Question

Difficulty: HardTransition Metals: General Properties, Catalytic Behavior, and Complex Ions

Chromium is a first-row transition element with an atomic number of 24. During a chemical reaction, a neutral chromium atom loses three electrons to form the chromium(III) cation, Cr3+Cr^{3+}. Which electronic structure correctly represents this Cr3+Cr^{3+} ion in its ground state?

  1. [Ar]3d3[Ar] 3d^3Answer
  2. B
    [Ar]3d14s2[Ar] 3d^1 4s^2
  3. C
    [Ar]3d24s1[Ar] 3d^2 4s^1
  4. D
    [Ar]3d4[Ar] 3d^4

Answer

The ground-state electronic configuration of the chromium(III) ion, Cr3+Cr^{3+}, is [Ar]3d3[Ar] 3d^3.
Neutral chromium has an electronic configuration of [Ar]3d54s1[Ar] 3d^5 4s^1 due to half-filled subshell stability. When forming a Cr3+Cr^{3+} cation, the atom loses three electrons: the single valence electron in the 4s4s orbital is lost first, followed by two electrons from the 3d3d subshell, leaving a final ground-state configuration of [Ar]3d3[Ar] 3d^3.

Step-by-Step Solution

1
Determine the ground-state electronic configuration of neutral chromium (CrCr, Z=24Z = 24).
Neutral chromium has the anomalous configuration [Ar]3d54s1[Ar] 3d^5 4s^1.
Chromium exhibits an exception to the standard Aufbau principle because a half-filled dd-subshell (3d53d^5) provides extra thermodynamic stability.
2
Apply the rule for cation formation in transition metals.
Electrons in the outermost principal quantum shell (4s4s) are removed prior to removing electrons from the inner (n1)d(n-1)d subshell (3d3d).
Once filled, the 3d3d orbitals experience greater nuclear attraction and drop lower in energy than the 4s4s orbital.
3
Deduct three electrons to account for the +3+3 charge of Cr3+Cr^{3+}.
Remove the single electron from 4s4s ([Ar]3d54s0[Ar] 3d^5 4s^0), then remove two electrons from 3d3d to obtain [Ar]3d3[Ar] 3d^3.
Removing 3 electrons total converts neutral CrCr into Cr3+Cr^{3+}.

Key Concept

Electronic Configuration of Transition Metal Cations
Estimated Time:2m 0s
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