Question

(i) Write the electronic configuration of the following ions: (a) Cr³⁺ (b) Cu²⁺.
(ii) Write the formula of chromate and dichromate ions.

Hint:

Remove 4s before 3d: Cr³⁺ is [Ar]3d³ and Cu²⁺ is [Ar]3d⁹. Chromate is CrO₄^2−, dichromate is Cr₂O₇^2−.

Answer 1

Step 1: Neutral atoms. Chromium (24) is an exception with configuration [Ar]3d⁵4s¹ (a half-filled 3d set is extra stable). Copper (29) is also an exception, [Ar]3d¹⁰4s¹ (a fully filled 3d set).
Step 2: Forming Cr³⁺. A 3+ charge means 24 − 3 = 21 electrons. Removing the 4s¹ and two 3d electrons leaves [Ar]3d³. So Cr³⁺ = [Ar]3d³, with 3 unpaired electrons.
Step 3: Forming Cu²⁺. A 2+ charge means 29 − 2 = 27 electrons. Removing 4s¹ and one 3d electron leaves [Ar]3d⁹. So Cu²⁺ = [Ar]3d⁹, with 1 unpaired electron.
Step 4: Oxo-anions of chromium. In the yellow chromate ion chromium is in the +6 state bonded to four oxygens: CrO₄^2−. In acid solution two chromate units share one oxygen to give the orange dichromate ion Cr₂O₇^2−.