JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

For low spin octahedral complexes, electrons fill the t_ 2g orbitals first before pairing or moving to e_g orbitals. The configuration depends on the number of d-electrons. For Cr^ 3+ : Electronic configuration is 3d^3. In an octahedral field, the configuration is t_ 2g ^3 e_g^0. Number of unpaired electrons (n) = 3. For Mn^ 3+ : Electronic configuration is 3d^4. For low spin, the configuration is t_ 2g ^4 e_g^0. Number of unpaired electrons (n) = 2. For Fe^ 3+ : Electronic configuration is 3d^5. For low spin, the configuration is t_ 2g ^5 e_g^0. Number of unpaired electrons (n) = 1. For Co^ 3+ : Electronic configuration is 3d^6. For low spin, the configuration is t_ 2g ^6 e_g^0. Number of unpaired electrons (n) = 0. Comparing the number of unpaired electrons: Cr^ 3+ (3) > Mn^ 3+ (2) > Fe^ 3+ (1) > Co^ 3+ (0). The correct decreasing order is Cr^ 3+ > Mn^ 3+ > Fe^ 3+ > Co^ 3+ .