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: Rule for writing ion configurations. First write the configuration of the neutral atom, then remove electrons for a cation. For d-block ions, electrons are removed from the outer 4s orbital first and then from 3d.
Step 2: Cr³⁺ (Z = 24). Neutral chromium: [Ar] 3d⁵ 4s¹. Removing 3 electrons (first the 4s electron, then two 3d electrons) gives:
Cr³⁺: [Ar] 3d³, i.e. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³.
Step 3: Cu²⁺ (Z = 29). Neutral copper: [Ar] 3d¹⁰ 4s¹. Removing 2 electrons (first 4s, then one 3d) gives:
Cu²⁺: [Ar] 3d⁹, i.e. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹.
Step 4: Formulae of the ions.
Chromate ion: CrO₄^2−
Dichromate ion: Cr₂O₇^2−