The d-electronic configuration of [CoCl4]2–in tetrahedral crystal field is emt₂n Sum of "m" and "num — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

**Step 1: Determine the electron configuration of Co** Cobalt (Co) has atomic number 27. Ground state configuration: [Ar] 3d⁷ 4s² **Step 2: Determine the configuration of Co²⁺ in [CoCl₄]²⁻** For the complex [CoCl₄]²⁻, cobalt is +2 oxidation state. Co²⁺ configuration: [Ar] 3d⁷ (loses 4s² electrons) So we have 7 d-electrons to distribute. **Step 3: Apply tetrahedral crystal field splitting** In tetrahedral geometry, the d-orbital splitting (from lowest to highest energy) is: - e orbital: lower energy (4 electrons capacity) - t₂ orbital: higher energy (6 electrons capacity) Order: e < t₂ (opposite of octahedral) **Step 4: Fill 7 d-electrons using Hund's rule** - e orbital: ↑↓ ↑↓ (4 electrons) - t₂ orbital: ↑ ↑ ↑ (3 electrons, one per orbital following Hund's rule) Configuration: e⁴t₂³ **Step 5: Identify m and unpaired electrons** - e⁴t₂³ means m = 3 - Unpaired electrons in t₂ = 3 **Step 6: Calculate sum** Sum of m and unpaired electrons = 3 + 3 = **6** Therefore, the answer is **6**.