )

Correct answer

2

Step-by-step 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−[Co(ox)_3]^{3-}:
Cobalt is in +3+3 oxidation state. Co3+Co^{3+} has a d6d^6 configuration.
Given Δ0≫P\Delta_0 \gg P, it is a low spin complex. The electrons fill the t2gt_{2g} level first.
Configuration: t2g6eg0t_{2g}^6 e_g^0.
CFSE for Co3+=6×(−0.4Δ0)+0×(0.6Δ0)=−2.4Δ0Co^{3+} = 6 \times (-0.4 \Delta_0) + 0 \times (0.6 \Delta_0) = -2.4 \Delta_0.
For [Cr(ox)3]3−[Cr(ox)_3]^{3-}:
Chromium is in +3+3 oxidation state. Cr3+Cr^{3+} has a d3d^3 configuration.
Configuration: t2g3eg0t_{2g}^3 e_g^0.
CFSE for Cr3+=3×(−0.4Δ0)+0×(0.6Δ0)=−1.2Δ0Cr^{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 Co3+=2.4Δ0Co^{3+} = 2.4 \Delta_0.
Magnitude of CFSE for Cr3+=1.2Δ0Cr^{3+} = 1.2 \Delta_0.
The ratio n=2.4Δ01.2Δ0=2n = \frac{2.4 \Delta_0}{1.2 \Delta_0} = 2.
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