Chemistry Q71 – JEE-MAIN 2026

Home Chemistry Coordination CompoundsQuestion #326

The total number of unpaired electrons present in the

d^{3}

d^{4}

(low spin)

d^{5}

(high spin),

d^{6}

(high spin) and

d^{7}

(low spin) octahedral complex systems is _______.

Enter Calculated Numerical Value:

Clue 1 of 3: Crystal Field Theory

Recall how d-orbitals split in an octahedral ligand field.

🧠 Full Solution Path

Step 1: Understand Octahedral Splitting and Spin States✦ Active

In an octahedral complex, the five d-orbitals split into a lower energy $t_{2 g}$ set (3 orbitals) and a higher energy $e_{g}$ set (2 orbitals). The number of unpaired electrons depends on the d-electron count and whether the complex is high spin or low spin.

Step 2: Calculate Unpaired Electrons for Each Configuration○ Expand

1. For $d^{3}$ : The configuration is $t_{2 g}^{3} e_{g}^{0}$ . All three electrons are unpaired. Number of unpaired electrons = 3.

2. For $d^{4}$ (low spin): Electrons pair up in $t_{2 g}$ first. The configuration is $t_{2 g}^{4} e_{g}^{0}$ . This results in one paired set and two unpaired electrons. Number of unpaired electrons = 2.

3. For $d^{5}$ (high spin): Electrons occupy $e_{g}$ before pairing in $t_{2 g}$ . The configuration is $t_{2 g}^{3} e_{g}^{2}$ . All five electrons are unpaired. Number of unpaired electrons = 5.

4. For $d^{6}$ (high spin): Electrons occupy $e_{g}$ before pairing in $t_{2 g}$ . The configuration is $t_{2 g}^{4} e_{g}^{2}$ . This results in one paired set in $t_{2 g}$ , two unpaired in $t_{2 g}$ , and two unpaired in $e_{g}$ . Number of unpaired electrons = 4.

5. For $d^{7}$ (low spin): Electrons pair up in $t_{2 g}$ first. The configuration is $t_{2 g}^{6} e_{g}^{1}$ . All $t_{2 g}$ electrons are paired, and there is one unpaired electron in $e_{g}$ . Number of unpaired electrons = 1.

💡 Teacher's Secret Hint

Remember Hund's rule and the effect of strong/weak field ligands on spin state.

Step 3: Sum the Unpaired Electrons○ Expand

Total number of unpaired electrons = $3 (d^{3}) + 2 (d^{4} low spin) + 5 (d^{5} high spin) + 4 (d^{6} high spin) + 1 (d^{7} low spin)$ . Total = $3 + 2 + 5 + 4 + 1 = 15$ .

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Chemistry Question 71 – JEE-MAIN 2026

Clue 1 of 3: Crystal Field Theory

🧠 Full Solution Path