JEE Main 2026 — Chemistry Inorganic Chemistry

In a tetrahedral crystal field, the d-orbitals split into a lower energy $e$e set and a higher energy $t_2$t2​ set. The energy of an electron in the $e$e orbital is $-0.6 \Delta_t$−0.6Δt​ and in the $t_2$t2​ orbital is $+0.4 \Delta_t$+0.4Δt​. Since $\Delta_t$Δt​ is generally small, tetrahedral complexes form high-spin configurations.

The Crystal Field Stabilization Energy (CFSE) is given by:

$$\text{CFSE} = (-0.6 n_e + 0.4 n_{t_2}) \Delta_t$$CFSE=(−0.6ne​+0.4nt2​​)Δt​

For $d^2$d2: The configuration is $e^2 t_2^0$e2t20​.

$\text{CFSE} = 2 \times (-0.6) + 0 = -1.2 \Delta_t$CFSE=2×(−0.6)+0=−1.2Δt​ (A $\rightarrow$→ III)

For $d^4$d4: The configuration is $e^2 t_2^2$e2t22​.

$\text{CFSE} = 2 \times (-0.6) + 2 \times (0.4) = -1.2 + 0.8 = -0.4 \Delta_t$CFSE=2×(−0.6)+2×(0.4)=−1.2+0.8=−0.4Δt​ (B $\rightarrow$→ IV)

For $d^6$d6: The configuration is $e^3 t_2^3$e3t23​.

$\text{CFSE} = 3 \times (-0.6) + 3 \times (0.4) = -1.8 + 1.2 = -0.6 \Delta_t$CFSE=3×(−0.6)+3×(0.4)=−1.8+1.2=−0.6Δt​ (C $\rightarrow$→ I)

For $d^8$d8: The configuration is $e^4 t_2^4$e4t24​.

$\text{CFSE} = 4 \times (-0.6) + 4 \times (0.4) = -2.4 + 1.6 = -0.8 \Delta_t$CFSE=4×(−0.6)+4×(0.4)=−2.4+1.6=−0.8Δt​ (D $\rightarrow$→ II)

Therefore, the correct match is A-III, B-IV, C-I, D-II.

Answer: A-III, B-IV, C-I, D-II