Multiple choice

Which metal ion represents zero value of CFSE in octahedral complexes if it associated with a weak field ligand?

  1. Fe(III)

  2. Fe(II)

  3. Cr(II)

  4. Mn(I)

  5. Mn(III)
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A Correct answer
Explanation

Correct answer. In weak field ligand, the filling of electrons are in accordance with Hund’s rule. The CFSE is given by the formula:  CFSE = value of t2g (n) + value of eg (n) Dq, Fe(III) corresponds to the d5 configuration. Now, CFSE for Fe(III) (d5) = -4 (3) + 6 (2) = -12 + 12 = 0 Dq CFSE for Fe(II) (d6) = -4 (4) + 6 (2) = -16 + 12 = -4 Dq CFSE for Cr(II) (d4) = -4 (3) + 6 (1) = -12 + 6 = -6 Dq CFSE for Mn(I) (d6) = -4 (4) + 6 (2) = -16 + 12 = -4 Dq CFSE for Mn(III) (d4) = -4 (3) + 6 (1) = -12 + 6 = -6 Dq Hence, the metal ion with zero CFSE is Fe(III) when it associated with a weak field ligand.

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