Match the LIST-I with LIST-II List-I Electronic configuration of tetrahedral metal ion A. d^2 B. d^4 C. d^6 D. d^8 List-II Crystal Field Stabilization Energy (Δt) I. -0.6 II. -0.8 III. -1.2 IV. -0.4 Choose the correct answer from the options given below:

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q61 – JEE-MAIN 2026

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Home Chemistry Coordination CompoundsQuestion #337

Match the LIST-I with LIST-II

\begin{array}{ll} List-I & List-II \\ Electronic configuration of & Crystal Field Stabilization \\ tetrahedral metal ion & Energy (Δ_{t}) \\ A. d^{2} & I. - 0.6 \\ B. d^{4} & II. - 0.8 \\ C. d^{6} & III. - 1.2 \\ D. d^{8} & IV. - 0.4 \end{array}

Choose the correct answer from the options given below:

Clue 1 of 3: Tetrahedral Field Splitting

Recall that in a tetrahedral crystal field, the d-orbitals split into a lower energy 'e' set and a higher energy 't2' set. Also, remember that tetrahedral complexes are almost always high-spin due to the small splitting energy (Δt).

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Ninja Strategy•Sequential Elimination

Verify the match for the first configuration (A) and the last (D) to quickly eliminate most incorrect options, then check one of the remaining configurations to find the answer.

🧠 Full Solution Path

Step 1: Understand d-orbital Splitting in Tetrahedral Fields✦ Active

In a tetrahedral crystal field, the d-orbitals split into two sets: a lower energy doubly degenerate set ( $e$ ) and a higher energy triply degenerate set ( $t_{2}$ ). The energy of the $e$ orbitals is $- 0.6 Δ_{t}$ and the energy of the $t_{2}$ orbitals is $+ 0.4 Δ_{t}$ relative to the barycenter. Tetrahedral complexes are high-spin, so electrons fill orbitals singly before pairing.

C F S E = [- 0.6 \times (e electrons) + 0.4 \times (t_{2} electrons)] Δ_{t}

Step 2: Calculate CFSE for Each Configuration○ Expand

We calculate the CFSE for each electronic configuration given in List-I:

A. $d^{2}$ : Configuration is $e^{2} t_{2}^{0}$ . CFSE = $[- 0.6 \times 2 + 0.4 \times 0] Δ_{t} = - 1.2 Δ_{t}$ .

B. $d^{4}$ : Configuration is $e^{2} t_{2}^{2}$ . CFSE = $[- 0.6 \times 2 + 0.4 \times 2] Δ_{t} = [- 1.2 + 0.8] Δ_{t} = - 0.4 Δ_{t}$ .

C. $d^{6}$ : Configuration is $e^{3} t_{2}^{3}$ . CFSE = $[- 0.6 \times 3 + 0.4 \times 3] Δ_{t} = [- 1.8 + 1.2] Δ_{t} = - 0.6 Δ_{t}$ .

D. $d^{8}$ : Configuration is $e^{4} t_{2}^{4}$ . CFSE = $[- 0.6 \times 4 + 0.4 \times 4] Δ_{t} = [- 2.4 + 1.6] Δ_{t} = - 0.8 Δ_{t}$ .

Step 3: Match the Lists and Select the Correct Option○ Expand

Based on the calculations, we can match the lists:

A ( $d^{2}$ ) $\to$ III ( $- 1.2$ ) B ( $d^{4}$ ) $\to$ IV ( $- 0.4$ ) C ( $d^{6}$ ) $\to$ I ( $- 0.6$ ) D ( $d^{8}$ ) $\to$ II ( $- 0.8$ )

The correct combination is A-III, B-IV, C-I, D-II, which corresponds to the third option.

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Chemistry Question 61 – JEE-MAIN 2026

Clue 1 of 3: Tetrahedral Field Splitting

🧠 Full Solution Path