![Q.8 Stability of [CrCl6]3− (X), [MnCl6]3− (Y) and [FeCl6]3− (Z) follows the order (Given: Atomic numbers o Cr = 24 Mn = 25 and Fe = 26) (A) X > Y > Z (B) X < Y < Z (C) Y < X < Z (D) X < Y = Z](https://www.letstalkacademy.com/wp-content/uploads/2026/01/slide_8-6.png)
Q.8 Stability of [CrCl6]3− (X), [MnCl6]3− (Y) and [FeCl6]3− (Z) follows the order (Given: Atomic numbers o Cr = 24 Mn = 25 and Fe = 26)
(A) X > Y > Z
(B) X < Y < Z
(C) Y < X < Z
(D) X < Y = Z
The stability of hexachlorocomplexes [CrCl₆]³⁻ (X), [MnCl₆]³⁻ (Y), and [FeCl₆]³⁻ (Z) follows the order X > Y > Z due to increasing number of unpaired electrons across Cr(III) (d³), Mn(III) (d⁴), and Fe(III) (d⁵) in high-spin octahedral fields with weak-field Cl⁻ ligands.
Correct Answer
(A) X > Y > Z
Key Factors
Stability decreases with more unpaired electrons because higher electron-electron repulsions in eg orbitals destabilize the complexes. All are d⁴–d⁵ high-spin with sp³d² hybridization.
Option Analysis
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(A) X > Y > Z: Correct. Cr(III) d³ (3 unpaired e⁻) has highest crystal field stabilization energy (CFSE = -0.4 Δo) and lowest repulsions; Mn(III) d⁴ high-spin (4 unpaired e⁻, CFSE = -0.6 Δo + P) less stable; Fe(III) d⁵ high-spin (5 unpaired e⁻, CFSE = 0 + 2P) least stable.
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(B) X < Y < Z: Incorrect. Reverses the trend; more unpaired electrons reduce stability due to eg repulsion.
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(C) Y < X < Z: Incorrect. Places Mn(III) most unstable (not true, as d⁴ better than d⁵) and Fe(III) too stable.
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(D) X < Y = Z: Incorrect. Mn(III) and Fe(III) differ (4 vs 5 unpaired e⁻); Cr(III) is most stable.
Why CrCl₆³⁻ Most Stable
Half-filled t₂g³ configuration in Cr(III) maximizes CFSE without eg electrons or Jahn-Teller distortion (unlike d⁴ Mn(III)). Fe(III) d⁵ has no CFSE gain.
