Question

Which of the following compounds has electrons symmetrically distributed in both \( t_{2g} \) and \( e_g \) orbitals?

• A. \( [CoF_6]^{3-} \)
• B. \( [Mn(CN)_6]^{4-} \)
• C. \( [Cr(NH_3)_6]^{3+} \)
• D. \( [FeCl_6]^{3-} \)

Hint:

High-spin \( d^5 \) configuration always gives symmetrical distribution because each d-orbital gets one electron before pairing occurs.

Answer 1

The Correct Option is D

Step 1: Understand the requirement.
Symmetrical distribution in both \( t_{2g} \) and \( e_g \) orbitals means equal or balanced filling of electrons in both sets of orbitals in an octahedral field.

Step 2: Find oxidation state and electronic configuration.
For \( [FeCl_6]^{3-} \):
Let oxidation state of Fe = \( x \)
\[ x + 6(-1) = -3 \]
\[ x = +3 \]
So, \( Fe^{3+} \) has configuration:
\[ 3d^5 \]

Step 3: Determine field strength of ligands.
\( Cl^- \) is a weak field ligand, so complex is high-spin.
Thus, electrons occupy orbitals according to Hund's rule.

Step 4: Fill orbitals.
For high-spin \( d^5 \) configuration:
\[ t_{2g}^3 e_g^2 \]
Each orbital has one electron, so electrons are symmetrically distributed.

Step 5: Check other options briefly.
(A) \( [CoF_6]^{3-} \): \( d^6 \), uneven distribution.
(B) \( [Mn(CN)_6]^{4-} \): strong field, low-spin, not symmetric in both sets.
(C) \( [Cr(NH_3)_6]^{3+} \): \( d^3 \), electrons only in \( t_{2g} \).

Step 6: Conclusion.
Only \( [FeCl_6]^{3-} \) has symmetrical distribution in both \( t_{2g} \) and \( e_g \) orbitals.
Therefore:
\[ \boxed{[FeCl_6]^{3-}} \]