What is CFSE of complex ion [FeF6]4- in terms of Dq ?

Question

Prashant Kotian · Accepted Answer

Correct answer: **ANSWER: 4**<BR>SOLUTION: FE2+ IS $D^6$. WITH THE WEAK FIELD F- LIGAND, IT ADOPTS A HIGH-SPIN $T_{2G}^4 E_G^2$ CONFIGURATION. CFSE = $(-0.4 \TIMES 4 + 0.6 \TIMES 2)\DELTA_O = -0.4 \DELTA_O = -4$ DQ. MOD VALUE IS 4.. Step 1: In [Co(CO3)(NH3)5]ClO4, the cobalt center is bonded to five neutral NH3 ligands and one anionic, monodentate carbonate (CO3^2-) ligand, resulting in a coordination number of 6. Step 2: The complex cation is [Co(CO3)(NH3)5]^+ with a net charge of +1. The oxidation state of Co is calculated as: x - 2 + 5(0) = +1 => x = +3, which leaves 6 d-electrons (3d^6 configuration). Step 3: Because NH3 is a strong-field ligand, Co^3+ (3d^6) undergoes spin-pairing to form a low-spin octahedral complex with 0 unpaired electrons. Row (a) correctly lists these properties.

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